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SOLUTION MANUAL FOR TABLE OF CONTENTS 1. Introduction. Engineering and Mechanics. Learning Mechanics. Fundamental Concepts. Units. Newtonian Gravitation. 2. Vectors. Vector Operations and Definitions. Scalars and Vectors. Rules for Manipulating Vectors. Cartesian Components. Components in Two Dimensions. Components in Three Dimensions. Products of Vectors. Dot Products. Cross Products. Mixed Triple Products. 3. Forces. Types of Forces. Equilibrium and Free-Body Diagrams. Two-Dimensional Force Systems. Three-Dimensional Force Systems. 4. Systems of Forces and Moments. Two-Dimensional Description of the Moment. The Moment Vector. Moment of a Force About a Line. Couples. Equivalent Systems. Representing Systems by Equivalent Systems. 5. Objects in Equilibrium. The Equilibrium Equations. Two-Dimensional Applications. Statically Indeterminate Objects. Three-Dimensional Applications. Two-Force and Three-Force. 6. Structures in Equilibrium. Trusses. The Method of Joints. The Method of Sections. Space Trusses. Frames and Machines. 7. Centroids and Centers of Mass 316. Centroids. Centroids of Areas. Centroids of Composite Areas. Distributed Loads. Centroids of Volumes and Lines. The Pappus-Guldinus Theorems. Centers of Mass. Definition of the Center of Mass. Centers of Mass of Objects. Centers of Mass of Composite Objects. 8. Moments of Inertia. Areas. Definitions. Parallel-Axis Theorems. Rotated and Principal Axes. Masses. Simple Objects. Parallel-Axis Theorem. 9. Friction. Theory of Dry Friction. Applications. 10. Internal Forces and Moments. Beams. Axial Force, Shear Force, and Bending Moment. Shear Force and Bending Moment Diagrams. Relations Between Distributed Load, Shear Force, and Bending Moment. Cables. Loads Distributed Uniformly Along Straight Lines. Loads Distributed Uniformly Along Cables. Discrete Loads. Liquids and Gasses. Pressure and the Center of Pressure. Pressure in a Stationary Liquid. 11. Virtual Work and Potential Energy. Virtual Work. Potential Energy. Problem 1.1 The value of is 3.14159265. . . If C is the circumference of a circle and r is its radius, determine the value of r/C to four significant digits. Problem 1.2 The base of natural logarithms is e D 2.718281828 . . . (a) (b) (c) Express e to five significant digits. Determine the value of e2 to five significant digits. Use the value of e you obtained in part (a) to determine the value of e2 to five significant digits. Solution: C D 2r ) r 1 D D 0.159154943. C 2 To four significant digits we have r D 0.1592 C Solution: The value of e is: e D 2.718281828 (a) To five significant figures e D 2.7183 (b) e2 to five significant figures is e2 D 7.3891 (c) Using the value from part (a) we find e2 D 7.3892 which is not correct in the fifth digit. [Part (c) demonstrates the hazard of using rounded-off values in calculations.] Problem 1.3 A machinist drills a circular hole in a panel with a nominal radius r D 5 mm. The actual radius of the hole is in the range r D 5 š 0.01 mm. (a) To what number of significant digits can you express the radius? (b) To what number of significant digits can you express the area of the hole? Solution: a) The radius is in the range r1 D 4.99 mm to r2 D 5.01 mm. These numbers are not equal at the level of three significant digits, but they are equal if they are rounded off to two significant digits. Two: r D 5.0 mm b) The area of the hole is in the range from A1 D r1 2 D 78.226 m2 to A2 D r2 2 D 78.854 m2 . These numbers are equal only if rounded to one significant digit: One: A D 80 mm2 Problem 1.4 The opening in the soccer goal is 24 ft wide and 8 ft high, so its area is 24 ft ð 8 ft D 192 ft2 . What is its area in m2 to three significant digits? Solution: A D 192 ft2 1m 3.281 ft 2 D 17.8 m2 A D 17.8 m2 Problem 1.5 The Burj Dubai, scheduled for completion in 2008, will be the world’s tallest building with a height of 705 m. The area of its ground footprint will be 8000 m2 . Convert its height and footprint area to U.S. customary units to three significant digits. Solution: 3.281 ft D 2.31 ð 103 ft 1m 3.218 ft 2 A D 8000 m2 D 8.61 ð 104 ft2 1m h D 705 m h D 2.31 ð 103 ft, A D 8.61 ð 104 ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 1 Problem 1.6 Suppose that you have just purchased a Ferrari F355 coupe and you want to know whether you can use your set of SAE (U.S. Customary Units) wrenches to work on it. You have wrenches with widths w D 1/4 in, 1/2 in, 3/4 in, and 1 in, and the car has nuts with dimensions n D 5 mm, 10 mm, 15 mm, 20 mm, and 25 mm. Defining a wrench to fit if w is no more than 2% larger than n, which of your wrenches can you use? Solution: Convert the metric size n to inches, and compute the percentage difference between the metric sized nut and the SAE wrench. The results are: 5 mm 1 inch 25.4 mm D 0.19685.. in, 0.19685 0.25 0.19685 100 D 27.0% 10 mm 15 mm n 20 mm 25 mm 1 inch 25.4 mm 1 inch 25.4 mm 1 inch 25.4 mm 1 inch 25.4 mm D 0.3937.. in, D 0.5905.. in, D 0.7874.. in, D 0.9843.. in, 0.3937 0.5 0.3937 0.5905 0.5 0.5905 100 D 27.0% 100 D C15.3% 0.7874 0.75 0.7874 0.9843 1.0 0.9843 100 D C4.7% 100 D 1.6% A negative percentage implies that the metric nut is smaller than the SAE wrench; a positive percentage means that the nut is larger then the wrench. Thus within the definition of the 2% fit, the 1 in wrench will fit the 25 mm nut. The other wrenches cannot be used. Problem 1.7 Suppose that the height of Mt. Everest is known to be between 29,032 ft and 29,034 ft. Based on this information, to how many significant digits can you express the height (a) in feet? (b) in meters?. Solution: a) h1 D 29032 ft h2 D 29034 ft The two heights are equal if rounded off to four significant digits. The fifth digit is not meaningful. Four: h D 29,030 ft b) In meters we have 1m D 8848.52 m h1 D 29032 ft 3.281 ft 1m D 8849.13 m h2 D 29034 ft 3.281 ft These two heights are equal if rounded off to three significant digits. The fourth digit is not meaningful. Three: h D 8850 m Problem 1.8 The maglev (magnetic levitation) train from Shanghai to the airport at Pudong reaches a speed of 430 km/h. Determine its speed (a) in mi/h; (b) ft/s. Problem 1.9 In the 2006 Winter Olympics, the men’s 15-km cross-country skiing race was won by Andrus Veerpalu of Estonia in a time of 38 minutes, 1.3 seconds. Determine his average speed (the distance traveled divided by the time required) to three significant digits (a) in km/h; (b) in mi/h. 2 Solution: a) v D 430 b) v D 430 km h 0.6214 mi 1 km D 267 mi/h v D 267 mi/h 1 ft 1h km 1000 m D 392 ft/s h 1 km 0.3048 m 3600 s v D 392 ft/s Solution: 15 km 1.3 38 C min 60 a) vD b) v D 23.7 km/h 60 min 1h 1 mi 1.609 km D 23.7 km/h v D 23.7 km/h D 14.7 mi/h v D 14.7 mi/h c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 1.10 The Porsche’s engine exerts 229 ft-lb (foot-pounds) of torque at 4600 rpm. Determine the value of the torque in N-m (Newton-meters). Solution: Problem 1.11 The kinetic energy of the man in Active Example 1.1 is defined by 12 mv2 , where m is his mass and v is his velocity. The man’s mass is 68 kg and he is moving at 6 m/s, so his kinetic energy is 1 2 2 2 2 (68 kg)(6 m/s) D 1224 kg-m /s . What is his kinetic energy in U.S. Customary units? Solution: T D 229 ft-lb 1N 0.2248 lb T D 1224 kg-m2 /s2 1m 3.281 ft 1 slug 14.59 kg D 310 N-m 1 ft 0.3048 m Solution: Use Table 1.2. The result is: Problem 1.13 A furlong per fortnight is a facetious unit of velocity, perhaps made up by a student as a satirical comment on the bewildering variety of units engineers must deal with. A furlong is 660 ft (1/8 mile). A fortnight is 2 weeks (14 nights). If you walk to class at 2 m/s, what is your speed in furlongs per fortnight to three significant digits? Solution: Solution: A D 200 mm2 280 mm120 mm D 20800 mm2 2 1m a) A D 20800 mm2 D 0.0208 m2 A D 0.0208 m2 1000 mm b) A D 20800 mm2 1 in 25.4 mm g D 9.81 m v D 12,000 D 903 slug-ft2 /s 1 ft 0.3048 m 1 ft 0.3048 m D 32.185 . . . 1 furlong 660 ft ft s2 3600 s hr D 32.2 24 hr 1 day ft s2 14 day 1 fortnight furlongs fortnight y 40 mm x 120 mm 40 mm 40 mm 200 mm 2 A D 32.2 in Problem 1.15 The cross-sectional area of the C12ð30 American Standard Channel steel beam is A D 8.81 in2 . What is its cross-sectional area in mm2 ? s2 v D 2 m/s 2 D 32.2 in2 2 T D 903 slug-ft2 /s Problem 1.12 The acceleration due to gravity at sea level in SI units is g D 9.81 m/s2 . By converting units, use this value to determine the acceleration due to gravity at sea level in U.S. Customary units. Problem 1.14 Determine the cross-sectional area of the beam (a) in m2 ; (b) in in2 . T D 310 N-m y A Solution: A D 8.81 in2 25.4 mm 1 in 2 D 5680 mm2 x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 3 Problem 1.16 A pressure transducer measures a value of 300 lb/in2 . Determine the value of the pressure in pascals. A pascal (Pa) is one newton per meter squared. Solution: Convert the units using Table 1.2 and the definition of the Pascal unit. The result: 300 lb in2 4.448 N 1 lb D 2.0683 . . . 106 12 in 1 ft N m2 2 1 ft 0.3048 m 2 D 2.07106 Pa Problem 1.17 A horsepower is 550 ft-lb/s. A watt is 1 N-m/s. Determine how many watts are generated by the engines of the passenger jet if they are producing 7000 horsepower. Solution: P D 7000 hp 550 ft-lb/s 1 hp 1m 3.281 ft 1N 0.2248 lb D 5.22 ð 106 W P D 5.22 ð 106 W Problem 1.18 Chapter 7 discusses distributed loads that are expressed in units of force per unit length. If the value of a distributed load is 400 N/m, what is its value in lb/ft?. Problem 1.19 The moment of inertia of the rectangular area about the x axis is given by the equation I D 13 bh3 . The dimensions of the area are b D 200 mm and h D 100 mm. Determine the value of I to four significant digits in terms of (a) mm4 ; (b) m4 ; (c) in4 . Solution: w D 400 N/m 0.2248 lb 1N 1m 3.281 ft D 27.4 lb/ft w D 27.4 lb/ft Solution: 1 200 mm100 mm3 D 66.7 ð 106 mm4 3 (a) ID (b) I D 66.7 ð 106 mm4 (c) I D 66.7 ð 106 mm4 y 1m 1000 mm 1 in 25.4 mm 4 D 66.7 ð 106 m4 4 D 160 in4 h x b Problem 1.20 In Example 1.3, instead of Einstein’s equation consider the equation L D mc, where the mass m is in kilograms and the velocity of light c is in meters per second. (a) What are the SI units of L? (b) If the value of L in SI units is 12, what is its value in U.S. Customary base units? 4 Solution: a) L D mc ) b) L D 12 kg-m/s Units L D kg-m/s 0.0685 slug 1 kg 3.281 ft 1m D 2.70 slug-ft/s L D 2.70 slug-ft/s c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 1.21 The equation D Solution: My I (a) is used in the mechanics of materials to determine normal stresses in beams. (a) (b) D (N-m)m My N D D 2 I m4 m D 2000 N-m0.1 m My D I 7 ð 105 m4 (b) When this equation is expressed in terms of SI base units, M is in newton-meters (N-m), y is in meters (m), and I is in meters to the fourth power (m4 ). What are the SI units of ? If M D 2000 N-m, y D 0.1 m, and I D 7 ð 105 m4 , what is the value of in U.S. Customary base units? D 59,700 1 lb 4.448 N 0.3048 m ft 2 lb ft2 Problem 1.22 The acceleration due to gravity on the Solution: a) The mass does surface of the moon is 1.62 m/s2 . (a) What would the on location. The mass in kg is not depend 14.59 kg mass of the C-clamp in Active Example 1.4 be on the surface mass D 0.397 kg D 0.397 kg 0.0272 slug 1 slug of the moon? (b) What would the weight of the C-clamp in newtons be on the surface of the moon? b) The weight on the surface of the moon is W D mg D 0.397 kg1.62 m/s2 D 0.643 N Problem 1.23 The 1 ft ð 1 ft ð 1 ft cube of iron weighs 490 lb at sea level. Determine the weight in newtons of a 1 m ð 1 m ð 1 m cube of the same material at sea level. Solution: The weight density is D The weight of the 1 m3 cube is: W D V D 490 lb 1 ft3 1 m3 W D 0.643N 1 ft 490 lb 1 ft3 1 ft 0.3048 m 3 1N 0.2248 lb 1 ft 1 ft D 77.0 kN Problem 1.24 The area of the Pacific Ocean is 64,186,000 square miles and its average depth is 12,925 ft. Assume that the weight per unit volume of ocean water is 64 lb/ft3 . Determine the mass of the Pacific Ocean (a) in slugs; (b) in kilograms Solution: The volume of the ocean is V D 64,186,000 mi2 12,925 ft 64 lb/ft3 32.2 ft/s2 2 D 2.312 ð 1019 ft3 (a) m D V D (b) m D 4.60 ð 1019 slugs 2.312 ð 1019 ft3 D 4.60 ð 1019 slugs Problem 1.25 The acceleration due to gravity at sea level is g D 9.81 m/s2 . The radius of the earth is 6370 km. The universal gravitational constant is G D 6.67 ð 1011 N-m2 /kg2 . Use this information to determine the mass of the earth. 5,280 ft 1 mi Solution: Use Eq (1.3) a D 14.59 kg 1 slug D 6.71 ð 1020 kg GmE . Solve for the mass, R2 m 2 9.81 m/s2 6370 km2 103 gR2 mE D D km G N-m2 11 6.6710 kg2 D 5.9679 . . . 1024 kg D 5.971024 kg c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5 Problem 1.26 A person weighs 180 lb at sea level. The radius of the earth is 3960 mi. What force is exerted on the person by the gravitational attraction of the earth if he is in a space station in orbit 200 mi above the surface of the earth? Solution: Use Eq (1.5). W D mg RE r 2 D WE g g RE RE C H 2 D WE 3960 3960 C 200 2 D 1800.90616 D 163 lb Problem 1.27 The acceleration due to gravity on the surface of the moon is 1.62 m/s2 . The moon’s radius is RM D 1738 km. (a) What is the weight in newtons on the surface of the moon of an object that has a mass of 10 kg? (b) Using the approach described in Example 1.5, determine the force exerted on the object by the gravity of the moon if the object is located 1738 km above the moon’s surface. Solution: Problem 1.28 If an object is near the surface of the earth, the variation of its weight with distance from the center of the earth can often be neglected. The acceleration due to gravity at sea level is g D 9.81 m/s2 . The radius of the earth is 6370 km. The weight of an object at sea level is mg, where m is its mass. At what height above the earth does the weight of the object decrease to 0.99 mg? Solution: Use a variation of Eq (1.5). W D mgM D 10 kg1.26 m/s2 D 12.6 N a) b) Adapting equation 1.4 we have aM D gM then F D maM D 10 kg1.62 m/s2 W D 12.6 N RM r 2 . The force is 1738 km 1738 km C 1738 km 2 D 4.05 N F D 4.05 N W D mg RE RE C h 2 D 0.99 mg Solve for the radial height, h D RE p 1 0.99 1 D 63701.0050378 1.0 D 32.09 . . . km D 32,100 m D 32.1 km Problem 1.29 The planet Neptune has an equatorial diameter of 49,532 km and its mass is 1.0247 ð 1026 kg. If the planet is modeled as a homogeneous sphere, what is the acceleration due to gravity at its surface? (The universal gravitational constant is G D 6.67 ð 1011 N-m2 /kg2 .) Solution: Problem 1.30 At a point between the earth and the moon, the magnitude of the force exerted on an object by the earth’s gravity equals the magnitude of the force exerted on the object by the moon’s gravity. What is the distance from the center of the earth to that point to three significant digits? The distance from the center of the earth to the center of the moon is 383,000 km, and the radius of the earth is 6370 km. The radius of the moon is 1738 km, and the acceleration due to gravity at its surface is 1.62 m/s2 . Solution: Let rEp be the distance from the Earth to the point where the gravitational accelerations are the same and let rMp be the distance from the Moon to that point. Then, rEp C rMp D rEM D 383,000 km. The fact that the gravitational attractions by the Earth and the Moon at this point are equal leads to the equation mN m mN mN D G 2 m ) gN D G 2 rN 2 r rN Note that the radius of Neptune is rN D 12 49,532 km D 24,766 km N-m2 1.0247 ð 1026 kg 1 km 2 Thus gN D 6.67 ð 1011 24766 km2 1000 m kg2 We have: W D G D 11.1 m/s2 gE RE rEp gN D 11.1 m/s2 2 D gM RM rMp 2 , where rEM D 383,000 km. Substituting the correct numerical values leads to the equation 9.81 m 1738 km 2 m 6370 km 2 D 1.62 , s2 rEp s2 rEM rEp where rEp is the only unknown. Solving, we get rEp D 344,770 km D 345,000 km. 6 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.1 In Active Example 2.1, suppose that the vectors U and V are reoriented as shown. The vector V is vertical. The magnitudes are jUj D 8 and jVj D 3. Graphically determine the magnitude of the vector U C 2V. 45⬚ U V Solution: Draw the vectors accurately and measure the resultant. R D jU C 2Vj D 5.7 R D 5.7 Problem 2.2 Suppose that the pylon in Example 2.2 is moved closer to the stadium so that the angle between the forces FAB and FAC is 50° . Draw a sketch of the new situation. The magnitudes of the forces are jFAB j D 100 kN and jFAC j D 60 kN. Graphically determine the magnitude and direction of the sum of the forces exerted on the pylon by the cables. Solution: Accurately draw the vectors and measure the magnitude and direction of the resultant jFAB C FAC j D 146 kN ˛ D 32° c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 7 Problem 2.3 The magnitude jFA j D 80 lb and the angle ˛ D 65° . The magnitude jFA C FB j D 120 lb. Graphically determine the magnitude of FB . FB FC  a FA Solution: Accurately draw the vectors and measure the magnitude of FB . jFB j D 62 lb Problem 2.4 The magnitudes jFA j D 40 N, jFB j D 50 N, and jFC j D 40 N. The angle ˛ D 50° and ˇ D 80° . Graphically determine the magnitude of FA C FB C FC . FB FC  a FA Solution: Accurately draw the vectors and measure the magnitude of FA C FB C FC . R D jFA C FB C FC j D 83 N 8 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.5 The magnitudes jFA j D jFB j D jFC j D 100 lb, and the angles ˛ D 30° . Graphically determine the value of the angle ˇ for which the magnitude jFA C FB C FC j is a minimum and the minimum value of jFA C FB C FC j. FB FC  a FA Solution: For a minimum, the vector FC must point back to the origin. R D jFA C FB C FC j D 93.2 lb ˇ D 165° Problem 2.6 The angle D 50° . Graphically determine the magnitude of the vector rAC . 150 mm 60 mm B rAB A rBC rAC C Solution: Draw the vectors accurately and then measure jrAC j. jrAC j D 181 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 9 Problem 2.7 The vectors FA and FB represent the forces exerted on the pulley by the belt. Their magnitudes are jFA j D 80 N and jFB j D 60 N. Graphically determine the magnitude of the total force the belt exerts on the pulley. Solution: Draw the vectors accurately and then measure jFA C FB j. jFA C FB j D 134 N FB 45⬚ FA 10⬚ Problem 2.8 The sum of the forces FA C FB C FC D 0. The magnitude jFA j D 100 N and the angle ˛ D 60° . Graphically determine the magnitudes jFB j and jFC j. Solution: Draw the vectors so that they add to zero. jFB j D 86.6 N, jFC j D 50.0 N FB 30 a FA FC Problem 2.9 The sum of the forces FA C FB C FC D 0. The magnitudes jFA j D 100 N and jFB j D 80 N. Graphically determine the magnitude jFC j and the angle ˛. FB 30 a FA FC Solution: Draw the vectors so that they add to zero. jFC j D 50.4 N, ˛ D 52.5° 10 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.10 The forces acting on the sailplane are represented by three vectors. The lift L and drag D are perpendicular. The magnitude of the weight W is 500 lb. The sum of the forces W C L C D D 0. Graphically determine the magnitudes of the lift and drag. L 25⬚ D W Solution: Draw the vectors so that they add to zero. Then measure the unknown magnitudes. jLj D 453 lb jDj D 211 lb Problem 2.11 A spherical storage tank is suspended from cables. The tank is subjected to three forces, the forces FA and FB exerted by the cables and its weight W. The weight of the tank is jWj D 600 lb. The vector sum of the forces acting on the tank equals zero. Graphically determine the magnitudes of FA and FB . FA 40˚ Solution: Draw the vectors so that they add to zero. Then measure the unknown magnitudes. jFA j D jFB j D 319 lb FB 20˚ 20˚ W c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 11 Problem 2.12 The rope ABC exerts forces FBA and FBC of equal magnitude on the block at B. The magnitude of the total force exerted on the block by the two forces is 200 lb. Graphically determine jFBA j. Solution: Draw the vectors accurately and then measure the unknown magnitudes. jFBA j D 174 lb FBC C 20⬚ B B FBA A Problem 2.13 Two snowcats tow an emergency shelter to a new location near McMurdo Station, Antarctica. (The top view is shown. The cables are horizontal.) The total force FA C FB exerted on the shelter is in the direction parallel to the line L and its magnitude is 400 lb. Graphically determine the magnitudes of FA and FB . Solution: Draw the vectors accurately and then measure the unknown magnitudes. jFA j D 203 lb jFB j D 311 lb L FA 30⬚ 50⬚ FB Top View Problem 2.14 A surveyor determines that the horizontal distance from A to B is 400 m and the horizontal distance from A to C is 600 m. Graphically determine the magnitude of the vector rBC and the angle ˛. Solution: Draw the vectors accurately and then measure the unknown magnitude and angle. jrBC j D 390 m ˛ D 21.2° North B a rBC C 60⬚ 20⬚ A 12 East c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.15 The vector r extends from point A to the midpoint between points B and C. Prove that C r D 12 rAB C rAC . rAC r rAB B A Solution: The proof is straightforward: C r D rAB C rBM , and r D rAC C rCM . rAC r Add the two equations and note that rBM C rCM D 0, since the two vectors are equal and opposite in direction. Thus 2r D rAC C rAB , or r D 1 2 rAC C rAB A M B rAB Problem 2.16 By drawing sketches of the vectors, explain why U C V C W D U C V C W. Solution: Additive associativity for vectors is usually given as an axiom in the theory of vector algebra, and of course axioms are not subject to proof. However we can by sketches show that associativity for vector addition is intuitively reasonable: Given the three vectors to be added, (a) shows the addition first of V C W, and then the addition of U. The result is the vector U C V C W. V U V+W (a) W U+[V+W] V (b) shows the addition of U C V, and then the addition of W, leading to the result U C V C W. U W The final vector in the two sketches is the same vector, illustrating that associativity of vector addition is intuitively reasonable. Problem 2.17 A force F D 40 i 20 j N. What is its magnitude jFj? U+V (b) [U+V]+W p Solution: jFj D 402 C 202 D 44.7 N Strategy: The magnitude of a vector in terms of its components is given by Eq. (2.8). Problem 2.18 An engineer estimating the components of a force F D Fx i C Fy j acting on a bridge abutment has determined that Fx D 130 MN, jFj D 165 MN, and Fy is negative. What is Fy ? Solution: jFj D jFy j D jFx j2 C jFy j2 jFj2 jFx j2 D 165 MN2 130 MN2 D 101.6 MN Fy D 102 MN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 13 Problem 2.19 A support is subjected to a force F D Fx i C 80j (N). If the support will safely support a force of 100 N, what is the allowable range of values of the component Fx ? Solution: Use the definition of magnitude in Eq. (2.8) and reduce algebraically. 100 ½ Fx 2 C 802 , from which 1002 802 ½ Fx 2 . Thus jFx j Problem 2.20 If FA D 600i 800j (kip) and FB D 200i 200j (kip), what is the magnitude of the force F D FA 2FB ? Solution: Take the scalar multiple of FB , add the components of the two forces as in Eq. (2.9), and use the definition of the magnitude. F D 600 2200i C 800 2200j D 200i 400j jFj D Problem 2.21 The forces acting on the sailplane are its weight W D 500jlb, the drag D D 200i C 100j(lb) and the lift L. The sum of the forces W C L C D D 0. Determine the components and the magnitude of L. p 3600, or 60 Fx C60 (N) 2002 C 4002 D 447.2 kip y L Solution: D L D W D D 500j 200i C 100j D 200i C 400jlb jLj D 200 lb2 C 400 lb2 D 447 lb W L D 200i C 400jlb, jLj D 447 lb x 14 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.22 Two perpendicular vectors U and V lie in the x-y plane. The vector U D 6i 8j and jVj D 20. What are the components of V? (Notice that this problem has two answers.) Solution: The two possible values of V are shown in the sketch. y The strategy is to (a) determine the unit vector associated with U, (b) express this vector in terms of an angle, (c) add š90° to this angle, (d) determine the two unit vectors perpendicular to U, and (e) calculate the components of the two possible values of V. The unit vector parallel to U is eU D 6i 62 C 82 8j D 0.6i 0.8j 2 6 C 82 V2 6 V1 U x 8 Expressed in terms of an angle, eU D i cos ˛ j sin ˛ D i cos53.1° j sin53.1° Add š90° to find the two unit vectors that are perpendicular to this unit vector: ep1 D i cos143.1° j sin143.1° D 0.8i 0.6j ep2 D i cos36.9° j sin36.9° D 0.8i C 0.6j Take the scalar multiple of these unit vectors to find the two vectors perpendicular to U. V1 D jVj0.8i 0.6j D 16i 12j. The components are Vx D 16, Vy D 12 V2 D jVj0.8i C 0.6j D 16i C 12j. The components are Vx D 16, Vy D 12 Problem 2.23 A fish exerts a 10-lb force on the line that is represented by the vector F. Express F in terms of components using the coordinate system shown. Solution: We can use similar triangles to determine the components of F. F D 10 lb y 11 7 p i p j D 5.37i 8.44j lb 72 C 112 72 C 112 F D 5.37i 8.44j lb 7 11 F x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 15 Problem 2.24 A man exerts a 60-lb force F to push a crate onto a truck. (a) Express F in terms of components using the coordinate system shown. (b) The weight of the crate is 100 lb. Determine the magnitude of the sum of the forces exerted by the man and the crate’s weight. Solution: (a) F D 60 lbcos 20° i C sin 20° j D 56.4i C 20.5j lb F D 56.4i C 20.5jlb (b) W D 100 lbj y F C W D 56.4i C [20.5 100]j lb D 56.4i 79.5j lb F 20⬚ jF C Wj D 56.4 lb2 C 79.5 lb2 D 97.4 lb jF C Wj D 97.4 lb x Problem 2.25 The missile’s engine exerts a 260-kN force F. (a) Express F in terms of components using the coordinate system shown. (b) The mass of the missile is 8800 kg. Determine the magnitude of the sum of the forces exerted by the engine and the missile’s weight. y F 3 4 Solution: (a) We can use similar triangles to determine the components of F. F D 260 kN p 3 iC p j D 208i C 156j kN 42 C 32 42 C 32 4 x F D 208i C 156j kN (b) The missile’s weight W can be expressed in component and then added to the force F. W D 8800 kg9.81 m/s2 j D 86.3 kNj F C W D 208i C [156 86.3]j kN D 208i 69.7j kN jF C Wj D 208 kN2 C 69.7 kN2 D 219 kN jF C Wj D 219 kN Problem 2.26 For the truss shown, express the position vector rAD from point A to point D in terms of components. Use your result to determine the distance from point A to point D. y rAD D 0 1.8 mi C 0.4 m 0.7 mj D 1.8i 0.3j m rAD D 1.8 m2 C 0.3 m2 D 1.825 m B A 0.6 m D 0.7 m 0.4 m C 0.6 m 16 Solution: Coordinates A(1.8, 0.7) m, D(0, 0.4) m x 1.2 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.27 The points A, B, . . . are the joints of the hexagonal structural element. Let rAB be the position vector from joint A to joint B, rAC the position vector from joint A to joint C, and so forth. Determine the components of the vectors rAC and rAF . Solution: Use the xy coordinate system shown and find the locations of C and F in those coordinates. The coordinates of the points in this system are the scalar components of the vectors rAC and rAF . For rAC , we have rAC D rAB C rBC D xB xA i C yB yA j y C xC xB i C yC yB j E D 2m rAC D 2m 0i C 0 0j C 2m cos 60° 0i or C 2m cos 60° 0j, F giving C rAC D 2m C 2m cos 60° i C 2m sin 60° j. For rAF , we have A B x rAF D xF xA i C yF yA j D 2m cos 60° xF 0i C 2m sin 60° 0j. Problem 2.28 For the hexagonal structural element in Problem 2.27, determine the components of the vector rAB rBC . Solution: rAB rBC . The angle between BC and the x-axis is 60° . rBC D 2 cos60° i C 2sin60° j m rBC D 1i C 1.73j m rAB rBC D 2i 1i 1.73j m rAB rBC D 1i 1.73j m Problem 2.29 The coordinates of point A are (1.8, 3.0) ft. The y coordinate of point B is 0.6 ft. The vector rAB has the same direction as the unit vector eAB D 0.616i 0.788j. What are the components of rAB ? y Solution: The vector rAB can be written two ways. A rAB rAB D jrAB j0.616i 0.788j D Bx Ax i C By Ay j Comparing the two expressions we have By Ay D 0.6 3.0ft D 0.788jrAB j jrAB j D B x 2.4 ft D 3.05 ft 0.788 Thus rAB D jrAB j0.616i 0.788j D 3.05 ft0.616i 0.788j D 1.88i 2.40j ft rAB D 1.88i 2.40j ft c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 17 y Problem 2.30 (a) Express the position vector from point A of the front-end loader to point B in terms of components. 98 in 45 in C (b) Express the position vector from point B to point C in terms of components. A 55 in (c) Use the results of (a) and (b) to determine the distance from point A to point C. B 50 in 35 in x 50 in Solution: The coordinates are A(50, 35); B(98, 50); C(45, 55). (a) The vector from point A to B: rAB D 98 50i C 50 35j D 48i C 15j (in) (b) The vector from point B to C is rBC D 45 98i C 55 50j D 53i C 5j (in). (c) The distance from A to C is the magnitude of the sum of the vectors, rAC D rAB C rBC D 48 53i C 15 C 5j D 5i C 20j. The distance from A to C is jrAC j D 18 52 C 202 D 20.62 in c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.31 In Active Example 2.3, the cable AB exerts a 900-N force on the top of the tower. Suppose that the attachment point B is moved in the horizontal direction farther from the tower, and assume that the magnitude of the force F the cable exerts on the top of the tower is proportional to the length of the cable. (a) What is the distance from the tower to point B if the magnitude of the force is 1000 N? (b) Express the 1000-N force F in terms of components using the coordinate system shown. A 80 m 40 m y Solution: In the new problem assume that point B is located a distance d away from the base. The lengths in the original problem and in the new problem are given by Loriginal D Lnew D (a) 40 m2 C 80 m2 D A Force exerted on the tower by cable AB 8000 m2 d2 C 80 m2 80 m F The force is proportional to the length. Therefore d2 C 80 m2 1000 N D 900 N 8000 m2 dD B 8000 m2 1000 N 900 N B x 40 m 2 80 m2 D 59.0 m d D 59.0 m (b) The force F is then F D 1000 N d d2 C 80 m2 i 80 m d2 C 80 m2 j D 593i 805j N F D 593i 805j N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 19 Problem 2.32 Determine the position vector rAB in terms of its components if (a) D 30° , (b) D 225° . y 150 mm 60 mm B rAB rBC C θ x A Solution: (a) y rAB D 60 cos30° i C 60 sin30° j, or 150 mm 60 mm rAB D 51.96i C 30j mm. And B (b) FAB rAB D 60 cos225° i C 60 sin225° j or A rAB D 42.4i 42.4j mm. θ FBC C F x Problem 2.33 In Example 2.4, the coordinates of the fixed point A are (17, 1) ft. The driver lowers the bed of the truck into a new position in which the coordinates of point B are (9, 3) ft. The magnitude of the force F exerted on the bed by the hydraulic cylinder when the bed is in the new position is 4800 lb. Draw a sketch of the new situation. Express F in terms of components. y B B 30⬚ F A 30⬚ A x Solution: D tan1 2 ft 8 ft D 14.04° F D 4800 lb cos i C sin j. F D 4660i C 1160j lb 20 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.34 A surveyor measures the location of point A and determines that rOA D 400i C 800j (m). He wants to determine the location of a point B so that jrAB j D 400 m and jrOA C rAB j D 1200 m. What are the cartesian coordinates of point B? B A N rAB rOA Proposed roadway x O Solution: Two possibilities are: The point B lies west of point A, or point B lies east of point A, as shown. The strategy is to determine the unknown angles ˛, ˇ, and . The magnitude of OA is jrOA j D B 4002 C 8002 D 894.4. α The angle ˇ is determined by tan ˇ D A B y α θ β 800 D 2, ˇ D 63.4° . 400 0 x The angle ˛ is determined from the cosine law: cos ˛ D 894.42 C 12002 4002 D 0.9689. 2894.41200 ˛ D 14.3° . The angle is D ˇ š ˛ D 49.12° , 77.74° . The two possible sets of coordinates of point B are rOB D 1200i cos 77.7 C j sin 77.7 D 254.67i C 1172.66j (m) rOB D 1200i cos 49.1 C j sin 49.1 D 785.33i C 907.34j (m) The two possibilities lead to B(254.7 m, 1172.7 m) or B(785.3 m, 907.3 m) Problem 2.35 The magnitude of the position vector rBA from point B to point A is 6 m and the magnitude of the position vector rCA from point C to point A is 4 m. What are the components of rBA ? y Thus rBA D xA 0i C yA 0j ) 6 m2 D xA 2 C yA 2 rCA D xA 3 mi C yA 0j ) 4 m2 D xA 3 m2 C yA 2 3m B Solution: The coordinates are: AxA , yA , B0, 0, C3 m, 0 x C Solving these two equations, we find xA D 4.833 m, yA D š3.555 m. We choose the “-” sign and find rBA D 4.83i 3.56j m A c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 21 Problem 2.36 In Problem 2.35, determine the components of a unit vector eCA that points from point C toward point A. Strategy: Determine the components of rCA and then divide the vector rCA by its magnitude. Solution: From the previous problem we have rCA D 1.83i 3.56j m, rCA D 1.832 C 3.562 m D 3.56 m Thus eCA D rCA D 0.458i 0.889j rCA Problem 2.37 The x and y coordinates of points A, B, and C of the sailboat are shown. (a) Determine the components of a unit vector that is parallel to the forestay AB and points from A toward B. (b) Determine the components of a unit vector that is parallel to the backstay BC and points from C toward B. Solution: rAB D xB xA i C yB yA j rCB D xB xC i C yC yB j Points are: A (0, 1.2), B (4, 13) and C (9, 1) Substituting, we get rAB D 4i C 11.8j m, jrAB j D 12.46 m y B (4, 13) m rCB D 5i C 12j m, jrCB j D 13 m The unit vectors are given by eAB D rAB rCB and eCB D jrAB j jrCB j Substituting, we get eAB D 0.321i C 0.947j eCB D 0.385i C 0.923j A (0, 1.2) m 22 C (9, 1) m x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.38 The length of the bar AB is 0.6 m. Determine the components of a unit vector eAB that points from point A toward point B. y B 0.4 m A 0.3 m x Solution: We need to find the coordinates of point Bx, y B We have the two equations m y m 0.6 0.4 0.3 m C x2 C y 2 D 0.6 m2 x 2 C y 2 D 0.4 m2 Solving we find x D 0.183 m, y D 0.356 m A 0.3 m O x Thus eAB D rAB 0.183 m [0.3 m]i C 0.356 mj D rAB 0.183 m C 0.3 m2 C 0.356 m2 D 0.806i C 0.593j y Problem 2.39 Determine the components of a unit vector that is parallel to the hydraulic actuator BC and points from B toward C. 1m D C Solution: Point B is at (0.75, 0) and point C is at (0, 0.6). The vector 1m rBC D xC xB i C yC yB j 0.6 m B A x rBC D 0 0.75i C 0.6 0j m rBC D 0.75i C 0.6j m jrBC j D eBC D 0.15 m 0.6 m Scoop 0.752 C 0.62 D 0.960 m 0.75 0.6 rBC D iC j jrBC j 0.96 0.96 eBC D 0.781i C 0.625j c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 23 Problem 2.40 The hydraulic actuator BC in Problem 2.39 exerts a 1.2-kN force F on the joint at C that is parallel to the actuator and points from B toward C. Determine the components of F. Solution: From the solution to Problem 2.39, eBC D 0.781i C 0.625j The vector F is given by F D jFjeBC F D 1.20.781i C 0.625j k Ð N F D 937i C 750j N Problem 2.41 A surveyor finds that the length of the line OA is 1500 m and the length of line OB is 2000 m. y N (a) Determine the components of the position vector from point A to point B. (b) Determine the components of a unit vector that points from point A toward point B. A Proposed bridge B Solution: We need to find the coordinates of points A and B rOA D 1500 cos 60° i C 1500 sin 60° j 60⬚ 30⬚ rOA D 750i C 1299j m Point A is at (750, 1299) (m) River x O rOB D 2000 cos 30° i C 2000 sin 30° j m rOB D 1732i C 1000j m Point B is at (1732, 1000) (m) (a) The vector from A to B is rAB D xB xA i C yB yA j rAB D 982i 299j m (b) The unit vector eAB is eAB D 982i 299j rAB D jrAB j 1026.6 eAB D 0.957i 0.291j 24 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.42 The magnitudes of the forces exerted by the cables are jT1 j D 2800 lb, jT2 D 3200 lb, jT3 j D 4000 lb, and jT4 j D 5000 lb. What is the magnitude of the total force exerted by the four cables? y T4 51⬚ T3 40⬚ T2 29⬚ T1 9⬚ x Solution: The x-component of the total force is Tx D jT1 j cos 9° C jT2 j cos 29° jT3 j cos 40° C jT4 j cos 51° Tx D 2800 lb cos 9° C 3200 lb cos 29° C 4000 lb cos 40° C 5000 lb cos 51° Tx D 11,800 lb The y-component of the total force is Ty D jT1 j sin 9° C jT2 j sin 29° C jT3 j sin 40° C jT4 j sin 51° Ty D 2800 lb sin 9° C 3200 lb sin 29° C 4000 lb sin 40° C 5000 lb sin 51° Ty D 8450 lb The magnitude of the total force is jTj D Tx 2 C Ty 2 D 11,800 lb2 C 8450 lb2 D 14,500 lb jTj D 14,500 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 25 Problem 2.43 The tensions in the four cables are equal: jT1 j D jT2 j D jT3 j D jT4 j D T. Determine the value of T so that the four cables exert a total force of 12,500-lb magnitude on the support. y T4 51⬚ T3 40⬚ T2 29⬚ T1 9⬚ x Solution: The x-component of the total force is Tx D T cos 9° C T cos 29° C T cos 40° C T cos 51° Tx D 3.26T The y-component of the total force is Ty D T sin 9° C T sin 29° C T sin 40° C T sin 51° Ty D 2.06T The magnitude of the total force is jTj D Tx 2 C Ty 2 D Solving for T we find 26 3.26T2 C 2.06T2 D 3.86T D 12,500 lb T D 3240 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.44 The rope ABC exerts forces FBA and FBC on the block at B. Their magnitudes are equal: jFBA j D jFBC j. The magnitude of the total force exerted on the block at B by the rope is jFBA C FBC j D 920 N. Determine jFBA j by expressing the forces FBA and FBC in terms of components. FBC C 20° B B FBA A Solution: FBC FBC D Fcos 20° i C sin 20° j 20° FBA D Fj FBC C FBA D Fcos 20° i C [sin 20° 1]j Therefore 920 N2 D F2 cos2 20° C [sin 20° 1]2 ) F D 802 N FBA Problem 2.45 The magnitude of the horizontal force F1 is 5 kN and F1 C F2 C F3 D 0. What are the magnitudes of F2 and F3 ? y F3 30˚ F1 Solution: Using components we have Fx : 5 kN C F2 cos 45° F3 cos 30° D 0 45˚ Fy : F2 sin 45° C F3 sin 30° D0 F2 Solving simultaneously yields: x ) F2 D 9.66 kN, F3 D 13.66 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 27 Problem 2.46 Four groups engage in a tug-of-war. The magnitudes of the forces exerted by groups B, C, and D are jFB j D 800 lb, jFC j D 1000 lb, jFD j D 900 lb. If the vector sum of the four forces equals zero, what are the magnitude of FA and the angle ˛? y FB FC 70° 30° Solution: The strategy is to use the angles and magnitudes to determine the force vector components, to solve for the unknown force FA and then take its magnitude. The force vectors are FB D 800i cos 110° C j sin 110° D 273.6i C 751.75j FC D 1000i cos 30° C j sin 30° D 866i C 500j α 20° FD FA x FD D 900i cos20° C j sin20° D 845.72i 307.8j FA D jFA ji cos180 C ˛ C j sin180 C ˛ D jFA ji cos ˛ j sin ˛ The sum vanishes: FA C FB C FC C FD D i1438.1 jFA j cos ˛ C j944 jFA j sin ˛ D 0 From which FA D 1438.1i C 944j. The magnitude is jFA j D 14382 C 9442 D 1720 lb The angle is: tan ˛ D 28 944 D 0.6565, or ˛ D 33.3° 1438 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.47 In Example 2.5, suppose that the attachment point of cable A is moved so that the angle between the cable and the wall increases from 40° to 55° . Draw a sketch showing the forces exerted on the hook by the two cables. If you want the total force FA C FB to have a magnitude of 200 lb and be in the direction perpendicular to the wall, what are the necessary magnitudes of FA and FB ? Solution: Let FA and FB be the magnitudes of FA and FB . The component of the total force parallel to the wall must be zero. And the sum of the components perpendicular to the wall must be 200 lb. A 40⬚ 20⬚ B FA cos 55° FB cos 20° D 0 FA sin 55° C FB sin 20° D 200 lb Solving we find FA 40⬚ FA D 195 lb FB D 119 lb 20⬚ FB c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 29 Problem 2.48 The bracket must support the two forces shown, where jF1 j D jF2 j D 2 kN. An engineer determines that the bracket will safely support a total force of magnitude 3.5 kN in any direction. Assume that 0 ˛ 90° . What is the safe range of the angle ˛? F2 α F1 F2 Solution: Fx : 2 kN C 2 kN cos ˛ D 2 kN1 C cos ˛ F1 β α α F1 + F2 Fy : 2 kN sin ˛ Thus the total force has a magnitude given by F D 2 kN p 1 C cos ˛2 C sin ˛2 D 2 kN 2 C 2 cos ˛ D 3.5 kN Thus when we are at the limits we have 2 C 2 cos ˛ D 3.5 kN 2 kN 2 D 49 17 ) cos ˛ D ) ˛ D 57.9° 16 32 In order to be safe we must have 57.9° ˛ 90° 30 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.49 The figure shows three forces acting on a joint of a structure. The magnitude of Fc is 60 kN, and FA C FB C FC D 0. What are the magnitudes of FA and FB ? y FC FB 15° x 40° FA Solution: We need to write each force in terms of its components. FA 195° FA D jFA j cos 40i C jFA j sin 40j kN 40° x FB D jFB j cos 195° i C jFB j sin 195j kN FC D jFC j cos 270° i C jFC j sin 270° j kN FB Thus FC D 60j kN 270° Since FA C FB C FC D 0, their components in each direction must also sum to zero. FC FAx C FBx C FCx D 0 FAy C FBy C FCy D 0 Thus, jFA j cos 40° C jFB j cos 195° C 0 D 0 jFA j sin 40° C jFB j sin 195° 60 kN D 0 Solving for jFA j and jFB j, we get jFA j D 137 kN, jFB j D 109 kN Problem 2.50 Four forces act on a beam. The vector sum of the forces is zero. The magnitudes jFB j D 10 kN and jFC j D 5 kN. Determine the magnitudes of FA and FD . FD 30° FA FB FC Solution: Use the angles and magnitudes to determine the vectors, and then solve for the unknowns. The vectors are: FA D jFA ji cos 30° C j sin 30° D 0.866jFA ji C 0.5jFA jj FB D 0i 10j, FC D 0i C 5j, FD D jFD ji C 0j. Take the sum of each component in the x- and y-directions: Fx D 0.866jFA j jFD ji D 0 and Fy D 0.5jFA j 10 5j D 0. From the second equation we get jFA j D 10 kN . Using this value in the first equation, we get jFD j D 8.7 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 31 Problem 2.51 Six forces act on a beam that forms part of a building’s frame. The vector sum of the forces is zero. The magnitudes jFB j D jFE j D 20 kN, jFC j D 16 kN, and jFD j D 9 kN. Determine the magnitudes of FA and FG . FA 70⬚ FC FD 40⬚ 50⬚ 40⬚ FB Solution: Write each force in terms of its magnitude and direction FG FE y as F D jFj cos i C jFj sin j where is measured counterclockwise from the Cx-axis. Thus, (all forces in kN) θ FA D jFA j cos 110° i C jFA j sin 110° j kN FB D 20 cos 270° i C 20 sin 270° j kN x FC D 16 cos 140° i C 16 sin 140° j kN FD D 9 cos 40° i C 9 sin 40° j kN FE D 20 cos 270° i C 20 sin 270° j kN FG D jFG j cos 50° i C jFG j sin 50° j kN We know that the x components and y components of the forces must add separately to zero. Thus FAx C FBx C FCx C FDx C FEx C FGx D 0 FAy C FBy C FCy C FDy C FEy C FGy D 0 jFA j cos 110° C 0 12.26 C 6.89 C 0 C jFG j cos 50° D 0 jFA j sin 110° 20 C 10.28 C 5.79 20 C jFG j sin 50° D 0 Solving, we get jFA j D 13.0 kN 32 jFG j D 15.3 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.52 The total weight of the man and parasail is jWj D 230 lb. The drag force D is perpendicular to the lift force L. If the vector sum of the three forces is zero, what are the magnitudes of L and D? y Solution: Let L and D be the magnitudes of the lift and drag forces. We can use similar triangles to express the vectors L and D in terms of components. Then the sum of the forces is zero. Breaking into components we have p L 5 p 2 2 22 C 52 5 22 C 52 L p L p 5 22 C 52 2 22 C 52 DD0 D 230 lb D 0 Solving we find D jDj D 85.4 lb, jLj D 214 lb x W Problem 2.53 The three forces acting on the car are shown. The force T is parallel to the x axis and the magnitude of the force W is 14 kN. If T C W C N D 0, what are the magnitudes of the forces T and N? Solution: Fx : T N sin 20° D 0 Fy : N cos 20° 14 kN D 0 Solving we find N D 14.90 N, T D 5.10 N 20⬚ y T W x 20⬚ N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 33 Problem 2.54 The cables A, B, and C help support a pillar that forms part of the supports of a structure. The magnitudes of the forces exerted by the cables are equal: jFA j D jFB j D jFC j. The magnitude of the vector sum of the three forces is 200 kN. What is jFA j? Solution: Use the angles and magnitudes to determine the vector components, take the sum, and solve for the unknown. The angles between each cable and the pillar are: A D tan1 B D tan1 FC FA FB D 33.7° , 8 D 53.1° 6 C D tan1 4m 6m 12 6 D 63.4° . 6m A B Measure the angles counterclockwise form the x-axis. The force vectors acting along the cables are: C FA D jFA ji cos 303.7° C j sin 303.7° D 0.5548jFA ji 0.8319jFA jj 4m 4m 4m FB D jFB ji cos 323.1° C j sin 323.1° D 0.7997jFB ji 0.6004jFB jj FC D jFC ji cos 333.4° C j sin 333.4° D 0.8944jFC ji0.4472jFC jj The sum of the forces are, noting that each is equal in magnitude, is F D 2.2489jFA ji 1.8795jFA jj. The magnitude of the sum is given by the problem: 200 D jFA j 2.24892 C 1.87952 D 2.931jFA j, from which jFA j D 68.24 kN Problem 2.55 The total force exerted on the top of the mast B by the sailboat’s forestay AB and backstay BC is 180i 820j (N). What are the magnitudes of the forces exerted at B by the cables AB and BC ? Solution: We first identify the forces: FAB D TAB 4.0 mi 11.8 mj 4.0 m2 C 11.8 m2 FBC D TBC y B (4, 13) m 5.0 mi 12.0 mj 5.0 m2 C 12.0 m2 Then if we add the force we find 4 5 TAB C p TBC D 180 N Fx : p 155.24 169 11.8 12 TAB p TBC D 820 N Fy : p 155.24 169 Solving simultaneously yields: ) TAB D 226 N, A (0, 1.2) m 34 C (9, 1) m TAC D 657 N x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.56 The structure shown forms part of a truss designed by an architectural engineer to support the roof of an orchestra shell. The members AB, AC, and AD exert forces FAB , FAC , and FAD on the joint A. The magnitude jFAB j D 4 kN. If the vector sum of the three forces equals zero, what are the magnitudes of FAC and FAD ? y B (– 4, 1) m FAB FAC (4, 2) m C x A FAD D (–2, – 3) m Solution: Determine the unit vectors parallel to each force: B 2 3 eAD D p iC p j D 0.5547i 0.8320j 22 C 32 22 C 32 C A 4 1 iC p j D 0.9701i C 0.2425j eAC D p 42 C 12 42 C 12 D 4 2 iC p j D 0.89443i C 0.4472j eAB D p 42 C 22 42 C 22 The forces are FAD D jFAD jeAD , FAC D jFAC jeAC , FAB D jFAB jeAB D 3.578i C 1.789j. Since the vector sum of the forces vanishes, the x- and y-components vanish separately: Fx D 0.5547jFAD j 0.9701jFAC j C 3.578i D 0, and Fy D 0.8320jFAD j C 0.2425jFAC j C 1.789j D 0 These simultaneous equations in two unknowns can be solved by any standard procedure. An HP-28S hand held calculator was used here: The results: jFAC j D 2.108 kN , jFAD j D 2.764 kN Problem 2.57 The distance s D 45 in. Solution: (a) (a) (b) Determine the unit vector eBA that points from B toward A. Use the unit vector you obtained in (a) to determine the coordinates of the collar C. The unit vector is the position vector from B to A divided by its magnitude rBA D [14 75]i C [45 12]jin D 61i C 33j in jrBA j D y eBA D A (14, 45) in 61 in2 C 33 in2 D 69.35 in 1 61i C 33j in D 0.880i C 0.476j 69.35 in eBA D 0.880i C 0.476j C s (b) B (75, 12) in x To find the coordinates of point C we will write a vector from the origin to point C. rC D rA C rAC D rA C seBA D 75i C 12j in C 45 in0.880i C 0.476j rC D 35.4i C 33.4j in Thus the coordinates of C are C 35.4, 33.4 in c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 35 Problem 2.58 In Problem 2.57, determine the x and y coordinates of the collar C as functions of the distance s. Solution: The coordinates of the point C are given by xC D xB C s0.880 and yC D yB C s0.476. Thus, the coordinates of point C are xC D 75 0.880s in and yC D 12 C 0.476s in. Note from the solution of Problem 2.57 above, 0 s 69.4 in. Problem 2.59 The position vector r goes from point A to a point on the straight line between B and C. Its magnitude is jrj D 6 ft. Express r in terms of scalar components. y B (7, 9) ft r A (3, 5) ft C (12, 3) ft x Solution: Determine the perpendicular vector to the line BC from y point A, and then use this perpendicular to determine the angular orientation of the vector r. The vectors are B[7,9] P rAB D 7 3i C 9 5j D 4i C 4j, jrAB j D 5.6568 r rAC D 12 3i C 3 5j D 9i 2j, jrAC j D 9.2195 rBC D 12 7i C 3 9j D 5i 6j, jrBC j D 7.8102 A[3,5] C[12,3] x The unit vector parallel to BC is eBC D rBC D 0.6402i 0.7682j D i cos 50.19° j sin 50.19° . jrBC j Add š90° to the angle to find the two possible perpendicular vectors: eAP1 D i cos 140.19° j sin 140.19° , or eAP2 D i cos 39.8° C j sin 39.8° . Choose the latter, since it points from A to the line. Given the triangle defined by vertices A, B, C, then the magnitude of the perpendicular corresponds to the altitude when the base is the line 2area . From geometry, the area of BC. The altitude is given by h D base a triangle with known sides is given by area D p ss jrBC js jrAC js jrAB j, where s is the semiperimeter, s D 12 jrAC j C jrAB j C jrBC j. Substituting values, s D 11.343, and area D 22.0 and the magnitude of the 222 D 5.6333. The angle between the perpendicular is jrAP j D 7.8102 5.6333 vector r and the perpendicular rAP is ˇ D cos1 D 20.1° . Thus 6 the angle between the vector r and the x-axis is ˛ D 39.8 š 20.1 D 59.1° or 19.7° . The first angle is ruled out because it causes the vector r to lie above the vector rAB , which is at a 45° angle relative to the x-axis. Thus: r D 6i cos 19.7° C j sin 19.7° D 5.65i C 2.02j 36 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.60 Let r be the position vector from point C to the point that is a distance s meters along the straight line between A and B. Express r in terms of components. (Your answer will be in terms of s). y B (10, 9) m s r Solution: First define the unit vector that points from A to B. A (3, 4) m rB/A D [10 3]i C [9 4]j m D 7i C 5j m jrB/A j D C (9, 3) m p 7 m2 C 5 m2 D 74 m x 1 eB/A D p 7i C 5j 74 Let P be the point that is a distance s along the line from A to B. The coordinates of point P are xp D 3 m C s yp D 4 m C s 7 p 74 5 p 74 D 3 C 0.814s m D 4 C 0.581s m. The vector r that points from C to P is then r D [3 C 0.814s 9]i C [4 C 0.581s 3]j m r D [0.814s 6]i C [0.581s C 1]j m Problem 2.61 A vector U D 3i 4j 12k. What is its magnitude? Strategy: The magnitude of a vector is given in terms of its components by Eq. (2.14). Problem 2.62 The vector e D 13 i C 23 j C ez k is a unit vector. Determine the component ez . (Notice that there are two answers.) Solution: Use definition given in Eq. (14). The vector magnitude is jUj D 32 C 42 C 122 D 13 Solution: eD 2 1 i C j C ez k ) 3 3 2 2 1 2 4 C C ez 2 D 1 ) e2 D 3 3 9 Thus ez D Problem 2.63 An engineer determines that an attachment point will be subjected to a force F D 20i C Fy j 45k kN. If the attachment point will safely support a force of 80-kN magnitude in any direction, what is the acceptable range of values for Fy ? y 2 3 or ez D 2 3 Solution: 802 ½ Fx2 C Fy2 C F2z 802 ½ 202 C Fy2 C 452 To find limits, use equality. Fy2LIMIT D 802 202 452 F Fy2LIMIT D 3975 Fy LIMIT D C63.0, 63.0 kN jFy LIMIT j 63.0 kN 63.0 kN Fy 63.0 kN z x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 37 Problem 2.64 A vector U D Ux i C Uy j C Uz k. Its magnitude is jUj D 30. Its components are related by the equations Uy D 2Ux and Uz D 4Uy . Determine the components. (Notice that there are two answers.) Solution: Substitute the relations between the components, deter- U D C3.61i C 23.61j C 423.61k mine the magnitude, and solve for the unknowns. Thus D 3.61i 7.22j 28.9k U D Ux i C 2Ux j C 42Ux k D Ux 1i 2j 8k where Ux can be factored out since it is a scalar. Take the magnitude, noting that the absolute value of jUx j must be taken: U D 3.61i C 23.61j C 423.61k D 3.61i C 7.22j C 28.9k p 30 D jUx j 12 C 22 C 82 D jUx j8.31. Solving, we get jUx j D 3.612, or Ux D š3.61. The two possible vectors are Problem 2.65 An object is acted upon by two forces F1 D 20i C 30j 24k (kN) and F2 D 60i C 20j C 40k (kN). What is the magnitude of the total force acting on the object? Solution: F1 D 20i C 30j 24k kN F2 D 60i C 20j C 40k kN F D F1 C F2 D 40i C 50j C 16k kN Thus FD Problem 2.66 Two vectors U D 3i 2j C 6k and V D 4i C 12j 3k. (a) Determine the magnitudes of U and V. (b) Determine the magnitude of the vector 3U C 2V. 40 kN2 C 50 kN2 C 16 kN2 D 66 kN Solution: The magnitudes: (a) jUj D p p 32 C 22 C 62 D 7 and jVj D 42 C 122 C 32 D 13 The resultant vector 3U C 2V D 9 C 8i C 6 C 24j C 18 6k D 17i C 18j C 12k (b) 38 The magnitude j3U C 2Vj D p 172 C 182 C 122 D 27.51 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.67 In Active Example 2.6, suppose that you want to redesign the truss, changing the position of point D so that the magnitude of the vector rCD from point C to point D is 3 m. To accomplish this, let the coordinates of point D be 2, yD , 1 m, and determine the value of yD so that jrCD j D 3 m. Draw a sketch of the truss with point D in its new position. What are the new directions cosines of rCD ? D (2, 3, 1) m rCD (4, 0, 0) m Solution: The vector rCD and the magnitude jrCD j are C rCD D [2 m 4 m]i C [yD 0]j C [1 m 0]k D 2 mi C yD j z C 1 mk (a) jrCD j D x 2 m2 C yCD 2 C 1 m2 D 3 m Solving we yCD D 2 m find yCD D 3 m2 2 m2 1 m2 D 2 m The new direction cosines of rCD . cos x D 2/3 D 0.667 cos y D 2/3 D 0.667 cos z D 1/3 D 0.333 Problem 2.68 A force vector is given in terms of its components by F D 10i 20j 20k (N). Solution: F D 10i 20j 20k N (a) (b) What are the direction cosines of F? Determine the components of a unit vector e that has the same direction as F. FD 10 N2 C 20 N2 C 20 N2 D 30 N cos x D (a) 10 N D 0.333, 30 N cos z D (b) cos y D 20 N D 0.667, 30 N 20 N D 0.667 30 N e D 0.333i 0.667j 0.667k c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 39 Problem 2.69 The cable exerts a force F on the hook at O whose magnitude is 200 N. The angle between the vector F and the x axis is 40° , and the angle between the vector F and the y axis is 70° . y 70° (a) What is the angle between the vector F and the z axis? (b) Express F in terms of components. Strategy: (a) Because you know the angles between the vector F and the x and y axes, you can use Eq. (2.16) to determine the angle between F and the z axis. (Observe from the figure that the angle between F and the z axis is clearly within the range 0 < z < 180° .) (b) The components of F can be obtained with Eqs. (2.15). F 40° x O z Solution: (a) cos 40° 2 C cos 70° 2 C cos z 2 D 1 ) z D 57.0° F D 200 Ncos 40° i C cos 70° j C cos 57.0° k (b) F D 153.2i C 68.4j C 108.8k N Problem 2.70 A unit vector has direction cosines cos x D 0.5 and cos y D 0.2. Its z component is positive. Express it in terms of components. Solution: Use Eq. (2.15) and (2.16). The third direction cosine is cos z D š 1 0.52 0.22 D C0.8426. The unit vector is u D 0.5i C 0.2j C 0.8426k Problem 2.71 The airplane’s engines exert a total thrust force T of 200-kN magnitude. The angle between T and the x axis is 120° , and the angle between T and the y axis is 130° . The z component of T is positive. (a) What is the angle between T and the z axis? (b) Express T in terms of components. l D cos 120° D 0.5, m D cos 130° D 0.6428 from which the z-direction cosine is n D cosz D š 1 0.52 0.64282 D C0.5804. Thus the angle between T and the z-axis is y y Solution: The x- and y-direction cosines are (a) z D cos1 0.5804 D 54.5° , and the thrust is T D 2000.5i 0.6428j C 0.5804k, or: 130⬚ x x (b) T D 100i 128.6j C 116.1k (kN) 120⬚ T z 40 z c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.72 Determine the components of the position vector rBD from point B to point D. Use your result to determine the distance from B to D. Solution: We have the following B5, 0, 3 m, C6, 0, 0 m, D4, 3, 1 m coordinates: A0, 0, 0, rBD D 4 m 5 mi C 3 m 0j C 1 m 3 mk y D (4, 3, 1) m rBD D i C 3j 2k m D 1 m2 C 3 m2 C 2 m2 D 3.74 m A C (6, 0, 0) m x z B (5, 0, 3) m Problem 2.73 What are the direction cosines of the position vector rBD from point B to point D? Solution: cos x D 1 m D 0.267, 3.74 m cos z D Problem 2.74 Determine the components of the unit vector eCD that points from point C toward point D. cos y D 3m D 0.802, 3.74 m 2 m D 0.535 3.74 m Solution: We have the following B5, 0, 3 m, C6, 0, 0 m, D4, 3, 1 m coordinates: A0, 0, 0, rCD D 4 m 6 mi C 3 m 0j C 1 m 0k D 2i C 3j C 1k rCD D 2 m2 C 3 m2 C 1 m2 D 3.74 m Thus eCD D 1 2i C 3j C k m D 0.535i C 0.802j C 0.267k 3.74 m Problem 2.75 What are the direction cosines of the unit vector eCD that points from point C toward point D? Solution: Using Problem 2.74 cos x D 0.535, cos y D 0.802, cos z D 0.267 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 41 Problem 2.76 In Example 2.7, suppose that caisson shifts on the ground to a new position. magnitude of the force F remains 600 lb. In the position, the angle between the force F and the x is 60° and the angle between F and the z axis is Express F in terms of components. the The new axis 70° . y 40⬚ F x 54⬚ Solution: We need to find the angle y between the force F and the y axis. We know that z cos2 x C cos2 y C cos2 z D 1 cos y D š 1 cos2 x cos2 z D š 1 cos2 60° cos2 70° D š0.7956 y D š cos1 0.7956 D 37.3° or 142.7° We will choose y D 37.3° because the picture shows the force pointing up. Now Fx D 600 lb cos 60° D 300 lb Fy D 600 lb cos 37.3° D 477 lb Fz D 600 lb cos 70° D 205 lb Thus F D 300i C 477j C 205k lb Problem 2.77 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector rA from the shuttle to satellite A has magnitude 2 km, and direction cosines cos x D 0.768, cos y D 0.384, cos z D 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos x D 0.743, cos y D 0.557, cos z D 0.371. What is the distance between the satellites? B rB x Solution: The two position vectors are: A y rA rA D 20.768iC 0.384jC 0.512k D 1.536i C 0.768j C 1.024k (km) rB D 40.743iC 0.557j 0.371k D 2.972i C 2.228j 1.484k (km) z The distance is the magnitude of the difference: jrA rB j D 1.5362.9272 C 0.7682.2282 C 1.0241.4842 D 3.24 (km) 42 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.78 Archaeologists measure a pre-Columbian ceremonial structure and obtain the dimensions shown. Determine (a) the magnitude and (b) the direction cosines of the position vector from point A to point B. y 4m 10 m 4m 10 m A 8m Solution: (a) B The coordinates are A (0, 16, 14) m and B (10, 8, 4) m. b rAB D [10 0]i C [8 16]j C [4 14]k m D 10i 8j 10k m jrAB j D p 102 C 82 C 102 m D 264 m D 16.2 m 8m z C x jrAB j D 16.2 m (b) 10 D 0.615 cos x D p 264 8 D 0.492 cos y D p 264 10 cos z D p D 0.615 264 Problem 2.79 Consider the structure described in Problem 2.78. After returning to the United States, an archaeologist discovers that a graduate student has erased the only data file containing the dimension b. But from recorded GPS data he is able to calculate that the distance from point B to point C is 16.61 m. y 10 m 4m 4m 10 m A 8m B (a) (b) What is the distance b? Determine the direction cosines of the position vector from B to C. b 8m z C x Solution: We have the coordinates B (10 m, 8 m, 4 m), C (10 m C b, 0 18 m). rBC D 10 m C b 10 mi C 0 8 mj C 18 m 4 mk rBC D bi C 8 mj C 14 mk (a) (b) We have 16.61 m2 D b2 C 8 m2 C 14 m2 ) b D 3.99 m The direction cosines of rBC are 3.99 m D 0.240 16.61 m 8 m cos y D D 0.482 16.61 m 14 m cos z D D 0.843 16.61 m cos x D c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 43 Problem 2.80 Observers at A and B use theodolites to measure the direction from their positions to a rocket in flight. If the coordinates of the rocket’s position at a given instant are (4, 4, 2) km, determine the direction cosines of the vectors rAR and rBR that the observers would measure at that instant. y rAR Solution: The vector rAR is given by rBR A rAR D 4i C 4j C 2k km x and the magnitude of rAR is given by jrAR j D z 42 C 42 C 22 B (5, 0, 2) km km D 6 km. The unit vector along AR is given by uAR D rAR /jrAR j. Thus, uAR D 0.667i C 0.667j C 0.333k and the direction cosines are cos x D 0.667, cos y D 0.667, and cos z D 0.333. The vector rBR is given by rBR D xR xB i C yR yB j C zR zB k km D 4 5i C 4 0j C 2 2k km and the magnitude of rBR is given by jrBR j D 12 C 42 C 02 km D 4.12 km. The unit vector along BR is given by eBR D rBR /jrBR j. Thus, uBR D 0.242i C 0.970j C 0k and the direction cosines are cos x D 0.242, cos y D 0.970, and cos z D 0.0. 44 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.81 In Problem 2.80, suppose that the coordinates of the rocket’s position are unknown. At a given instant, the person at A determines that the direction cosines of rAR are cos x D 0.535, cos y D 0.802, and cos z D 0.267, and the person at B determines that the direction cosines of rBR are cos x D 0.576, cos y D 0.798, and cos z D 0.177. What are the coordinates of the rocket’s position at that instant. Solution: The vector from A to B is given by rAB D xB xA i C yB yA j C zB zA k or Similarly, the vector along BR, uBR D 0.576i C 0.798 0.177k. From the diagram in the problem statement, we see that rAR D rAB C rBR . Using the unit vectors, the vectors rAR and rBR can be written as rAB D 5 0i C 0 0j C 2 0k D 5i C 2k km. The magnitude of rAB is given by jrAB j D The unit vector along AB, uAB , is given by rAR D 0.535rAR i C 0.802rAR j C 0.267rAR k, and 52 C 22 D 5.39 km. uAB D rAB /jrAB j D 0.928i C 0j C 0.371k km. The unit vector along the line AR, uAR D cos x i C cos y j C cos z k D 0.535i C 0.802j C 0.267k. rBR D 0.576rBR i C 0.798rBR j 0.177rBR k. Substituting into the vector addition rAR D rAB C rBR and equating components, we get, in the x direction, 0.535rAR D 0.576rBR , and in the y direction, 0.802rAR D 0.798rBR . Solving, we get that rAR D 4.489 km. Calculating the components, we get rAR D rAR eAR D 0.5354.489i C 0.8024.489j C 0.2674.489k. Hence, the coordinates of the rocket, R, are (2.40, 3.60, 1.20) km. Problem 2.82* The height of Mount Everest was originally measured by a surveyor in the following way. He first measured the altitudes of two points and the horizontal distance between them. For example, suppose that the points A and B are 3000 m above sea level and are 10,000 m apart. He then used a theodolite to measure the direction cosines of the vector rAP from point A to the top of the mountain P and the vector rBP from point B to P. Suppose that the direction cosines of rAP are cos x D 0.5179, cos y D 0.6906, and cos z D 0.5048, and the direction cosines of rBP are cos x D 0.3743, cos y D 0.7486, and cos z D 0.5472. Using this data, determine the height of Mount Everest above sea level. z P y B x A Solution: We have the following coordinates A0, 0, 3000 m, B10, 000, 0, 3000 m, Px, y, z Then rAP D xi C yj C z 3000 mk D rAP 0.5179i C 0.6906j C 0.5048k rBP D x 10,000 mi C yj C z 3000 mk D rBP 0.3743i C 0.7486j C 0.5472k Equating components gives us five equations (one redundant) which we can solve for the five unknowns. x D rAP 0.5179 y D rAP 0.6906 z 3000 m D rAP 0.5048 ) z D 8848 m x 10000 m D rBP 0.7486 y D rBP 0.5472 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 45 y Problem 2.83 The distance from point O to point A is 20 ft. The straight line AB is parallel to the y axis, and point B is in the x-z plane. Express the vector rOA in terms of scalar components. A rOA Strategy: You can resolve rOA into a vector from O to B and a vector from B to A. You can then resolve the vector form O to B into vector components parallel to the x and z axes. See Example 2.8. O x 30° 60° B z Solution: See Example 2.8. The length BA is, from the right triangle The vector rOA is given by rOA D rOB C rBA , from which OAB, rOA D 15i C 10j C 8.66k (ft) jrAB j D jrOA j sin 30° D 200.5 D 10 ft. Similarly, the length OB is A jrOB j D jrOA j cos 30° D 200.866 D 17.32 ft The vector rOB can be resolved into components along the axes by the right triangles OBP and OBQ and the condition that it lies in the x-z plane. Hence, rOB D jrOB ji cos 30° C j cos 90° C k cos 60° rOA y 30° O x Q z P 60° B or rOB D 15i C 0j C 8.66k. The vector rBA can be resolved into components from the condition that it is parallel to the y-axis. This vector is rBA D jrBA ji cos 90° C j cos 0° C k cos 90° D 0i C 10j C 0k. y Problem 2.84 The magnitudes of the two force vectors are jFA j D 140 lb and jFB j D 100 lb. Determine the magFB nitude of the sum of the forces FA C FB . FA Solution: We have the vectors 60⬚ FA D 140 lb[cos 40° sin 50° ]i C [sin 40° ]j C [cos 40° cos 50° ]k 30⬚ 40⬚ x 50⬚ FA D 82.2i C 90.0j C 68.9k lb z FB D 100 lb[ cos 60° sin 30° ]i C [sin 60° ]j C [cos 60° cos 30° ]k FB D 25.0i C 86.6j C 43.3k lb Adding and taking the magnitude we have FA C FB D 57.2i C 176.6j C 112.2k lb jFA C FB j D 57.2 lb2 C 176.6 lb2 C 112.2 lb2 D 217 lb jFA C FB j D 217 lb 46 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.85 Determine the direction cosines of the vectors FA and FB . FB Solution: We have the vectors FA FA D 140 lb[cos 40° sin 50° ]i C [sin 40° ]j C [cos 40° cos 50° ]k FA D 82.2i C 90.0j C 68.9k lb 60⬚ FB D 100 lb[ cos 60° sin 30° ]i C [sin 60° ]j C [cos 60° cos 30° ]k 30⬚ FB D 25.0i C 86.6j C 43.3k lb 40⬚ x 50⬚ z The direction cosines for FA are cos x D 82.2 lb 90.0 lb D 0.587, cos y D D 0.643, 140 lb 140 lb cos z D 68.9 lb D 0.492 140 lb The direction cosines for FB are cos x D 25.0 lb 86.6 lb D 0.250, cos y D D 0.866, 100 lb 100 lb cos z D 43.3 lb D 0.433 100 lb FA : cos x D 0.587, cos y D 0.643, cos z D 0.492 FB : cos x D 0.250, cos y D 0.866, cos z D 0.433 Problem 2.86 In Example 2.8, suppose that a change in the wind causes a change in the position of the balloon and increases the magnitude of the force F exerted on the hook at O to 900 N. In the new position, the angle between the vector component Fh and F is 35° , and the angle between the vector components Fh and Fz is 40° . Draw a sketch showing the relationship of these angles to the components of F. Express F in terms of its components. y B F O O z x A Solution: We have jFy j D 900 N sin 35° D 516 N jFh j D 900 N cos 35° D 737 N jFx j D jFh j sin 40° D 474 N jFz j D jFh j cos 40° D 565 N Thus F D 474i C 516j C 565k N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 47 Problem 2.87 An engineer calculates that the magnitude of the axial force in one of the beams of a geodesic dome is jPj D 7.65 kN. The cartesian coordinates of the endpoints A and B of the straight beam are (12.4, 22.0, 18.4) m and (9.2, 24.4, 15.6) m, respectively. Express the force P in terms of scalar components. Solution: The components of the position vector from B to A are rBA D xA xB i C yA yB j C zA zB k D 12.4 C 9.2i C 22.0 24.4j C 18.4 C 15.6k D 3.2i 2.4j 2.8k m. B P A Dividing this vector by its magnitude, we obtain a unit vector that points from B toward A: eBA D 0.655i 0.492j 0.573k. Therefore P D jPjeBA D 7.65 eBA D 5.01i 3.76j 4.39k kN. y Problem 2.88 The cable BC exerts an 8-kN force F on the bar AB at B. B (5, 6, 1) m (a) Determine the components of a unit vector that points from B toward point C. (b) Express F in terms of components. F A Solution: (a) eBC D x xC xB i C yC yB j C zC zB k rBC D jrBC j xC xB 2 C yC yB 2 C zC zB 2 C (3, 0, 4) m z 2i 6j C 3k 2 6 3 eBC D p D i jC k 7 7 7 22 C 62 C 32 eBC D 0.286i 0.857j C 0.429k (b) 48 F D jFjeBC D 8eBC D 2.29i 6.86j C 3.43k kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.89 A cable extends from point C to point E. It exerts a 50-lb force T on plate C that is directed along the line from C to E. Express T in terms of components. 6 ft E A D z 4 ft T 2 ft 20⬚ B x C 4 ft y Solution: Find the unit vector eCE and multiply it times the magnitude of the force to get the vector in component form, eCE D 6 ft xE xC i C yE yC j C zE zC k rCE D jrCE j xE xC 2 C yE yC 2 C zE zC 2 A E The coordinates of point C are 4, 4 sin 20° , 4 cos 20° or 4, 1.37, 3.76 (ft) The coordinates of point E are (0, 2, 6) (ft) D T 2 ft eCE 0 4i C 2 1.37j C 6 3.76k p D 42 C 3.372 C 2.242 x 4 ft T z B 20° C 4 ft eCE D 0.703i C 0.592j C 0.394k T D 50eCE lb T D 35.2i C 29.6j C 19.7k lb Problem 2.90 In Example 2.9, suppose that the metal loop at A is moved upward so that the vertical distance to A increases from 7 ft to 8 ft. As a result, the magnitudes of the forces FAB and FAC increase to jFAB j D jFAC j D 240 lb. What is the magnitude of the total force F D FAB C FAC exerted on the loop by the rope? Solution: The new coordinates of point A are (6, 8, 0) ft. The position vectors are rAB D 4i 8j C 4k ft rAC D 4i 8j C 6k ft The forces are 6 ft A FAB D 240 lb rAB D 98.0i 196j C 98.0k lb jrAB j FAC D 240 lb rAC D 89.1i 178j C 134.0k lb jrAC j 7 ft 4 ft 6 ft B The sum of the forces is C 2 ft F D FAB C FAC D 8.85i 374j C 232k lb 10 ft The magnitude is jFj D 440 lb y 6 ft A FAB FAC 7 ft 4 ft 2 ft x B 6 ft C 10 ft z c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 49 y Problem 2.91 The cable AB exerts a 200-lb force FAB at point A that is directed along the line from A to B. Express FAB in terms of components. 8 ft C 8 ft Solution: The coordinates of B are B(0,6,8). The position vector from A to B is 6 ft B x rAB D 0 6i C 6 0j C 8 10k D 6i C 6j 2k The magnitude is jrAB j D p 62 C 62 C 22 D 8.718 ft. FAB The unit vector is uAB D 6 2 6 iC j k 8.718 8.718 8.718 z FAC A (6, 0, 10) ft or uAB D 0.6882i C 0.6882j 0.2294k. FAB D jFAB juAB D 2000.6882i C 0.6882j 0.2294k The components of the force are FAB D jFAB juAB D 2000.6882i C 0.6882j 0.2294k or FAB D 137.6i C 137.6j 45.9k Problem 2.92 Consider the cables and wall described in Problem 2.91. Cable AB exerts a 200-lb force FAB at point A that is directed along the line from A to B. The cable AC exerts a 100-lb force FAC at point A that is directed along the line from A to C. Determine the magnitude of the total force exerted at point A by the two cables. Solution: Refer to the figure in Problem 2.91. From Problem 2.91 the force FAB is The force is FAC D jFAC juAC D 100uAC D 16.9i C 50.7j 84.5k. FAB D 137.6i C 137.6j 45.9k The resultant of the two forces is The coordinates of C are C(8,6,0). The position vector from A to C is FR D FAB C FAC D 137.6 C 16.9i C 137.6 C 50.7j rAC D 8 6i C 6 0j C 0 10k D 2i C 6j 10k. The magnitude is jrAC j D The unit vector is uAC D C 84.5 45.9k. p 22 C 62 C 102 D 11.83 ft. 6 10 2 iC j k D 0.1691i C 0.5072j 0.8453k. 11.83 11.83 11.83 FR D 120.7i C 188.3j 130.4k. The magnitude is jFR j D 50 p 120.72 C 188.32 C 130.42 D 259.0 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.93 The 70-m-tall tower is supported by three cables that exert forces FAB , FAC , and FAD on it. The magnitude of each force is 2 kN. Express the total force exerted on the tower by the three cables in terms of components. A y FAD A FAB FAC D Solution: The coordinates of the points are A (0, 70, 0), B (40, 0, 0), C (40, 0, 40) D (60, 0, 60). 60 m 60 m B The position vectors corresponding to the cables are: x 40 m rAD D 60 0i C 0 70j C 60 0k C 40 m 40 m z rAD D 60i 70k 60k rAC D 40 0i C 0 70j C 40 0k rAC D 40i 70j C 40k rAB D 40 0i C 0 70j C 0 0k rAB D 40i 70j C 0k The unit vectors corresponding to these position vectors are: uAD D 60 70 60 rAD D i j k jrAD j 110 110 110 D 0.5455i 0.6364j 0.5455k uAC D 40 70 40 rAC D i jC k jrAC j 90 90 90 D 0.4444i 0.7778j C 0.4444k uAB D 40 70 rAB D i j C 0k D 0.4963i 0.8685j C 0k jrAB j 80.6 80.6 The forces are: FAB D jFAB juAB D 0.9926i 1.737j C 0k FAC D jFAC juAC D 0.8888i 1.5556j C 0.8888 FAD D jFAD juAD D 1.0910i 1.2728j 1.0910k The resultant force exerted on the tower by the cables is: FR D FAB C FAC C FAD D 0.9875i 4.5648j 0.2020k kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 51 Problem 2.94 Consider the tower described in Problem 2.93. The magnitude of the force FAB is 2 kN. The x and z components of the vector sum of the forces exerted on the tower by the three cables are zero. What are the magnitudes of FAC and FAD ? Solution: From the solution of Problem 2.93, the unit vectors are: uAC D 40 70 40 rAC D i jC k jrAC j 90 90 90 Taking the sum of the forces: FR D FAB C FAC C FAD D 0.9926 0.4444jFAC j 0.5455jFAD ji C 1.737 0.7778jFAC j 0.6364jFAD jj D 0.4444i 0.7778j C 0.4444k uAD D 60 70 60 rAD D i j jrAD j 110 110 110 C 0.4444jFAC j 0.5455jFAD jk The sum of the x- and z-components vanishes, hence the set of simultaneous equations: D 0.5455i 0.6364j 0.5455k From the solution of Problem 2.93 the force FAB is FAB D jFAB juAB D 0.9926i 1.737j C 0k The forces FAC and FAD are: FAC D jFAC juAC D jFAC j0.4444i 0.7778j C 0.4444k FAD D jFAD juAD D jFAD j0.5455i 0.6364j 0.5455k 0.4444jFAC j C 0.5455jFAD j D 0.9926 and 0.4444jFAC j 0.5455jFAD j D 0 These can be solved by means of standard algorithms, or by the use of commercial packages such as TK Solver Plus or Mathcad. Here a hand held calculator was used to obtain the solution: jFAC j D 1.1163 kN Problem 2.95 In Example 2.10, suppose that the distance from point C to the collar A is increased from 0.2 m to 0.3 m, and the magnitude of the force T increases to 60 N. Express T in terms of its components. jFAD j D 0.9096 kN y 0.15 m 0.4 m B C Solution: The position vector from C to A is now rCA D 0.3 meCD D 0.137i 0.205j C 0.171km The position vector form the origin to A is T A 0.2 m 0.5 m rOA D rOC C rCA D 0.4i C 0.3j m C 0.137i 0.205j C 0.171k m O rOA D 0.263i C 0.0949j C 0.171k m The coordinates of A are (0.263, 0.0949, 0.171) m. The position vector from A to B is 0.3 m x D 0.25 m 0.2 m z rAB D [0 0.263]i C [0.5 0.0949]j C [0.15 0.171]k m rAB D 0.263i C 0.405j 0.209k m The force T is T D 60 N rAB D 32.7i C 50.3j 2.60k N jrAB j T D 32.7i C 50.3j 2.60k N 52 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.96 The cable AB exerts a 32-lb force T on the collar at A. Express T in terms of components. 4 ft B Solution: The coordinates of point B are B (0, 7, 4). The vector T 6 ft position of B is rOB D 0i C 7j C 4k. The vector from point A to point B is given by 7 ft rAB D rOB rOA . A x From Problem 2.95, rOA D 2.67i C 2.33j C 2.67k. Thus 4 ft rAB D 0 2.67i C 7 2.33j C 4 2.67j 4 ft z rAB D 2.67i C 4.67j C 1.33k. The magnitude is jrAB j D p 2.672 C 4.672 C 1.332 D 5.54 ft. The unit vector pointing from A to B is uAB D rAB D 0.4819i C 0.8429j C 0.2401k jrAB j The force T is given by TAB D jTAB juAB D 32uAB D 15.4i C 27.0j C 7.7k (lb) Problem 2.97 The circular bar has a 4-m radius and lies in the x-y plane. Express the position vector from point B to the collar at A in terms of components. y Solution: From the figure, the point B is at (0, 4, 3) m. The coordinates of point A are determined by the radius of the circular bar and the angle shown in the figure. The vector from the origin to A is rOA D 4 cos20° i C 4 sin20° j m. Thus, the coordinates of point A are (3.76, 1.37, 0) m. The vector from B to A is given by rBA D xA xB i C yA yB j C zA zB k D 3.76i 2.63j 3k m. Finally, the scalar components of the vector from B to A are (3.76, 2.63, 3) m. 3m B A 4m 20° 4m x z c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 53 Problem 2.98 The cable AB in Problem 2.97 exerts a 60-N force T on the collar at A that is directed along the line from A toward B. Express T in terms of components. Solution: We know rBA D 3.76i 2.63j 3k m from Problem 2.97. The unit vector uAB D rBA /jrBA j. The unit vector is uAB D 0.686i C 0.480j C 0.547k. Hence, the force vector T is given by T D jTj0.686iC 0.480jC 0.547k N D 41.1i C 28.8j C 32.8k N Problem 2.99 In Active Example 2.11, suppose that the vector V is changed to V D 4i 6j 10k. (a) (b) What is the value of UžV? What is the angle between U and V when they are placed tail to tail? Solution: From Active Example 2.4 we have the expression for U. Thus U D 6i 5j 3k, V D 4i 6k 10k U Ð V D 64 C 56 C 310 D 84 cos D 84 UÐV D D 0.814 jVjjVj 62 C 52 C 32 42 C 62 C 102 D cos1 0.814 D 35.5° a U Ð V D 84, b D 35.5° Problem 2.100 In Example 2.12, suppose that the coordinates of point B are changed to (6, 4, 4) m. What is the angle between the lines AB and AC? Solution: Using the new coordinates we have rAB D 2i C j C 2k m, jrAB j D 3 m rAC D 4i C 5j C 2k m, jrAC j D 6.71 m y C (8, 8, 4) m u A (4, 3, 2) m B (6, 1, ⫺2) m x cos D 24 C 15 C 22 m2 rAB Ð rAC D D 0.845 jrAB jjrAC j 3 m6.71 m D cos1 0.845 D 32.4° D 32.4° z Problem 2.101 What is the dot product of the position vector r D 10i C 25j (m) and the force vector Solution: Use Eq. (2.23). F Ð r D 30010 C 25025 C 3000 D 3250 N-m F D 300i C 250j C 300k N? Problem 2.102 Suppose that the dot product of two vectors U and V is U Ð V D 0. If jUj 6D 0, what do you know about the vector V? 54 Solution: Either jVj D 0 or V ? U c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.103 Two perpendicular vectors are given in terms of their components by Solution: When the vectors are perpendicular, U Ð V 0. Thus U D Ux i 4j C 6k U Ð V D Ux Vx C Uy Vy C Uz Vz D 0 and V D 3i C 2j 3k. Use the dot product to determine the component Ux . D 3Ux C 42 C 63 D 0 3Ux D 26 Ux D 8.67 Problem 2.104 Three vectors Solution: For mutually perpendicular vectors, we have three equations, i.e., U D Ux i C 3j C 2k UÐVD0 V D 3i C Vy j C 3k W D 2i C 4j C Wz k are mutually perpendicular. Use the dot product to determine the components Ux , Vy , and Wz . UÐWD0 VÐWD0 Thus 3Ux C 3Vy C 6 D 0 3 Eqns 2Ux C 12 C 2Wz D 0 3 Unknowns C6 C 4Vy C 3Wz D 0 Solving, we get Ux Vy Wz Problem 2.105 The magnitudes jUj D 10 and jVj D 20. (a) (b) Use the definition of the dot product to determine U Ð V. Use Eq. (2.23) to obtain U Ð V. Solution: (a) D 2.857 D 0.857 D 3.143 y V U 45⬚ The definition of the dot product (Eq. (2.18)) is 30⬚ x U Ð V D jUjjVj cos . Thus U Ð V D 1020 cos45° 30° D 193.2 (b) The components of U and V are U D 10i cos 45° C j sin 45° D 7.07i C 7.07j V D 20i cos 30° C j sin 30° D 17.32i C 10j From Eq. (2.23) U Ð V D 7.0717.32 C 7.0710 D 193.2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 55 Problem 2.106 By evaluating the dot product U Ð V, prove the identity cos1 2 D cos 1 cos 2 C sin 1 sin 2 . y Strategy: Evaluate the dot product both by using Eq. (2.18) and by using Eq. (2.23). U V u1 Solution: The strategy is to use the definition Eq. (2.18) and the u2 x Eq. (2.23). From Eq. (2.18) and the figure, U Ð V D jUjjVj cos1 2 . From Eq. (2.23) and the figure, U D jUji cos 1 C j sin 2 , V D jVji cos 2 C j sin 2 , and the dot product is U Ð V D jUjjVjcos 1 cos 2 C sin 1 sin 2 . Equating the two results: U Ð V D jUjjVj cos1 2 D jUjjVjcos 1 cos 2 C sin 1 sin 2 , from which if jUj 6D 0 and jVj 6D 0, it follows that cos1 2 D cos 1 cos 2 C sin 1 sin 2 , Q.E.D. Problem 2.107 Use the dot product to determine the angle between the forestay (cable AB) and the backstay (cable BC). y B (4, 13) m Solution: The unit vector from B to A is eBA D rBA D 0.321i 0.947j jrBA j The unit vector from B to C is eBC D rBC D 0.385i 0.923j jrBC j From the definition of the dot product, eBA Ð eBC D 1 Ð 1 Ð cos , where is the angle between BA and BC. Thus A (0, 1.2) m C (9, 1) m x cos D 0.3210.385 C 0.9470.923 cos D 0.750 D 41.3° 56 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.108 Determine the angle between the lines AB and AC (a) by using the law of cosines (see Appendix A); (b) by using the dot product. B (4, 3, ⫺1) m Solution: (a) A We have the distances: AB D AC D BC D 42 C 32 C 12 m D x u p 26 m (5, ⫺1, 3) m C z p 52 C 12 C 32 m D 35 m 5 42 C 1 32 C 3 C 12 m D p 33 m The law of cosines gives BC2 D AB2 C AC2 2ABAC cos cos D (b) AB2 C AC2 BC2 D 0.464 2ABAC ) D 62.3° Using the dot product rAB D 4i C 3j k m, rAC D 5i j C 3k m rAB Ð rAC D 4 m5 m C 3 m1 m C 1 m3 m D 14 m2 rAB Ð rAC D ABAC cos Therefore 14 m2 p D 0.464 ) D 62.3° cos D p 26 m 35 m Problem 2.109 The ship O measures the positions of the ship A and the airplane B and obtains the coordinates shown. What is the angle between the lines of sight OA and OB? y B (4, 4, ⫺4) km Solution: From the coordinates, the position vectors are: u x rOA D 6i C 0j C 3k and rOB D 4i C 4j 4k O The dot product: rOA Ð rOB D 64 C 04 C 34 D 12 The magnitudes: jrOA j D p 62 C 02 C 32 D 6.71 km and A z (6, 0, 3) km p jrOA j D 42 C 42 C 42 D 6.93 km. rOA Ð rOB D 0.2581, from which D š75° . jrOA jjrOB j From the problem and the construction, only the positive angle makes sense, hence D 75° From Eq. (2.24) cos D c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 57 Problem 2.110 Astronauts on the space shuttle use radar to determine the magnitudes and direction cosines of the position vectors of two satellites A and B. The vector rA from the shuttle to satellite A has magnitude 2 km and direction cosines cos x D 0.768, cos y D 0.384, cos z D 0.512. The vector rB from the shuttle to satellite B has magnitude 4 km and direction cosines cos x D 0.743, cos y D 0.557, cos z D 0.371. What is the angle between the vectors rA and rB ? B rB x θ Solution: The direction cosines of the vectors along rA and rB are the components of the unit vectors in these directions (i.e., uA D cos x i C cos y j C cos z k, where the direction cosines are those for rA ). Thus, through the definition of the dot product, we can find an expression for the cosine of the angle between rA and rB . y rA A z cos D cos xA cos xB C cos yA cos yB C cos zA cos zB . Evaluation of the relation yields cos D 0.594 ) D 53.5° . Problem 2.111 In Example 2.13, if you shift your position and the coordinates of point A where you apply the 50-N force become (8, 3, 3) m, what is the vector component of F parallel to the cable OB? y A (6, 6, –3) m F Solution: We use the following vectors to define the force F. O rOA D 8i C 3j 3k m eOA z x (10, ⫺2, 3) m B rOA D 0.833i C 0.331j 0.331k D jrOA j F D 50 NeOA D 44.2i C 16.6j 16.6k N Now we need the unit vector eOB . rOB D 10i 2j C 3k m eOB D rOB D 0.941i 0.188j C 0.282k jrOB j To find the vector component parallel to OB we use the dot product in the following manner F Ð eOB D 44.2 N0.941 C 16.6 N0.188 C 16.6 N0.282 D 33.8 N Fp D F Ð eOB eOB D 33.8 N0.941i 0.188j C 0.282k Fp D 31.8i 6.35j C 9.53k N 58 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.112 The person exerts a force F D 60i 40j (N) on the handle of the exercise machine. Use Eq. (2.26) to determine the vector component of F that is parallel to the line from the origin O to where the person grips the handle. r D 250i C 200j 150k mm, jrj D 354 mm To produce the unit vector that is parallel to this line we divide by the magnitude 150 mm y Solution: The vector r from the O to where the person grips the handle is eD r 250i C 200j 150k mm D D 0.707i C 0.566j 0.424k jrj 354 mm Using Eq. (2.26), we find that the vector component parallel to the line is F O Fp D e Ð Fe D [0.70760 N C 0.56640 N]0.707i 200 mm z C 0.566j 0.424k 250 mm Fp D 14.0i C 11.2j C 8.4k N x Problem 2.113 At the instant shown, the Harrier’s thrust vector is T D 17,000i C 68,000j 8,000k (N) and its velocity vector is v D 7.3i C 1.8j 0.6k (m/s). The quantity P D jTp jjvj, where Tp is the vector component of T parallel to v, is the power currently being transferred to the airplane by its engine. Determine the value of P. Solution: y v T T D 17,000i C 68,000j 8,000k N v D 7.3i C 1.8j 0.6k m/s x Power D T Ð v D 17,000 N7.3 m/s C 68,000 N1.8 m/s C 8,000 N0.6 m/s Power D 251,000 Nm/s D 251 kW c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 59 Problem 2.114 Cables extend from A to B and from A to C. The cable AC exerts a 1000-lb force F at A. y A (a) What is the angle between the cables AB and AC? (b) Determine the vector component of F parallel to the cable AB. (0, 7, 0) ft F x Solution: Use Eq. (2.24) to solve. (a) From the coordinates of the points, the position vectors are: rAB D 0 0i C 0 7j C 10 0k B (0, 0, 10) ft z C (14, 0, 14) ft rAB D 0i 7j C 10k rAC D 14 0i C 0 7j C 14 0k rAC D 14i 7j C 14k The magnitudes are: jrAB j D p 72 C 102 D 12.2 (ft) and jrAB j D p 142 C 72 C 142 D 21. The dot product is given by rAB Ð rAC D 140 C 77 C 1014 D 189. The angle is given by cos D 189 D 0.7377, 12.221 from which D š42.5° . From the construction: D C42.5° (b) The unit vector associated with AB is eAB D rAB D 0i 0.5738j C 0.8197k. jrAB j The unit vector associated with AC is eAC D rAC jrAC j D 0.6667i 0.3333j C 0.6667k. Thus the force vector along AC is FAC D jFjeAC D 666.7i 333.3j C 666.7k. The component of this force parallel to AB is FAC Ð eAB eAB D 737.5eAB D 0i 422.8j C 604.5k (lb) 60 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.115 Consider the cables AB and AC shown in Problem 2.114. Let rAB be the position vector from point A to point B. Determine the vector component of rAB parallel to the cable AC. Solution: From Problem 2.114, rAB D 0i 7j C 10k, and eAC D 0.6667i 0.3333j C 0.6667k. Thus rAB Ð eAC D 9, and rAB Ð eAC eAC D 6i 3j C 6k ft. Problem 2.116 The force F D 10i C 12j 6k N. Determine the vector components of F parallel and normal to line OA. y A (0, 6, 4) m Solution: Find eOA F O rOA D jrOA j x Then z FP D F Ð eOA eOA and FN D F FP eOA D 0i C 6j C 4k 6j C 4k p D p 52 62 C 42 eOA D 6 4 jC k D 0.832j C 0.555k 7.21 7.21 FP D [10i C 12j 6k Ð 0.832j C 0.555k]eOA FP D [6.656]eOA D 0i C 5.54j C 3.69k N FN D F FP FN D 10i C 12 5.54j C 6 3.69k FN D 10i C 6.46j 9.69k N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 61 Problem 2.117 The rope AB exerts a 50-N force T on collar A. Determine the vector component of T parallel to the bar CD. y 0.15 m Solution: We have the following vectors 0.4 m B C rCD D 0.2i 0.3j C 0.25k m eCD D T rCD D 0.456i 0.684j C 0.570k jrCD j 0.2 m 0.3 m A 0.5 m rOB D 0.5j C 0.15k m O x 0.25 m D rOC D 0.4i C 0.3j m 0.2 m rOA D rOC C 0.2 meCD D 0.309i C 0.163j C 0.114k m z rAB D rOB rOA D 0.309i C 0.337j C 0.036k m eAB D rAB D 0.674i C 0.735j C 0.079k jrAB j We can now write the force T and determine the vector component parallel to CD. T D 50 NeAB D 33.7i C 36.7j C 3.93k N Tp D eCD Ð TeCD D 3.43i C 5.14j 4.29k N Tp D 3.43i C 5.14j 4.29k N Problem 2.118 In Problem 2.117, determine the vector component of T normal to the bar CD. y 0.15 m Solution: From Problem 2.117 we have 0.4 m B C T D 33.7i C 36.7j C 3.93k N T Tp D 3.43i C 5.14j 4.29k N A The normal component is then 0.5 m O Tn D T T p x D Tn D 37.1i C 31.6j C 8.22k N 62 0.2 m 0.3 m 0.25 m 0.2 m z c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.119 The disk A is at the midpoint of the sloped surface. The string from A to B exerts a 0.2-lb force F on the disk. If you express F in terms of vector components parallel and normal to the sloped surface, what is the component normal to the surface? B (0, 6, 0) ft F 2 ft A x 8 ft 10 ft z 2 Solution: Consider a line on the sloped surface from A perpendicular to the surface. (see the diagram above) By SIMILAR triangles we see that one such vector is rN D 8j C 2k. Let us find the component of F parallel to this line. The unit vector in the direction normal to the surface is eN D y 8 rN 8j C 2k D 0.970j C 0.243k D p jrN j 82 C 22 2 The unit vector eAB can be found by z xB xA i C yB yA j C zB zA h eAB D xB xA 2 C yB yA 2 C zB zA 2 8 Point B is at (0, 6, 0) (ft) and A is at (5, 1, 4) (ft). Substituting, we get eAB D 0.615i C 0.615j 0.492k Now F D jFjeAB D 0.2eAB F D 0.123i C 0.123j 0.0984k lb The component of F normal to the surface is the component parallel to the unit vector eN . FNORMAL D F Ð eN eN D 0.955eN FNORMAL D 0i C 0.0927j C 0.0232k lb Problem 2.120 In Problem 2.119, what is the vector component of F parallel to the surface? Solution: From the solution to Problem 2.119, Thus F D 0.123i C 0.123j 0.0984k lb and Fparallel D F FNORMAL FNORMAL D 0i C 0.0927j C 0.0232k lb Substituting, we get The component parallel to the surface and the component normal to the surface add to give FF D FNORMAL C Fparallel . Fparallel D 0.1231i C 0.0304j 0.1216k lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 63 Problem 2.121 An astronaut in a maneuvering unit approaches a space station. At the present instant, the station informs him that his position relative to the origin of the station’s coordinate system is rG D 50i C 80j C 180k (m) and his velocity is v D 2.2j 3.6k (m/s). The position of the airlock is rA D 12i C 20k (m). Determine the angle between his velocity vector and the line from his position to the airlock’s position. Solution: Points G and A are located at G: (50, 80, 180) m and A: (12, 0, 20) m. The vector rGA is rGA D xA xG i C yA yG j C zA zG k D 12 50i C 0 80j C 20 180k m. The dot product between v and rGA is v ž rGA D jvjjrGA j cos D vx xGA C vy yGA C vz zGA , where is the angle between v and rGA . Substituting in the numerical values, we get D 19.7° . y G A z x Problem 2.122 In Problem 2.121, determine the vector component of the astronaut’s velocity parallel to the line from his position to the airlock’s position. Solution: The coordinates are A (12, 0, 20) m, G (50, 80, 180) m. Therefore rGA D 62i 80j 160k m eGA D rGA D 0.327i 0.423j 0.845k jrGA j The velocity is given as v D 2.2j 3.6k m/s The vector component parallel to the line is now vp D eGA Ð veGA D [0.4232.2 C 0.8453.6]eGA vp D 1.30i 1.68j 3.36k m/s 64 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.123 Point P is at longitude 30° W and latitude 45° N on the Atlantic Ocean between Nova Scotia and France. Point Q is at longitude 60° E and latitude 20° N in the Arabian Sea. Use the dot product to determine the shortest distance along the surface of the earth from P to Q in terms of the radius of the earth RE . y N P Q Strategy: Use the dot product to detrmine the angle between the lines OP and OQ; then use the definition of an angle in radians to determine the distance along the surface of the earth from P to Q. 45⬚ z 20⬚ O 30⬚ 60⬚ G Equator x Solution: The distance is the product of the angle and the radius of the sphere, d D RE , where is in radian measure. From Eqs. (2.18) and (2.24), the angular separation of P and Q is given by cos D PÐQ jPjjQj . The strategy is to determine the angle in terms of the latitude and longitude of the two points. Drop a vertical line from each point P and Q to b and c on the equatorial plane. The vector position of P is the sum of the two vectors: P D rOB C rBP . The vector rOB D jrOB ji cos P C 0j C k sin P . From geometry, the magnitude is jrOB j D RE cos P . The vector rBP D jrBP j0i C 1j C 0k. From geometry, the magnitude is jrBP j D RE sin P . Substitute and reduce to obtain: The dot product is P Ð Q D RE2 cosP Q cos P cos Q C sin P sin Q Substitute: cos D PÐQ D cosP Q cos P cos Q C sin P sin Q jPjjQj Substitute P D C30° , Q D 60° , p D C45° , Q D C20° , to obtain cos D 0.2418, or D 1.326 radians. Thus the distance is d D 1.326RE y P D rOB C rBP D RE i cos P cos P C j sin P C k sin P cos P . N P θ A similar argument for the point Q yields 45° Q D rOC C rCQ D RE i cos Q cos Q C j sin Q C k sin Q cos Q Q RE b 30° 60° Using the identity cos2 ˇ C sin2 ˇ D 1, the magnitudes are x 20° c G jPj D jQj D RE Problem 2.124 In Active Example 2.14, suppose that the vector V is changed to V D 4i 6j 10k. (a) Determine the cross product U ð V. (b) Use the dot product to prove that U ð V is perpendicular to V. Solution: We have U D 6i 5j k, V D 4k 6j 10k (a) i U ð V D 6 4 j 5 6 k 1 D 44i C 56j 16k 10 U ð V D 44i C 56j 16k (b) U ð V Ð V D 444 C 566 C 1610 D 0 ) U ð V ? V c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 65 Problem 2.125 Two vectors U D 3i C 2j and V D 2i C 4j. (a) What is the cross product U ð V? (b) What is the cross product V ð U? Solution: Use Eq. (2.34) and expand into 2 by 2 determinants. i U ð V D 3 2 j 2 4 k 0 D i20 40 j30 20 0 j 4 2 k 0 D i40 20 j20 30 0 C k34 22 D 8k i V ð U D 2 3 C k22 34 D 8k y Problem 2.126 The two segments of the L-shaped bar are parallel to the x and z axes. The rope AB exerts a force of magnitude jFj D 500 lb on the bar at A. Determine the cross product rCA ð F, where rCA is the position vector form point C to point A. 4 ft C 5 ft Solution: We need to determine the force F in terms of its components. The vector from A to B is used to define F. 4 ft rAB D 2i 4j k ft A x F 2i 4j k rAB D 500 lb F D 500 lb jrAB j 22 C 42 C 12 F D 218i 436j 109k lb B z (6, 0, 4) ft Also we have rCA D 4i C 5k ft Therefore i j k rCA ð F D 4 0 5 D 2180i C 1530j 1750k ft-lb 218 436 109 rCA ð F D 2180i C 1530j 1750k ft-lb 66 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.127 The two segments of the L-shaped bar are parallel to the x and z axes. The rope AB exerts a force of magnitude jFj D 500 lb on the bar at A. Determine the cross product rCB ð F, where rCB is the position vector form point C to point B. Compare your answers to the answer to Problem 2.126. 4 ft C 5 ft Solution: We need to determine the force F in terms of its compo4 ft nents. The vector from A to B is used to define F. A rAB D 2i 4j k ft F D 500 lb x F 2i 4j k rAB D 500 lb jrAB j 22 C 42 C 12 B z F D 218i 436j 109k lb (6, 0, 4) ft Also we have rCB D 6i 4j C 4k ft Therefore i j k rCB ð F D 6 4 4 D 2180i C 1530j 1750k ft-lb 218 436 109 rCB ð F D 2180i C 1530j 1750k ft-lb The answer is the same for 2.126 and 2.127 because the position vectors just point to different points along the line of action of the force. Problem 2.128 Suppose that the cross product of two vectors U and V is U ð V D 0. If jUj 6D 0, what do you know about the vector V? Solution: Either V D 0 or VjjU Problem 2.129 The cross product of two vectors U and V is U ð V D 30i C 40k. The vector V D 4i 2j C 3k. The vector U D 4i C Uy j C Uz k. Determine Uy and Uz . Solution: From the given information we have i U ð V D 4 4 j Uy 2 k Uz 3 D 3Uy C 2Uz i C 4Uz 12j C 8 4Uy k U ð V D 30i C 40k Equating the components we have 3Uy C 2Uz D 30, 4Uz 12 D 0, 8 4Uy D 40. Solving any two of these three redundant equations gives Uy D 12, Uz D 3. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 67 Problem 2.130 The magnitudes jUj D 10 and jVj D 20. y V (a) Use the definition of the cross product to determine U ð V. (b) Use the definition of the cross product to determine V ð U. (c) Use Eq. (2.34) to determine U ð V. (d) U 30° 45° x Use Eq. (2.34) to determine V ð U. Solution: From Eq. (228) U ð V D jUjjVj sin e. From the sketch, the positive z-axis is out of the paper. For U ð V, e D 1k (points into the paper); for V ð U, e D C1k (points out of the paper). The angle D 15° , hence (a) U ð V D 10200.2588e D 51.8e D 51.8k. Similarly, (b) V ð U D 51.8e D 51.8k (c) The two vectors are: U D 10i cos 45° C j sin 45 D 7.07i C 0.707j, V D 20i cos 30° C j sin 30° D 17.32i C 10j i U ð V D 7.07 17.32 j k 7.07 0 D i0 j0 C k70.7 122.45 10 0 D 51.8k i (d) V ð U D 17.32 7.07 j k 10 0 D i0 j0 C k122.45 70.7 7.07 0 D 51.8k Problem 2.131 The force F D 10i 4j (N). Determine the cross product rAB ð F. y (6, 3, 0) m A rA B x z (6, 0, 4) m B F Solution: The position vector is y A (6, 3, 0) rAB D 6 6i C 0 3j C 4 0k D 0i 3j C 4k The cross product: i j k rAB ð F D 0 3 4 D i16 j40 C k30 10 4 0 rA B x z B (6, 0, 4) F D 16i C 40j C 30k (N-m) 68 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.132 By evaluating the cross product U ð V, prove the identity sin1 2 D sin 1 cos 2 cos 1 sin 2 . y U V θ1 θ2 Solution: Assume that both U and V lie in the x-y plane. The strategy is to use the definition of the cross product (Eq. 2.28) and the Eq. (2.34), and equate the two. From Eq. (2.28) U ð V D jUjjVj sin1 2 e. Since the positive z-axis is out of the paper, and e points into the paper, then e D k. Take the dot product of both sides with e, and note that k Ð k D 1. Thus sin1 2 D U ð V Ð k jUjjVj x y U V θ1 θ2 x The vectors are: U D jUji cos 1 C j sin 2 , and V D jVji cos 2 C j sin 2 . The cross product is i U ð V D jUj cos 1 jVj cos 2 j jUj sin 1 jVj sin 2 k 0 0 D i0 j0 C kjUjjVjcos 1 sin 2 cos 2 sin 1 Substitute into the definition to obtain: sin1 2 D sin 1 cos 2 cos 1 sin 2 . Q.E.D. Problem 2.133 In Example 2.15, what is the minimum distance from point B to the line OA? y Solution: Let be the angle between rOA and rOB . Then the minimum distance is d D jrOB j sin Using the cross product, we have B (6, 6, ⫺3) m jrOA ð rOB j D jrOA jjrOB j sin D jrOA jd ) d D jrOA ð rOB j jrOA j We have O z x A (10, ⫺2, 3) m rOA D 10i 2j C 3k m rOB D 6i C 6j 3k m rOA ð rOB i D 10 6 j k 2 3 D 12i C 48j C 72k m2 6 3 Thus dD 12 m2 C 48 m2 2 C 72 m2 2 D 8.22 m 10 m2 C 2 m2 C 3 m2 d D 8.22 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 69 Problem 2.134 (a) What is the cross product rOA ð rOB ? (b) Determine a unit vector e that is perpendicular to rOA and rOB . y B ( 4, 4, –4) m Solution: The two radius vectors are rOB rOB D 4i C 4j 4k, rOA D 6i 2j C 3k (a) The cross product is i D 6 4 rOA ð rOB O j 2 4 k 3 D i8 12 j24 12 4 x rOA A (6, –2, 3) m z C k24 C 8 D 4i C 36j C 32k m2 The magnitude is jrOA ð rOB j D (b) p 42 C 362 C 322 D 48.33 m2 The unit vector is eDš rOA ð rOB jrOA ð rOB j D š0.0828i C 0.7448j C 0.6621k (Two vectors.) Problem 2.135 For the points O, A, and B in Problem 2.134, use the cross product to determine the length of the shortest straight line from point B to the straight line that passes through points O and A. Solution: (The magnitude of C is 338.3) rOA D 6i 2j C 3k (m) rOB D 4i C 4j 4k m We now want to find the length of the projection, P, of line OB in direction ec . P D rOB Ð eC rOA ð rOB D C D 4i C 4j 4k Ð eC (C is ? to both rOA and rOB ) i C D 6 4 j 2 4 C8 12i k 3 D C12 C 24j C24 C 8k 4 P D 6.90 m y B ( 4, 4, –4) m C D 4i C 36j C 32k C is ? to both rOA and rOB . Any line ? to the plane formed by C and rOA will be parallel to the line BP on the diagram. C ð rOA is such a line. We then need to find the component of rOB in this direction and compute its magnitude. C ð rOA i D 4 6 j C36 2 rOB k 32 3 C D 172i C 204j 208k O x rOA P z A(6, –2, 3) m The unit vector in the direction of C is eC D 70 C D 0.508i C 0.603j 0.614k jCj c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.136 The cable BC exerts a 1000-lb force F on the hook at B. Determine rAB ð F. y Solution: The coordinates of points A, B, and C are A (16, 0, 12), B (4, 6, 0), C (4, 0, 8). The position vectors are B F 6 ft rAB rOA D 16i C 0j C 12k, rOB D 4i C 6j C 0k, rOC D 4i C 0j C 8k. x 8 ft The force F acts along the unit vector eBC D rAC 4 ft rBC rOC rOB rAB D D jrBC j jrOC rOB j jrAB j 4 ft p 62 C 82 D 10. Thus y eBC D 0i 0.6j C 0.8k, and F D jFjeBC D 0i 600j C 800k (lb). B The vector 6 ft 8 ft C Thus the cross product is k 12 D 2400i C 9600j C 7200k (ft-lb) 800 4 ft 4 ft A 12 ft y Problem 2.137 The force vector F points along the straight line from point A to point B. Its magnitude is jFj D 20 N. The coordinates of points A and B are xA D 6 m, yA D 8 m, zA D 4 m and xB D 8 m, yB D 1 m, zB D 2 m. (a) (b) r x rAB D 4 16i C 6 0j C 0 12k D 12i C 6j 12k i j 6 rAB ð F D 12 0 600 A 12 ft z Noting rOC rOB D 4 4i C 0 6j C 8 0k D 0i 6j C 8k jrOC rOB j D C A F B rA Express the vector F in terms of its components. Use Eq. (2.34) to determine the cross products rA ð F and rB ð F. rB x z Solution: We have rA D 6i C 8j C 4k m, rB D 8i C j 2k m, F D 20 N (a) 8 6 mi C 1 8 mj C 2 4 mk 2 m2 C 7 m2 C 6 m2 20 N D p 2i 7j 6k 89 20 N rA ð F D p 89 (b) i 6 m 2 j 8m 7 k 4 m 6 D 42.4i C 93.3j 123.0k Nm i j k 20 N rB ð F D p 8 m 1 m 2 m 89 2 7 6 D 42.4i C 93.3j 123.0k Nm Note that both cross products give the same result (as they must). c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 71 Problem 2.138 The rope AB exerts a 50-N force T on the collar at A. Let rCA be the position vector from point C to point A. Determine the cross product rCA ð T. y 0.15 m 0.4 m B C Solution: We define the appropriate vectors. T rCD D 0.2i 0.3j C 0.25k m A 0.2 m 0.3 m 0.5 m rCA rCD D 0.091i 0.137j C 0.114k m D 0.2 m jrCD j O x D 0.25 m 0.2 m rOB D 0.5j C 0.15k m z rOC D 0.4i C 0.3j m rAB D rOB rOC C rCA D 0.61i 1.22j 0.305k m T D 50 N rAB D 33.7i C 36.7j C 3.93k N jrAB j Now take the cross product i j rCA ð T D 0.091 0.137 33.7 36.7 l 0.114 D 4.72i 3.48j C 7.96k N-m 3.93 rCA ð T D 47.2i 3.48j C 7.96k N-m Problem 2.139 In Example 2.16, suppose that the attachment point E is moved to the location (0.3, 0.3, 0) m and the magnitude of T increases to 600 N. What is the magnitude of the component of T perpendicular to the door? Solution: We first develop the force T. rCE D 0.3i C 0.1j m T D 600 N E (0.2, 0.4, ⫺0.1) m From Example 2.16 we know that the unit vector perpendicular to the door is e D 0.358i C 0.894j C 0.268k y T D The magnitude of the force perpendicular to the door (parallel to e) is then C (0, 0.2, 0) m A (0.5, 0, 0) m B (0.35, 0, 0.2) m rCE D 569i C 190j N jrCE j x jTn j D T Ð e D 569 N0.358 C 190 N0.894 D 373 N jTn j D 373 N z 72 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.140 The bar AB is 6 m long and is perpendicular to the bars AC and AD. Use the cross product to determine the coordinates xB , yB , zB of point B. B Solution: The strategy is to determine the unit vector perpendicular to both AC and AD, and then determine the coordinates that will agree with the magnitude of AB. The position vectors are: (0, 3, 0) m (xB, yB, zB) A rOA D 0i C 3j C 0k, rOD D 0i C 0j C 3k, and rOC D 4i C 0j C 0k. The vectors collinear with the bars are: rAD D 0 0i C 0 3j C 3 0k D 0i 3j C 3k, rAC D 4 0i C 0 3j C 0 0k D 4i 3j C 0k. The vector collinear with rAB is i j k R D rAD ð rAC D 0 3 3 D 9i C 12j C 12k 4 3 0 C D (0, 0, 3) m x (4, 0, 0) m z The magnitude jRj D 19.21 (m). The unit vector is eAB D R D 0.4685i C 0.6247j C 0.6247k. jRj Thus the vector collinear with AB is rAB D 6eAB D C2.811i C 3.75j C 3.75k. Using the coordinates of point A: xB D 2.81 C 0 D 2.81 (m) yB D 3.75 C 3 D 6.75 (m) zB D 3.75 C 0 D 3.75 (m) Problem 2.141* Determine the minimum distance from point P to the plane defined by the three points A, B, and C. y B (0, 5, 0) m P (9, 6, 5) m Solution: The strategy is to find the unit vector perpendicular to the plane. The projection of this unit vector on the vector OP: rOP Ð e is the distance from the origin to P along the perpendicular to the plane. The projection on e of any vector into the plane (rOA Ð e, rOB Ð e, or rOC Ð e) is the distance from the origin to the plane along this same perpendicular. Thus the distance of P from the plane is A (3, 0, 0) m C d D rOP Ð e rOA Ð e. The position vectors are: rOA D 3i, rOB D 5j, rOC D 4k and rOP D 9i C 6j C 5k. The unit vector perpendicular to the plane is found from the cross product of any two vectors lying in the plane. Noting: rBC D rOC rOB D 5j C 4k, and rBA D rOA rOB D 3i 5j. The cross product: rBC ð rBA i D 0 3 (0, 0, 4) m z y P[9,6,5] j k 5 4 D 20i C 12j C 15k. 5 0 The magnitude is jrBC ð rBA j D 27.73, thus the unit vector is e D 0.7212i C 0.4327j C 0.5409k. The distance of point P from the plane is d D rOP Ð e rOA Ð e D 11.792 2.164 D 9.63 m. The second term is the distance of the plane from the origin; the vectors rOB , or rOC could have been used instead of rOA . x B[0,5,0] x O A[3,0,0] z C[0,0,4] c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 73 y Problem 2.142* The force vector F points along the straight line from point A to point B. Use Eqs. (2.28)–(2.31) to prove that A F rB ð F D rA ð F. rA Strategy: Let rAB be the position vector from point A to point B. Express rB in terms of of rA and rAB . Notice that the vectors rAB and F are parallel. B rB x z Solution: We have rB D rA C rAB . Therefore rB ð F D rA C rAB ð F D rA ð F C rAB ð F The last term is zero since rAB jjF. Therefore rB ð F D rA ð F Problem 2.143 For the vectors U D 6i C 2j 4k, V D 2i C 7j, and W D 3i C 2k, evaluate the following mixed triple products: (a) U Ð V ð W; (b) W Ð V ð U; (c) V Ð W ð U. Solution: Use Eq. (2.36). 6 (a) U Ð V ð W D 2 3 2 7 0 4 0 2 D 614 24 C 421 D 160 3 (b) W Ð V ð U D 2 6 0 7 2 2 0 4 D 328 0 C 24 42 D 160 2 (c) V Ð W ð U D 3 6 7 0 2 0 2 4 D 24 712 12 C 0 D 160 74 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.144 Use the mixed triple product to calculate the volume of the parallelepiped. y (140, 90, 30) mm (200, 0, 0) mm x (160, 0, 100) mm z Solution: We are given the coordinates of point D. From the geometry, we need to locate points A and C. The key to doing this is to note that the length of side OD is 200 mm and that side OD is the x axis. Sides OD, AE, and CG are parallel to the x axis and the coordinates of the point pairs (O and D), (A and E), and (C and D) differ only by 200 mm in the x coordinate. Thus, the coordinates of point A are (60, 90, 30) mm and the coordinates of point C are (40, 0, 100) mm. Thus, the vectors rOA , rOD , and rOC are rOD D 200i mm, rOA D 60i C 90j C 30k mm, and rOC D 40i C 0j C 100k mm. The mixed triple product of the three vectors is the volume of the parallelepiped. The volume is 60 rOA Ð rOC ð rOD D 40 200 90 0 0 30 100 0 y (140, 90, 30) mm E A B F O D G C x (200, 0, 0) mm (160, 0, 100) mm z D 600 C 90200100 C 300 mm3 D 1,800,000 mm3 Problem 2.145 By using Eqs. (2.23) and (2.34), show that Ux Uy Uz U Ð V ð W D Vx Vy Vz W W W x y z . Solution: One strategy is to expand the determinant in terms of its components, take the dot product, and then collapse the expansion. Eq. (2.23) is an expansion of the dot product: Eq. (2.23): U Ð V D UX VX C UY VY C UZ VZ . Eq. (2.34) is the determinant representation of the cross product: i Eq. (2.34) U ð V D UX VX j UY VY k UZ VZ U Q Ð P D QX Y VY For notational convenience, write P D U ð V. Expand the determinant about its first row: U P D i Y VY UX UZ j VX VZ UX UZ C k VX VZ Since the two-by-two determinants are scalars, this can be written in the form: P D iPX C jPY C kPZ where the scalars PX , PY , and PZ are the two-by-two determinants. Apply Eq. (2.23) to the dot product of a vector Q with P. Thus Q Ð P D QX PX C QY PY C QZ PZ . Substitute PX , PY , and PZ into this dot product UZ VZ UX UZ Q Y VZ VX UX UZ C Q z VZ VX UZ VZ But this expression can be collapsed into a three-by-three determinant directly, thus: QX Q Ð U ð V D UX VX QY UY VY QZ UZ . This completes the demonstration. VZ c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 75 Problem 2.146 The vectors U D i C UY j C 4k, V D 2i C j 2k, and W D 3i C j 2k are coplanar (they lie in the same plane). What is the component Uy ? Solution: Since the non-zero vectors are coplanar, the cross product of any two will produce a vector perpendicular to the plane, and the dot product with the third will vanish, by definition of the dot product. Thus U Ð V ð W D 0, for example. 1 UY 1 U Ð V ð W D 2 3 1 4 2 2 D 12 C 2 UY 4 6 C 42 C 3 D C10UY C 20 D 0 Thus UY D 2 Problem 2.147 The magnitude of F is 8 kN. Express F in terms of scalar components. Solution: The unit vector collinear with the force F is developed as follows: The collinear vector is r D 7 3i C 2 7j D 4i 5j y The magnitude: jrj D (3, 7) m eD F p 42 C 52 D 6.403 m. The unit vector is r D 0.6247i 0.7809j. The force vector is jrj F D jFje D 4.998i 6.247j D 5i 6.25j (kN) (7, 2) m x Problem 2.148 The magnitude of the vertical force W is 600 lb, and the magnitude of the force B is 1500 lb. Given that A C B C W D 0, determine the magnitude of the force A and the angle ˛. B Solution: The strategy is to use the condition of force balance to determine the unknowns. The weight vector is W D 600j. The vector B is W 50° α A B D 1500i cos 50° C j sin 50° D 964.2i C 1149.1j The vector A is A D jAji cos180 C ˛ C j sin180 C ˛ A D jAji cos ˛ j sin ˛. The forces balance, hence A C B C W D 0, or 964.2 jAj cos ˛i D 0, and 1149.1 600 jAj sin ˛j D 0. Thus jAj cos ˛ D 964.2, and jAj sin ˛ D 549.1. Take the ratio of the two equations to obtain tan ˛ D 0.5695, or ˛ D 29.7° . Substitute this angle to solve: jAj D 1110 lb 76 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.149 The magnitude of the vertical force vector A is 200 lb. If A C B C C D 0, what are the magnitudes of the force vectors B and C? Solution: The strategy is to express the forces in terms of scalar components, and then solve the force balance equations for the unknowns. C D jCji cos ˛ j sin ˛, where tan ˛ D 100 in. 70 in. 50 in. 50 D 0.7143, or ˛ D 35.5° . 70 C E B D A F Thus C D jCj0.8137i 0.5812j. Similarly, B D CjBji, and A D C200j. The force balance equation is A C B C C D 0. Substituting, 0.8137jCj C jBji D 0, and 0.5812jCj C 200j D 0. Solving, jCj D 344.1 lb, jBj D 280 lb Problem 2.150 The magnitude of the horizontal force vector D in Problem 2.149 is 280 lb. If D C E C F D 0, what are the magnitudes of the force vectors E and F? Solution: The strategy is to express the force vectors in terms of scalar components, and then solve the force balance equation for the unknowns. The force vectors are: E D jEji cos ˇ j sin ˇ, where tan ˇ D 50 D 0.5, or ˇ D 26.6° . 100 Thus E D jEj0.8944i 0.4472j D D 280i, and F D jFjj. The force balance equation is D C E C F D 0. Substitute and resolve into two equations: 0.8944jEj 280i D 0, and 0.4472jEj C jFjj D 0. Solve: jEj D 313.1 lb, jFj D 140 lb Problem 2.151 What are the direction cosines of F? y Refer to this diagram when solving Problems 2.151– 2.157. A (4, 4, 2) ft Solution: Use the definition of the direction cosines and the ensuing discussion. p The magnitude of F: jFj D 202 C 102 C 102 D 24.5. F ⫽ 20i ⫹ 10j ⫺ 10k (lb) u B (8, 1, ⫺2) ft x z Fx 20 D D 0.8165, The direction cosines are cos x D jFj 24.5 cos y D Fy 10 D D 0.4082 jFj 24.5 cos z D 10 Fz D D 0.4082 jFj 24.5 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 77 Problem 2.152 Determine the scalar components of a unit vector parallel to line AB that points from A toward B. Solution: Use the definition of the unit vector, we get The position vectors are: rA D 4i C 4j C 2k, rB D 8i C 1j 2k. The vector from A to B is rAB D 8 p4i C 1 4j C 2 2k D 4i 3j 4k. The magnitude: jrAB j D 42 C 32 C 42 D 6.4. The unit vector is eAB D Problem 2.153 What is the angle between the line AB and the force F? rAB 4 3 4 D i j k D 0.6247i 0.4685j 0.6247k jrAB j 6.4 6.4 6.4 Solution: Use the definition of the dot product Eq. (2.18), and Eq. (2.24): cos D rAB Ð F . jrAB jjFj From the solution to Problem 2.130, the vector parallel to AB is rAB D 4i 3j 4k, with a magnitude jrAB j D 6.4. From Problem 2.151, the force is F D 20i C 10j 10k, with a magnitude of jFj D 24.5. The dot product is rAB Ð F D 420 C 310 C 410 D 90. Substi90 tuting, cos D D 0.574, D 55° 6.424.5 Problem 2.154 Determine the vector component of F that is parallel to the line AB. Solution: Use the definition in Eq. (2.26): UP D e Ð Ue, where e is parallel to a line L. From Problem 2.152 the unit vector parallel to line AB is eAB D 0.6247i 0.4688j 0.6247k. The dot product is e Ð F D 0.624720 C 0.468810 C 0.624710 D 14.053. The parallel vector is e Ð Fe D 14.053e D 8.78i 6.59j 8.78k (lb) Problem 2.155 Determine the vector component of F that is normal to the line AB. Solution: Use the Eq. (2.27) and the solution to Problem 2.154. FN D F FP D 20 8.78i C 10 C 6.59j C 10 C 8.78k D 11.22i C 16.59j 1.22k (lb) Problem 2.156 Determine the vector rBA ð F, where rBA is the position vector from B to A. Solution: Use the definition in Eq. (2.34). Noting rBA D rAB , from Problem 2.155 rBA D 4i C 3j C 4k. The cross product is i rBA ð F D 4 20 j 3 10 k 4 D 30 40i 40 80j 10 C 40 60 D 70i C 40j 100k (ft-lb) 78 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.157 (a) Write the position vector rAB from point A to point B in terms of components. y A (4, 4, 2) ft (b) A vector R has magnitude jRj D 200 lb and is parallel to the line from A to B. Write R in terms of components. F ⫽ 20i ⫹ 10j ⫺ 10k (lb) u B (8, 1, ⫺2) ft x Solution: (a) rAB D [8 4]i C [1 4]j C [2 2]k ft z rAB D 4i 3j 4k ft (b) R D 200 N rAB D 125i 93.7j 125k N jrAB j R D 125i 96.3j 125k N y Problem 2.158 The rope exerts a force of magnitude jFj D 200 lb on the top of the pole at B. (a) (b) Determine the position vector Determine the position vector vector rAB ð F, where rAB is the from A to B. vector rAC ð F, where rAC is the from A to C. F A Solution: The strategy is to define the unit vector pointing from B to A, express the force in terms of this unit vector, and take the cross product of the position vectors with this force. The position vectors rAB D 5i C 6j C 1k, rAC D 3i C 0j C 4k, B (5, 6, 1) ft x C (3, 0, 4) ft z rBC D 3 5i C 0 6j C 4 1k D 2i 6j C 3k. The magnitude jrBC j D eBC D p 22 C 62 C 32 D 7. The unit vector is rBC D 0.2857i 0.8571j C 0.4286k. jrBC j The force vector is F D jFjeBC D 200eBC D 57.14i 171.42j C 85.72k. The cross products: i rAB ð F D 5 57.14 j 6 171.42 k 1 85.72 D 685.74i 485.74j 514.26k D 685.7i 485.7j 514.3k (ft-lb) i rAC ð F D 3 57.14 j 0 171.42 k 4 85.72 D 685.68i 485.72j 514.26k D 685.7i 485.7j 514.3k (ft-lb) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 79 Problem 2.159 The pole supporting the sign is parallel to the x axis and is 6 ft long. Point A is contained in the y –z plane. (a) Express the vector r in terms of components. (b) What are the direction cosines of r? Solution: The vector r is r D jrjsin 45° i C cos 45° sin 60° j C cos 45° cos 60° k The length of the pole is the x component of r. Therefore y jrj sin 45° D 6 ft ) jrj D A (a) Bedford Falls (b) r D 6.00i C 5.20j C 3.00k ft The direction cosines are cos x D r 45⬚ 6 ft D 8.49 ft sin 45° rx ry rz D 0.707, cos y D D 0.612, cos z D D 0.354 jrj jrj jrj cos x D 0.707, cos y D 0.612, cos z D 0.354 60⬚ x O z Problem 2.160 The z component of the force F is 80 lb. (a) Express F in terms of components. (b) what are the angles x , y , and z between F and the positive coordinate axes? y F Solution: We can write the force as x 60⬚ We know that the z component is 80 lb. Therefore jFj cos 20° cos 60° (a) (b) 20⬚ O F D jFjcos 20° sin 60° i C sin 20° j C cos 20° cos 60° k A D 80 lb ) jFj D 170 lb F D 139i C 58.2j C 80k lb z The direction cosines can be found: 139 D 35.5° x D cos1 170 y D cos1 z D cos1 58.2 170 80 170 D 70.0° D 62.0° x D 35.5° , y D 70.0° , z D 62.0° 80 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 2.161 The magnitude of the force vector FB is 2 kN. Express it in terms of scalar components. F D (4, 3, 1) m FC FA FB A z C x (6, 0, 0) m B Solution: The strategy is to determine the unit vector collinear with FB and then express the force in terms of this unit vector. F y The radius vector collinear with FB is D (4,3,1) rBD D 4 5i C 3 0j C 1 3k or rBD D 1i C 3j 2k. FA The magnitude is p jrBD j D 12 C 32 C 22 D 3.74. The unit vector is eBD (5, 0, 3) m FC A z x C(6,0,0) FB B (5,0,3) rBD D D 0.2673i C 0.8018j 0.5345k jrBD j The force is FB D jFB jeBD D 2eBD (kN) FB D 0.5345i C 1.6036j 1.0693k D 0.53i C 1.60j 1.07k (kN) Problem 2.162 The magnitude of the vertical force vector F in Problem 2.161 is 6 kN. Determine the vector components of F parallel and normal to the line from B to D. Solution: The projection of the force F onto the line from B to D is FP D F Ð eBD eBD . The vertical force has the component F D 6j (kN). From Problem 2.139, the unit vector pointing from B to D is eBD D 0.2673i C 0.8018j 0.5345k. The dot product is F Ð eBD D 4.813. Thus the component parallel to the line BD is FP D 4.813eBD D C1.29i 3.86j C 2.57k (kN). The component perpendicular to the line is: FN D F FP . Thus FN D 1.29i 2.14j 2.57k (kN) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 81 Problem 2.163 The magnitude of the vertical force vector F in Problem 2.161 is 6 kN. Given that F C FA C FB C FC D 0, what are the magnitudes of FA , FB , and FC ? Solution: The strategy is to expand the forces into scalar components, and then use the force balance equation to solve for the unknowns. The unit vectors are used to expand the forces into scalar components. The position vectors, magnitudes, and unit vectors are: rAD D 4i C 3j C 1k, jrAD j D The forces are: FA D jFA jeAD , FB D jFB jeBD , FC D jFC jeCD , F D 6j (kN). Substituting into the force balance equation p 26 D 5.1, F C FA C FB C FC D 0, eAD D 0.7845i C 0.5883j C 0.1961k. rBD D 1i C 3j 2k, jrBD j D 0.7843jFA j 0.2674jFB j 0.5348jFC ji D 0 p 14 D 3.74, 0.5882jFA j C 0.8021jFB j C 0.8021jFC j 6j eBD D 0.2673i C 0.8018j 0.5345k. rCD D 2i C 3j C 1k, jrCD j D p D 00.1961jFA j 0.5348jFB j C 0.2674jFC jk D 0 14 D 3.74, These simple simultaneous equations can be solved a standard method (e.g., Gauss elimination) or, conveniently, by using a commercial package, such as TK Solver, Mathcad, or other. An HP-28S hand held calculator was used here: jFA j D 2.83 (kN), jFB j D 2.49 (kN), jFC j D 2.91 (kN) eCD D 0.5345i C 0.8018j C 0.2673k Problem 2.164 The magnitude of the vertical force W is 160 N. The direction cosines of the position vector from A to B are cos x D 0.500, cos y D 0.866, and cos z D 0, and the direction cosines of the position vector from B to C are cos x D 0.707, cos y D 0.619, and cos z D 0.342. Point G is the midpoint of the line from B to C. Determine the vector rAG ð W, where rAG is the position vector from A to G. Solution: Express the position vectors in terms of scalar components, calculate rAG , and take the cross product. The position vectors are: rAB D 0.6.5i C 0.866j C 0k rAB D 0.3i C 0.5196j C 0k, rBG D 0.30.707i C 0.619j 0.342k, rBG D 0.2121i C 0.1857j 0.1026k. rAG D rAB C rBG D 0.5121i C 0.7053j 0.1026k. W D 160j y 0 60 mm C i rAG ð W D 0.5121 0 j 0.7053 160 k 0.1026 0 G D 16.44i C 0j 81.95k D 16.4i C 0j 82k (N m) B W 600 mm 600 mm C 600 mm G B W A A z 82 x x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 2.165 The rope CE exerts a 500-N force T on the hinged door. E (0.2, 0.4, ⫺0.1) m y (a) (b) Express T in terms of components. Determine the vector component of T parallel to the line from point A to point B. T D C (0, 0.2, 0) m Solution: We have A (0.5, 0, 0) m rCE D 0.2i C 0.2j 0.1k m T D 500 N (a) (b) x B (0.35, 0, 0.2) m rCE D 333i C 333j 167k N jrCE j z T D 333i C 333j 167k N We define the unit vector in the direction of AB and then use this vector to find the component parallel to AB. rAB D 0.15i C 0.2k m eAB D rAB D 0.6i C 0.8k jrAB j Tp D eAB Ð TeAB D [0.6][333 N] C [0.8][167 N]0.6i C 0.8k Tp D 200i 267k N Problem 2.166 In Problem 2.165, let rBC be the position vector from point B to point C. Determine the cross product rBC ð T. E (0.2, 0.4, ⫺0.1) m y T Solution: From Problem 2.165 we know that D C (0, 0.2, 0) m A (0.5, 0, 0) m T D 333i C 333j 167k N x B (0.35, 0, 0.2) m The vector rBC is rBC D 035i C 0.2j 0.2k m The cross product is i j rBC ð T D 0.35 0.2 333 333 z k 0.2 D 33.3i 125j 183k Nm 137 rBC ð T D 33.3i 125j 183k Nm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 83 Problem 3.1 In Active Example 3.1, suppose that the angle between the ramp supporting the car is increased from 20° to 30° . Draw the free-body diagram of the car showing the new geometry. Suppose that the cable from A to B must exert a 1900-lb horizontal force on the car to hold it in place. Determine the car’s weight in pounds. A B 20⬚ Solution: The free-body diagram is shown to the right. Applying the equilibrium equations Fx : T N sin 30° D 0, Fy : N cos 30° mg D 0 Setting T D 1900 lb and solving yields N D 3800 lb, mg D 3290 lb Problem 3.2 The ring weighs 5 lb and is in equilibrium. The force F1 D 4.5 lb. Determine the force F2 and the angle ˛. y F2 a F1 30⬚ x Solution: The free-body diagram is shown below the drawing. The equilibrium equations are Fx : F1 cos 30° F2 cos ˛ D 0 Fy : F1 sin 30° C F2 sin ˛ 5 lb D 0 We can write these equations as F2 sin ˛ D 5 lb F1 sin 30° F2 cos ˛ D F1 cos 30° Dividing these equations and using the known value for F1 we have. tan ˛ D F2 D 5 lb 4.5 lb sin 30° D 0.706 ) ˛ D 35.2° 4.5 lb cos 30° 4.5 lb cos 30° D 4.77 lb cos ˛ F2 D 4.77 lb, ˛ D 35.2° 84 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.3 In Example 3.2, suppose that the attachment point C is moved to the right and cable AC is extended so that the angle between cable AC and the ceiling decreases from 45° to 35° . The angle between cable AB and the ceiling remains 60° . What are the tensions in cables AB and AC? B 60⬚ 45⬚ C A Solution: The free-body diagram is shown below the picture. The equilibrium equations are: Fx : TAC cos 35° TAB cos 60° D 0 Fy : TAC sin 35° C TAB sin 60° 1962 N D 0 Solving we find TAB D 1610 N, TAC D 985 N Problem 3.4 The 200-kg engine block is suspended by the cables AB and AC. The angle ˛ D 40° . The freebody diagram obtained by isolating the part of the system within the dashed line is shown. Determine the forces TAB and TAC . y TAB B TAC C a A a A x (200 kg) (9.81 m/s2) Solution: TAB TAC ˛ D 40° Fx : TAC cos ˛ TAB cos ˛ D 0 α α Fy : TAC sin ˛ C TAB sin ˛ 1962 N D 0 Solving: TAB D TAC D 1.526 kN 1962 Ν c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 85 Problem 3.5 A heavy rope used as a mooring line for a cruise ship sags as shown. If the mass of the rope is 90 kg, what are the tensions in the rope at A and B? 55⬚ A B 40⬚ Solution: The free-body diagram is shown. The equilibrium equations are Fx : TB cos 40° TA cos 55° D 0 Fy : TB sin 40° C TA sin 55° 909.81 N D 0 Solving: TA D 679 N, TB D 508 N Problem 3.6 A physiologist estimates that the masseter muscle of a predator, Martes, is capable of exerting a force M as large as 900 N. Assume that the jaw is in equilibrium and determine the necessary force T that the temporalis muscle exerts and the force P exerted on the object being bitten. 22⬚ T P M Solution: The equilibrium equations are Fx : T cos 22° M cos 36° D 0 36⬚ Fy : T sin 22° C M sin 36° P D 0 Setting M D 900 N, and solving, we find T D 785 N, P D 823 N 86 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.7 The two springs are identical, with unstretched lengths 250 mm and spring constants k D 1200 N/m. (a) Draw the free-body diagram of block A. (b) Draw the free-body diagram of block B. (c) What are the masses of the two blocks? 300 mm A 280 mm B Solution: The tension in the upper spring acts on block A in the positive Y direction, Solve the spring force-deflection equation for the tension in the upper spring. Apply the equilibrium conditions to block A. Repeat the steps for block B. 300 mm N 0.3 m 0.25 mj D 0i C 60j N TUA D 0i C 1200 m Similarly, the tension in the lower spring acts on block A in the negative Y direction TLA D 0i 1200 A 280 mm N m 0.28 m 0.25 mj D 0i 36j N B The weight is WA D 0i jWA jj The equilibrium conditions are FD Fx C Fy D 0, Tension, upper spring F D WA C TUA C TLA D 0 A Collect and combine like terms in i, j Solve Fy D jWA j C 60 36j D 0 Tension, lower spring Weight, mass A jWA j D 60 36 D 24 N The mass of A is mA D 24 N jWL j D D 2.45 kg jgj 9.81 m/s2 The free body diagram for block B is shown. The tension in the lower spring TLB D 0i C 36j The weight: WB D 0i jWB jj Apply the equilibrium conditions to block B. Tension, lower spring y B x Weight, mass B F D WB C TLB D 0 Collect and combine like terms in i, j: Solve: Fy D jWB j C 36j D 0 jWB j D 36 N The mass of B is given by mB D 36 N jWB j D D 3.67 kg jgj 9.81 m/s2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 87 Problem 3.8 The two springs in Problem 3.7 are identical, with unstretched lengths of 250 mm. Suppose that their spring constant k is unknown and the sum of the masses of blocks A and B is 10 kg. Determine the value of k and the masses of the two blocks. Solution: All of the forces are in the vertical direction so we will use scalar equations. First, consider the upper spring supporting both masses (10 kg total mass). The equation of equilibrium for block the entire assembly supported by the upper spring is A is TUA mA C mB g D 0, where TUA D kU 0.25 N. The equation of equilibrium for block B is TUB mB g D 0, where TUB D kL 0.25 N. The equation of equilibrium for block A alone is TUA C TLA mA g D 0 where TLA D TUB . Using g D 9.81 m/s2 , and solving simultaneously, we get k D 1962 N/m, mA D 4 kg, and mB D 6 kg . Problem 3.9 The inclined surface is smooth (Remember that “smooth” means that friction is negligble). The two springs are identical, with unstretched lengths of 250 mm and spring constants k D 1200 N/m. What are the masses of blocks A and B? 300 mm A 280 mm B 30⬚ Solution: F1 D 1200 N/m0.3 0.25m D 60 N mAg F1 F2 D 1200 N/m0.28 0.25m D 36 N F2 FB &: F2 C mB g sin 30° D 0 m Bg F2 FA &: F1 C F2 C mA g sin 30° D 0 NA Solving: mA D 4.89 kg, mB D 7.34 kg NB 88 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.10 The mass of the crane is 20,000 kg. The crane’s cable is attached to a caisson whose mass is 400 kg. The tension in the cable is 1 kN. (a) (b) Determine the friction forces level ground. Determine the friction forces level ground. magnitudes of the normal and exerted on the crane by the 45° magnitudes of the normal and exerted on the caisson by the Strategy: To do part (a), draw the free-body diagram of the crane and the part of its cable within the dashed line. Solution: (a) 45° Fy : Ncrane 196.2 kN 1 kN sin 45° D0 196.2 kN 1 kN Fx : Fcrane C 1 kN cos 45° D 0 y Ncrane D 196.9 kN, Fcrane D 0.707 kN (b) Fy : Ncaisson 3.924 kN C 1 kN sin 45° D 0 x Fcrane Fx : 1 kN cos 45° C Fcaisson D 0 Ncrane Ncaisson D 3.22 kN, Fcaisson D 0.707 kN 1 kN 3.924 kN 45° Fcaisson Ncaisson Problem 3.11 The inclined surface is smooth. The 100-kg crate is held stationary by a force T applied to the cable. Solution: (a) The FBD T (a) (b) Draw the free-body diagram of the crate. Determine the force T. 981 Ν T Ν 60° 60⬚ (b) F -: T 981 N sin 60° D 0 T D 850 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 89 Problem 3.12 The 1200-kg car is stationary on the sloping road. If ˛ D 20° , what are the magnitudes of the total normal and friction forces exerted on the car’s tires by the road? (b) The car can remain stationary only if the total friction force necessary for equilibrium is not greater than 0.6 times the total normal force. What is the largest angle ˛ for which the car can remain stationary? (a) a Solution: 11.772 kN (a) ˛D 20° F% : N 11.772 kN cos ˛ D 0 F- : F 11.772 kN sin ˛ D 0 N D 11.06 kN, F D 4.03 kN (b) α F D 0.6 N F F% : N 11.772 kN cos ˛ D 0 ) ˛ D 31.0° N F- : F 11.772 kN sin ˛ D 0 Problem 3.13 The 100-lb crate is in equilibrium on the smooth surface. The spring constant is k D 400 lb/ft. Let S be the stretch of the spring. Obtain an equation for S (in feet) as a function of the angle ˛. a Solution: The free-body diagram is shown. The equilibrium equation in the direction parallel to the inclined surface is kS 100 lb sin ˛ D 0 Solving for S and using the given value for k we find S D 0.25 ft sin ˛ 90 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.14 A 600-lb box is held in place on the smooth bed of the dump truck by the rope AB. (a) If ˛ D 25° , what is the tension in the rope? (b) If the rope will safely support a tension of 400 lb, what is the maximum allowable value of ˛? B A α Solution: Isolate the box. Resolve the forces into scalar components, and solve the equilibrium equations. A B The external forces are the weight, the tension in the rope, and the normal force exerted by the surface. The angle between the x axis and the weight vector is 90 ˛ (or 270 C ˛). The weight vector is α W D jWji sin ˛ j cos ˛ D 600i sin ˛ j cos ˛ The projections of the rope tension and the normal force are y T D jTx ji C 0j N D 0i C jNy jj T The equilibrium conditions are x FDWCNCTD0 Substitute, and collect like terms N W α Fx D 600 sin ˛ jTx ji D 0 Fy D 600 cos ˛ C jNy jj D 0 Solve for the unknown tension when For ˛ D 25° jTx j D 600 sin ˛ D 253.6 lb. For a tension of 400 lb, (600 sin ˛ 400 D 0. Solve for the unknown angle sin ˛ D 400 D 0.667 or ˛ D 41.84° 600 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 91 A Problem 3.15 The 80-lb box is held in place on the smooth inclined surface by the rope AB. Determine the tension in the rope and the normal force exerted on the box by the inclined surface. 30⬚ B 50⬚ Solution: The equilibrium equations (in terms of a coordinate system with the x axis parallel to the inclined surface) are Fx : 80 lb sin 50° T cos 50 D 0 Fx : N 80 lb cos 50° T sin 50 D 0 Solving: T D 95.34 lb, N D 124 lb Problem 3.16 The 1360-kg car and the 2100-kg tow truck are stationary. The muddy surface on which the car’s tires rest exerts negligible friction forces on them. What is the tension in the tow cable? 18⬚ 10⬚ 26⬚ Solution: FBD of the car being towed F- : T cos 8° 13.34 kN sin 26° D 0 13.34 kN T 18° T D 5.91 kN 26° N 92 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.17 Each box weighs 40 lb. The angles are measured relative to the horizontal. The surfaces are smooth. Determine the tension in the rope A and the normal force exerted on box B by the inclined surface. A B C 70⬚ 45⬚ D 20⬚ Solution: The free-body diagrams are shown. The equilibrium equations for box D are Fx : 40 lb sin 20° TC cos 25° D 0 Fy : ND 40 lb cos 20° C TC sin 25° D 0 The equilibrium equations for box B are Fx : 40 lb sin 70° C TC cos 25° TA D 0 Fy : NB 40 lb cos 70° C TC sin 25° D 0 Solving these four equations yields: TA D 51.2 lb, TC D 15.1 lb, NB D 7.30 lb, ND D 31.2 lb Thus TA D 51.2 lb, NB D 7.30 lb Problem 3.18 A 10-kg painting is hung with a wire supported by a nail. The length of the wire is 1.3 m. (a) (b) What is the tension in the wire? What is the magnitude of the force exerted on the nail by the wire? 1.2 m Solution: (a) Fy : 98.1 N 2 98.1 N 5 TD0 13 T D 128 N T (b) T 5 Force D 98.1 N 12 12 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 93 Problem 3.19 A 10-kg painting is hung with a wire supported by two nails. The length of the wire is 1.3 m. (a) What is the tension in the wire? (b) What is the magnitude of the force exerted on each nail by the wire? (Assume that the tension is the same in each part of the wire.) 0.4 m 0.4 m 0.4 m Compare your answers to the answers to Problem 3.18. T Solution: (a) Examine the point on the left where the wire is attached to the picture. This point supports half of the weight R 27.3° Fy : T sin 27.3° 49.05 N D 0 T D 107 N (b) 49.05 N Examine one of the nails Fx : Rx T cos 27.3° C T D 0 Ry Fy : Ry T sin 27.3° D 0 Rx 27.3° RD Rx 2 C Ry 2 T T R D 50.5 N Problem 3.20 Assume that the 150-lb climber is in equilibrium. What are the tensions in the rope on the left and right sides? 15⬚ 14⬚ y Solution: Fx D TR cos15° TL cos14° D 0 Fy D TR sin15° C TL sin14° 150 D 0 14° TR TL 15° Solving, we get TL D 299 lb, TR D 300 lb x 150 lb 94 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.21 If the mass of the climber shown in Problem 3.20 is 80 kg, what are the tensions in the rope on the left and right sides? y Solution: Fx D TR cos15° TL cos14° D 0 Fy D TR sin15° C TL sin14° mg D 0 TR TL 14° 15° Solving, we get x TL D 1.56 kN, TR D 1.57 kN mg = (80) (9.81) N Problem 3.22 The construction worker exerts a 20-lb force on the rope to hold the crate in equilibrium in the position shown. What is the weight of the crate? 5⬚ 30⬚ Solution: The free-body diagram is shown. The equilibrium equations for the part of the rope system where the three ropes are joined are Fx : 20 lb cos 30° T sin 5° D 0 Fy : 20 lb sin 30° C T cos 5° W D 0 Solving yields W D 188 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 95 Problem 3.23 A construction worker on the moon, where the acceleration due to gravity is 1.62 m/s2 , holds the same crate described in Problem 3.22 in the position shown. What force must she exert on the cable to hold the crate ub equilibrium (a) in newtons; (b) in pounds? 5⬚ 30⬚ Solution: The free-body diagram is shown. From Problem 3.22 we know that the weight is W D 188 lb. Therefore its mass is mD 188 lb D 5.84 slug 32.2 ft/s2 m D 5.84 slug 14.59 kg slug D 85.2 kg The equilibrium equations for the part of the rope system where the three ropes are joined are Fx : F cos 30° T sin 5° D 0 Fy : F sin 30° C T cos 5° mgm D 0 where gm D 1.62 m/s2 . Solving yields F D 3.30 lb D 14.7 N a F D 14.7 N, b F D 3.30 lb 96 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.24 The person wants to cause the 200-lb crate to start sliding toward the right. To achieve this, the horizontal component of the force exerted on the crate by the rope must equal 0.35 times the normal force exerted on the crate by the floor. In Fig.a, the person pulls on the rope in the direction shown. In Fig.b, the person attaches the rope to a support as shown and pulls upward on the rope. What is the magnitude of the force he must exert on the rope in each case? 20⬚ (a) 10⬚ (b) Solution: The friction force Ffr is given by Ffr D 0.35N (a) For equilibrium we have Fx : T cos 20° 0.35N D 0 Fy : T sin 20° 200 lb C N D 0 Solving: T D 66.1 lb (b) The person exerts the force F. Using the free-body diagram of the crate and of the point on the rope where the person grabs the rope, we find Fx : TL 0.35N D 0 Fy : N 200 lb D 0 Fx : TL C TR cos 10° D 0 Fy : F TR sin 10° D 0 Solving we find F D 12.3 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 97 Problem 3.25 A traffic engineer wants to suspend a 200-lb traffic light above the center of the two right lanes of a four-lane thoroughfare as shown. Determine the tensions in the cables AB and BC. Solution: 80 ft 20 ft A C 6 2 Fx : p TAB C p TBC D 0 37 5 1 1 Fy : p TAB C p TBC 200 lb D 0 37 5 Solving: TAB D 304 lb, TBC D 335 lb 10 ft B TBC 30 ft 6 1 1 TAB 2 200 lb Problem 3.26 Cable AB is 3 m long and cable BC is 4 m long. The mass of the suspended object is 350 kg. Determine the tensions in cables AB and BC. 5m A C B Solution: TAB TAC 3 4 Fx : TAB C TBC D 0 5 5 4 4 3 Fy : TAB C TBC 3.43 kN D 0 5 5 4 3 3 TAB D 2.75 kN, TBC D 2.06 kN 3.43 kN 98 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.27 In Problem 3.26, the length of cable AB is adjustable. If you don’t want the tension in either cable AB or cable BC to exceed 3 kN, what is the minimum acceptable length of cable AB? Solution: Consider the geometry: x 5−x We have the constraints LAB 2 D x 2 C y 2 , 4 m2 D 5 m x2 C y 2 y LAB 4m These constraint imply yD 10 mx x2 9 m2 TAB L D 10 mx 9 m2 Now draw the FBD and write the equations in terms of x x 5x TBC D 0 Fx : p TAB C 4 10x 9 Fy : TBC y 4 y x 5−x p p 10x x2 9 10x x2 9 p TBC 3.43 kN D 0 TAB C 4 10x 9 If we set TAB D 3 kN and solve for x we find x D 1.535, TBC D 2.11 kN < 3 kN 3.43 kN Using this value for x we find that LAB D 2.52 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 99 Problem 3.28 What are the tensions in the upper and lower cables? (Your answers will be in terms of W. Neglect the weight of the pulley.) 45° 30° W Solution: Isolate the weight. The frictionless pulley changes the TU direction but not the magnitude of the tension. The angle between the right hand upper cable and the x axis is ˛, hence TU β TUR D jTU ji cos ˛ C j sin ˛. α y The angle between the positive x and the left hand upper pulley is 180° ˇ, hence TUL D jTU ji cos180 ˇ C j sin180 ˇ TL W x D jTU ji cos ˇ C j sin ˇ. The lower cable exerts a force: TL D jTL ji C 0j The weight: W D 0i jWjj The equilibrium conditions are F D W C TUL C TUR C TL D 0 Substitute and collect like terms, Fx D jTU j cos ˇ C jTU j cos ˛ jTL ji D 0 Fy D jTU j sin ˛ C jTU j sin ˇ jWjj D 0. Solve: jTU j D jWj sin ˛ C sin ˇ , jTL j D jTU jcos ˛ cos ˇ. From which For jTL j D jWj cos ˛ cos ˇ sin ˛ C sin ˇ . ˛ D 30° and ˇ D 45° jTU j D 0.828jWj, jTL j D 0.132jWj 100 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.29 Two tow trucks lift a 660-lb motorcycle out of a ravine following an accident. If the motorcycle is in equilibrium in the position shown, what are the tensions in cables AB and AC? (36, 36) ft y (12, 32) ft C B (26, 16) ft A x Solution: The angles are ˛ D tan1 ˇ D tan1 32 16 26 12 36 16 36 26 D 48.8° D 63.4° Now from equilibrium we have Fx : TAB cos ˛ C TAC cos ˇ D 0 Fy : TAB sin ˛ C TAC sin ˇ 660 lb D 0 Solving yields TAB D 319 lb, TAC D 470 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 101 Problem 3.30 An astronaut candidate conducts experiments on an airbearing platform. While she carries out calibrations, the platform is held in place by the horizontal tethers AB, AC, and AD. The forces exerted by the tethers are the only horizontal forces acting on the platform. If the tension in tether AC is 2 N, what are the tensions in the other two tethers? TOP VIEW D 4.0 m A 3.5 m B C 3.0 m Solution: Isolate the platform. The angles ˛ and ˇ are tan ˛ D Also, tan ˇ D 1.5 3.5 3.0 3.5 D 0.429, ˛ D 23.2° . B 3.0 m A D 0.857, ˇ D 40.6° . C 3.5 m 4.0 m y B x TAB D jTAB ji cos180° ˇ C j sin180° ˇ β α TAB D jTAB ji cos ˇ C j sin ˇ. The angle between the tether AC and the positive x axis is 180° C ˛. The tension is D jTAC ji cos ˛ j sin ˛. C Solve: jTAB j D The tether AD is aligned with the positive x axis, TAD D jTAD ji C 0j. The equilibrium condition: F D TAD C TAB C TAC D 0. Substitute and collect like terms, 102 D A TAC D jTAC ji cos180° C ˛ C j sin180° C ˛ D 1.5 m The angle between the tether AB and the positive x axis is 180° ˇ, hence 1.5 m jTAD j D sin ˛ sin ˇ jTAC j, jTAC j sin˛ C ˇ sin ˇ . For jTAC j D 2 N, ˛ D 23.2° and ˇ D 40.6° , jTAB j D 1.21 N, jTAD j D 2.76 N Fx D jTAB j cos ˇ jTAC j cos ˛ C jTAD ji D 0, Fy D jTAB j sin ˇ jTAC j sin ˛j D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.31 The bucket contains concrete and weighs 5800 lb. What are the tensions in the cables AB and AC? (5, 34) ft y B C (12, 16) ft (20, 34) ft A x Solution: The angles are ˛ D tan1 ˇ D tan1 34 16 12 5 34 16 20 12 D 68.7° D 66.0° Now from equilibrium we have Fx : TAB cos ˛ C TAC cos ˇ D 0 Fy : TAB sin ˛ C TAC sin ˇ 660 lb D 0 Solving yields TAB D 319 lb, TAC D 470 lb Problem 3.32 The slider A is in equilibrium and the bar is smooth. What is the mass of the slider? 20⬚ 200 N A 45⬚ Solution: The pulley does not change the tension in the rope that passes over it. There is no friction between the slider and the bar. y T = 200 N 20° Eqns. of Equilibrium: Fx D T sin 20° C N cos 45° D 0 T D 200 N Fy D N sin 45° C T cos 20° mg D 0 g D 9.81 m/s2 x Substituting for T and g, we have two eqns in two unknowns (N and m). Solving, we get N D 96.7 N, m D 12.2 kg. N 45° mg = (9.81) g c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 103 Problem 3.33 The 20-kg mass is suspended from three cables. Cable AC is equipped with a turnbuckle so that its tension can be adjusted and a strain gauge that allows its tension to be measured. If the tension in cable AC is 40 N, what are the tensions in cables AB and AD? 0.4 m 0.4 m B 0.48 m C D 0.64 m A Solution: TAC TAB 5 TAC D 40 N 5 5 11 TAD D 0 Fx : p TAB C p TAC C p 89 89 185 8 11 5 TAD 8 8 8 8 8 TAD 196.2 N D 0 Fy : p TAB C p TAC C p 89 89 185 Solving: TAB D 144.1 N, TAD D 68.2 N 196.2 N Problem 3.34 The structural joint is in equilibrium. If FA D 1000 lb and FD D 5000 lb, what are FB and FC ? FC 80⬚ FB 65⬚ 35⬚ FA FD Solution: The equilibrium equations are Fx : FD FC cos 65° FB cos 35° FA D 0 Fy : FC sin 65° C FB sin 35° D 0 Solving yields FB D 3680 lb, FC D 2330 lb 104 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.35 The collar A slides on the smooth vertical bar. The masses mA D 20 kg and mB D 10 kg. When h D 0.1 m, the spring is unstretched. When the system is in equilibrium, h D 0.3 m. Determine the spring constant k. 0.25 m h A B k Solution: The triangles formed by the rope segments and the horizontal line level with A can be used to determine the lengths Lu and Ls . The equations are Lu D 0.252 C 0.12 and Ls D 0.252 C 0.32 . The stretch in the spring when in equilibrium is given by υ D Ls Lu . Carrying out the calculations, we get Lu D 0.269 m, Ls D 0.391 m, and υ D 0.121 m. The angle, , between the rope at A and the horizontal when the system is in equilibrium is given by tan D 0.3/0.25, or D 50.2° . From the free body diagram for mass A, we get two equilibrium equations. They are and T NA A mA g Fx D NA C T cos D 0 T Fy D T sin mA g D 0. We have two equations in two unknowns and can solve. We get NA D 163.5 N and T D 255.4 N. Now we go to the free body diagram for B, where the equation of equilibrium is T mB g kυ D 0. This equation has only one unknown. Solving, we get k D 1297 N/m Lu 0.1 m B mBg Kδ 0.25 m Ls Lu 0.3 m 0.25 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 105 Problem 3.36* Suppose that you want to design a cable system to suspend an object of weight W from the ceiling. The two wires must be identical, and the dimension b is fixed. The ratio of the tension T in each wire to its cross-sectional area A must equal a specified value T/A D . The “cost” of your design ispthe total volume of material in the two wires, V D 2A b2 C h2 . Determine the value of h that minimizes the cost. b b h W Solution: From the equation T T θ θ Fy D 2T sin W D 0, we obtain T D p W W b2 C h2 D . 2 sin 2h Since T/A D , A D p W b2 C h2 T D 2h W p Wb2 C h2 . and the “cost” is V D 2A b2 C h2 D h To determine the value of h that minimizes V, we set dV W b2 C h2 D C2 D0 dh h2 and solve for h, obtaining h D b. 106 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.37 The system of cables suspends a 1000-lb bank of lights above a movie set. Determine the tensions in cables AB, CD, and CE. 20 ft 18 ft B D Solution: Isolate juncture A, and solve the equilibrium equations. C Repeat for the cable juncture C. E The angle between the cable AC and the positive x axis is ˛. The tension in AC is TAC D jTAC ji cos ˛ C j sin ˛ 45° 30° A The angle between the x axis and AB is 180° ˇ. The tension is TAB D jTAB ji cos180 ˇ C j sin180 ˇ TAB D i cos ˇ C j sin ˇ. The weight is W D 0i jWjj. The equilibrium conditions are Solve: jTCE j D jTCA j cos ˛, F D 0 D W C TAB C TAC D 0. jTCD j D jTCA j sin ˛; Substitute and collect like terms, for jTCA j D 732 lb and ˛ D 30° , Fx D jTAC j cos ˛ jTAB j cos ˇi D 0 jTAB j D 896.6 lb, Fy D jTAB j sin ˇ C jTAC j sin ˛ jWjj D 0. jTCE j D 634 lb, Solving, we get jTAB j D cos ˛ cos ˇ and jTAC j jTAC j D jWj cos ˇ sin˛ C ˇ jTCD j D 366 lb , jWj D 1000 lb, and ˛ D 30° , ˇ D 45° jTAC j D 1000 jTAB j D 732 0.7071 0.9659 0.866 0.7071 B C A β α D 732.05 lb y x W D 896.5 lb Isolate juncture C. The angle between the positive x axis and the cable CA is 180° ˛. The tension is D TCA D jTCA ji cos180° C ˛ C j sin180° C ˛, C 90° E or TCA D jTCA ji cos ˛ j sin ˛. The tension in the cable CE is α y A TCE D ijTCE j C 0j. x The tension in the cable CD is TCD D 0i C jjTCD j. The equilibrium conditions are F D 0 D TCA C TCE C TCD D 0 Substitute t and collect like terms, Fx D jTCE j jTCA j cos ˛i D 0, Fy D jTCD j jTCA j sin ˛j D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 107 Problem 3.38 Consider the 1000-lb bank of lights in Problem 3.37. A technician changes the position of the lights by removing the cable CE. What is the tension in cable AB after the change? Solution: The original configuration in Problem 3.35 is used to solve for the dimensions and the angles. Isolate the juncture A, and solve the equilibrium conditions. 18 ft 20 ft D B C The lengths are calculated as follows: The vertical interior distance in the triangle is 20 ft, since the angle is 45 deg. and the base and altitude of a 45 deg triangle are equal. The length AB is given by α β A AB D 20 ft D 28.284 ft. cos 45° The length AC is given by AC D 18 ft D 20.785 ft. cos 30° 38 B The altitude of the triangle for which AC is the hypotenuse is 18 tan 30° D 10.392 ft. The distance CD is given by 20 10.392 D 9.608 ft. D β α β 20.784+9.608 = 30.392 α 28.284 The distance AD is given by A AD D AC C CD D 20.784 C 9.608 D 30.392 B The new angles are given by the cosine law AB2 D 382 C AD2 D β 238AD cos ˛. A α Reduce and solve: cos ˛ D C 30.3922 28.2842 23830.392 cos ˇ D 382 28.2842 C 382 30.3922 228.28438 y D 0.6787, ˛ D 47.23° . D 0.6142, ˇ D 52.1° . Isolate the juncture A. The angle between the cable AD and the positive x axis is ˛. The tension is: Solve: jTAB j D and jTAD j D TAD D jTAD ji cos ˛ C j sin ˛. The angle between x and the cable AB is 180° ˇ. The tension is TAB D jTAB ji cos ˇ C j sin ˇ. The weight is W D 0i jWjj F D 0 D W C TAB C TAD D 0. cos ˛ cos ˇ jTAD j, jWj cos ˇ sin˛ C ˇ . For jWj D 1000 lb, and ˛ D 51.2° , ˇ D 47.2° jTAD j D 1000 The equilibrium conditions are x W 0.6142 0.989 jTAB j D 622.3 0.6787 0.6142 D 621.03 lb, D 687.9 lb Substitute and collect like terms, 108 Fx D jTAD j cos ˛ jTAB j cos ˇi D 0, Fy D jTAB j sin ˇ C jTAD j sin ˛ jWjj D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.39 While working on another exhibit, a curator at the Smithsonian Institution pulls the suspended Voyager aircraft to one side by attaching three horizontal cables as shown. The mass of the aircraft is 1250 kg. Determine the tensions in the cable segments AB, BC, and CD. D C B 30° 50° Solution: Isolate each cable juncture, beginning with A and solve the equilibrium equations at each juncture. The angle between the cable AB and the positive x axis is ˛ D 70° ; the tension in cable AB is TAB D jTAB ji cos ˛ C j sin ˛. The weight is W D 0i jWjj. The tension in cable AT is T D jTji C 0j. The equilibrium conditions are A 70° F D W C T C TAB D 0. Substitute and collect like terms Fx jTAB j cos ˛ jTji D 0, Fy D jTAB j sin ˛ jWjj D 0. y Solve: the tension in cable AB is jTAB j D For jWj D 1250 kg 9.81 jTAB j D 12262.5 0.94 m s2 jWj . sin ˛ D 12262.5 N and ˛ D 70° B x α A T D 13049.5 N W Isolate juncture B. The angles are ˛ D 50° , ˇ D 70° , and the tension cable BC is TBC D jTBC ji cos ˛ C j sin ˛. The angle between the cable BA and the positive x axis is 180 C ˇ; the tension is y C x TBA D jTBA ji cos180 C ˇ C j sin180 C ˇ The tension in the left horizontal cable is T D jTji C 0j. The equilibrium conditions are β A F D TBA C TBC C T D 0. Substitute and collect like terms α B T D jTBA ji cos ˇ j sin ˇ y T Fy D jTBC j sin ˛ jTBA j sin ˇj D 0. Solve: jTBC j D sin ˇ sin ˛ D x Fx D jTBC j cos ˛ jTBA j cos ˇ jTji D 0 α C β jTBA j. B For jTBA j D 13049.5 N, and ˛ D 50° , ˇ D 70° , jTBC j D 13049.5 0.9397 0.7660 D 16007.6 N Isolate the cable juncture C. The angles are ˛ D 30° , ˇ D 50° . By symmetry with the cable juncture B above, the tension in cable CD is jTCD j D sin ˇ sin ˛ jTCB j. Substitute: jTCD j D 16007.6 0.7660 0.5 D 24525.0 N. This completes the problem solution. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 109 Problem 3.40 A truck dealer wants to suspend a 4000kg truck as shown for advertising. The distance b D 15 m, and the sum of the lengths of the cables AB and BC is 42 m. Points A and C are at the same height. What are the tensions in the cables? 40 m b A C B Solution: Determine the dimensions and angles of the cables. Iso- 15 m late the cable juncture B, and solve the equilibrium conditions. The dimensions of the triangles formed by the cables: L D 25 m, L b A b D 15 m, 25 m β AB C BC D S D 42 m. C α Subdivide into two right triangles with a common side of unknown length. Let the unknown length of this common side be d, then by the Pythagorean Theorem b2 C d2 D AB2 , L2 C d2 D BC2 . B y Subtract the first equation from the second to eliminate the unknown d, L 2 b2 D BC2 AB2 . A α B β C Note that BC2 AB2 D BC ABBC C AB. W Substitute and reduce to the pair of simultaneous equations in the unknowns x BC AB D L 2 b2 S Solve: 2 1 L b2 CS 2 S BC D D , BC C AB D S Substitute and collect like terms 2 1 25 152 C 42 D 25.762 m 2 42 Fx D jTBC j cos ˛ jTBA j cos ˇi D 0, Fy D jTBC j sin ˛ C jTBA j sin ˇ jWjj D 0 and AB D S BC D 42 25.762 D 16.238 m. Solve: jTBC j D The interior angles are found from the cosine law: cos ˛ D cos ˇ D L C b2 C BC2 2L C bBC AB2 L C b2 C AB2 BC2 2L C bAB and jTBA j D D 0.9704 ˛ D 13.97° cos ˇ cos ˛ jTBA j, jWj cos ˛ sin˛ C ˇ . For jWj D 40009.81 D 39240 N, D 0.9238 ˇ D 22.52° Isolate cable juncture B. The angle between BC and the positive x axis is ˛; the tension is and ˛ D 13.97° , ˇ D 22.52° , jTBA j D 64033 D 64 kN, jTBC j D 60953 D 61 kN TBC D jTBC ji cos ˛ C j sin ˛ The angle between BA and the positive x axis is 180° ˇ; the tension is TBA D jTBA ji cos180 ˇ C j sin180 ˇ D jTBA ji cos ˇ C j sin ˇ. The weight is W D 0i jWjj. The equilibrium conditions are 110 F D W C TBA C TBC D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.41 The distance h D 12 in, and the tension in cable AD is 200 lb. What are the tensions in cables AB and AC? B 12 in. A D C 12 in. h 8 in. 12 in. Solution: Isolated the cable juncture. From the sketch, the angles are found from tan ˛ D tan ˇ D 8 12 4 12 D 0.667 8 in. y B 12 in α ˛ D 33.7° 8 in A D D 0.333 ˇ D 18.4° β 4 in C The angle between the cable AB and the positive x axis is 180° ˛, the tension in AB is: x TAB D jTAB ji cos180 ˛ C j sin180 ˛ TAB D jTAB ji cos ˛ C j sin ˛. The angle between AC and the positive x axis is 180 C ˇ. The tension is TAC D jTAC ji cos180 C ˇ C j sin180 C ˇ TAC D jTAC ji cos ˇ j sin ˇ. The tension in the cable AD is TAD D jTAD ji C 0j. The equilibrium conditions are F D TAC C TAB C TAD D 0. Substitute and collect like terms, Fx D jTAB j cos ˛ jTAC j cos ˇ C jTAD ji D 0 Fy D jTAB j sin ˛ jTAC j sin ˇj D 0. Solve: jTAB j D and jTAC j D sin ˇ sin ˛ jTAC j, sin ˛ sin˛ C ˇ jTAD j. For jTAD j D 200 lb, ˛ D 33.7° , ˇ D 18.4° jTAC j D 140.6 lb, jTAB j D 80.1 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 111 Problem 3.42 You are designing a cable system to support a suspended object of weight W. Because your design requires points A and B to be placed as shown, you have no control over the angle ˛, but you can choose the angle ˇ by placing point C wherever you wish. Show that to minimize the tensions in cables AB and BC, you must choose ˇ D ˛ if the angle ˛ ½ 45° . B A Strategy: Draw a diagram of the sum of the forces exerted by the three cables at A. W Solution: Draw the free body diagram of the knot at point A. Then y TAC draw the force triangle involving the three forces. Remember that ˛ is fixed and the force W has both fixed magnitude and direction. From the force triangle, we see that the force TAC can be smaller than TAB for a large range of values for ˇ. By inspection, we see that the minimum simultaneous values for TAC and TAB occur when the two forces are equal. This occurs when ˛ D ˇ. Note: this does not happen when ˛ < 45° . TAB and B α x A W In this case, we solved the problem without writing the equations of equilibrium. For reference, these equations are: C β α B Possible locations for C lie on line C? C? α TAB Fx D TAB cos ˛ C TAC cos ˇ D 0 Fy D TAB sin ˛ C TAC sin ˇ W D 0. Candidate β W Candidate values for TAC Fixed direction for line AB Problem 3.43* The length of the cable ABC is 1.4 m. The 2-kN force is applied to a small pulley. The system is stationary. What is the tension in the cable? 1m A C B 0.75 m 15⬚ Solution: Examine the geometry h2 C 0.75 m2 C tan ˛ D ) 0.75 m 2 kN 0.25 m β α h2 C 0.25 m2 D 1.4 m h h h , tan ˇ D 0.75 m 0.25 m h D 0.458 m, ˛ D 31.39° , ˇ D 61.35° Now draw a FBD and solve for the tension. We can use either of the equilibrium equations Fx : T cos ˛ C T cos ˇ C 2 kN sin 15° D 0 T T β α T D 1.38 kN 2 kN 15° 112 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.44 The masses m1 D 12 kg and m2 D 6 kg are suspended by the cable system shown. The cable BC is horizontal. Determine the angle ˛ and the tensions in the cables AB, BC, and CD. TAB A D α TBC α B B C 70⬚ m1 m2 117.7 N Solution: We have 4 unknowns and 4 equations TCD FBx : TAB cos ˛ C TBC D 0 FBy : TAB sin ˛ 117.7 N D 0 FCx : TBC C TCD cos 70° D 0 70° TBC C FCy : TCD sin 70° 58.86 N D 0 Solving we find ˛ D 79.7° , TAB D 119.7 N, TBC D 21.4 N, TCD D 62.6 N Problem 3.45 The weights W1 D 50 lb and W2 are suspended by the cable system shown. Determine the weight W2 and the tensions in the cables AB, BC, and CD. 58.86 N 30 in 30 in 30 in A D 16 in 20 in C B W2 W1 Solution: We have 4 unknowns and 4 equilibrium equations to use 3 15 TBC D 0 FBx : p TAB C p 229 13 TAB 2 15 2 3 2 2 TBC 50 lb D 0 FBy : p TAB C p 229 13 TBC B 15 15 TBC C TCD D 0 FCx : p 17 229 50 lb 2 8 TBC C TCD W2 D 0 FCy : p 17 229 TCD W2 D 25 lb, TAB D 75.1 lb 8 ) C TBC D 63.1 lb, TCD D 70.8 lb TBC 15 15 2 W2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 113 Problem 3.46 In the system shown in Problem 3.45, assume that W2 D W1 /2. If you don’t want the tension anywhere in the supporting cable to exceed 200 lb, what is the largest acceptable value of W1 ? TAB Solution: 3 15 TBC D 0 FBx : p TAB C p 229 13 2 15 2 3 2 2 TBC W1 D 0 FBy : p TAB C p 229 13 TBC B 15 15 TBC C TCD D 0 FCx : p 17 229 W1 2 8 W1 TCD D0 TBC C FCy : p 17 2 229 TCD TAB D 1.502W1 , TBC D 1.262W1 , TCD D 1.417W1 8 C AB is the critical cable 200 lb D 1.502W1 ) W1 D 133.2 lb TBC 15 2 15 W2 = W1/2 114 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.47 The hydraulic cylinder is subjected to three forces. An 8-kN force is exerted on the cylinder at B that is parallel to the cylinder and points from B toward C. The link AC exerts a force at C that is parallel to the line from A to C. The link CD exerts a force at C that is parallel to the line from C to D. (a) (b) Draw the free-body diagram of the cylinder. (The cylinder’s weight is negligible). Determine the magnitudes of the forces exerted by the links AC and CD. Solution: From the figure, if C is at the origin, then points A, B, and D are located at 1m D C Hydraulic cylinder 1m 0.6 m B A 0.15 m 0.6 m Scoop y A0.15, 0.6 B0.75, 0.6 D FCD D1.00, 0.4 and forces FCA , FBC , and FCD are parallel to CA, BC, and CD, respectively. C x FBC We need to write unit vectors in the three force directions and express the forces in terms of magnitudes and unit vectors. The unit vectors are given by eCA D rCA D 0.243i 0.970j jrCA j eCB D rCB D 0.781i 0.625j jrCB j eCD D rCD D 0.928i C 0.371j jrCD j FCA A B Now we write the forces in terms of magnitudes and unit vectors. We can write FBC as FCB D 8eCB kN or as FCB D 8eCB kN (because we were told it was directed from B toward C and had a magnitude of 8 kN. Either way, we must end up with FCB D 6.25i C 5.00j kN Similarly, FCA D 0.243FCA i 0.970FCA j FCD D 0.928FCD i C 0.371FCD j For equilibrium, FCA C FCB C FCD D 0 In component form, this gives Fx D 0.243FCA C 0.928FCD 6.25 (kN) D 0 F D 0.970F C 0.371F C 5.00 (kN) D 0 y CA CD Solving, we get FCA D 7.02 kN, FCD D 4.89 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 115 Problem 3.48 The 50-lb cylinder rests on two smooth surfaces. (a) (b) Draw the free-body diagram of the cylinder. If ˛ D 30° , what are the magnitudes of the forces exerted on the cylinder by the left and right surfaces? α Solution: Isolate the cylinder. (a) The free body diagram of the 45° y isolated cylinder is shown. (b) The forces acting are the weight and the normal forces exerted by the surfaces. The angle between the normal force on the right and the x axis is 90 C ˇ. The normal force is β α NL NR D jNR ji cos90 C ˇ C j sin90 C ˇ NR W x NR D jNR ji sin ˇ C j cos ˇ. The angle between the positive x axis and the left hand force is normal 90 ˛; the normal force is NL D jNL ji sin ˛ C j cos ˛. The weight is W D 0i jWjj. The equilibrium conditions are jNR j D F D W C NR C NL D 0. and jNL j D Substitute and collect like terms, Solve: sin ˛ sin ˇ jNL j, jWj sin ˇ sin˛ C ˇ . For jWj D 50 lb, and ˛ D 30° , ˇ D 45° , the normal forces are Fx D jNR j sin ˇ C jNL j sin ˛i D 0, jNL j D 36.6 lb, jNR j D 25.9 lb Fy D jNR j cos ˇ C jNL j cos ˛ jWjj D 0. Problem 3.49 For the 50-lb cylinder in Problem 3.48, obtain an equation for the force exerted on the cylinder by the left surface in terms of the angle ˛ in two ways: (a) using a coordinate system with the y axis vertical, (b) using a coordinate system with the y axis parallel to the right surface. y Solution: The solution for Part (a) is given in Problem 3.48 (see free body diagram). jNR j D sin ˛ sin ˇ jNL j jNL j D jWj sin ˇ sin˛ C ˇ . Part (b): The isolated cylinder with the coordinate system is shown. The angle between the right hand normal force and the positive x axis is 180° . The normal force: NR D jNR ji C 0j. The angle between the left hand normal force and the positive x is 180 ˛ C ˇ. The normal force is NL D jNL ji cos˛ C ˇ C j sin˛ C ˇ. The angle between the weight vector and the positive x axis is ˇ. The weight vector is W D jWji cos ˇ j sin ˇ. The equilibrium conditions are NR NL W x Substitute and collect like terms, Fx D jNR j jNL j cos˛ C ˇ C jWj cos ˇi D 0, Fy D jNL j sin˛ C ˇ jWj sin ˇj D 0. F D W C NR C NL D 0. Solve: 116 β α jNL j D jWj sin ˇ sin˛ C ˇ c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.50 The two springs are identical, with unstretched length 0.4 m. When the 50-kg mass is suspended at B, the length of each spring increases to 0.6 m. What is the spring constant k? 0.6 m A C k k B Solution: F F F D k0.6 m 0.4 m Fy : 2F sin 60° 490.5 N D 0 60° 60° k D 1416 N/m 490.5 N Problem 3.51 The cable AB is 0.5 m in length. The unstretched length of the spring is 0.4 m. When the 50-kg mass is suspended at B, the length of the spring increases to 0.45 m. What is the spring constant k? 0.7 m A C k B Solution: The Geometry 0.7 m Law of Cosines and Law of Sines φ θ 0.72 D 0.52 C 0.452 20.50.45 cos ˇ sin sin sin ˇ D D 0.45 m 0.5 m 0.7 m 0.5 m 0.45 m β ˇ D 94.8° , D 39.8° D 45.4° Now do the statics TAB F F D k0.45 m 0.4 m Fx : TAB cos C F cos D 0 θ φ Fy : TAB sin C F sin 490.5 N D 0 Solving: k D 7560 N/m 490.5 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 117 Problem 3.52 The small sphere of mass m is attached to a string of length L and rests on the smooth surface of a fixed sphere of radius R. The center of the sphere is directly below the point where the string is attached. Obtain an equation for the tension in the string in terms of m, L, h, and R. L h m R Solution: From the geometry we have cos D x RChy , sin D L L cos D x y , sin D R R Thus the equilibrium equations can be written x x Fx : T C N D 0 L R Fy : RChy y T C N mg D 0 L R Solving, we find T D mgL RCh Problem 3.53 The inclined surface is smooth. Determine the force T that must be exerted on the cable to hold the 100-kg crate in equilibrium and compare your answer to the answer of Problem 3.11. T 60⬚ 3T Solution: 981 N F- : 3 T 981 N sin 60° D 0 T D 283 N N 60° 118 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.54 In Example 3.3, suppose that the mass of the suspended object is mA and the masses of the pulleys are mB D 0.3mA , mC D 0.2mA , and mD D 0.2mA . Show that the force T necessary for the system to be in equilibrium is 0.275mA g. D C B T A Solution: From the free-body diagram of pulley C TD TD 2T mC g D 0 ) TD D 2T C mC g Then from the free-body diagram of pulley B D C mg T C T C 2T C mC g mB g mA g D 0 T Thus TD TD C 1 mA C mB mC g 4 T D 0.275mA g TD ⫽ 2T ⫹ mg B 1 mA C 0.3mA 0.2mA g D 0.275mA g 4 T (a) T T T A B mg mAg (b) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 119 Problem 3.55 The mass of each pulley of the system is m and the mass of the suspended object A is mA . Determine the force T necessary for the system to be in equilibrium. Solution: Draw free body diagrams of each pulley and the object A. Each pulley and the object A must be in equilibrium. The weights of the pulleys and object A are W D mg and WA D mA g. The equilibrium equations for the weight A, the lower pulley, second pulley, third pulley, and the top pulley are, respectively, B WA D 0, 2C B W D 0, 2D C W D 0, 2T D W D 0, and FS 2T W D 0. Begin with the first equation and solve for B, substitute for B in the second equation and solve for C, substitute for C in the third equation and solve for D, and substitute for D in the fourth equation and solve for T, to get T in terms of W and WA . The result is T A Fs W T W B D WA , DD CD WA W C , 2 2 3W WA 7W WA C , and T D C , 4 4 8 8 or in terms of the masses, TD 120 g mA C 7m. 8 T T T W D D D C W C C B B WA c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.56 The suspended mass m1 D 50 kg. Neglecting the masses of the pulleys, determine the value of the mass m2 necessary for the system to be in equilibrium. A B C m2 m1 Solution: T1 T T FC : T1 C 2m2 g m1 g D 0 FB : T1 2m2 g D 0 C m2 D m1 D 12.5 kg 4 T1 m1 g B T = m2 g T c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 121 Problem 3.57 The boy is lifting himself using the block and tackle shown. If the weight of the block and tackle is negligible, and the combined weight of the boy and the beam he is sitting on is 120 lb, what force does he have to exert on the rope to raise himself at a constant rate? (Neglect the deviation of the ropes from the vertical.) Solution: A free-body diagram can be obtained by cutting the four ropes between the two pulleys of the block and tackle and the rope the boy is holding. The tension has the same value T in all five of these ropes. So the upward force on the free-body diagram is 5T and the downward force is the 120-lb weight. Therefore the force the boy must exert is T D 120 lb/5 D 24 lb T D 24 lb 122 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.58 Pulley systems containing one, two, and three pulleys are shown. Neglecting the weights of the pulleys, determine the force T required to support the weight W in each case. T T T W (a) One pulley W (b) Two pulleys W (c) Three pulleys Solution: (a) (b) (b) For two pulleys T Fy : 2T W D 0 ) T D Fupper : 2T T1 D 0 Flower : 2T1 W D 0 T1 W TD 4 (c) T W 2 T1 Fupper : 2T T1 D 0 Fmiddle : 2T1 T2 D 0 W Flower : 2T2 W D 0 (c) For three pulleys TD T W 8 T (a) For one pulleys T T T1 T1 W T2 T2 T2 W c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 123 Problem 3.59 Problem 3.58 shows pulley systems containing one, two, and three pulleys. The number of pulleys in the type of system shown could obviously be extended to an arbitrary number N. Solution: By extrapolation of the previous problem Neglecting the weights of the pulleys, determine the force T required to support the weight W as a function of the number of pulleys N in the system. (b) Using the result of part (a), determine the force T required to support the weight W for a system with 10 pulleys. (a) TD W 2N (b) TD W 1024 (a) Problem 3.60 A 14,000-kg airplane is in steady flight in the vertical plane. The flight path angle is D 10° , the angle of attack is ˛ D 4° , and the thrust force exerted by the engine is T D 60 kN. What are the magnitudes of the lift and drag forces acting on the airplane? (See Example 3.4). Solution: Let us draw a more detailed free body diagram to see the angles involved more clearly. Then we will write the equations of equilibrium and solve them. y L W D mg D 14,0009.81 N The equilibrium equations are x Fx D T cos ˛ D W sin D 0 Fy D T sin ˛ C L W cos D 0 α α = 4° γ = 10° T γ D W T D 60 kN D 60000 N Solving, we get γ D D 36.0 kN, L D 131.1 kN y Path x T γ L α D Horizon W 124 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.61 An airplane is in steady flight, the angle of attack ˛ D 0, the thrust-to-drag ratio T/D D 2, and the lift-to-drag ratio L/D D 4. What is the flight path angle ? (See Example 3.4). Solution: Use the same strategy as in Problem 3.52. The angle between the thrust vector and the positive x axis is ˛, T D jTji cos ˛ C j sin ˛ The lift vector: L D 0i C jLjj The drag: D D jDji C 0j. The angle between the weight vector and the positive x axis is 270 ; W D jWji sin j cos . The equilibrium conditions are F D T C L C D C W D 0. Substitute and collect like terms and Fx D jTj cos ˛ jDj jWj sin i D 0, Fy D jTj sin ˛ C jLj jWj cos j D 0 Solve the equations for the terms in : jWj sin D jTj cos ˛ jDj, and jWj cos D jTj sin ˛ C jLj. Take the ratio of the two equations tan D jTj cos ˛ jDj jTj sin ˛ C jLj . Divide top and bottom on the right by jDj. For ˛ D 0, jLj jTj D 2, D 4, jDj jDj tan D 21 4 D 1 or D 14° 4 y Path T L α D x γ Horizontal W c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 125 Problem 3.62 An airplane glides in steady flight (T D 0), and its lift-to-drag ratio is L/D D 4. (a) What is the flight path angle ? (b) If the airplane glides from an altitude of 1000 m to zero altitude, what horizontal distance does it travel? (See Example 3.4.) Solution: See Example 3.4. The angle between the thrust vector y and the positive x axis is ˛: Path x T T D jTji cos ˛ C j sin ˛. L α The lift vector: L D 0i C jLjj. The drag: D D jDji C 0j. The angle between the weight vector and the positive x axis is 270 : γ Horizontal D W W D jWji sin j cos . The equilibrium conditions are F D T C L C D C W D 0. γ Substitute and collect like terms: 1 km Fx D jTj cos ˛ jDj jWj sin i D 0 γ Fy D jTj sin ˛ C jLj jWj cos j D 0 h Solve the equations for the terms in , jWj sin D jTj cos ˛ jDj, and jWj cos D jTj sin ˛ C jLj Part (a): Take the ratio of the two equilibrium equations: tan D jTj cos ˛ jDj jTj sin ˛ C jLj . Divide top and bottom on the right by jDj. For ˛ D 0, jTj D 0, jLj D 4, jDj tan D 1 4 D 14° Part (b): The flight path angle is a negative angle measured from the horizontal, hence from the equality of opposite interior angles the angle is also the positive elevation angle of the airplane measured at the point of landing. tan D 126 1 , h hD 1 1 D D 4 km 1 tan 4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.63 In Active Example 3.5, suppose that the attachment point B is moved to the point (5,0,0) m. What are the tensions in cables AB, AC, and AD? C (⫺2, 0, ⫺2) m y x B (4, 0, 2) m (⫺3, 0, 3) m D Solution: The position vector from point A to point B can be used to write the force TAB . A (0, ⫺4, 0) m z rAB D 5i C 4j m TAB D TAB rAB D TAB 0.781i C 0.625j jrAB j 100 kg Using the other forces from Active Example 3.5, we have Fx : 0.781TAB 0.408TAC 0.514TAD D 0 Fy : 0.625TAB C 0.816TAC C 0.686TAD 981 N D 0 Fz : 0.408TAC C 0.514TAD D 0 Solving yields TAB D 509 N, TAC D 487 N, TAD D 386 N y Problem 3.64 The force F D 800i C 200j (lb) acts at point A where the cables AB, AC, and AD are joined. What are the tensions in the three cables? F D (0, 6, 0) ft A (12, 4, 2) ft C B (0, 4, 6) ft (6, 0, 0) ft x z Solution: We first write the position vectors rAB D 6i 4j 2k ft rAC D 12i C 6k ft rAD D 12i C 2j 2k ft Now we can use these vectors to define the force vectors TAB D TAB rAB D TAB 0.802i 0.535j 0.267k jrAB j TAC D TAC rAC D TAC 0.949i C 0.316k jrAC j TAD D TAD rAD D TAD 0.973i C 0.162j 0.162k jrAD j The equilibrium equations are then Fx : 0.802TAB 0.949TAC 0.973TAD C 800 lb D 0 Fy : 0.535TAB C 0.162TAD C 200 lb D 0 Fz : 0.267TAB C 0.316TAC 0.162TAD D 0 Solving, we find TAB D 405 lb, TAC D 395 lb, TAD D 103 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 127 y Problem 3.65 Suppose that you want to apply a 1000-lb force F at point A in a direction such that the resulting tensions in cables AB, AC, and AD are equal. Determine the components of F. F D (0, 6, 0) ft A (12, 4, 2) ft C B (0, 4, 6) ft (6, 0, 0) ft x z Solution: We first write the position vectors rAB D 6i 4j 2k ft rAC D 12i C 6k ft rAD D 12i C 2j 2k ft Now we can use these vectors to define the force vectors TAB D T rAB D T0.802i 0.535j 0.267k jrAB j TAC D T rAC D T0.949i C 0.316k jrAC j TAD D T rAD D T0.973i C 0.162j 0.162k jrAD j The force F can be written F D Fx i C Fy j C Fz k The equilibrium equations are then Fx : 0.802T 0.949T 0.973T C Fx D 0 ) Fx D 2.72T Fy : 0.535T C 0.162T C Fy D 0 ) Fy D 0.732T Fz : 0.267T C 0.316T 0.162T C Fz D 0 ) Fz D 0.113T We also have the constraint equation ) T D 363 lb Fx 2 C Fy 2 C Fz 2 D 1000 lb Solving, we find Fx D 990 lb, Fy D 135 lb, Fz D 41.2 lb 128 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.66 The 10-lb metal disk A is supported by the smooth inclined surface and the strings AB and AC. The disk is located at coordinates (5,1,4) ft. What are the tensions in the strings? y B (0, 6, 0) ft C (8, 4, 0) ft Solution: The position vectors are 2 ft rAB D 5i C 5j 4k ft A x 8 ft rAC D 3i C 3j 4k ft z 10 ft The angle ˛ between the inclined surface the horizontal is ˛ D tan1 2/8 D 14.0° We identify the following force: TAB D TAB rAB D TAB 0.615i C 0.615j 0.492k jrAB j TAC D TAC rAC D TAC 0.514i C 0.514j 0.686k jrAC j N D Ncos ˛j C sin ˛k D N0.970j C 0.243k W D 10 lbj The equilibrium equations are then Fx : 0.615TAB C 0.514TAC D 0 Fy : 0.615TAB C 0.514TAC C 0.970N 10 lb D 0 Fz : 0.492TAB 0.686TAC C 0.243N D 0 Solving, we find N D 8.35 lb TAB D 1.54 lb, TAC D 1.85 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 129 Problem 3.67 The bulldozer exerts a force F D 2i (kip) at A. What are the tensions in cables AB, AC, and AD? y 6 ft C 8 ft 2 ft Solution: Isolate the cable juncture. Express the tensions in terms of unit vectors. Solve the equilibrium equations. The coordinates of points A, B, C, D are: A8, 0, 0, B0, 3, 8, C0, 2, 6, B A 3 ft D0, 4, 0. D z The radius vectors for these points are 4 ft rA D 8i C 0j C 0k, rB D 0i C 3j C 8k, rC D 0i C 2j 6k, rD D 0i C 4j C 0k. 8 ft x By definition, the unit vector parallel to the tension in cable AB is eAB D rB r A . jrB rA j Carrying out the operations for each of the cables, the results are: eAB D 0.6835i C 0.2563j C 0.6835k, eAC D 0.7845i C 0.1961j 0.5883k, eAD D 0.8944i 0.4472j C 0k. The tensions in the cables are expressed in terms of the unit vectors, TAB D jTAB jeAB , TAC D jTAC jeAC , TAD D jTAD jeAD . The external force acting on the juncture is F D 2000i C 0j C 0k. The equilibrium conditions are F D 0 D TAB C TAC C TAD C F D 0. Substitute the vectors into the equilibrium conditions: Fx D 0.6835jTAB j 0.7845jTAC j 0.8944jTAD jC2000i D 0 Fy D 0.2563jTAB j C 0.1961jTAC j 0.4472jTAD jj D 0 Fz D 0.6835jTAB j 0.5883jTAC j C 0jTAD jk D 0 The commercial program TK Solver Plus was used to solve these equations. The results are jTAB j D 780.31 lb , 130 jTAC j D 906.49 lb , jTAD j D 844.74 lb . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 3.68 Prior to its launch, a balloon carrying a set of experiments to high altitude is held in place by groups of student volunteers holding the tethers at B, C, and D. The mass of the balloon, experiments package, and the gas it contains is 90 kg, and the buoyancy force on the balloon is 1000 N. The supervising professor conservatively estimates that each student can exert at least a 40-N tension on the tether for the necessary length of time. Based on this estimate, what minimum numbers of students are needed at B, C, and D? A (0, 8, 0) m C (10,0, –12) m D (–16, 0, 4) m x B (16, 0, 16) m z Solution: 1000 N Fy D 1000 909.81 T D 0 T D 117.1 N (90) g A0, 8, 0 B16, 0, 16 T C10, 0, 12 D16, 0, 4 We need to write unit vectors eAB , eAC , and eAD . y T eAB D 0.667i 0.333j C 0.667k (0, 8, 0) eAC D 0.570i 0.456j 0.684k A FAC eAD D 0.873i 0.436j C 0.218k FAD We now write the forces in terms of magnitudes and unit vectors FAB D 0.667FAB i 0.333FAB j C 0.667FAB k FAC D 0.570FAC i 0.456FAC j 0.684FAC k FAD D 0.873FAD i 0.436FAC j C 0.218FAC k T D 117.1j (N) C (10, 0, −12) m D x (−16, 0, 4) z B (16, 0, 16) m The equations of equilibrium are Fx D 0.667FAB C 0.570FAC 0.873FAD D 0 Fy D 0.333FAB 0.456FAC 0.436FAC C 117.1 D 0 Fz D 0.667FAB 0.684FAC C 0.218FAC D 0 Solving, we get FAB D 64.8 N ¾ 2 students FAC D 99.8 N ¾ 3 students FAD D 114.6 N ¾ 3 students c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 131 Problem 3.69 The 20-kg mass is suspended by cables attached to three vertical 2-m posts. Point A is at (0, 1.2, 0) m. Determine the tensions in cables AB, AC, and AD. y C B D A 1m 1m 2m 0.3 m x z Solution: Points A, B, C, and D are located at A0, 1.2, 0, B0.3, 2, 1, C0, 2, 1, D2, 2, 0 y C FAC B FAB FAD Write the unit vectors eAB , eAC , eAD D A eAB D 0.228i C 0.608j C 0.760k W eAC D 0i C 0.625j 0.781k eAD D 0.928i C 0.371j C 0k z (20) (9.81) N x The forces are FAB D 0.228FAB i C 0.608FAB j C 0.760FAB k FAC D 0FAC i C 0.625FAC j 0.781FAC k FAD D 0.928FAD i C 0.371FAD j C 0k W D 209.81j The equations of equilibrium are Fx D 0.228FAB C 0 C 0.928FAD D 0 Fy D 0.608FAB C 0.625FAC C 0.371FAD 209.81 D 0 Fz D 0.760FAB 0.781FAC C 0 D 0 We have 3 eqns in 3 unknowns solving, we get FAB D 150.0 N FAC D 146.1 N FAD D 36.9 N 132 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.70 The weight of the horizontal wall section is W D 20,000 lb. Determine the tensions in the cables AB, AC, and AD. Solution: Set the coordinate origin at A with axes as shown. The upward force, T, at point A will be equal to the weight, W, since the cable at A supports the entire wall. The upward force at A is T D W k. From the figure, the coordinates of the points in feet are A D 10 ft A4, 6, 10, B0, 0, 0, C12, 0, 0, and 7 ft D4, 14, 0. The three unit vectors are of the form 6 ft C B 4 ft xI xA i C yI yA j C zI zA k , eAI D xI xA 2 C yI yA 2 C zI zA 2 14 ft 8 ft W where I takes on the values B, C, and D. The denominators of the unit vectors are the distances AB, AC, and AD, respectively. Substitution of the coordinates of the points yields the following unit vectors: T z eAB D 0.324i 0.487j 0.811k, A y TD eAC D 0.566i 0.424j 0.707k, 10 ft TB D 7 ft TC and eAD D 0i C 0.625j 0.781k. 6 ft The forces are TAB D TAB eAB , 14 ft 4 ft TAC D TAC eAC , and TAD D TAD eAD . C B X 8 ft W The equilibrium equation for the knot at point A is T C TAB C TAC C TAD D 0. From the vector equilibrium equation, write the scalar equilibrium equations in the x, y, and z directions. We get three linear equations in three unknowns. Solving these equations simultaneously, we get TAB D 9393 lb, TAC D 5387 lb, and TAD D 10,977 lb A D 10 ft 6 ft C B 4 ft 8 ft 7 ft 14 ft W c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 133 Problem 3.71 The car in Fig. a and the pallet supporting it weigh 3000 lb. They are supported by four cables AB, AC, AD, and AE. The locations of the attachment points on the pallet are shown in Fig. b. The tensions in cables AB and AE are equal. Determine the tensions in the cables. y A (0, 10, 0) ft Solution: Isolate the knot at A. Let TAB , TAC , TAD and TAE be the forces exerted by the tensions in the cables. The force exerted by the vertical cable is (3000 lb)j. We first find the position vectors and then express all of the forces as vectors. rAB D 5i 10j C 5k ft rAC D 6i 10j 5k ft E C rAD D 8i 10j 4k ft z rAE D 6i 10j C 5k ft B TAB rAB D TAB 0.408i 0.816j C 0.408k D TAB jrAB j x (a) TAC D TAC rAC D TAC 0.473i 0.788j 0.394k jrAC j TAD rAD D TAD D TAD 0.596i 0.745j 0.298k jrAD j TAE rAE D TAE 0.473i 0.788j C 0.394k D TAE jrAE j 8 ft 6 ft C D 5 ft 4 ft x The equilibrium equations are 5 ft Fx : 0.408TAB C 0.473TAC 0.596TAD 0.473TAE D 0 E Fy : 0.816TAB 0.788TAC 0.745TAD 0.788TAE C 3000 lb D 0 B 6 ft 5 ft z Fz : 0.408TAB 0.394TAC 0.298TAD C 0.394TAE D 0 (b) Solving, we find TAB D 896 lb, TAC D 1186 lb, TAD D 843 lb, TAE D 896 lb 134 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.72 The 680-kg load suspended from the helicopter is in equilibrium. The aerodynamic drag force on the load is horizontal. The y axis is vertical, and cable OA lies in the x-y plane. Determine the magnitude of the drag force and the tension in cable OA. y A 10° O x B C D y Solution: TOA Fx D TOA sin 10° D D 0, Fy D TOA cos 10° 6809.81 D 0. Solving, we obtain D D 1176 N, TOA D 6774 N. 10° D x (680) (9.81) N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 135 Problem 3.73 In Problem 3.72, the coordinates of the three cable attachment points B, C, and D are (3.3, 4.5, 0) m, (1.1, 5.3, 1) m, and (1.6, 5.4, 1) m, respectively. What are the tensions in cables OB, OC, and OD? Solution: The position vectors from O to pts B, C, and D are rOB D 3.3i 4.5j (m), rOC D 1.1i 5.3j C k (m), rOD D 1.6i 5.4j k (m). Dividing by the magnitudes, we obtain the unit vectors eOB D 0.591i 0.806j, eOC D 0.200i 0.963j C 0.182k, eOD D 0.280i 0.944j 0.175k. Using these unit vectors, we obtain the equilibrium equations Fx D TOA sin 10° 0.591TOB C 0.200TOC C 0.280TOD D 0, Fy D TOA cos 10° 0.806TOB 0.963TOC 0.944TOD D 0, Fz D 0.182TOC 0.175TOD D 0. From the solution of Problem 3.72, TOA D 6774 N. Solving these equations, we obtain TOB D 3.60 kN, TOC D 1.94 kN, TOD D 2.02 kN. y TOA 10° x TOB TOC 136 TOD c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.74 If the mass of the bar AB is negligible compared to the mass of the suspended object E, the bar exerts a force on the “ball” at B that points from A toward B. The mass of the object E is 200 kg. The y-axis points upward. Determine the tensions in the cables BC and CD. y (0, 4, ⫺3) m C B (4, 3, 1) m D (0, 5, 5) m Strategy: Draw a free-body diagram of the ball at B. (The weight of the ball is negligible.) x A E z Solution: FAB D FAB 4i 3j k p 26 , TBC D TBC 4i C j 4k p 33 , The forces TBD D TBD 4i C 2j C 4k 6 , W D 200 kg9.81 m/s2 j The equilibrium equations 4 4 4 Fx : p FAB p TBC TBD D 0 6 26 33 3 1 2 Fy : p FAB C p TBC C TBD 1962 N D 0 6 26 33 1 4 4 Fz : p FAB p TBC C TBD D 0 6 26 33 TBC D 1610 N ) TBD D 1009 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 137 Problem 3.75* The 3400-lb car is at rest on the plane surface. The unit vector en D 0.456i C 0.570j C 0.684k is perpendicular to the surface. Determine the magnitudes of the total normal force N and the total friction force f exerted on the surface by the car’s wheels. y en Solution: The forces on the car are its weight, the normal force, and the friction force. The normal force is in the direction of the unit vector, so it can be written N D Nen D N0.456i C 0.570j C 0.684k The equilibrium equation is x z Nen C f 3400 lbj D 0 The friction force f is perpendicular to N, so we can eliminate the friction force from the equilibrium equation by taking the dot product of the equation with en . Nen C f 3400 lbj Ð en D N 3400 lbj Ð en D 0 N D 3400 lb0.57 D 1940 lb Now we can solve for the friction force f. f D 3400 lbj Nen D 3400 lbj 1940 lb0.456i C 0.570j C 0.684k f D 884i C 2300j 1330k lb jfj D 884 lb2 C 2300 lb2 C 1330 lb2 D 2790 lb jNj D 1940 lb, jfj D 2790 lb 138 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 3.76 The system shown anchors a stanchion of a cable-suspended roof. If the tension in cable AB is 900 kN, what are the tensions in cables EF and EG? G (0, 1.4, ⫺1.2) m Solution: Using the coordinates for the points we find E F rBA D [3.4 2i C 1 1j C 0 0k] m (2, 1, 0) m (1, 1.2, 0) m rBA D 1.4 mi eBA rBA Di D jrBA j (3.4, 1, 0) m B (0, 1.4, 1.2) m A (2.2, 0, ⫺1) m Using the same procedure we find the other unit vectors that we need. D eBC D 0.140i 0.700j C 0.700k z x (2.2, 0, 1) m C eBD D 0.140i 0.700j C 0.700k eBE D 0.981i C 0.196j eEG D 0.635i C 0.127j 0.762k eEF D 0.635i C 0.127j C 0.762k We can now write the equilibrium equations for the connections at B and E. 900 kNeBA C TBC eBC C TBD eBD C TBE eBE D 0, TBE eBE C TEF eEF C TEG eEG 0 Breaking these equations into components, we have the following six equations to solve for five unknows (one of the equations is redundant). 900 kN C TBC 0.140 C TBD 0.140 C TBE 0.981 D 0 TBC 0.700 C TBD 0.700 C TBE 0.196 D 0 TBC 0.700 C TBD 0.700 D 0 TBE 0.981 C TEG 0.635 C TEF 0.635 D 0 TBE 0.196 C TEG 0.127 C TEF 0.127 D 0 TEG 0.726 C TEF 0.726 D 0 Solving, we find TBC D TBD D 134 kN, TBE D 956 kN TEF D TEG D 738 kN y Problem 3.77* The cables of the system will each safely support a tension of 1500 kN. Based on this criterion, what is the largest safe value of the tension in cable AB? Solution: From Problem 3.76 we know that if the tension in AB is 900 kN, then the largest force in the system occurs in cable BE and that tension is 956 kN. To solve this problem, we can just scale the results from Problem 3.76 1500 kN 900 kN TAB D 1410 kN TAB D 956 kN G (0, 1.4, ⫺1.2) m F E (2, 1, 0) m (1, 1.2, 0) m (3.4, 1, 0) m B (0, 1.4, 1.2) m A (2.2, 0, ⫺1) m D z C x (2.2, 0, 1) m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 139 y Problem 3.78 The 200-kg slider at A is held in place on the smooth vertical bar by the cable AB. 2m (a) Determine the tension in the cable. (b) Determine the force exerted on the slider by the bar. B A 5m 2m x 2m z Solution: The coordinates of the points A, B are A2, 2, 0, B0, 5, 2. The vector positions rA D 2i C 2j C 0k, T rB D 0i C 5j C 2k N The equilibrium conditions are: F D T C N C W D 0. W Eliminate the slider bar normal force as follows: The bar is parallel to the y axis, hence the unit vector parallel to the bar is eB D 0i C 1j C 0k. The dot product of the unit vector and the normal force vanishes: eB Ð N D 0. Take the dot product of eB with the equilibrium conditions: eB Ð N D 0. eB Ð F D eB Ð T C eB Ð W D 0. The weight is eB Ð W D 1j Ð jjWj D jWj D 2009.81 D 1962 N. Note: For this specific configuration, the problem can be solved without eliminating the slider bar normal force, since it does not appear in the y-component of the equilibrium equation (the slider bar is parallel to the y-axis). However, in the general case, the slider bar will not be parallel to an axis, and the unknown normal force will be projected onto all components of the equilibrium equations (see Problem 3.79 below). In this general situation, it will be necessary to eliminate the slider bar normal force by some procedure equivalent to that used above. End Note. The unit vector parallel to the cable is by definition, eAB D rB r A . jrB rA j Substitute the vectors and carry out the operation: eAB D 0.4851i C 0.7278j C 0.4851k. (a) The tension in the cable is T D jTjeAB . Substitute into the modified equilibrium condition eB F D 0.7276jTj 1962 D 0. Solve: jTj D 2696.5 N from which the tension vector is T D jTjeAB D 1308i C 1962j C 1308k. (b) The equilibrium conditions are F D 0 D T C N C W D 1308i C 1308k C N D 0. Solve for the normal force: N D 1308i 1308k. The magnitude is jNj D 1850 N. 140 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.79 In Example 3.6, suppose that the cable AC is replaced by a longer one so that the distance from point B to the slider C increases from 6 ft to 8 ft. Determine the tension in the cable. Solution: The vector from B to C is now y B 4 ft A 6 ft rBC D 8 ft eBD rBC D 8 ft 7 4 4 i jC k 9 9 9 7 ft C rBC D 3.56i 6.22j C 3.56k ft O x We can now find the unit vector form C to A. rCA D rOA rOB C rBC D [7j C 4k z 4 ft D 4 ft f7j C 3.56i 6.22j C 3.56kg] ft rCA D 3.56i C 6.22j C 0.444k ft eCA D rCA D 0.495i C 0.867j C 0.0619k jrCA j Using N to stand for the normal force between the bar and the slider, we can write the equilibrium equation: TeCA C N 100 lbj D 0 We can use the dot product to eliminate N from the equation [TeCA C N 100 lbj] Ð eBD D TeCA Ð eBD 100 lbj Ð eBD D 0 T 7 4 4 [0.495] C [0.867] C [0.0619] 100 lb0.778 D 0 9 9 9 T0.867 C 77.8 lb D 0 ) T D 89.8 lb T D 89.8 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 141 Problem 3.80 The cable AB keeps the 8-kg collar A in place on the smooth bar CD. The y axis points upward. What is the tension in the cable? y 0.15 m 0.4 m B Solution: We develop the following position vectors and unit C vectors rCD D 0.2i 0.3j C 0.25k m 0.2 m 0.3 m A 0.5 m eCD D rCD D 0.456i 0.684j C 0.570k jrCD j O x 0.25 m D 0.2 m rCA D 0.2 meCD D 0.091i 0.137j C 0.114k m z rAB D rOB rOC C rCA rAB D [0.5j C 0.15k f0.4i C 0.3jg C f0.091i 0.137j C 0.114kg] m rAB D 0.309i C 0.337j C 0.036k m eAB D rAB D 0.674i C 0.735j C 0.079k jrAB j We can now write the equilibrium equation for the slider using N to stand for the normal force between the slider and the bar CD. TeAB C N 8 kg9.81 m/s2 j D 0 To eliminate the normal force N we take a dot product with eCD . [TeAB C N 8 kg9.81 m/s2 j] Ð eCD D 0 TeAB Ð eCD 78.5 Nj Ð eCD D 0 T[0.674][0.456] C [0.735][0.684] C [0.079][0.570] 78.5 N0.684 D 0 T0.150 C 53.6 N D 0 T D 357 N Problem 3.81 Determine the magnitude of the normal force exerted on the collar A by the smooth bar. y 0.15 m Solution: From Problem 3.81 we have 0.4 m B C eAB D 0.674i C 0.735j C 0.079k T D 357 N 0.2 m 0.3 m A 0.5 m The equilibrium equation is O TeAB C N 78.5 Nj D 0 D We can now solve for the normal force N. x 0.25 m 0.2 m z N D 78.5 Nj 357 N0.674i C 0.735j C 0.079k N D 240i 184j 28.1k N The magnitude of N is jNj D 240 N2 C 184 N2 C 28.1 N2 jNj D 304 N 142 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 3.82* The 10-kg collar A and 20-kg collar B are held in place on the smooth bars by the 3-m cable from A to B and the force F acting on A. The force F is parallel to the bar. Determine F. (0, 5, 0) m (0, 3, 0) m F Solution: The geometry is the first part of the Problem. To ease our work, let us name the points C, D, E, and G as shown in the figure. The unit vectors from C to D and from E to G are essential to the location of points A and B. The diagram shown contains two free bodies plus the pertinent geometry. The unit vectors from C to D and from E to G are given by A 3m B (4, 0, 0) m eCD D erCDx i C eCDy j C eCDz k, (0, 0, 4) m z and eEG D erEGx i C eEGy j C eEGz k. y Using the coordinates of points C, D, E, and G from the picture, the unit vectors are D (0, 5, 0) m eCD D 0.625i C 0.781j C 0k, yA D yC C CAeCDy , B mBg z yB D yA C ABeABy , A TAB NB and zA D zC C CAeCDz , xB D xA C ABeABx , NA TAB The location of point A is given by where CA D 3 m. From these equations, we find that the location of point A is given by A (2.13, 2.34, 0) m. Once we know the location of point A, we can proceed to find the location of point B. We have two ways to determine the location of B. First, B is 3 m from point A along the line AB (which we do not know). Also, B lies on the line EG. The equations for the location of point B based on line AB are: F G (0, 3, 0) m and eEG D 0i C 0.6j C 0.8k. xA D xC C CAeCDx , mAg 3m C (4, 0, 0) m We now have two fewer equation than unknowns. Fortunately, there are two conditions we have not yet invoked. The bars at A and B are smooth. This means that the normal force on each bar can have no component along that bar. This can be expressed by using the dot product of the normal force and the unit vector along the bar. The two conditions are NA Ð eCD D NAx eCDx C NAy eCDy C NAz eCDz D 0 The equations based on line EG are: for slider A and yB D yE C EBeEGy , and zB D zE C EBeEGz . We have six new equations in the three coordinates of B and the distance EB. Some of the information in the equations is redundant. However, we can solve for EB (and the coordinates of B). We get that the length EB is 2.56 m and that point B is located at (0, 1.53, 1.96) m. We next write equilibrium equations for bodies A and B. From the free body diagram for A, we get NAx C TAB eABx C FeCDx D 0, x E (0, 0, 4) m and zB D zA C ABeABz . xB D xE C EBeEGx , x NB Ð eEG D NBx eEGx C NBy eEGy C NBz eEGz D 0. Solving the eight equations in the eight unknowns, we obtain F D 36.6 N . Other values obtained in the solution are EB D 2.56 m, NAx D 145 N, NBx D 122 N, NAy D 116 N, NBy D 150 N, NAz D 112 N, and NBz D 112 N. NAy C TAB eABy C FeCDy mA g D 0, and NAz C TAB eABz C FeCDz D 0. From the free body diagram for B, we get NBx TAB eABx D 0, Nby TAB eABy mB g D 0, and NBz TAB eABz D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 143 Problem 3.83 The 100-lb crate is held in place on the smooth surface by the rope AB. Determine the tension in the rope and the magnitude of the normal force exerted on the crate by the surface. A 45⬚ B 30⬚ Solution: The free-body diagram is sketched. The equilibrium equations are Fx - : T cos45° 30° 100 lb sin 30° D 0 F% : T sin45° 30° 100 lb cos 30° C N D 0 Solving, we find T D 51.8 lb, N D 73.2 lb Problem 3.84 The system shown is called Russell’s traction. If the sum of the downward forces exerted at A and B by the patient’s leg is 32.2 lb, what is the weight W? y 60⬚ 20⬚ 25⬚ B A W x Solution: The force in the cable is W everywhere. The free-body diagram of the leg is shown. The downward force is given, but the horizontal force FH is unknown. The equilibrium equation in the vertical direction is Fy : W sin 25° C W sin 60° 32.2 lb D 0 Thus WD 32.2 lb sin 25° C sin 60° W D 25.0 lb 144 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.85 The 400-lb engine block is suspended by the cables AB and AC. If you don’t want either TAB or TAC to exceed 400 lb, what is the smallest acceptable value of the angle ˛? y TAB B TAC C a A a A x 400 lb Solution: The equilibrium equations are Fx : TAB cos ˛ C TAC cos ˛ D 0 Fy : TAB sin ˛ C TAC sin ˛ 400 lb D 0 Solving, we find TAB D TAC D 400 lb 2 sin ˛ If we limit the tensions to 400 lb, we have 400 lb D 400 lb 1 ) sin ˛ D ) ˛ D 30° 2 sin ˛ 2 Problem 3.86 The cable AB is horizontal, and the box on the right weighs 100 lb. The surface are smooth. (a) What is the tension in the cable? (b) What is the weight of the box on the left? A B 20⬚ 40⬚ Solution: We have the following equilibrium equations FyB : NB cos 40° 100 lb D 0 FxB : NB sin 40° T D 0 FxA : T NA sin 20° D 0 FyA : NA cos 20° WA D 0 Solving these equations sequentially, we find NB D 131 lb, T D 83.9 lb NA D 245 lb, WA D 230.5 lb Thus we have T D 83.9 lb, WA D 230.5 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 145 Problem 3.87 Assume that the forces exerted on the 170-lb climber by the slanted walls of the “chimney” are perpendicular to the walls. If he is in equilibrium and is exerting a 160-lb force on the rope, what are the magnitudes of the forces exerted on him by the left and right walls? 10⬚ 4⬚ 3⬚ Solution: The forces in the free-body diagram are in the directions shown on the figure. The equilibrium equations are: Fx : T sin 10° C NL cos 4° NR cos 3° D 0 Fy : T cos 10° 170 lb C NL sin 40° C NR sin 3° D 0 where T D 160 lb. Solving we find NL D 114 lb, NR D 85.8 lb Left Wall: 114 lb Right Wall: 85.8 lb 146 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.88 The mass of the suspended object A is mA and the masses of the pulleys are negligible. Determine the force T necessary for the system to be in equilibrium. T A Solution: Break the system into four free body diagrams as shown. Carefully label the forces to ensure that the tension in any single cord is uniform. The equations of equilibrium for the four objects, starting with the leftmost pulley and moving clockwise, are: S 3T D 0, R 3S D 0, F F 3R D 0, R and 2T C 2S C 2R mA g D 0. We want to eliminate S, R, and F from our result and find T in terms of mA and g. From the first two equations, we get S D 3T, and R D 3S D 9T. Substituting these into the last equilibrium equation results in 2T C 23T C 29T D mA g. R R R S S S Solving, we get T D mA g/26 . S T T S S R R T T T A mAg Note: We did not have to solve for F to find the appropriate value of T. The final equation would give us the value of F in terms of mA and g. We would get F D 27mA g/26. If we then drew a free body diagram of the entire assembly, the equation of equilibrium would be F T mA g D 0. Substituting in the known values for T and F, we see that this equation is also satisfied. Checking the equilibrium solution by using the “extra” free body diagram is often a good procedure. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 147 Problem 3.89 The assembly A, including the pulley, weighs 60 lb. What force F is necessary for the system to be in equilibrium? F A Solution: From the free body diagram of the assembly A, we have 3F 60 D 0, or F D 20 lb F F F F F F F 60 lb. 148 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.90 The mass of block A is 42 kg, and the mass of block B is 50 kg. The surfaces are smooth. If the blocks are in equilibrium, what is the force F? B F 45⬚ A 20⬚ Solution: Isolate the top block. Solve the equilibrium equations. The weight is. The angle between the normal force N1 and the positive x axis is. The normal force is. The force N2 is. The equilibrium conditions are from which Solve: y B N2 F D N1 C N2 C W D 0 N1 Fx D 0.7071jN1 j jN2 ji D 0 Fy D 0.7071jN1 j 490.5j D 0. N1 D 693.7 N, W α x y N1 jN2 j D 490.5 N Isolate the bottom block. The weight is F β α A W D 0i jWjj D 0i 429.81j D 0i 412.02j (N). The angle between the normal force N1 and the positive x axis is 270° 45° D 225° . x N3 W The normal force: N1 D jN1 ji cos 225° C j sin 225° D jN1 j0.7071i 0.7071j. The angle between the normal force N3 and the positive x-axis is 90° 20° D 70° . The normal force is N1 D jN3 ji cos 70° C j sin 70° D jN3 j0.3420i C 0.9397j. The force is . . . F D jFji C 0j. The equilibrium conditions are F D W C N1 C N3 C F D 0, from which: Fx D 0.7071jN1 j C 0.3420jN3 j C jFji D 0 Fy D 0.7071jN1 j C 0.9397jN3 j 412j D 0 For jN1 j D 693.7 N from above: jFj D 162 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 149 Problem 3.91 The climber A is being helped up an icy slope by two friends. His mass is 80 kg, and the direction cosines of the force exerted on him by the slope are cos x D 0.286, cos y D 0.429, cos z D 0.857. The y axis is vertical. If the climber is in equilibrium in the position shown, what are the tensions in the ropes AB and AC and the magnitude of the force exerted on him by the slope? y B (2, 2, 0) m B dinates of the end points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. The rope tensions, the normal force, and the weight act on the climber. The coordinates of points A, B, C are given by the problem, A3, 0, 4, B2, 2, 0, C5, 2, 1. rB D 2i C 2j C 0k, C A The vector locations of the points A, B, C are: rA D 3i C 0j C 4k, x A (3, 0, 4) m z Solution: Get the unit vectors parallel to the ropes using the coor- C (5, 2, ⫺1) m W N rC D 5i C 2j 1k. Substitute and collect like terms, The unit vector parallel to the tension acting between the points A, B in the direction of B is eAB D rB r A jrB rA j The unit vectors are eAB D 0.2182i C 0.4364j 0.8729k, Fx D 0.2182jTAB j C 0.3482jTAC j 0.286jNji D 0 Fy D 0.4364jTAB j C 0.3482jTAC j C 0.429jNj 784.8j D 0 Fz D 0.8729jTAB j C 0.8704jTAC j 0.857jNjk D 0 We have three linear equations in three unknowns. The solution is: eAC D 0.3482i C 0.3482j 0.8704k, jTAB j D 100.7 N , jTAC j D 889.0 N , jNj D 1005.5 N . and eN D 0.286i C 0.429j C 0.857k. where the last was given by the problem statement. The forces are expressed in terms of the unit vectors, TAB D jTAB jeAB , TAC D jTAC jeAC , N D jNjeN . The weight is W D 0i jWjj C 0k D 0i 809.81j C 0k 0i 784.8j C 0k. The equilibrium conditions are 150 F D 0 D TAB C TAC C N C W D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.92 Consider the climber A being helped by his friends in Problem 3.91. To try to make the tensions in the ropes more equal, the friend at B moves to the position (4, 2, 0) m. What are the new tensions in the ropes AB and AC and the magnitude of the force exerted on the climber by the slope? Solution: Get the unit vectors parallel to the ropes using the coordinates of the end points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. The coordinates of points A, B, C are A3, 0, 4, B4, 2, 0, C5, 2, 1. The vector locations of the points A, B, C are: rA D 3i C 0j C 4k, rB D 4i C 2j C 0k, rC D 5i C 2j 1k. The unit vectors are eAB D C0.2182i C 0.4364j 0.8729k, eAC D C0.3482i C 0.3482j 0.8704k, eN D 0.286i C 0.429j C 0.857k. where the last was given by the problem statement. The forces are expressed in terms of the unit vectors, TAB D jTAB jeAB , TAC D jTAC jeAC , N D jNjeN . The weight is W D 0i jWjj C 0k D 0i 809.81j C 0k 0i 784.8j C 0k. The equilibrium conditions are F D 0 D TAB C TAC C N C W D 0. Substitute and collect like terms, Fx D C0.281jTAB j C 0.3482jTAC j 0.286jNji D 0 Fy D 0.4364jTAB j C 0.3482jTAC j C 0.429jNj 784.8j D 0 Fz D 0.8729jTAB j C 0.8704jTAC j 0.857jNjk D 0 The HP-28S hand held calculator was used to solve these simultaneous equations. The solution is: jTAB j D 420.5 N , jTAC j D 532.7 N , jNj D 969.3 N . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 151 Problem 3.93 A climber helps his friend up an icy slope. His friend is hauling a box of supplies. If the mass of the friend is 90 kg and the mass of the supplies is 22 kg, what are the tensions in the ropes AB and CD? Assume that the slope is smooth. That is, only normal forces are exerted on the man and the box by the slope. A 20⬚ B C 40⬚ Solution: Isolate the box. The weight vector is 75 W2 D 229.81j D 215.8j (N). D The angle between the normal force and the positive x axis is 90° 60° D 30° . 60⬚ The normal force is NB D jNB j0.866i 0.5j. The angle between the rope CD and the positive x axis is 180° 75° D 105° ; the tension is: T y β T2 D jT2 ji cos 105° C j sin 105° D jT2 j0.2588i C 0.9659j The equilibrium conditions are N Fx D 0.866jNB j C 0.2588jT2 ji D 0, α x W Fy D 0.5jNB j C 0.9659jT2 j 215.8j D 0. Solve: NB D 57.8 N, jT2 j D 193.5 N. y T1 Isolate the friend. The weight is 20° W D 909.81j D 882.9j (N). The angle between the normal force and the positive x axis is 40° D 50° . The normal force is: 90° 40° N 75° T2 W x NF D jNF j0.6428i C 0.7660j. The angle between the lower rope and the x axis is 75° ; the tension is T2 D jT2 j0.2588i C 0.9659j. The angle between the tension in the upper rope and the positive x axis is 180° 20° D 160° , the tension is T1 D jT1 j0.9397i C 0.3420j. The equilibrium conditions are F D W C T1 C T2 C NF D 0. From which: Fx D 0.6428jNF j C 0.2588jT2 j 0.9397jT1 ji D 0 Fy D 0.7660jNF j 0.9659jT2 j C 0.3420jT1 j 882.9j D 0 Solve, for jT2 j D 193.5 N. The result: jNF j D 1051.6 N , 152 jT1 j D 772.6 N . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.94 The 2800-lb car is moving at constant speed on a road with the slope shown. The aerodynamic forces on the car the drag D D 270 lb, which is parallel to the road, and the lift L D 120 lb, which is perpendicular to the road. Determine the magnitudes of the total normal and friction forces exerted on the car by the road. L D 15⬚ Solution: The free-body diagram is shown. If we write the equilibrium equations parallel and perpendicular to the road, we have: F- : N 2800 lb cos 15° C 120 lb D 0 F% : f 270 lb 2800 lb sin 15° D 0 Solving, we find N D 2580 lb, f D 995 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 153 Problem 3.95 An engineer doing preliminary design studies for a new radio telescope envisions a triangular receiving platform suspended by cables from three equally spaced 40-m towers. The receiving platform has a mass of 20 Mg (megagrams) and is 10 m below the tops of the towers. What tension would the cables be subjected to? TOP VIEW Solution: Isolate the platform. Choose a coordinate system with the origin at the center of the platform, with the z axis vertical, and the x,y axes as shown. Express the tensions in terms of unit vectors, and solve the equilibrium conditions. The cable connections at the platform are labeled a, b, c, and the cable connections at the towers are labeled A, B, C. The horizontal distance from the origin (center of the platform) to any tower is given by LD 20 m 65 m C 65 D 37.5 m. 2 sin60 z c The coordinates of points A, B, C are b a A37.5, 0, 10, B A y x B37.5 cos120° , 37.5 sin120° .10, C37.5 cos240° , 37.5 sin240° , 10, The vector locations are: The tensions in the cables are expressed in terms of the unit vectors, rA D 37.5i C 0j C 10k, rB D 18.764i C 32.5j C 10k, rC D 18.764i C 32.5j C 10k. The distance from the origin to any cable connection on the platform is dD 20 D 11.547 m. 2 sin60° The coordinates of the cable connections are a11.547, 0, 0, b11.547 cos120° , 11547 sin120° , 0, TaA D jTaA jeaA , TcC D jTcC jecC . The weight is W D 0i 0j 200009.81k D 0i C 0j 196200k. The equilibrium conditions are F D 0 D TaA C TbB C TcC C W D 0, from which: c11.547 cos240° , 11.547 sin240° , 0. The vector locations of these points are, ra D 11.547i C 0j C 0k, TbB D jTbB jebB , Fx D 0.9333jTaA j 0.4666jTbB j 0.4666jTcC ji D 0 Fy D 0jTaA j C 0.8082jTbB j 0.8082jTcC jj D 0 Fz D 0.3592jTaA j 0.3592jTbB j rb D 5.774i C 10j C 0k, C 0.3592jTcC 196200jk D 0 rc D 5.774i C 10j C 0k. The unit vector parallel to the tension acting between the points A, a in the direction of A is by definition eaA r A ra D . jrA ra Perform this operation for each of the unit vectors to obtain eaA D C0.9333i C 0j 0.3592k The commercial package TK Solver Plus was used to solve these equations. The results: jTaA j D 182.1 kN , Fz D 3jTj sin 196200 D 0, where D tan1 10 37.5 11.547 D 21.07° , from which each tension is jTj D 182.1 kN. 154 jTcC j D 182.1 kN . Check: For this geometry, where from symmetry all cable tensions may be assumed to be the same, only the z-component of the equilibrium equations is required: ebB D 0.4667i C 0.8082j 0.3592k ecC D 0.4667i C 0.8082j C 0.3592k jTbB j D 182.1 kN , check. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.96 To support the tent, the tension in the rope AB must be 35 lb. What are the tensions in the ropes AC, AD, and AE? y Solution: We develop the following position vectors rAB D 2i ft rAC D 6i C j 3k ft (0, 5, 0) ft rAD D 6i C 2j C 3k ft C (0, 6, 6) ft rAE D 3i 4j ft (6, 4, 3) ft A B (8, 4, 3) ft D x If we divide by the respective magnitudes we can develop the unit vectors that are parallel to these position vectors. E (3, 0, 3) ft eAB D 1.00i eAC D 0.885i C 0.147j 0.442k z eAD D 0.857i C 0.286j C 0.429k eAE D 6.00i 0.800j The equilibrium equation is TAB eAB C TAC eAC C TAD eAD C TAE eAE D 0. If we break this up into components, we have Fx : TAB 0.885TAC 0.857TAD 0.600TAE D 0 Fy : 0.147TAC C 0.286TAD 0.800TAE D 0 Fz : 0.442TAC C 0.429TAD D 0 If we set TAB D 35 lb, we cans solve for the other tensions. The result is TAC D 16.7 lb, TAD D 17.2 lb, TAE D 9.21 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 155 y Problem 3.97 Cable AB is attached to the top of the vertical 3-m post, and its tension is 50 kN. What are the tensions in cables AO, AC, and AD? 5m 5m C D Solution: Get the unit vectors parallel to the cables using the coordinates of the end points. Express the tensions in terms of these unit vectors, and solve the equilibrium conditions. The coordinates of points A, B, C, D, O are found from the problem sketch: The coordinates of the points are A6, 2, 0, B12, 3, 0, C0, 8, 5, D0, 4, 5, O0, 0, 0. 4m 8m (6, 2, 0) m The vector locations of these points are: rA D 6i C 2j C 0k, rB D 12i C 3j C 0k, O rC D 0i C 8j C 5k, B A z 3m 12 m rD D 0i C 4j 5k, rO D 0i C 0j C 0k. x The unit vector parallel to the tension acting between the points A, B in the direction of B is by definition eAB D y rB r A . jrB rA j 5m 5m Perform this for each of the unit vectors D 4m C eAB D C0.9864i C 0.1644j C 0k eAC D 0.6092i C 0.6092j C 0.5077k 8m O (6, 2, 0) m eAD D 0.7442i C 0.2481j 0.6202k A eAO D 0.9487i 0.3162j C 0k The tensions in the cables are expressed in terms of the unit vectors, TAB D jTAB jeAB D 50eAB , TAD D jTAD jeAD , TAC D jTAC jeAC , TAO D jTAO jeAO . The equilibrium conditions are F D 0 D TAB C TAC C TAD C TAO D 0. Substitute and collect like terms, Fx D 0.986450 0.6092jTAC j 0.7422jTAD j 0.9487jTAO ji D 0 Fy D 0.164450 C 0.6092jTAC j C 0.2481jTAD j 0.3162jTAO jj D 0 Fz D C0.5077jTAC j 0.6202jTAD jk D 0. This set of simultaneous equations in the unknown forces may be solved using any of several standard algorithms. The results are: jTAO j D 43.3 kN, 156 jTAC j D 6.8 kN, jTAD j D 5.5 kN. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 3.98* The 1350-kg car is at rest on a plane surface with its brakes locked. The unit vector en D 0.231i C 0.923j C 0.308k is perpendicular to the surface. The y axis points upward. The direction cosines of the cable from A to B are cos x D 0.816, cos y D 0.408, cos z D 0.408, and the tension in the cable is 1.2 kN. Determine the magnitudes of the normal and friction forces the car’s wheels exert on the surface. y en B ep x z Solution: Assume that all forces act at the center of mass of the car. The vector equation of equilibrium for the car is y en "car" FN TAB FS C TAB C W D 0. Writing these forces in terms of components, we have x W D mgj D 13509.81 D 13240j N, FS D FSx i C FSy j C FSz k, FS F W z and TAB D TAB eAB , where eAB D cos x i C cos y j C cos z k D 0.816i C 0.408j 0.408k. Substituting these values into the equations of equilibrium and solving for the unknown components of FS , we get three scalar equations of equilibrium. These are: FSx TABx D 0, FSy TABy W D 0, and FSz TABz D 0. Substituting in the numbers and solving, we get FSx D 979.2 N, FSy D 12, 754 N, and FSz D 489.6 N. The next step is to find the component of FS normal to the surface. This component is given by FN D FN Ð en D FSx eny C FSx eny C FSz enz . Substitution yields FN D 12149 N . From its components, the magnitude of FS is FS D 12800 N. Using the Pythagorean theorem, the friction force is fD F2S F2N D 4033 N. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 157 Problem 3.99* The brakes of the car in Problem 3.98 are released, and the car is held in place on the plane surface by the cable AB. The car’s front wheels are aligned so that the tires exert no friction forces parallel to the car’s longitudinal axis. The unit vector ep D 0.941i C 0.131j C 0.314k is parallel to the plane surface and aligned with the car’s longitudinal axis. What is the tension in the cable? Solution: Only the cable and the car’s weight exert forces in the direction parallel to ep . Therefore ep Ð T mgj D 0: 0.941i C 0.131j C 0.314k Ð [T0.816i C 0.408j 0.408k mgj] D 0, 0.9410.816T C 0.1310.408T mg C 0.3140.408T D 0. Solving, we obtain T D 2.50 kN. 158 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.1 In Active Example 4.1, the 40-kN force points 30° above the horizontal. Suppose that the force points 30° below the horizontal instead. Draw a sketch of the beam with the new orientation of the force. What is the moment of the force about point A? 40 kN 30° A 6m Solution: The perpendicular distance from A to the line of action of the force is unchanged D D 6 m sin 30° D 3 m The magnitude of the moment is therefore unchanged M D 3 m40 kN D 120 kN-m However, with its new orientation, the force would tend to cause the beam to rotate about A in the clockwise direction. The moment is clockwise M D 120 kN-m clockwise Problem 4.2 The mass m1 D 20 kg. The magnitude of the total moment about B due to the forces exerted on bar AB by the weights of the two suspended masses is 170 N-m. What is the magnitude of the total moment due to the forces about point A? 0.35 m 0.35 m 0.35 m A B m1 m2 Solution: The total moment about B is MB D m2 9.81 m/s2 0.35 m C 20 kg9.81 m/s2 0.7 m D 170 N-m Solving, we find m2 D 9.51 kg The moment about A is then jMA j D 20 kg9.81 m/s2 0.35 m C 9.51 kg9.81 m/s2 0.7 m jMA j D 134 N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 159 Problem 4.3 The wheels of the overhead crane exert downward forces on the horizontal I-beam at B and C. If the force at B is 40 kip and the force at C is 44 kip, determine the sum of the moments of the forces on the beam about (a) point A, (b) point D. 10 ft 25 ft B A 15 ft C D Solution: Use 2-dimensional moment strategy: determine normal distance to line of action D; calculate magnitude DF; determine sign. Add moments. (a) The normal distances from A to the lines of action are DAB D 10 ft, and DAC D 35 ft. The moments are clockwise (negative). Hence, (b) 10 ft 25 ft 15 ft A D B C MA D 1040 3544 D 1940 ft-kip . The normal distances from D to the lines of action are DDB D 40 ft, and DDC D 15 ft. The actions are positive; hence MD D C4040 C 1544 D 2260 ft-kip Problem 4.4 What force F applied to the pliers is required to exert a 4 N-m moment about the center of the bolt at P? Solution: MP D 4 N-m D F0.165 m sin 42° ) F D 4 N-m 0.165 m sin 42° D 36.2 N P F 165 mm 42⬚ 160 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.5 Two forces of equal magnitude F are applied to the wrench as shown. If a 50 N-m moment is required to loosen the nut, what is the necessary value of F? Solution: 300 mm 380 mm F Mnut center D F cos 30° 0.3 m C F cos 20° 0.38 m F 30⬚ D 50 N-m 50 N-m FD D 81.1 N 0.3 m cos 30° C 0.38 m cos 20° Problem 4.6 The force F D 8 kN. What is the moment of the force about point P? 20⬚ F F y (3, 7) m Solution: The angle between the force F and the x axis is F ˛ D tan1 5/4 D 51.3° Q (8, 5) m The force can then be written P (3, 2) m F D 8 kNcos ˛i sin ˛j D 5.00i 6.25j kN (7, 2) m x The line of action of the j component passes through P, so it exerts no moment about P. The moment of the i component about P is clockwise, and its magnitude is MP D 5 m5.00 kN D 25.0 kN-m MP D 25.0 kN-m clockwise Problem 4.7 If the magnitude of the moment due to the force F about Q is 30 kN-m, what is F? y (3, 7) m F Solution: The angle between the force F and the x axis is Q (8, 5) m ˛ D tan1 5/4 D 51.3° The force can then be written F D Fcos ˛i sin ˛j D F0.625i 0.781j P (3, 2) m (7, 2) m x Treating counterclockwise moment as positive, the total moment about point Q is MQ D 0.781F5 m 0.625F2 m D 30 kN-m Solving, we find F D 11.3 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 161 Problem 4.8 The support at the left end of the beam will fail if the moment about A of the 15-kN force F exceeds 18 kN-m. Based on this criterion, what is the largest allowable length of the beam? F 30° B A 25° Solution: MA D L Ð F sin 30° D L 15 2 30° F = 15 kN 30° MA D 7.5 L kN Ð m L set MA D MAmax D 18 kN Ð m D 7.5 Lmax Lmax D 2.4 m 25° Problem 4.9 The length of the bar AP is 650 mm. The radius of the pulley is 120 mm. Equal forces T D 50 N are applied to the ends of the cable. What is the sum of the moments of the forces (a) about A; (b) about P. 45⬚ A 30⬚ T T Solution: (a) MA D 50 N0.12 m 50 N0.12 m D 0 MA D 0 (b) P 45⬚ MP D 50 N0.12 m 50 N cos 30° 0.65 m sin 45° C 0.12 m cos 30° 50 N sin 30° 0.65 m cos 45° C 0.12 m sin 30° MP D 31.4 N-m 162 or MP D 31.4 N-m CW c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.10 The force F D 12 kN. A structural engineer determines that the magnitude of the moment due to F about P should not exceed 5 kN-m. What is the acceptable range of the angle ˛? Assume that 0 ˛ 90° . F α 1m P 2m Solution: We have the moment about P 12 kN MP D 12 kN sin ˛2 m 12 kN cos ˛1 m α MP D 122 sin ˛ cos ˛ kN-m The moment must not exceed 5 kN-m 1m Thus 5 kN-m ½ j122 sin ˛ cos ˛jkN-m P The limits occur when 122 sin ˛ cos ˛ D 5 ) ˛ D 37.3 2m 122 sin ˛ cos ˛ D 5 ) ˛ D 15.83° So we must have 15.83° ˛ 37.3° c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 163 Problem 4.11 The length of bar AB is 350 mm. The moments exerted about points B and C by the vertical force F are MB D 1.75 kN-m and MC D 4.20 kN-m. Determine the force F and the length of bar AC. B 30° C 20° A Solution: We have 1.75 kN-m D F0.35 m sin 30° ) F D 10 kN F 4.20 kN-m D FLAC cos 20° ) LAC D 0.447 m In summary F D 10 kN, LAC D 447 mm B C 30° 20° F d1 30° 50 0.3 m 0.450 m 20° d2 164 600 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.12 In Example 4.2, suppose that the 2-kN force points upward instead of downward. Draw a sketch of the machine part showing the orientations of the forces. What is the sum of the moments of the forces about the origin O? 4 kN 30⬚ 2 kN 300 mm O 3 kN 5 kN 300 mm 400 mm Solution: If the 2-kN force points upward, the magnitude of its moment about O does not change, but the direction of the moment changes from clockwise to counterclockwise. Treating counterclockwise moments as positive, the moment due to the 2-kN force is 0.3 m2 kN D 0.6 kN-m The moments due to the other forces do not change, so the sum of the moments of the four forces is MO D 0.6 1.039 C 1.400 kN-m MO D 0.961 kN-m Problem 4.13 Two equal and opposite forces act on the beam. Determine the sum of the moments of the two forces (a) about point P; (b) about point Q; (c) about the point with coordinates x D 7 m, y D 5 m. y 40 N P 30⬚ 40 N 2m Solution: 30⬚ Q 2m y 40 N 40 N MP D 40 N cos 30° 2 m C 40 N cos 30° 4 m x (a) D 69.3 N-m CCW 30° 30° (b) MQ D 40 N cos 30° 2 m D 69.3 N-m CCW M D 40 N sin 30° 5 m C 40 N cos 30° 5 m (c) x P 2m 2m Q 40 N sin 30° 5 m 40 N cos 30° 3 m D 69.3 N-m CCW c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 165 Problem 4.14 The moment exerted about point E by the weight is 299 in-lb. What moment does the weight exert about point S? S 13 in. 30° 12 E 40° in. Solution: The key is the geometry From trigonometry, cos 40° D Thus d1 D 12 in cos 30° d1 D and d2 d1 , cos 30° D 13 in 12 in d1 S 30° 13 in 12 i 10.3900 n W 40° d2 D 13 in cos 40° d2 E d2 D 9.9600 We are given that 299 in-lb D d2 W D 9.96 W W D 30.0 lb Now, Ms D d1 C d2 W Ms D 20.3530.0 Ms D 611 in-lb clockwise Problem 4.15 The magnitudes of the forces exerted on the pillar at D by the cables A, B, and C are equal: FA D FB D FC . The magnitude of the total moment about E due to the forces exerted by the three cables at D is 1350kN-m. What is FA ? D FC D FA FB 6m A Solution: The angles between the three cables and the pillar are ˛A D tan1 4/6 D 33.7° B C E 4m 4m 4m ˛B D tan1 8/6 D 53.1° ˛C D tan1 12/6 D 63.4° The vertical components of each force at point D exert no moment about E. Noting that FA D FB D FC , the magnitude of the moment about E due to the horizontal components is ME D FA sin ˛A C sin ˛B C sin ˛C 6 m D 1350 kN-m Solving for FA yields FA D 100 kN 166 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.16 Three forces act on the piping. Determine the sum of the moments of the three forces about point P. 2 kN Solution: MP D 4 kN0.2 m C 2 kN0.6 m 2 kN cos 20° 0.2 m C 2 kN sin 20° 0.4 m D 10.18 kN-m 20⬚ MP D 0.298 kN-m CCW 2 kN 4 kN 0.2 m 2 kN 20° 4 kN P 0.2 m 0.2 m 0.2 m P 0.2 m Problem 4.17 The forces F1 D 30 N, F2 D 80 N, and F3 D 40 N. What is the sum of the moments of the forces about point A? Solution: The moment about point A due to F1 is zero. Treating counterclockwise moments as positive the sum of the moments is 2 kN 0.2 m 0.2 m 0.2 m y F3 A C 30⬚ F1 2m MA D F3 sin 30° 8 m C F2 cos 45° 2 m B 45⬚ F2 MA D 273 N-m counterclockwise 8m Problem 4.18 The force F1 D 30 N. The vector sum of the forces is zero. What is the sum of the moments of the forces about point A? x y F3 A 30⬚ C F1 2m Solution: The sums of the forces in the x and y directions equal zero: Fx : F1 C F2 cos 45° F3 cos 30° D 0 B 45⬚ F2 8m Fy : F2 sin 45° C F3 sin 30° x D0 Setting F1 D 30 N and solving yields F2 D 58.0 N, F3 D 82.0 N. The sum of the moments about point A is MA D F2 sin 30° 8 m C F2 cos 45° 2 m D 410 N-m MA D 410 N-m counterclockwise c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 167 Problem 4.19 The forces FA D 30 lb, FB D 40 lb, FC D 20 lb, and FD D 30 lb. What is the sum of the moments of the forces about the origin of the coordinate system? y FD 30⬚ FA FB Solution: The moment about the origin due to FA and FD is zero. Treating counterclockwise moments as positive, the sum of the moments is M D FB 6 ft C FC 10 ft 6 ft x FC 4 ft D 40 lb6 ft C 20 lb10 ft D 40 ft-lb M D 40 ft-lb clockwise y Problem 4.20 The force FA D 30 lb. The vector sum of the forces on the beam is zero, and the sum of the moments of the forces about the origin of the coordinate system is zero. (a) (b) FD 30⬚ FA FB Determine the forces FB , FC , and FD . Determine the sum of the moments of the forces about the right end of the beam. 6 ft x FC 4 ft Solution: (a) The sum of the forces and the sum of the moments equals zero Fx : FA cos 30° FD D 0 Fy : FA sin 30° FB C FC D 0 Morigin : FB 6 ft C FC 10 ft D 0 Setting FA D 30 lb and solving yields (b) FB D 37.5 lb, FC D 22.5 lb, FD D 26.0 lb The sum of the moments about the right end is MRight End : FB 4 ft FA sin 30° 10 ft D 37.5 lb4 ft 30 lb10 ft D0 168 MRight End D0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.21 Three forces act on the car. The sum of the forces is zero and the sum of the moments of the forces about point P is zero. (a) (b) 3 ft 6 ft Determine the forces A and B. Determine the sum of the moments of the forces about point Q. x B P Q 2800 lb A Solution: 6 ft (a) Fy : A C B 2800 lb D 0 MP : 2800 lb6 ft C A9 ft D 0 Solving we find A D 1867 lb, B D 933 lb (b) 3 ft Q P 2800 lb MQ D 2800 lb3 ft B9 ft D 0 B A MQ D 0 80 lb Problem 4.22 Five forces act on the piping. The vector sum of the forces is zero and the sum of the moments of the forces about point P is zero. (a) (b) 45⬚ y Determine the forces A, B, and C. Determine the sum of the moments of the forces about point Q. 2 ft 20 lb Q x A P C 2 ft B 2 ft 2 ft Solution: 80 lb The conditions given in the problem are: Fx : A C 80 lb cos 45° D 0 45° y 20 lb 2 ft (a) Fy : B C 20 lb C 80 lb sin 45° D 0 P MP : 20 lb2 ft C6 ft 80 lb cos 45° 2 ft Q x A 2 ft C 80 lb sin 45° 4 ft D 0 Solving we have B 2 ft 2 ft C A D 56.6 lb, B D 24.4 lb, C D 12.19 lb MQ : 80 lb cos 45° 2 ft 80 lb sin 45° 2 ft (b) C20 lb4 ft C B6 ft D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 169 Problem 4.23 In Example 4.3, suppose that the attachment point B is moved upward and the cable is lengthened so that the vertical distance from C to B is 9 ft. (the positions of points C and A are unchanged.) Draw a sketch of the system with the cable in its new position. What is the tension in the cable? B A 7 ft W 4 ft C 2 ft 2 ft Solution: The angle ˛ between the cable AB and the horizontal is. ˛ D tan1 5/4 D 51.3° The sum of the moments about C is MC : W2 ft C T cos ˛4 ft C T sin ˛4 ft D 0 Solving yields T D 106.7 lb Problem 4.24 The tension in the cable is the same on both sides of the pulley. The sum of the moments about point A due to the 800-lb force and the forces exerted on the bar by the cable at B and C is zero. What is the tension in the cable? Solution: Let T be the tension in the cable. The sum of the moments about A is MA : T30 in C T sin 30° 90 in 800 lb60in D 0 A B 30 30 in 30 in C 800 lb 30 in Solving yields T D 640 lb 170 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.25 The 160-N weights of the arms AB and BC of the robotic manipulator act at their midpoints. Determine the sum of the moments of the three weights about A. 150 600 mm mm 40 N C 20° B m 0m Solution: The strategy is to find the perpendicular distance from 40° 60 160 N the points to the line of action of the forces, and determine the sum of the moments, using the appropriate sign of the action. The distance from A to the action line of the weight of the arm AB is: A 160 N dAB D 0.300 cos 40° D 0.2298 m The distance from A to the action line of the weight of the arm BC is dBC D 0.600cos 40° C 0.300cos 20° D 0.7415 m. The distance from A to the line of action of the force is dF D 0.600cos 40° C 0.600cos 20° C 0.150cos 20° D 1.1644 m. The sum of the moments about A is MA D dAB 160 dBC 160 dF 40 D 202 N-m Problem 4.26 The space shuttle’s attitude thrusters exert two forces of magnitude F D 7.70 kN. What moment do the thrusters exert about the center of mass G? 2.2 m 2.2 m F F G 5° 18 m Solution: The key to this problem is getting the geometry correct. The simplest way to do this is to break each force into components parallel and perpendicular to the axis of the shuttle and then to sum the moments of the components. (This will become much easier in the next section) 6° 12 m F sin 6° F sin 5° 5˚ 6°c 2.2 m 18 m F cos 5° FRONT 2.2 m 12 m F cos 6° REAR CMFRONTý D 18F sin 5° 2.2F cos 5° CMREARý D 2.2F cos 6° 12F sin 6° CMTOTAL D MFRONT C MREAR CMTOTAL D 4.80 C 7.19 N-m CMTOTAL D 2.39 N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 171 Problem 4.27 The force F exerts a 200 ft-lb counterclockwise moment about A and a 100 ft-lb clockwise moment about B. What are F and ? y A (–5, 5) ft F θ (4, 3) ft x B (3, – 4) ft Solution: The strategy is to resolve F into x- and y-components, and compute the perpendicular distance to each component from A and B. The components of F are: F D iFX C jFY . The vector from A to the point of application is: rAF D 4 5i C 3 5j D 9i 2j. The perpendicular distances are dAX D 9 ft, and dAY D 2 ft, and the actions are positive. The moment about A is MA D 9FY C 2FX D 200 ft-lb. The vector from B to the point of application is rBF D 4 3i C 3 4j D 1i C 7j; the distances dBX D 1 ft and dBY D 7 ft, the action of FY is positive and the action of FX is negative. The moment about B is MB D 1FY 7FX D 100 ft-lb. The two simultaneous equations have solution: FY D 18.46 lb and FX D 16.92 lb. Take the ratio to find the angle: D tan1 FY FX D tan1 18.46 16.92 y A (–5, 5) ft F θ (4, 3) ft x B (3, –4) ft D tan1 1.091 D 47.5° . From the Pythagorean theorem jFj D 172 F2Y C F2X D p 18.462 C 16.922 D 25.04 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.28 Five forces act on a link in the gearshifting mechanism of a lawn mower. The vector sum of the five forces on the bar is zero. The sum of their moments about the point where the forces Ax and Ay act is zero. (a) Determine the forces Ax , Ay , and B. (b) Determine the sum of the moments of the forces about the point where the force B acts. Ay Ax 25 kN 20° 650 mm 450 mm 30 kN 45° B 650 mm 350 mm Solution: The strategy is to resolve the forces into x- and y-components, determine the perpendicular distances from B to the line of action, determine the sign of the action, and compute the moments. The angles are measured counterclockwise from the x axis. The forces are F2 D 30i cos 135° C j sin 135° D 21.21i C 21.21j F1 D 25i cos 20° C j sin 20° D 23.50i C 8.55j. (a) The sum of the forces is F D A C B C F1 C F2 D 0. Substituting: and FX D AX C BX C 23.5 21.2i D 0, FY D AY C 21.2 C 8.55j D 0. Solve the second equation: AY D 29.76 kN. The distances of the forces from A are: the triangle has equal base and altitude, hence the angle is 45° , so that the line of action of F1 passes through A. The distance to the line of action of B is 0.65 m, with a positive action. The distance to the line of action of the y-component of F2 is 0.650 C 0.350 D 1 m, and the action is positive. The distance to the line of action of the x-component of F2 is 0.650 0.450 D 0.200 m, and the action is positive. The moment about A is MA D 8.551 C 23.50.2 C BX 0.65 D 0. Solve: BX D 20.38 kN. Substitute into the force equation to obtain AX D 18.09 kN (b) The distance from B to the line of action of the y-component of F1 is 0.350 m, and the action is negative. The distance from B to the line of action of AX is 0.650 m and the action is negative. The distance from B to the line of action of AY is 1 m and the action is positive. The distance from B to the line of action of the x-component of F2 is 0.450 m and the action is negative. The sum of the moments about B: MB D 0.35021.21 0.65018.09 C 129.76 0.45023.5 D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 173 F Problem 4.29 Five forces act on a model truss built by a civil engineering student as part of a design project. The dimensions are b D 300 mm and h D 400 mm; F D 100 N. The sum of the moments of the forces about the point where Ax and Ay act is zero. If the weight of the truss is negligible, what is the force B? F 60° 60° h Ax Ay b b b b b b B Solution: The x- and y-components of the force F are F D jFji cos 60° C j sin 60° D jFj0.5i C 0.866j. The distance from A to the x-component is h and the action is positive. The distances to the y-component are 3b and 5b. The distance to B is 6b. The sum of the moments about A is MA D 2jFj0.5h 3bjFj0.866 5bjFj0.866 C 6bB D 0. Substitute and solve: B D 1.6784jFj D 93.2 N 1.8 Problem 4.30 Consider the truss shown in Problem 4.29. The dimensions are b D 3 ft and h D 4 ft; F D 300 lb. The vector sum of the forces acting on the truss is zero, and the sum of the moments of the forces about the point where Ax and Ay act is zero. (a) (b) Determine the forces Ax , Ay , and B. Determine the sum of the moments of the forces about the point where the force B acts. Solution: The forces are resolved into x- and y-components: Solve the first: Ax D 300 lb. The distance from point A to the x-components of the forces is h, and the action is positive. The distances between the point A and the lines of action of the ycomponents of the forces are 3b and 5b. The actions are negative. The distance to the line of action of the force B is 6b. The action is positive. The sum of moments about point A is F D 300i cos 60° C j sin 60° D 150i 259.8j. (a) The sum of the forces: F D 2F C A C B D 0. MA D 2150 h 3b259.8 5b259.8 C 6b B D 0. The x- and y-components: Substitute and solve: B D 279.7 lb. Substitute this value into the force equation and solve: Ax D 519.6 279.7 D 239.9 lb Fx D Ax 300i D 0, (b) Fy D 519.6 C Ay C Bj D 0. The distances from B and the line of action of AY is 6b and the action is negative. The distance between B and the x-component of the forces is h and the action is positive. The distance between B and the y-components of the forces is b and 3b, and the action is positive. The sum of the moments about B: 174 MB D 6b239.9 C 2150 h C b259.8 C 3b259.8 D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.31 The mass m D 70 kg. What is the moment about A due to the force exerted on the beam at B by the cable? B A 45° 30° 3m m Solution: The strategy is to resolve the force at B into components parallel to and normal to the beam, and solve for the moment using the normal component of the force. The force at B is to be determined from the equilibrium conditions on the cable juncture O. Angles are measured from the positive x axis. The forces at the cable juncture are: FOB FOC O FOB D jFOB ji cos 150° C j sin 150° D jFOB j0.866i C 0.5j W FOC D jFOC ji cos 45° C j sin 45° D jFOC j0.707i C 0.707j. W D 709.810i 1j D 686.7j (N). The equilibrium conditions are: Fx D 0.866jFOB j C 0.7070jFOC ji D 0 FY D 0.500jFOB j C .707jFOC j 686.7j D 0. Solve: jFOB j D 502.70 N. This is used to resolve the cable tension at B: FB D 502.7i cos 330° C j sin 330° D 435.4i 251.4j. The distance from A to the action line of the y-component at B is 3 m, and the action is negative. The x-component at passes through A, so that the action line distance is zero. The moment at A is MA D 3251.4 D 754.0 N-m Problem 4.32 The weights W1 and W2 are suspended by the cable system shown. The weight W1 D 12 lb. The cable BC is horizontal. Determine the moment about point P due to the force exerted on the vertical post at D by the cable CD. A D 50⬚ W1 B 6 ft C W2 P Solution: Isolate part of the cable system near point B. The equilibrium equations are Fx : TBC TAC cos 50° D 0 Fy : TAB 12 lb D 0 Solving yields TAB D 15.7 lb, TBC D 10.1 lb Let ˛ be the angle between the cable CD and the horizontal. The magnitude of the moment about P due to the force exerted at D by cable CD is M D TCD cos ˛6 ft Isolate part of the cable system near point C. From the equilibrium equation Fx : TCD cos ˛ TBC D 0 ) TCD cos ˛ D TBC D 10.1 lb Thus M D 10.1 lb6 ft M D 60.4 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 175 Problem 4.33 The bar AB exerts a force at B that helps support the vertical retaining wall. The force is parallel to the bar. The civil engineer wants the bar to exert a 38 kN-m moment about O. What is the magnitude of the force the bar must exert? B 4m A 1m O 1m 3m Solution: The strategy is to resolve the force at B into components FB parallel to and normal to the wall, determine the perpendicular distance from O to the line of action, and compute the moment about O in terms of the magnitude of the force exerted by the bar. B By inspection, the bar forms a 3, 4, 5 triangle. The angle the bar makes with the horizontal is cos D 35 D 0.600, and sin D 45 D 0.800. The force at B is FB D jFB j0.600i C 0.800j. The perpendicular distance from O to the line of action of the x-component is 4 C 1 D 5 m, and the action is positive. The distance from O to the line of action of the y-component is 1 m, and the action is positive. The moment about O is MO D 50.600jFB j C 10.800jFB j D 3.8jFB j D 38 kN, from which jFB j D 10 kN 4m A θ O 1m 1m 3m Problem 4.34 A contestant in a fly-casting contest snags his line in some grass. If the tension in the line is 5 lb, what moment does the force exerted on the rod by the line exert about point H, where he holds the rod? H Solution: The strategy is to resolve the line tension into a compo- 6 ft nent normal to the rod; use the length from H to tip as the perpendicular distance; determine the sign of the action, and compute the moment. 4 ft The line and rod form two right triangles, as shown in the sketch. The angles are: ˛ D tan1 2 D 15.95° 7 ˇ D tan1 7 ft 6 15 α D 21.8° . β α 2 ft 7 ft The angle between the perpendicular distance line and the fishing line is D ˛ C ˇ D 37.7° . The force normal to the p distance line is F D 5sin 37.7° D 3.061 lb. The distance is d D 22 C 72 D 7.28 ft, and the action is negative. The moment about H is MH D 7.283.061 D 22.3 ft-lb Check: The tension can be resolved into x and y components, 15 ft 6 ft 15 ft β Fx D F cos ˇ D 4.642 lb, Fy D F sin ˇ D 1.857 lb. The moment is M D 2Fx C 7Fy D 22.28 D 22.3 ft-lb. check. 176 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. A Problem 4.35 The cables AB and AC help support the tower. The tension in cable AB is 5 kN. The points A, B, C, and O are contained in the same vertical plane. (a) What is the moment about O due to the force exerted on the tower by cable AB? (b) If the sum of the moments about O due to the forces exerted on the tower by the two cables is zero, what is the tension in cable AC? Solution: The strategy is to resolve the cable tensions into components normal to the vertical line through OA; use the height of the tower as the perpendicular distance; determine the sign of the action, and compute the moments. (a) (b) 20 m 60° 45° C O A B FN 60° FN A 45° The component normal to the line OA is FBN D 5cos 60° D 2.5 kN. The action is negative. The moment about O is MOA D 2.520 D 50 kN-m By a similar process, the normal component of the tension in the cable AC is FCN D jFC j cos 45° D 0.707jFC j. The action is positive. If the sum of the moments is zero, MO D 0.70720jFC j 50 D 0, from which jFC j D 50 kN m D 3.54 kN 0.70720 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 177 Problem 4.36 The cable from B to A (the sailboat’s forestay) exerts a 230-N force at B. The cable from B to C (the backstay) exerts a 660-N force at B. The bottom of the sailboat’s mast is located at x D 4 m, y D 0. What is the sum of the moments about the bottom of the mast due to the forces exerted at B by the forestay and backstay? Solution: Triangle ABP tan ˛ D Triangle BCQ tan ˇ D y 4 , ˛ D 18.73° 11.8 5 , ˇ D 22.62° 12 CMO D 13230 sin ˛ 13660 sin ˇ B (4,13) m CMO D 2340 N-m B (4,13) 230 N 660 N β α A (0,1.2) m C (9,1) m x P A (0,1.2) C (9,1) O (4,0) Q 660 sin β 230 sin α α β 13 m O 178 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.37 The cable AB exerts a 290-kN force on the building crane’s boom at B. The cable AC exerts a 148-kN force on the boom at C. Determine the sum of the moments about P due to the forces the cables exert on the boom. A B 8m C G Boom P 16 m 38 m 56 m Solution: A 56 8 290 8 8 290 kN56 m p 148 kN16 m MP D p 3200 320 kN 8 kN 16 14 40 m B D 3.36 MNm 8 C 16 m P MP D 3.36 MN-m CW Problem 4.38 The mass of the building crane’s boom in Problem 4.37 is 9000 kg. Its weight acts at G. The sum of the moments about P due to the boom’s weight, the force exerted at B by the cable AB, and the force exerted at C by the cable AC is zero. Assume that the tensions in cables AB and AC are equal. Determine the tension in the cables. Solution: A 56 8 8 8 8 TAB 56 m p TAC 16 m MP D p 3200 320 C 9000 kg9.81 m/s2 38 m D 0 using TAB D TAC we solve and find T AB B 18 m C 16 TA 22 m C 16 m P 9000 kg (9.81 m/s2) TAB D TAC D 223 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 179 Problem 4.39 The mass of the luggage carrier and the suitcase combined is 12 kg. Their weight acts at A. The sum of the moments about the origin of the coordinate system due to the weight acting at A and the vertical force F applied to the handle of the luggage carrier is zero. Determine the force F (a) if ˛ D 30° ; (b) if ˛ D 50° . F y 0.28 m Solution: O is the origin of the coordinate system MO D F1.2 m cos ˛ 12 kg9.81 m/s2 0.28 cos ˛ 0.14 sin ˛ D 0 Solving we find x FD 12 kg9.81 m/s2 0.28 cos ˛ 0.14 sin ˛ 1.2 m cos ˛ (a) For ˛ D 30° We find F D 19.54 N (b) For ˛ D 50° We find F D 11.10 N 0.14 m 1.2 m A a C Problem 4.40 The hydraulic cylinder BC exerts a 300-kN force on the boom of the crane at C. The force is parallel to the cylinder. What is the moment of the force about A? Solution: The strategy is to resolve the force exerted by the hydraulic cylinder into the normal component about the crane; determine the distance; determine the sign of the action, and compute the moment. Two right triangles are constructed: The angle formed by the hydraulic cylinder with the horizontal is ˇ D tan1 2.4 1.2 D 63.43° . The angle formed by the crane with the horizontal is C ˛ D tan1 1.4 3 D 25.02° . A 2.4 m 1m B 1.8 m 1.2 m 7m The angle between the hydraulic cylinder and the crane is D ˇ ˛ D 38.42° . The normal component of the force is: Fp N D 300sin 38.42° D 186.42 kN. The distance from point A is d D 1.42 C 32 D 3.31 m. The action is positive. The moment about A is MO D C3.31186.42 D 617.15 kN-m Check: The force exerted by the actuator can be resolved into x- and y-components, Fx D F cos ˇ D 134.16 kN, Fy D F sin ˇ D 268.33 kN. The moment about the point A is M D 1.4Fx C 3.0 Fy D 617.15 kN m. check. β α 1.2 m α 3m 2.4 m β 1.4 m 180 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.41 The hydraulic piston AB exerts a 400-lb force on the ladder at B in the direction parallel to the piston. The sum of the moments about C due to the force exerted on the ladder by the piston and the weight W of the ladder is zero. What is the weight of the ladder? 6 ft W Solution: The angle between the piston AB and the horizontal is 3 ft ˛ D tan1 3/6 D 26.6° A B C The sum of the counterclockwise moment about C is MC : W6 ft 400 lb cos ˛3 ft 400 lb sin ˛3 ft D 0 6 ft 3 ft Solving yields W D 268 lb Problem 4.42 The hydraulic cylinder exerts an 8-kN force at B that is parallel to the cylinder and points from C toward B. Determine the moments of the force about points A and D. 1m D C Hydraulic cylinder 1m 0.6 m B A 0.6 m Scoop 0.15 m Solution: Use x, y coords with origin A. We need the unit vector from C to B, eCB . From the geometry, eCB D 0.780i 0.625j 5.00 kN 6.25 kN C (−0.15, + 0.6) The force FCB is given by FCB D 0.7808i 0.6258j kN 0.6 m FCB D 6.25i 5.00j kN For the moments about A and D, treat the components of FCB as two separate forces. 0.15 m CMA D 5, 000.15 0.66.25 kN Ð m A (0 , 0) CMA D 3.00 kN Ð m 5.0 kN m D For the moment about D C 0,4 m MD D 5 kN1 m C 6.25 kN0.4 m C 6.25 kN CMD D 7.5 kN Ð m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 181 Problem 4.43 The structure shown in the diagram is one of the two identical structures that support the scoop of the excavator. The bar BC exerts a 700-N force at C that points from C toward B. What is the moment of this force about K? 320 mm C Shaft 100 mm Scoop 260 mm H B 180 260 mm mm J D 160 mm L K 380 mm 1040 mm 1120 mm Solution: 320 320 700 N0.52 m D 353 Nm MK D p 108800 80 700 N 520 mm MK D 353 Nm CW K Problem 4.44 In the structure shown in Problem 4.43, the bar BC exerts a force at C that points from C toward B. The hydraulic cylinder DH exerts a 1550-N force at D that points from D toward H. The sum of the moments of these two forces about K is zero. What is the magnitude of the force that bar BC exerts at C? Solution: 320 80 260 mm 320 1120 1550 N0.26 m p F0.52 D 0 MK D p 1264400 108800 Solving we find BC 1120 F D 796 N 100 1550 N 260 mm K 182 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.45 In Active Example 4.4, what is the moment of F about the origin of the coordinate system? C (7, 7, 0) ft A (0, 6, 5) ft Solution: The vector from the origin to point B is F x r D 11i C 4k ft B (11, 0, 4) ft From Active Example 4.4 we know that the force F is z F D 40i C 70j 40k lb The moment of F about the origin is i j k M D r ð F D 11 0 4 D 280i C 280j C 770k ft-lb 40 70 40 M D 280i C 280j C 770k ft-lb Problem 4.46 Use Eq. (4.2) to determine the moment of the 80-N force about the origin O letting r be the vector (a) from O to A; (b) from O to B. Solution: (a) MO D rOA ð F y D 6i ð 80j D 480k N-m. 80j (N) B MO D rOB ð F (b) (6, 4, 0) m D 6i C 4j ð 80j D 480k N-m. O x A (6, 0, 0) m Problem 4.47 A bioengineer studying an injury sustained in throwing the javelin estimates that the magnitude of the maximum force exerted was jFj D 360 N and the perpendicular distance from O to the line of action of F was 550 mm. The vector F and point O are contained in the xy plane. Express the moment of F about the shoulder joint at O as a vector. Solution: The magnitude of the moment is jFj0.55 m D 360 N 0.55 m D 198 N-m. The moment vector is perpendicular to the xy plane, and the right-hand rule indicates it points in the positive z direction. Therefore MO D 198k N-m. y F y 550 mm F O O x x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 183 y Problem 4.48 Use Eq. (4.2) to determine the moment of the 100-kN force (a) about A, (b) about B. A Solution: (a) The coordinates of A are (0,6,0). The coordinates of the point of application of the force are (8,0,0). The position vector from A to the point of application of the force is rAF D 8 0i C 0 6j D 8i 6j. The force is F D 100j (kN). The cross product is i rAF ð F D 8 0 100j (kN) 6m j k 6 0 D 800k (kN-m) 100 0 B x 8m 12 m (b) The coordinates of B are (12,0,0). The position vector from B to the point of application of the force is rBF D 8 12 i D 4i. The cross product is: i rBF ð F D 4 0 j k 0 0 D 400k (kN-m) 100 0 Problem 4.49 The cable AB exerts a 200-N force on the support at A that points from A toward B. Use Eq. (4.2) to determine the moment of this force about point P in two ways: (a) letting r be the vector from P to A; (b) letting r be the vector from P to B. y P (0.9, 0.8) m (0.3, 0.5) m Solution: First we express the force as a vector. The force points in the same direction as the position vector AB. AB D 1 0.3 mi C 0.2 0.5 mj D 0.7i 0.3j m jABj D p 0.7 m2 C 0.3 m2 D 0.58 m A B (1, 0.2) m x 200 N 0.7i 0.3j FD p 0.58 (a) 200 N 0.7i 0.3j MP D PA ð F D 0.6 mi 0.3 mj ð p 0.58 Carrying out the cross product we find MP D 102.4 N-mk (b) 200 N 0.7i 0.3j MP D PB ð F D 0.1 mi 0.6 mj ð p 0.58 Carrying out the cross product we find MP D 102.4 N-mk 184 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.50 The line of action of F is contained in the xy plane. The moment of F about O is 140k (Nm), and the moment of F about A is 280k (N-m). What are the components of F? y A (0, 7, 0) m F (5, 3, 0) m Solution: The strategy is to find the moments in terms of the components of F and solve the resulting simultaneous equations. The position vector from O to the point of application is rOF D 5i C 3j. The position vector from A to the point of application is rAF D 5 0i C 3 7j D 5i 4j. The cross products: i rOF ð F D 5 FX j 3 FY k 0 D 5FY 3FX k D 140k, and 0 i rAF ð F D 5 FX j 4 FY k 0 D 5FY C 4FX k D 280k. 0 Take the dot product of both sides with k to eliminate k. The simultaneous equations are: x O 5FY 3FX D 140, 5FY C 4FX D 280. Solving: FY D 40, FX D 20, from which F D 20i C 40j (N) y A (0,7,0) F (5,3,0) x O Problem 4.51 Use Eq. (4.2) to determine the sum of the moments of the three forces (a) about A, (b) about B. y 6 kN 3 kN Solution: (a) 3 kN B A x MA D 0.2i ð 3j C 0.4i ð 6j C 0.6i ð 3j 0.2 m 0.2 m 0.2 m 0.2 m D O. (b) MB D 0.2i ð 3j C 0.4i ð 6j C 0.6i ð 3j D O. Problem 4.52 Three forces are applied to the plate. Use Eq. (4.2) to determine the sum of the moments of the three forces about the origin O. Solution: The position vectors from O to the points of application of the forces are: rO1 D 3j, F1 D 200i; rO2 D 10i, F2 D 500j; rO3 D 6i C 6j, F3 D 200i. y 200 lb 3 ft 200 lb 3 ft The sum of the moments about O is i MO D 0 200 j k i 3 0 C 10 0 0 0 j k i 0 0 C 6 500 0 200 j k 6 0 lb 0 0 O x 6 ft 4 ft 500 lb D 600k 5000k 1200k D 5600k ft-lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 185 Problem 4.53 Three forces act on the plate. Use Eq. (4.2) to determine the sum of the moments of the three forces about point P. y 4 kN 45⬚ Solution: 3 kN r1 D 0.12i C 0.08j m, F1 D 4 cos 45° i C 4 sin 45° j kN 30⬚ 0.18 m P 0.10 m r2 D 0.16i m, F2 D 3 cos 30° i C 3 sin 30° j kN r3 D 0.16i 0.1j m, F3 D 12 cos 20° i 12 sin 20° j kN 20⬚ 0.12 m 12 kN 0.28 m MP D r1 ð F1 C r2 ð F2 C r3 ð F3 x MP D 0.145 kN-mk D 145 N-mk Problem 4.54 (a) Determine the magnitude of the moment of the 150-N force about A by calculating the perpendicular distance from A to the line of action of the force. y (0, 6, 0) m 150k (N) (b) Use Eq. (4.2) to determine the magnitude of the moment of the 150-N force about A. A Solution: (a) x (6, 0, 0) m The perpendicular from A to the line of action of the force lies in the xy plane d p D z 62 C 62 D 8.485 m jMj D dF D 8.485150 D 1270 N-m (b) M D 6i C 6j ð 150k D 900j C 900i N-m p jMj D 186 9002 C 9002 D 1270 N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.55 (a) Determine the magnitude of the moment of the 600-N force about A by calculating the perpendicular distance from A to the line of action of the force. A (0.6, 0.5, 0.4) m (b) Use Eq. (4.2) to determine the magnitude of the moment of the 600-N force about A. Solution: (a) x 0.8 m Choose some point Px, 0, 0.8 m. on the line of action of the force. The distance from A to P is then d D x 0.6 m2 C 0 0.5 m2 C 0.8 m 0.4 m2 600i (N) z The perpendicular distance is the shortest distance d which occurs when x D 0.6 m. We have d D 0.6403 m. Thus the magnitude of the moment is M D 600 N0.6403 m D 384 N-m (b) Define the point on the end of the rod to be B. Then AB D 0.6i 0.5j C 0.4k m we have M D AB ð F D 0.6i 0.5j C 0.4k m ð 600 Ni M D 240j C 300k N-m Thus the magnitude is MD 240 Nm2 C 300 Nm2 D 384 N-m y Problem 4.56 what is the magnitude of the moment of F about point B? A (4, 4, 2) ft Solution: The position vector from B to A is B (8, 1, ⫺2) ft x rBA D [4 8i C 4 1j C 2 2k] ft z rBA D 4i C 3j C 4k ft The moment of F about B i MB D rBA ð F D 4 20 F ⫽ 20i ⫹ 10j ⫺ 10k (lb) is j 3 10 k 4 D 70i C 40j 100k ft-lb 10 Its magnitude is jMB j D 70 ft-lb2 C 40 ft-lb2 C 100 ft-lb2 D 128 ft-lb jMB j D 128 ft-lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 187 Problem 4.57 In Example 4.5, suppose that the attachment point C is moved to the location (8,2,0) m and the tension in cable AC changes to 25 kN. What is the sum of the moments about O due to the forces excerted on the attachment point A by the two cables? Solution: The position vector from A to C is y C (6, 3, 0) m B (0, 4, 8) m O x A (4, 0, 6) m rAC D [8 4i C 2 0j C 0 6k] m z rAC D 4i C 2j 6k The force exerted at A by cable AC can be written FAC D 25 kN rAC D 13.4i C 6.68j 20.0k kN jrAC j The total force exerted at A by the two cables is F D FAB C FAC D 6.70i C 13.3j 16.7k kN The moment about O is i j 0 MO D rAB ð F D 4 6.70 13.3 k 6 D 80.1i C 107j C 53.4k kN-m 16.7 MO D 80.1i C 107j C 53.4k kN-m y Problem 4.58 The rope exerts a force of magnitude jFj D 200 lb on the top of the pole at B. Determine the magnitude of the moment of F about A. Solution: The position vector from B to C is B (5, 6, 1) ft F rBC D [3 5i C 0 6j C 4 1k] ft A x rBC D 2i 6j C 3k ft C (3, 0, 4) ft The force F can be written F D 200 lb rBC D 57.1i 171j C 85.7k lb jrBC j The moment of F about A is i j MA D rAB ð F D 5 6 57.1 171 z k 1 85.7 D 686i 486j 514k ft-lb Its magnitude is jMA j D 686 ft-lb2 C 486 ft-lb2 C 514 ft-lb2 D 985 ft-lb jMA j D 985 ft-lb 188 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.59 The force F D 30i C 20j 10k (N). (a) Determine the magnitude of the moment of F about A. (b) Suppose that you can change the direction of F while keeping its magnitude constant, and you want to choose a direction that maximizes the moment of F about A. What is the magnitude of the resulting maximum moment? Solution: The vector from A to the point of application of F is F A (8, 2, – 4) m (4, 3, 3) m x z r D 4i 1j 7k m and p jrj D (a) 42 C 12 C 72 D 8.12 m The moment of F about A is MA i j D r ð F D 4 1 30 20 jMA j D (b) k 7 D 150i 170j C 110k N-m 10 p 1502 C 1702 C 1102 D 252 N-m The maximum moment occurs when r ? F. In this case jMAmax j D jrjjFj Hence, we need jFj. p jFj D 302 C 202 C 102 D 37.4 N Thus, jMAmax j D 8.1237.4 D 304 N-m Problem 4.60 The direction cosines of the force F are cos x D 0.818, cos y D 0.182, and cos z D 0.545. The support of the beam at O will fail if the magnitude of the moment of F about O exceeds 100 kN-m. Determine the magnitude of the largest force F that can safely be applied to the beam. y z O F 3m Solution: The strategy is to determine the perpendicular distance x from O to the action line of F, and to calculate the largest magnitude of F from MO D DjFj. The position vector from O to the point of application of F is rOF D 3i (m). Resolve the position vector into components parallel and normal to F. The component parallel to F is rP D rOF Ð eF eF , where the unit vector eF parallel to F is eF D i cos X C j cos Y C k cos Z D 0.818i C 0.182j 0.545k. The dot product is rOF Ð eF D 2.454. The parallel component is rP D 2.007i C 0.4466j 1.3374k. The component normal to F is rN D rOF rP D 3 2i 0.4466j C 1.3374k. The magnitude p of the normal component is the perpendicular distance: D D 12 C 0.44662 C 1.3372 D 1.7283 m. The maximum moment allowed is MO D 1.7283jFj D 100 kN-m, from which jFj D 100 kN-m D 57.86 ¾ D 58 kN 1.7283 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 189 Problem 4.61 The force F exerted on the grip of the exercise machine points in the direction of the unit vector e D 23 i 23 j C 13 k and its magnitude is 120 N. Determine the magnitude of the moment of F about the origin O. 150 mm y F Solution: The vector from O to the point of application of the O 200 mm force is z r D 0.25i C 0.2j 0.15k m 250 mm x and the force is F D jFje or F D 80i 80j C 40k N. The moment of F about O is i j MO D r ð F D 0.25 0.2 80 80 k 0.15 N-m 40 or MO D 4i 22j 36k N-m and p jMO j D 42 C 222 C 362 N-m jMO j D 42.4 N-m Problem 4.62 The force F in Problem 4.61 points in the direction of the unit vector e D 23 i 23 j C 13 k. The support at O will safely support a moment of 560 N-m magnitude. (a) Based on this criterion, what is the largest safe magnitude of F? (b) If the force F may be exerted in any direction, what is its largest safe magnitude? Solution: See the figure of Problem 4.61. The moment in Problem 4.61 can be written as i j MO D 0.25 0.2 2F 2F 3 3 k 0.15 where F D jFj C 31 F MO D 0.0333i 0.1833j 0.3kF And the magnitude of MO is p jMO j D 0.03332 C 0.18332 C 0.32 F jMO j D 0.353 F 190 If we set jMO j D 560 N-m, we can solve for jFmax j 560 D 0.353jFmax j jFmax j D 1586 N (b) If F can be in any direction, then the worst case is when r ? F. The moment in this case is jMO j D jrjjFworst j jrj D p 0.252 C 0.22 C 0.152 D 0.3536 m 560 D 0.3536jFWORST j jFworst j D 1584 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.63 A civil engineer in Boulder, Colorado estimates that under the severest expected Chinook winds, the total force on the highway sign will be F D 2.8i 1.8j (kN). Let MO be the moment due to F about the base O of the cylindrical column supporting the sign. The y component of MO is called the torsion exerted on the cylindrical column at the base, and the component of MO parallel to the xz plane is called the bending moment. Determine the magnitudes of the torsion and bending moment. y F 8m 8m O x Solution: The total moment is z M D 8j C 8k m ð 2.8i 1.8j kN D 14.4i C 22.4j 22.4k kN-m We now identify Torsion D My D 22.4 kN-m Bending moment D Mx 2 C Mz 2 D 14.4 kNm2 C 22.4 kNm2 D 26.6 kN-m Problem 4.64 The weights of the arms OA and AB of the robotic manipulator act at their midpoints. The direction cosines of the centerline of arm OA are cos x D 0.500, cos y D 0.866, and cos z D 0, and the direction cosines of the centerline of arm AB are cos x D 0.707, cos y D 0.619, and cos z D 0.342. What is the sum of the moments about O due to the two forces? Solution: By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are e1 D 0.5i C 0.866j, and e2 D 0.707i C 0.619j 0.342k. The position vectors of the midpoints of the arms are r1 D 0.3e1 D 0.30.5i C 0.866j D 0.15i C 0.2598j r2 D 0.6e1 C 0.3e2 D 0.512i C 0.7053j 0.1026k. The sum of moments is y 0 60 mm B 160 N M D r1 ð W 1 C r 2 ð W 2 i j k i D 0.15 0.2598 0 C 0.512 0 200 0 0 j 0.7053 160 k 0.1026 0 D 16.42i 111.92k (N-m) 600 mm A 200 N O z x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 191 Problem 4.65 The tension in cable AB is 100 lb. If you want the magnitude of the moment about the base O of the tree due to the forces exerted on the tree by the two ropes to be 1500 ft-lb, what is the necessary tension in rope AC ? Solution: We have the forces 100 lb TAC 8j C 10k, F2 D p 14i 8j C 14k F1 D p 164 456 Thus the total moment is M D 8 ftj ð F1 C F2 D 625 ft lb C 5.24 ft TAC i y 5.24 ftTAC K The magnitude squared is then 625 ft lb C 5.24 ft TAC 2 C 5.24 ft TAC 2 D 1500 ft lb2 Solving we find TAC D 134 lb (0, 8, 0) ft A x O B (0, 0, 10) ft (14, 0, 14) ft C z Problem 4.66* A force F acts at the top end A of the pole. Its magnitude is jFj D 6 kN and its x component is Fx D 4 kN. The coordinates of point A are shown. Determine the components of F so that the magnitude of the moment due to F about the base P of the pole is as large as possible. (There are two answers.) Solution: The force is given by F D 4 kNi C Fy j C Fz k. Since the magnitude is constrained we must have 4 kN2 C Fy 2 C Fz 2 D 6 kN2 ) Fz D 20 kN2 Fy 2 Thus we will use (suppressing the units) y FD F 4i C Fy j C 20 Fy 2 k A (4, 3, ⫺2) m The moment is now given by M D 4i C 3j 2k ð F M D 2Fy C 3 20 Fy 2 i 8 C 4 20 Fy 2 j C 12 C 4Fy k The magnitude is P x M2 D 708 5Fy 2 C 64 20 Fy 2 C 12Fy 8 C 20 Fy 2 To maximize this quantity we solve z dM2 D 0 for the critical values dFy of Fy . There are three solutions Fy D 4.00, 3.72, 3.38. The first and third solutions produce the same maximum moment. The second answer corresponds to a local minimum and is therefore discarded. So the force that produces the largest moment is F D 4i 4j C 2k 192 or F D 4i 3.38j C 2.92k c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.67 The force F D 5i (kN) acts on the ring A where the cables AB, AC, and AD are joined. What is the sum of the moments about point D due to the force F and the three forces exerted on the ring by the cables? D (0, 6, 0) m A Strategy: The ring is in equilibrium. Use what you know about the four forces acting on it. C B Solution: The vector from D to A is rDA D 12i 2j C 2k m. The sum of the moments about point D is given by F (12, 4, 2) m (6, 0, 0) m x (0, 4, 6) m z FAD A F MD D rDA ð FAD C rDA ð FAC C rDA ð FAB C rDA ð F MD D rDA ð FAD C FAC C FAB C F FAC FAB However, we are given that ring A is in equilibrium and this implies that FAD C FAC C FAB C F D O D 0 Thus, MD D rDA ð O D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 193 Problem 4.68 In Problem 4.67, determine the moment about point D due to the force exerted on the ring A by the cable AB. Solution: We need to write the forces as magnitudes times the D(0, 6, 0) appropriate unit vectors, write the equilibrium equations for A in component form, and then solve the resulting three equations for the three unknown magnitudes. The unit vectors are of the form FAD FAC eAP D A(12, 4, 2) m F = 5i (kN) xP xA i C yP yA j C zP zA k jrAP j C(0, 4, 6) m Where P takes on values B, C, and D B(6, 0, 0) m Calculating the unit vectors, we get e D 0.802i 0.535j 0.267k AB eAC D 0.949i C 0j C 0.316k eAD D 0.973i C 0.162j 0.162k From equilibrium, we have FAB eAB C FAC eAC C FAD eAD C 5i kN D 0 In component form, we get i: 0.802FAB 0.949FAC 0.973FAD C 5 D 0 j: 0.535FAB C 0FAC C 0.162FAD D 0 k: 0.267FAB C 0.316FAC 0.162FAD D 0 Solving, we get FAB D 779.5 N, FAC D 1976 N FAD D 2569 N The vector from D to A is rDA D 12i 2j C 2k m The force FAB is given by FAB D FAB eAB FAB D 0.625i 0.417j 0.208k kN The moment about D is given by MD D rDA ð FAB i D 12 0.625 j 2 0.417 k 2 0.208 MD D 1.25i C 1.25j 6.25k kN-m 194 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.69 The tower is 70 m tall. The tensions in cables AB, AC, and AD are 4 kN, 2 kN, and 2 kN, respectively. Determine the sum of the moments about the origin O due to the forces exerted by the cables at point A. y A D 35 m B 35 m 40 m C O x 40 m 40 m z Solution: The coordinates of the points are A (0, 70, 0), B (40, 0, 0), C (40, 0, 40) D(35, 0, 35). The position vectors corresponding to the cables are: rAD D 35 0i C 0 70j C 35 0k rAD D 35i 70k 35k rAC D 40 0i C 0 70j C 40 0k rAC D 40i 70j C 40k The sum of the forces acting at A are TA D 0.2792i 6.6615j C 0.07239k (kN-m) The position vector of A is rOA D 70j. The moment about O is M D rOA ð TA i M D 0 0.2792 j 70 6.6615 k 0 0.07239 D 700.07239i j0 k700.2792 D 5.067i 19.54k rAB D 40 0i C 0 70j C 0 0k rAB D 40i 70j C 0k The unit vectors corresponding to these position vectors are: eAD D rAD 35 70 35 D i j jrAD j 85.73 85.73 85.73 D 0.4082i 0.8165j 0.4082k eAC D rAC 40 70 40 D i jC k jrAC j 90 90 90 D 0.4444i 0.7778j C 0.4444k eAB D 40 70 rAB D i j C 0k D 0.4962i 0.8682j C 0k jrAB j 80.6 80.6 The forces at point A are TAB D 4eAB D 1.9846i 3.4729j C 0k TAC D 2eAB D 0.8889i 1.5556j C 0.8889k TAD D 2eAD D 0.8165i 1.6330j 0.8165k. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 195 Problem 4.70 Consider the 70-m tower in Problem 4.69. Suppose that the tension in cable AB is 4 kN, and you want to adjust the tensions in cables AC and AD so that the sum of the moments about the origin O due to the forces exerted by the cables at point A is zero. Determine the tensions. Solution: From Varignon’s theorem, the moment is zero only if the resultant of the forces normal to the vector rOA is zero. From Problem 4.69 the unit vectors are: eAD 35 70 35 rAD D i j D jrAD j 85.73 85.73 85.73 The tensions are TAB D 4eAB , TAC D jTAC jeAC , and TAD D jTAD jeAD . The components normal to rOA are D 0.4082i 0.8165j 0.4082k eAC D 40 70 40 rAC D i jC k jrAC j 90 90 90 D 0.4444i 0.7778j C 0.4444k eAB D FX D 0.4082jTAD j 0.4444jTAC j C 1.9846i D 0 FZ D 0.4082jTAD j C 0.4444jTAC jk D 0. The HP-28S calculator was used to solve these equations: jTAC j D 2.23 kN, jTAD j D 2.43 kN rAB 40 70 D i j C 0k D 0.4963i 0.8685j C 0k jrAB j 80.6 80.6 Problem 4.71 The tension in cable AB is 150 N. The tension in cable AC is 100 N. Determine the sum of the moments about D due to the forces exerted on the wall by the cables. y 5m 5m B C Solution: The coordinates of the points A, B, C are A (8, 0, 0), B (0, 4, 5), C (0, 8, 5), D(0, 0, 5). The point A is the intersection of the lines of action of the forces. The position vector DA is 4m 8m 8m D rDA D 8i C 0j 5k. A z x The position vectors AB and AC are rAB D 8i C 4j 5k, rAB D rAC D 8i C 8j C 5k, rAC D p 82 C 42 C 52 D 10.247 m. p 82 C 82 C 52 D 12.369 m. The unit vectors parallel to the cables are: eAB D 0.7807i C 0.3904j 0.4879k, eAC D 0.6468i C 0.6468j C 0.4042k. i MD D 8 181.79 j 0 123.24 k 5 D 123.245i C32.77 8C32.77 5181.79j C 8123.24k MD D 616.2i 117.11j 985.9k (N-m) (Note: An alternate method of solution is to express the moment in terms of the sum: MD D rDC ð TC C rDB ð TB . The tensions are y TAB D 150eAB D 117.11i C 58.56j 73.19k, 5m TAC D 100eAC D 64.68i C 64.68j C 40.42k. The sum of the forces exerted by the wall on A is 5m B 4m C TA D 181.79i C 123.24j 32.77k. The force exerted on the wall by the cables is TA . The moment about D is MD D rDA ð TA , 196 A 8m z D 8m F x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.72 Consider the wall shown in Problem 4.71. The total force exerted by the two cables in the direction perpendicular to the wall is 2 kN. The magnitude of the sum of the moments about D due to the forces exerted on the wall by the cables is 18 kN-m. What are the tensions in the cables? Solution: From the solution of Problem 4.71, we have rDA D 8i C 0j 5k. Forces in both cables pass through point A and we can use this vector to determine moments of both forces about D. The position vectors AB and AC are p rAB D 8i C 4j 5k, jrAB j D rAC D 8i C 8j C 5k, jrAC j D 82 C 42 C 52 D 10.247 m. p 82 C 82 C 52 D 12.369 m. The unit vectors parallel to the cables are: eAB D 0.7807i C 0.3904j 0.4879k, eAC D 0.6468i C 0.6468j C 0.4042k. The tensions are TBA D TBA eAB D TBA 0.7807i C 0.3904j 0.4879k, and TCA D TCA eAC D TCA 0.6468i C 0.6468j C 0.4042k. The sum of the forces exerted by the cables perpendicular to the wall is given by TPerpendicular D TAB 0.7807 C TAC 0.6468 D 2 kN. The moments of these two forces about D are given by MD D rDA ð TCA C rDA ð TBA D rDA ð TCA C TBA . The sum of the two forces is given by i 8 MD D TCA C TCB X j 0 TCA C TCB Y k . 5 TCA C TCB Z This expression can be expanded to yield MD D 5TCA C TCB Y i C [8TCA C TCB Z 5TCA C TCB X ]j C 8TCA C TCB Y k. The magnitude of this vector is given as 18 kN-m. Thus, we obtain the relation jMD j D 25TCA C TCB 2Y C [8TCA C TCB Z D 18 kN-m. 5TCA C TCB X ]2 C 64TCA C TCB 2Y We now have two equations in the two tensions in the cables. Either algebraic substitution or a numerical solver can be used to give TBA D 1.596 kN, and TCA D 1.166 kN. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 197 Problem 4.73 The tension in the cable BD is 1 kN. As a result, cable BD exerts a 1-kN force on the “ball” at B that points from B toward D. Determine the moment of this force about point A. Solution: We have the force and position vectors FD 1 kN 4i C 2j C 4k, r D AB D 4i C 3j C k m 6 The moment is then y M D r ð F D 1.667i 3.33j C 3.33k kN-m C (0, 4, ⫺3) m B (4, 3, 1) m D (0, 5, 5) m x A E z Problem 4.74* Suppose that the mass of the suspended object E in Problem 4.73 is 100 kg and the mass of the bar AB is 20 kg. Assume that the weight of the bar acts at its midpoint. By using the fact that the sum of the moments about point A due to the weight of the bar and the forces exerted on the “ball” at B by the three cables BC, BD, and BE is zero, determine the tensions in the cables BC and BD. Solution: We have the following forces applied at point B. F1 D 100 kg9.81 m/s2 j, F3 D TBC F2 D p 4i C j 4k, 33 TBD 4i C 2j C 4k 6 In addition we have the weight of the bar F4 D 20 kg9.81 m/s2 j The moment around point A is MA D 4i C 3j C k m ð F1 C F2 C F3 C 2i C 1.5j C 0.5k m ð F4 D 0 Carrying out the cross products and breaking into components we find Mx D 1079 2.26TBC C 1.667TBD D 0 My D 2.089TBC 3.333TBD D 0 Mz D 4316 C 2.785TBC C 3.333TBD D 0 Only two of these three equations are independent. Solving we find TBC D 886 N, TBD D 555 N 198 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.75 The 200-kg slider at A is held in place on the smooth vertical bar by the cable AB. Determine the moment about the bottom of the bar (point C with coordinates x D 2 m, y D z D 0) due to the force exerted on the slider by the cable. y 2m B A 5m 2m 2m C x z Solution: The slider is in equilibrium. The smooth bar exerts no vertical forces on the slider. Hence, the vertical component of FAB supports the weight of the slider. FAB The unit vector from A to B is determined from the coordinates of points A and B A2, 2, 0, B0, 5, 2 m Thus, H rAB D 2i C 3j C 2k m eAB D 0.485i C 0.728j C 0.485k and −mg j FAB D FAB eAB The horizontal force exerted by the bar on the slider is H D Hx i C H z k Equilibrium requires H C FAB mgj D 0 i: Hx 0.485FAB D 0 m D 200 kg j: 0.728FAB mg D 0 g D 9.81 m/s2 k: Hz C 0.485FAB D 0 Solving, we get FAB D 2697N D 2, 70 kN Hx D 1308N D 1.31 kN Hz D 1308N D 1.31 kN rCA D 2j m FAB D FAB eAB FAB D 1308i C 1962j C 1308k N i Mc D 0 1308 j 2 1962 k 0 1308 Mc D 2616i C 0j C 2616k N-m Mc D 2.62i C 2.62i kN-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 199 y Problem 4.76 To evaluate the adequacy of the design of the vertical steel post, you must determine the moment about the bottom of the post due to the force exerted on the post at B by the cable AB. A calibrated strain gauge mounted on cable AC indicates that the tension in cable AC is 22 kN. What is the moment? 5m 5m C D 4m 8m (6, 2, 0) m B A O z 3m 12 m x D(0, 4, −5) m Solution: To find the moment, we must find the force in cable AB. In order to do this, we must find the forces in cables AO and AD also. This requires that we solve the equilibrium problem at A. Our first task is to write unit vectors eAB , eAO , eAC , and eAD . Each will be of the form eAi xi xA i C yi yA j C zi zA k D xi xA 2 C yi yA 2 C zi zA 2 where i takes on the values B, C, D, and O. We get eAB D 0.986i C 0.164j C 0k eAC D 0.609i C 0.609j C 0.508k eAD D 0.744i C 0.248j 0.620k C (0, 8, 5) m TAD TAC A (6, 2, 0) m TAO B(12, 3, 0) m TAB O(0, 0, 0) m In component form, T e C TAC eACx C TAD eADx C TAO eAOx D 0 AB ABx TAB eABy C TAC eACy C TAD eADy C TAO eAOy D 0 TAB eABz C TAC eACz C TAD eADz C TAO eAOz D 0 We know TAC D 22 kN. Substituting this in, we have 3 eqns in 3 unknowns. Solving, we get eAO D 0.949i 0.316j C 0k We now write the forces as TAB D 163.05 kN, TAD D 18.01 kN TAO D 141.28 kN We now know that TAB is given as TAB D TAB eAB TAB D TAB eAB D 160.8i C 26.8j kN TAC D TAC eAC and that the force acting at B is TAB . TAD D TAD eAD The moment about the bottom of the post is given by TAO D TAO eAO MBOTTOM D r ð TAB D 3j ð TAB We then sum the forces and write the equilibrium equations in component form. For equilibrium at A, 200 FA D 0 Solving, we get MBOTTOM D 482k kN-m FA D TAB C TAC C TAD C TAO D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.77 The force F D 20i C 40j 10k (N). Use both of the procedures described in Example 4.7 to determine the moment due to F about the z axis. y F Solution: First Method: We can use Eqs. (4.5) and (4.6) r D 8i m x (8, 0, 0) m F D 20i C 40j 10k N Mzaxis D [k Ð r ð F]k 0 jMzaxis j D k Ð r ð F D 8 m 20 N 01 0 40 N 0 D 320 N-m 10 N z Mzaxis D 320 N-mk Second Method: The y-component of the force is perpendicular to the plane containing the z axis and the position vector r. The perpendicular distance from the z axis to the y-component of the force is 8 m. Therefore jMzaxis j D 40 N8 m D 320 N-m Using the right-hand rule we see that the moment is about the Cz axis. Thus Mzaxis D 320 N-mk Problem 4.78 Use Eqs. (4.5) and (4.6) to determine the moment of the 20-N force about (a) the x axis, (b) the y axis, (c) the z axis. (First see if you can write down the results without using the equations.) Solution: The force is parallel to the z axis. The perpendicular distance from the x axis to the line of action of the force is 4 m. The perpendicular distance frompthe y axis p is 7 m and the perpendicular distance from the z axis is 42 C 72 D 65 m. By inspection, the moment about the x axis is y Mx D 420i (N-m) (7, 4, 0) m Mx D 80i N-m By inspection, the moment about the y axis is My D 720j N-m 20 k (N) x My D 140j (N-m) z By inspection, the moment about the z axis is zero since F is parallel to the z axis. Mz D 0 N-m Now for the calculations using (4.5) and (4.6) ML D [e Ð r ð F]e 1 Mx D 7 0 0 4 0 0 0 i D 80i N-m 20 0 My D 7 0 1 4 0 0 0 j D 140j N-m 20 0 Mz D 7 0 0 4 0 1 0 k D 0k N-m 20 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 201 y Problem 4.79 Three forces parallel to the y axis act on the rectangular plate. Use Eqs. (4.5) and (4.6) to determine the sum of the moments of the forces about the x axis. (First see if you can write down the result without using the equations.) 3 kN x 2 kN 6 kN 600 mm 900 mm z Solution: By inspection, the 3 kN force has no moment about the M6 x axis since it acts through the x axis. The perpendicular distances of the other two forces from the x axis is 0.6 m. The H 2 kN force has a positive moment and the 6 kN force has a negative about the x axis. Mx D [20.6 60.6]i kN Mx D 2.4i kN kN 1 D 0 0 Mx D M3 0 0 6 kN 0 .6 i D 3.6i kN 0 C M2 kN C M6 kN Mx D 0 C 1.2i 3.6i kN Mx D 2.4i kN Calculating the result: M3 M2 kN kN 1 D 0 0 0 0 3 0 0 i D 0i kN 0 1 D 0 0 0 0 2 0 .6 i D 1.2i kN 0 Problem 4.80 Consider the rectangular plate shown in Problem 4.79. The three forces are parallel to the y axis. Determine the sum of the moments of the forces (a) about the y axis, (b) about the z axis. Solution: (a) The magnitude of the moments about the y axis is M D eY Ð r ð F. The position vectors of the three forces are given in the solution to Problem 4.79. The magnitude for each force is: 0 1 0 eY Ð r ð F D 0.9 0 0 D 0, 0 3 0 0 1 eY Ð r ð F D 0.9 0 0 6 0 0.6 D 0, 0 0 eY Ð r ð F D 0 0 0 0.6 D 0 0 1 0 2 Thus the moment about the y axis is zero, since the magnitude of each moment is zero. 202 (b) The magnitude of each moment about the z axis is 0 1 eZ Ð r ð F D 0.9 0 0 3 0 0 D 2.7, 0 0 0 eZ Ð r ð F D 0.9 0 0 C 6 1 0.6 D 5.4, 0 0 eZ Ð r ð F D 0 0 0 0 2 1 0.6 D 0. 0 Thus the moment about the z axis is MZ D 2.7eZ C 5.4eZ D 2.7k (kN-m) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.81 The person exerts a force F D 0.2i 0.4j C 1.2k (lb) on the gate at C. Point C lies in the xy plane. What moment does the person exert about the gate’s hinge axis, which is coincident with the y axis? y A C Solution: 3.5 ft M D [e Ð r ð F]e e D j, 0 MY D 2 .2 r D 2i ft, F is given x 1 0 0 0 j D 2.4j ft-lb .4 1.2 B 2 ft y Problem 4.82 Four forces act on the plate. Their components are FB FA D 2i C 4j C 2k (kN), FA FB D 3j 3k (kN), x FD FC FC D 2j C 3k (kN), FD D 2i C 6j C 4k (kN). Determine the sum of the moments of the forces (a) about the x axis; (b) about the z axis. z 2m 3m Solution: Note that FA acts at the origin so no moment is generated about the origin. For the other forces we have i j k j k i 0 2m 0 0 C3 m MO D 3 m 0 2 kN 3 kN 3 kN 3 kN 0 i C 0 2 kN j 0 6 kN k 2m 4 kN MO D 16i C 4j C 15k kN-m Now we find Mx D MO Ð i D 16 kN-m, Mz D MO Ð k D 15 kN-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 203 y Problem 4.83 The force F D 30i C 20j 10k (lb). (a) What is the moment of F about the y axis? (b) Suppose that you keep the magnitude of F fixed, but you change its direction so as to make the moment of F about the y axis as large as possible. What is the magnitude of the resulting moment? F (4, 2, 2) ft Solution: x z (a) My D j Ð [4i C 2j C 2k ft ð 30i C 20j 10k lb] 0 My D 4 ft 30 lb 2 ft 2 ft D 100 ft lb 20 lb 10 lb 1 0 ) My D 100 ft-lbj (b) p p Mymax D Fd D 302 C 202 C 102 lb 42 C 22 ft D 167.3 ft-lb Note that d is the distance from the y axis, not the distance from the origin. Problem 4.84 The moment of the force F shown in Problem 4.83 about the x axis is 80i (ft-lb), the moment about the y axis is zero, and the moment about the z axis is 160k (ft-lb). If Fy D 80 lb, what are Fx and Fz ? Solution: The magnitudes of the moments: eX e ž r ð F D rX FX eY rY FY eZ rZ , FZ 0 eZ Ð r ð F D 4 FX 0 2 80 1 2 D 320 2FX D 160 FZ Solve: FX D 80 lb, FZ D 40 lb, from which the force vector is F D 80i C 80j C 40k 204 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.85 The robotic manipulator is stationary. The weights of the arms AB and BC act at their midpoints. The direction cosines of the centerline of arm AB are cos x D 0.500, cos y D 0.866, cos z D 0, and the direction cosines of the centerline of arm BC are cos x D 0.707, cos y D 0.619, cos z D 0.342. What total moment is exerted about the z axis by the weights of the arms? m y 0m C 60 160 N 600 mm B Solution: The unit vectors along AB and AC are of the form 200 N e D cos x i C cos y j C cos z k. The unit vectors are A eAB D 0.500i C 0.866j C 0k and eBC D 0.707i C 0.619j 0.342k. z The vector to point G at the center of arm AB is x rAG D 3000.500i C 0.866j C 0k D 150i C 259.8j C 0k mm, and the vector from A to the point H at the center of arm BC is given by rAH D rAB C rBH D 600eAB C 300eBC D 512.1i C 705.3j 102.6k mm. The weight vectors acting at G and H are WG D 200j N, and WH D 160j N. The moment vectors of these forces about the z axis are of the form eX e ž r ð F D rX FX ey rY FY ez rZ . FZ Here, WG and WH take on the role of F, and e D k. Substituting into the form for the moment of the force at G, we get 0 e ž r ð F D 0.150 0 0 1 0.260 0 D 30 N-m. 200 0 Similarly, for the moment of the force at H, we get 0 0 e ž r ð F D 0.512 0.705 0 160 1 0.103 D 81.9 N-m. 0 The total moment about the z axis is the sum of the two moments. Hence, Mz axis D 111.9 N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 205 Problem 4.86 In Problem 4.85, what total moment is exerted about the x axis by the weights of the arms? Solution: The solution is identical to that of Problem 4.85 except that e D i. Substituting into the form for the moment of the force at G, we get 1 0 0 e Ð r ð F D 0.150 0.260 0 D 0 N-m. 0 200 0 Similarly, for the moment of the force at H, we get 1 0 e Ð r ð F D 0.512 0.705 0 160 0 0.103 D 16.4 N-m. 0 The total moment about the x axis is the sum of the two moments. Hence, Mx axis D 16.4 N-m Problem 4.87 In Active Example 4.6, suppose that the force changes to F D 2i C 3j C 6k (kN). Determine the magnitude of the moment of the force about the axis of the bar BC. y C (0, 4, 0) m F ⫽ ⫺2i ⫹ 6j ⫹ 3k (kN) A (4, 2, 2) m Solution: We have the following vectors rBA D 4i C 2j 1k m x B z (0, 0, 3) m F D 2i C 3j C 6k kN rBC D 4j 3k m eBC D rBC D 0.8j 0.6k jrBC j The moment of F about the axis of the bar is 0 0.8 0.6 jMBC j D eBC Ð r ð F D 4 2 1 D 27.2 kN-m 2 3 6 Thus MBC D 27.2 kN-meBC , jMBC j D 27.2 kN-m 206 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.88 Determine the moment of the 20-N force about the line AB. Use Eqs. (4.5) and (4.6), letting the unit vector e point (a) from A toward B, (b) from B toward A. y A (0, 5, 0) m (7, 4, 0) m 20k (N) B (– 4, 0, 0) m x z Solution: First, we need the unit vector Using eAB xB xA i C yB yA j C zB zA k eAB D xB xA 2 C yB yA 2 C zB zA 2 0.625 ML D 7 0 0.781 0 1 0 0.625i 0.781j 0 20 eAB D 0.625i 0.781j D eBA ML D 76.1i 95.1j N-m Now, the moment of the 20k (N) force about AB is given as ex ML D rx Fx ey ry Fy ez rz e Fz Using eBA 0.625 ML D 7 0 where e is eAB or eBA 0.781 0 1 0 0.625i C 0.781j 0 20 For this problem, r must go from line AB to the point of application of the force. Let us use point A. ML D 76.1i 95.1j N-m r D 7 0i C 4 5j C 0 0k m Ł Results are the same r D 7i 1j C 0k m Problem 4.89 The force F D 10i C 5j 5k (kip). Determine the moment of F about the line AB. Draw a sketch to indicate the sense of the moment. y B (6, 6, 0) ft Solution: The moment of F about pt. A is MA D 6i ð F i j D 6 0 10 5 F k 0 5 A x (6, 0, 0) ft z D 30j 30k ft-kip. y The unit vector j is parallel to line AB, so the moment about AB is (6, 6, 0) ft MAB D j Ð MA j B D 30j ft-kip. F y B z x A (6, 0, 0) ft −30j (ft-kip) Direction of moment x A z c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 207 y Problem 4.90 The force F D 10i C 12j 6k (N). What is the moment of F about the line OA? Draw a sketch to indicate the sense of the moment. Solution: The strategy is to determine a unit vector parallel to OA and to use this to determine the moment about OA. The vector parallel to OA is rOA D 6j C 4k. The magnitude: F. The unit vector parallel to OA is eOA D 0.8321j C 0.5547k. The vector from O to the point of application of F is rOF D 8i C 6k. The magnitude of the moment about OA is 0 jMO j D eOA Ð rOF ð F D 8 10 A (0, 6, 4) m x (8, 0, 6) m 0.5547 6 6 0.8321 0 12 F O z D 89.8614 C 53.251 D 143.1 N-m. The moment about OA is MOA D jMOA jeOA D 119.1j C 79.4k (N-m). The sense of the moment is in the direction of the curled fingers of the right hand when the thumb is parallel to OA, pointing to A. Problem 4.91 The tension in the cable AB is 1 kN. Determine the moment about the x axis due to the force exerted on the hatch by the cable at point B. Draw a sketch to indicate the sense of the moment. Solution: The vector parallel to BA is y A (400, 300, 0) mm x rBA D 0.4 1i C 0.3j 0.6k D 0.6i C 0.3j 0.6k. 600 mm B The unit vector parallel to BA is 1000 mm z eBA D 0.6667i C 0.3333j 0.6667k. The moment about O is i MO D rOB ð T D 1 0.6667 j 0 0.3333 k 0.6 0.66667 MO D 0.2i C 0.2667j C 0.3333k. The magnitude is jMX j D eX Ð MO D 0.2 kN-m. The moment is MX D 0.2i kN-m. The sense is clockwise when viewed along the x axis toward the origin. 208 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.92 Determine the moment of the force applied at D about the straight line through the hinges A and B. (The line through A and B lies in the yz plane.) 6 ft 20i – 60j (lb) Solution: From the figure, we see that the unit vector along the line from A toward B is given by eAB D sin 20° j C cos 20° k. The position vector is rAD D 4i ft, and the force vector is as shown in the figure. The moment vector of a force about an axis is of the form eX e ž r ð F D rX FX ey rY FY E A D 4 ft 2 ft ez rZ . FZ x B z 20° C 4 ft For this case, 0 sin 20° e ž r ð F D 4 0 20 60 cos 20° 0 D 240 cos 20° ft-lb 0 D 225.5 ft-lb. The negative sign is because the moment is opposite in direction to the unit vector from A to B. Problem 4.93 In Problem 4.92, the tension in the cable CE is 160 lb. Determine the moment of the force exerted by the cable on the hatch at C about the straight line through the hinges A and B. Solution: From the figure, we see that the unit vector along the line from A toward B is given by eAB D sin 20° j C cos 20° k. The position vector is rBC D 4i ft. The coordinates of point C are (4, 4 sin 20° , 4 cos 20° ). The unit vector along CE is 0.703i C 0.592j C 0.394k and the force vector is as shown acting at point D. The moment vector is a force about an axis is of the form eX e ž r ð F D rX FX ey rY FY ez rZ . FZ For this case, rCE D 4i C 3.368j C 2.242k TCE D 160eCE D 112.488i C 94.715j C 63.049k 0 sin 20° e ž r ð F D 4 0 112.488 94.715 cos 20° 0 D 240 cos 20° ft-lb 63.049 D 701 ft-lbs. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 209 Problem 4.94 The coordinates of A are (2.4, 0, 0.6) m, and the coordinates of B are (2.2, 0.7, 1.2) m. The force exerted at B by the sailboat’s main sheet AB is 130 N. Determine the moment of the force about the centerline of the mast (the y axis). Draw a sketch to indicate the sense of the moment. y x B Solution: The position vectors: rOA D 2.4i 0.6k (m), rOB D 2.2i C 0.7j 1.2k (m), A rBA D 2.4 C 2.2i C 0 0.7j C 0.6 C 1.2k (m) z D 0.2i 0.7j C 0.6k (m). The magnitude is jrBA j D 0.9434 m. The unit vector parallel to BA is eBA D 0.2120i 0.7420j C 0.6360k. The tension is TBA D 130eBA . The moment of TBA about the origin is MO D rOB ð TBA or i D 2.2 27.56 j 0.7 96.46 k 1.2 , 82.68 MO D 57.88i C 214.97j C 231.5k. The magnitude of the moment about the y axis is jMY j D eY Ð MO D 214.97 N-m. The moment is MY D eY 214.97 D 214.97j N-m. 210 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.95 The tension in cable AB is 200 lb. Determine the moments about each of the coordinate axes due to the force exerted on point B by the cable. Draw sketches to indicate the senses of the moments. y A (2, 5, –2) ft x z B (10, –2, 3) ft y Solution: The position vector from B to A is 443 ft-lb rBA D 2 10i C [5 2]j C 2 3k D 8i C 7j 5k ft, 187 ft-lb So the force exerted on B is F D 200 x rBA D 136.2i C 119.2j 85.1k lb. jrBA j The moment of F about the origin O is i rOB ð F D 10 136.2 j 2 119.2 k 3 85.1 919 ft-b z D 187i C 443j C 919k ft-lb. The moments about the x, y, and z axes are [rOB ð F Ð i]i D 187i ft-lb, [rOB ð F Ð j]j D 443j ft-lb, [rOB ð F Ð k]k D 919k ft-lb. Problem 4.96 The total force exerted on the blades of the turbine by the steam nozzle is F D 20i 120j C 100k (N), and it effectively acts at the point (100, 80, 300) mm. What moment is exerted about the axis of the turbine (the x axis)? y Fixed Rotating x Solution: The moment about the origin is i MO D 0.1 20 j 0.08 120 k 0.3 100 D 44.0i 4.0j 13.6k N-m. The moment about the x axis is z MO Ð ii D 44.0i N-m. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 211 Problem 4.97 The pneumatic support AB holds a trunk lid in place. It exerts a 35-N force on the fixture at B that points in the direction from A toward B. Determine the magnitude of the moment of the force about the hinge axis of the lid, which is the z axis. Solution: The vector from A to B is rAB D [60 480i C 100 40j C 30 40k] mm rAB D 420i C 140j 70k mm y The 35-N force can be written F D 35 N rAB D 32.8i C 10.9j 5.47k N jrAB j The moment about point O is i j MO D rOB ð F D 60 100 32.8 10.9 B (60, 100, ⫺30) mm k 30 5.47 O D 219i C 1310j C 3940k N-mm The magnitude of the moment about the z axis is z Mz D MO Ð k D 3940 N-mm D 3.94 N-m x Mz D 3.94 N-m 212 A (480, ⫺40, 40) mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.98 The tension in cable AB is 80 lb. What is the moment about the line CD due to the force exerted by the cable on the wall at B? y 8 ft 3 ft Solution: The strategy is to find the moment about the point C exerted by the force at B, and then to find the component of that moment acting along the line CD. The coordinates of the points B, C, D are B (8, 6, 0), C (3, 6, 0), D(3, 0, 0). The position vectors are: rOB D 8i C 6j, rOC D 3i C 6j, rOD D 3i. The vector parallel to CD is rCD D rOD rOC D 6j. The unit vector parallel to CD is eCD D 1j. The vector from point C to B is rCB D rOB rOC D 5i. B The position vector of A is rOA D 6i C 10k. The vector parallel to BA is rBA D rOA rOB D 2i 6j C 10k. The magnitude is jrBA j D 11.832 ft. The unit vector parallel to BA is C 6 ft eBA D 0.1690i 0.5071j C 0.8452k. The tension acting at B is x D TBA D 80eBA D 13.52i 40.57j C 67.62k. The magnitude of the moment about CD due to the tension acting at B is z A (6, 0, 10) ft 0 jMCD j D eCD Ð rCB ð TBA D 5 13.52 1 0 40.57 0 0 67.62 D 338.1 (ft lb). The moment about CD is MCD D 338.1eCD D 338.1j (ft lb). The sense of the moment is along the curled fingers of the right hand when the thumb is parallel to CD, pointing toward D. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 213 Problem 4.99 The magnitude of the force F is 0.2 N and its direction cosines are cos x D 0.727, cos y D 0.364, and cos z D 0.582. Determine the magnitude of the moment of F about the axis AB of the spool. Solution: We have rAB D 0.3i 0.1j 0.4k m, rAB D 0.32 C 0.12 C 0.42 m D p 0.26 m y 1 0.3i 0.1j 0.4k eAB D p 0.26 B F D 0.2 N0.727i 0.364j C 0.582k (200, 400, 0) mm rAP D 0.26i 0.025j 0.11k m (160, 475, 290) mm Now the magnitude of the moment about the spool axis AB is P A F (⫺100, 500, 400) mm x MAB 0.2 N D p 0.26 0.3 0.26 m 0.727 0.4 0.11 m D 0.0146 N-m 0.582 0.1 0.025 m 0.364 z Problem 4.100 A motorist applies the two forces shown to loosen a lug nut. The direction cosines of 4 3 F are cos x D 13 , cos y D 12 , and cos z D 13 . If the 13 magnitude of the moment about the x axis must be 32 ftlb to loosen the nut, what is the magnitude of the forces the motorist must apply? y Solution: The unit vectors for the forces are the direction cosines. The position vector of the force F is rOF D 1.333k ft. The magnitude of the moment due to F is 1 0 jMOF j D eX Ð rOF ð F D 0.3077F 0 0 0.9231F 0 1.333 0.2308F jMOF j D 1.230F ft lb. The magnitude of the moment due to F is –F F jMOF j D eX Ð rOF ð F 1 D 0 .3077F z 0 0 0.9231F 0 1.333 D 1.230F ft lb. 0.2308F The total moment about the x axis is 16 in 16 in x MX D 1.230Fi C 1.230Fi D 2.46Fi, from which, for a total magnitude of 32 ft lb, the force to be applied is FD 214 32 D 13 lb 2.46 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.101 The tension in cable AB is 2 kN. What is the magnitude of the moment about the shaft CD due to the force exerted by the cable at A? Draw a sketch to indicate the sense of the moment about the shaft. 2m C A Solution: The strategy is to determine the moment about C due to A, and determine the component parallel to CD. The moment is determined from the distance CA and the components of the tension, which is to be found from the magnitude of the tension and the unit vector parallel to AB. The coordinates of the points A, B, C, and D are: A (2, 2, 0), B (3, 0, 1), C (0, 2, 0), and D (0,0,0). The unit vector parallel to CD is by inspection eCD D 1j. The position vectors parallel to DC, DA, and DB: rDC D 2j, rDA D 2i C 2j, rDB D 3i C 1k. 2m The vector parallel to CA is rCA D 2i. The vector parallel to AB is rAB D rDB rDA D 1i 2j C 1k. D B 1m The magnitude: jrAB j D 2.4495 m. The unit vector parallel to AB is 3m eAB D 0.4082i 0.8165j C 0.4082k. The tension is TAB D 2eAB D 0.8165i 1.633j C 0.8165k. The magnitude of the moment about CD is 0 jMCD j D eCD Ð rCA ð TAB D 2 0.8164 1 0 1.633 0 0 0.8165 D 1.633 kN-m. The moment about CD is MCD D eCD jMCD j D 1.633j (kN-m). The sense is in the direction of the curled fingers of the right hand when the thumb is parallel to DC, pointed toward D. Problem 4.102 The axis of the car’s wheel passes through the origin of the coordinate system and its direction cosines are cos x D 0.940, cos y D 0, cos z D 0.342. The force exerted on the tire by the road effectively acts at the point x D 0, y D 0.36 m, z D 0 and has components F D 720i C 3660j C 1240k (N). What is the moment of F about the wheel’s axis? Solution: We have to determine the moment about the axle where a unit vector along the axle is e D cos x i C cos y j C cos z k e D 0.940i C 0j C 0.342k The vector from the origin to the point of contact with the road is r D 0i 0.36j C 0k m The force exerted at the point of contact is F D 720i C 3660j C 1240k N y The moment of the force F about the axle is MAXLE D [e Ð r ð F]e x MAXLE z 0.940 D 0 720 0 0.36 C3660 0.342 0 0.940i C 0.342k N-m C1240 MAXLE D 508.260.940i C 0.342k N-m MAXLE D 478i 174k N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 215 Problem 4.103 The direction cosines of the centerline OA are cos x D 0.500, cos y D 0.866, and cos z D 0, and the direction cosines of the line AG are cos x D 0.707, cos y D 0.619, and cos z D 0.342. What is the moment about OA due to the 250-N weight? Draw a sketch to indicate the sense of the moment about the shaft. Solution: By definition, the direction cosines are the scalar components of the unit vectors. Thus the unit vectors are e1 D 0.5i C 0.866j, and e2 D 0.707i C 0.619j 0.341k. The force is W D 250j (N). The position vector of the 250 N weight is rW D 0.600e1 C 0.750e2 D 0.8303i C 0.9839j 0.2565k The moment about OA is G m MOA D eOA eOA Ð rW ð W m 50 y 0.5 D 0.8303 0 7 250 N 0.866 0.9839 250 0 0.2565 e1 D 32.06e1 0 D 16i 27.77j (N-m) A 600 mm The moment is anti parallel to the unit vector parallel to OA, with the sense of the moment in the direction of the curled fingers when the thumb of the right hand is directed oppositely to the direction of the unit vector. O z x Problem 4.104 The radius of the steering wheel is 200 mm. The distance from O to C is 1 m. The center C of the steering wheel lies in the x y plane. The driver exerts a force F D 10i C 10j 5k (N) on the wheel at A. If the angle ˛ D 0, what is the magnitude of the moment about the shaft OC? Draw a sketch to indicate the sense of the moment about the shaft. y Solution: The strategy is to determine the moment about C, and then determine its component about OC. The radius vectors parallel to OC and CA are: rOC D 1i cos 20° C j sin 20° C 0k D 0.9397i C 0.3420j. The line from C to the x axis is perpendicular to OC since it lies in the plane of the steering wheel. The unit vector from C to the x axis is eCX D i cos20 90 C j sin20 90 D 0.3420i 0.9397j, where the angle is measured positive counterclockwise from the x axis. The vector parallel to CA is F C A O z rCA D 0.2eCX D C0.0684i 0.1879j (m). The magnitude of the moment about OC 20° α 0.9397 jMOC j D eOC Ð rCA ð F D 0.0684 10 0.3420 0 0.1879 0 10 5 x D 0.9998 D 1 N-m. The sense of the moment is in the direction of the curled fingers of the right hand if the thumb is parallel to OC, pointing from O to C. 216 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.105* The magnitude of the force F is 10 N. Suppose that you want to choose the direction of the force F so that the magnitude of its moment about the line L is a maximum. Determine the components of F and the magnitude of its moment about L. (There are two solutions for F.) y Solution: The moment of the general force F D Fx i C Fy j C Fz k about the line is developed by eBA D 1 3i C 6j 6k D i C 2j 2k, 9 3 rBP D 12i C 2j 2k m, MBA D eBA Ð rBP ð F A (3, 8, 0) m This expression simplifies to MBA D We also have the constraint that 10 N2 D Fx 2 C Fy 2 C Fz 2 L F Since Fx does not contribute to the moment we set it equal to zero. Solving the constraint equation for Fz and substituting this into the expression for the moment we find B (0, 2, 6) m P (12, 4, 4) m MBA D 22 Fy š 3 100 Fy 2 . ) x z 22 m Fy C Fz 3 dMBA D0 dFy p p ) Fy D š5 2N ) Fz D š5 2 We thus have two answers: F D 7.07j C 7.07k N or F D 7.07j C 7.07k c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 217 Problem 4.106 The weight W causes a tension of 100 lb in cable CD. If d D 2 ft, what is the moment about the z axis due to the force exerted by the cable CD at point C? y (12, 10, 0) ft (0, 3, 0) ft W Solution: The strategy is to use the unit vector parallel to the bar to locate point C relative to the origin, and then use this location to find the unit vector parallel to the cable CD. With the tension resolved into components about the origin, the moment about the origin can be resolved into components along the z axis. Denote the top of the bar by T and the bottom of the bar by B. The position vectors of the ends of the bar are: D C d x z (3, 0, 10) ft rOB D 3i C 0j C 10k, rOT D 12i C 10j C 0k. The vector from the bottom to the top of the bar is rBT D rOT rOB D 9i C 10j 10k. The magnitude: p jrBT j D 92 C 102 C 102 D 16.763 ft. The unit vector parallel to the bar, pointing toward the top, is eBT D 0.5369i C 0.5965j 0.5965k. The position vector of the point C relative to the bottom of the bar is rBC D 2eBT D 1.074i C 1.193j 1.193k. The position vector of point C relative to the origin is rOC D rOB C rBC D 4.074i C 1.193j C 8.807k. The position vector of point D is rOD D 0i C 3j C 0k. The vector parallel to CD is rCD D rOD rOC D 4.074i C 1.807j 8.807k. The magnitude is jrCD j D p 4.0742 C 1.8072 C 8.8072 D 9.87 ft. The unit vector parallel to CD is eCD D 0.4127i C 0.1831j 0.8923k. The tension is TCD D 100eCD D 41.27i C 18.31j 89.23k lb. The magnitude of the moment about the z axis is 0 jMO j D eZ Ð rOC ð TCD D 4.074 41.27 0 1.193 18.31 1 8.807 89.23 D 123.83 ft lb 218 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.107* The y axis points upward. The weight of the 4-kg rectangular plate acts at the midpoint G of the plate. The sum of the moments about the straight line through the supports A and B due to the weight of the plate and the force exerted on the plate by the cable CD is zero. What is the tension in the cable? Solution: Note (150, 152.5, 195). that the coordinates of point G are We calculate the moment about the line BA due to the two forces as follows. eBA D 0.1i C 0.07j 0.36k p 0.1445 y r1 D 0.2i 0.125j C 0.03k m, A (100, 500, 700) mm (100, 250, 0) mm F1 D TCD D 0.1i C 0.445j C 0.31k p 0.304125 r2 D 0.15i 0.0275j 0.165k m, F2 D 4 kg9.81 m/s2 j G x B MBA D eBA Ð r1 ð F1 C r2 ð F2 The moment reduces to (0, 180, 360) mm C MBA D 3.871 N-m 0.17793 mTCD D 0 ) TCD D 21.8 N (200, 55, 390) mm z Problem 4.108 In Active Example 4.9, suppose that the point of application of the force F is moved from (8, 3, 0) m to (8, 8, 0) m. Draw a sketch showing the new position of the force. From your sketch, will the moment due to the couple be clockwise or counterclockwise? Calculate the moment due to the couple. Represent the moment by its magnitude and a circular arrow indicating its direction. y ⴚF (6, 6, 0) m (8, 3, 0) m F x Solution: From Active Example 4.9 we know that F D 10i 4j N From the sketch, it is evident that the moment will be clockwise. The moment due to the couple is the sum of the moments of the two forces about any point. If we determine the sum of the moments about the point of application of one of the forces, the moment due to that force is zero and we only need to determine the moment due to the other force. Let us determine the moment about the point of application of the force F. The vector from the point of application of F to the point of application of the force -F is r D [6 8i C 6 8j] m D 2i 2j m The sum of the moments of the two forces is i j k M D r ð F D 2 2 0 D 28k N-m 10 4 0 The magnitude of the moment is 28 N-m. Pointing the thumb of the right hand into the page, the right-hand rule indicates that the moment is clockwise. M D 28 N-m clockwise c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 219 Problem 4.109 The forces are contained in the xy plane. Solution: The right hand force is F D [1000 lb]cos 60° i sin 60° j (a) (b) Determine the moment of the couple and represent it as shown in Fig. 4.28c. What is the sum of the moments of the two forces about the point (10, 40, 20) ft? y F D C500i 867j lb. The vector from the x intercept of the left force to that of the right force is r D 40i ft. The moment is MC D r ð F 1000 lb 1000 lb 60° MC D 40i ð 500i 867j ft-lb 60° MC D 34700 ft-lb k x 20 ft 20 ft or Problem 4.110 The moment of the couple is 600 k (N-m). What is the angle ˛? MC D 34700 ft-lb) clockwise Solution: M D 100 N cos ˛4 m C 100 N sin ˛5 m D 600 N-m y Solving yields two answers: ˛ D 30.9° a or ˛ D 71.8° (0, 4) m 100 N 100 N a x (5, 0) m Problem 4.111 Point P is contained in the xy plane, jFj D 100 N, and the moment of the couple is 500k (N-m). What are the coordinates of P? Solution: The force is F D 100i cos30° C j sin30° D 86.6i 50j. Let r be the distance OP. The vector parallel to OP is y P r D ri cos 70° C j sin 70° D r0.3420i C 0.9397j. 30° F The moment is –F 70° x i M D r ð F D 0.3420r 86.6 From which, r D j 0.9397r 50.0 k 0 D 98.48rk. 0 500 D 5.077 m. From above, 98.48 r D 5.0770.3420i C 0.9397j. The coordinates of P are x D 5.0770.3420 D 1.74 m, y D 5.0770.9397 D 4.77 m 220 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.112 Three forces of equal magnitude are applied parallel to the sides of an equilateral triangle. (a) Show that the sum of the moments of the forces is the same about any point. (b) Determine the magnitude of the sum of the moments. F L F Solution: (a) (b) F Resolving one of the forces into vector components parallel to the other two forces results in two equal and opposite forces with the same line of action and one couple. Therefore the moment due to the forces is the same about any point. Determine the moment about one of the vertices of the triangle. A vertex lies on the line of action of two of the forces, so the moment due to them is zero. The perpendicular distance to the line of action of the third force is L cos 30° , so the magnitude of the moment due to the three force is M D FL cos 30° Problem 4.113 In Example 4.10, suppose that the 200 ft-lb couple is counterclockwise instead of clockwise. Draw a sketch of the beam showing the forces and couple acting on it. What are the forces A and B? y 200 ft-lb A B 4 ft 4 ft x Solution: In Example 4.10 we are given that the sum of the forces is zero and the sum of the moments is zero. Thus Fy D A C B D 0 MA D B 4 ft C 200 ft-lb D 0 Solving we find A D 50 lb, B D 50 lb Problem 4.114 The moments of two couples are shown. What is the sum of the moments about point P? y 50 ft-lb P x (– 4, 0, 0) ft 10 ft-lb Solution: The moment of a couple is the same anywhere in the plane. Hence the sum about the point P is M D 50k C 10k D 40k ft lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 221 Problem 4.115 Determine the sum of the moments exerted on the plate by the two couples. y Solution: The moment due to the 30 lb couple, which acts in a clockwise direction is M30 D 330k D 90k ft lb. 30 lb The moment due to the 20 lb couple, which acts in a counterclockwise direction, is 3 ft 30 lb M20 D 920k D 180k ft lb. 2 ft The sum of the moments is x 20 lb 20 lb 5 ft 4 ft M D 90k C 180k D C90k ft lb. The sum of the moments is the same anywhere on the plate. Problem 4.116 Determine the sum of the moments exerted about A by the couple and the two forces. 100 lb Solution: Let the x axis point to the right and the y axis point upward in the plane of the page. The moments of the forces are 400 lb M100 D 3i ð 100j D 300k (ft-lb), 900 ft-lb and A 3 ft B 4 ft 3 ft M400 D 7i ð 400j D 2800k (ft-lb). The moment of the couple is MC D 900k (ft-lb). Summing the moments, we get 4 ft MTotal D 2200k (ft-lb) Problem 4.117 Determine the sum of the moments exerted about A by the couple and the two forces. Solution: MA D 0.2i ð 200j C 0.4i C 0.2j 100 N 30° ð 86.7i C 50j C 300k N-m 200 N 0.2 m A MA D 40k C 2.66k C 300k N-m MA D 262.7k N-m ' 263k N-m 300 N-m 0.2 m 0.2 m 0.2 m Problem 4.118 The sum of the moments about point A due to the forces and couples acting on the bar is zero. Solution: (a) MA D 20 kN-m 2 kN5 m 4 kN3 m (a) What is the magnitude of the couple C? (b) Determine the sum of the moments about point B due to the forces and couples acting on the bar. 3 kN8 C C D 0 C D 26 kN-m B 4 kN 3m 20 kN-m MB D 3 kN3 m 4 kN3 m 5 kN5 m (b) C 20 kN-m C 26 kN-m D 0 C A 4 kN 2 kN 5 kN 5m 222 3 kN 3m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.119 In Example 4.11, suppose that instead of acting in the positive z direction, the upper 20-N force acts in the positive x axis direction. Instead of acting in the negative z axis direction, let the lower 20-N force act in the negative x axis direction. Draw a sketch of the pipe showing the forces acting on it. Determine the sum of the moments exerted on the pipe by the two couples. y 20 N 30 N 30 N 2m 4m 4m x 60⬚ 20 N 60⬚ z Solution: The magnitude of the moment of the 20-N couple is unchanged, 2 m20 N D 40 N-m. The direction of the moment vector is perpendicular to eh x-y plane, and the right-hand rule indicates that it points in the negative z axis direction. The moment of the 20-N couple is (40 N-m) k. The sum of the moments exerted on the pipe by the two couples is M D 40 N-m k C 30 N cos 60° 4 m j 30 N sin 60° 4 m k M D 60j 144k N-m Problem 4.120 (a) What is the moment of the couple? (b) Determine the perpendicular distance between the lines of action of the two forces. y (0, 4, 0) m ⫺2i ⫹ 2j ⫹ k (kN) Solution: 2i ⫺ 2j ⫺ k (kN) x M D 4j 5k m ð 2i 2j k kN (a) D 14i 10j 8k kN-m (b) MD FD (0, 0, 5) m 142 C 102 C 82 kN-m D 18.97 kN-m z 22 C 22 C 12 kN D 3 kN M D Fd ) d D M 18.97 kN-m D D 6.32 m F 3 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 223 Problem 4.121 Determine the sum of the moments exerted on the plate by the three couples. (The 80-lb forces are contained in the xz plane.) y 3 ft 20 lb 3 ft 20 lb 40 lb x 8 ft 40 lb z 60° 60° 80 lb Solution: The moments of two of the couples can be determined The moment is from inspection: i j M3 D r3 ð F3 D 6 0 69.282 0 M1 D 320k D 60k ft lb. M2 D 840j D 320j ft lb The forces in the 3rd couple are resolved: 80 lb k 0 D 240j. 40 The sum of the moments due to the couples: M D 60k C 320j 240j D 80j 60k ft lb F D 80i sin 60° C k cos 60° D 69.282i C 40k The two forces in the third couple are separated by the vector r3 D 6i C 8k 8k D 6i Problem 4.122 What is the magnitude of the sum of the moments exerted on the T-shaped structure by the two couples? y 3 ft 50i + 20j – 10k (lb) 3 ft 50j (lb) 3 ft z –50j (lb) 3 ft x –50i – 20j + 10k (lb) Solution: The moment of the 50 lb couple can be determined by y 3 ft inspection: F 3 ft 50j (lb) M1 D 503k D 150k ft lb. The vector separating the other two force is r D 6k. The moment is i M2 D r ð F D 0 50 j 0 20 k 6 D 120i C 300j. 10 3 ft x –F –50j (lb) 3 ft z The sum of the moments is M D 120i C 300j 150k. The magnitude is p jMj D 224 1202 C 3002 C 1502 D 356.23 ft lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.123 The tension in cables AB and CD is 500 N. (a) (b) A (0, 2, 0) m Show that the two forces exerted by the cables on the rectangular hatch at B and C form a couple. What is the moment exerted on the plate by the cables? 3m B z x 3m C D Solution: One condition for a couple is that the sum of a pair of forces vanish; another is for a non-zero moment to be the same anywhere. The first condition is demonstrated by determining the unit vectors parallel to the action lines of the forces. The vector position of point B is rB D 3i m. The vector position of point A is rA D 2j. The vector parallel to cable AB is rBA D rA rB D 3i C 2j. (6, –2, 3) m The moment about the origin is MO D rB rC ð TAB D rCB ð TAB , which is identical with the above expression for the moment. Let rPC and rPB be the distances to points C and B from an arbitrary point P on the plate. Then MP D rPB rPC ð TAB D rCB ð TAB which is identical to the above expression. Thus the moment is the same everywhere on the plate, and the forces form a couple. The magnitude is: jrAB j D p 32 C 22 D 3.606 m. The unit vector: eAB D rAB D 0.8321i C 0.5547j. jrAB j The tension is TAB D jTAB jeAB D 416.05i C 277.35j. The vector position of points C and D are: rC D 3i C 3k, rD D 6i 2j C 3k. The vector parallel to the cable CD is rCD D rD rC D 3i 2j. The magnitude is jrCD j D 3.606 m, and the unit vector parallel to the cable CD is eCD D C0.8321i 0.5547j. The magnitude of the tension in the two cables is the same, and eBA D eCD , hence the sum of the tensions vanish on the plate. The second condition is demonstrated by determining the moment at any point on the plate. By inspection, the distance between the action lines of the forces is rCB D rB rC D 3i 3i 3k D 3k. The moment is M D rCB ð TAB i D 0 416.05 j k 0 3 277.35 0 D 832.05i 1248.15j (N-m). c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 225 Problem 4.124 The cables AB and CD exert a couple on the vertical pipe. The tension in each cable is 8 kN. Determine the magnitude of the moment the cables exert on the pipe. (⫺1.6, 2.2, ⫺1.2) m y D C (0.2, 1.6, ⫺0.2) m Solution: FAB D 8 kN A 1.4i 0.6j C 1.0k p , rDB D 3.2i 2.2j C 2.4k m 3.32 (0.2, 0.6, 0.2) m M D rBD ð FAB D 3.34i C 0.702j C 5.09k kN-m x ) M D 6.13 kN-m B z Problem 4.125 The bar is loaded by the forces y FB FB D 2i C 6j C 3k (kN), A FC D i 2j C 2k (kN), and the couple (1.6, 0, 1.2) m B MC C x 1m z 1m FC MC D 2i C j 2k (kN-m). Determine the sum of the moments of the two forces and the couple about A. Solution: The moments of the two forces about A are given by MFB D 1i ð 2i C 6j C 3k (kN-m) D 0i 3j C 6k (kN-m) and MFC D 2i ð 1i 2j C 2k (kN-m) D 0i 4j 4k (kN-m). Adding these two moments and MC D 2i C 1j 2k (kN-m), we get MTOTAL D 2i 6j C 0k (kN-m) Problem 4.126 In Problem 4.125, the forces Solution: From the solution to Problem 4.125, the sum of the moments of the two forces about A is FB D 2i C 6j C 3k (kN), MForces D 0i 7j C 2k (kN-m). FC D i 2j C 2k (kN), The required moment, MC , must be the negative of this sum. and the couple Thus MCy D 7 (kN-m), and MCz D 2 (kN-m). MC D MCy j C MCz k (kN-m). Determine the values for MCy and MCz , so that the sum of the moments of the two forces and the couple about A is zero. 226 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.127 Two wrenches are used to tighten an elbow fitting. The force F D 10k (lb) on the right wrench is applied at (6, 5, 3) in, and the force F on the left wrench is applied at (4, 5, 3) in. (a) Determine the moment about the x axis due to the force exerted on the right wrench. (b) Determine the moment of the couple formed by the forces exerted on the two wrenches. (c) Based on the results of (a) and (b), explain why two wrenches are used. z x F –F Solution: The position vector of the force on the right wrench is rR D 6i 5j 3k. The magnitude of the moment about the x axis is 1 jMR j D eX Ð rR ð F D 6 0 (a) 0 0 5 3 D 50 in lb 0 10 from which MXL D 50i in lb, which is opposite in direction and equal in magnitude to the moment exerted on the x axis by the right wrench. The left wrench force is applied 2 in nearer the origin than the right wrench force, hence the moment must be absorbed by the space between, where it is wanted. The moment about the x axis is MR D jMR jeX D 50i (in lb). (b) The moment of the couple is i MC D rR rL ð FR D 2 0 (c) j 0 0 k 6 D 20j in lb 10 The objective is to apply a moment to the elbow relative to connecting pipe, and zero resultant moment to the pipe itself. A resultant moment about the x axis will affect the joint at the origin. However the use of two wrenches results in a net zero moment about the x axis the moment is absorbed at the juncture of the elbow and the pipe. This is demonstrated by calculating the moment about the x axis due to the left wrench: 1 jMX j D eX Ð rL ð FL D 4 0 0 5 0 0 3 D 50 in lb 10 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 227 Problem 4.128 Two systems of forces act on the beam. Are they equivalent? Strategy: Check the two conditions for equivalence. The sums of the forces must be equal, and the sums of the moments about an arbitrary point must be equal. System 1 y 100 N x 50 N 1m 1m System 2 y 50 N x 2m Solution: The strategy is to check the two conditions for equivalence: (a) the sums of the forces must be equal and (b) the sums of the moments about an arbitrary point must be equal. The sums of the forces of the two systems: FX D 0, (both systems) and FY1 D 100j C 50j D 50j (N) FY2 D 50j (N). The sums of the forces are equal. The sums of the moments about the left end are: M1 D 1100k D 100k (N-m) M2 D 250k D 100k (N-m). The sums of the moments about the left end are equal. Choose any point P at the same distance r D xi from the left end on each beam. The sums of the moments about the point P are M1 D 50x C 100x 1k D 50x 100k (N-m) M2 D 502 xk D 50x 100k (N-m). Thus the sums of the moments about any point on the beam are equal for the two sets of forces; the systems are equivalent. Yes 228 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.129 Two systems of forces and moments act on the beam. Are they equivalent? System 1 y 20 lb 50 ft-lb 10 lb x Solution: The sums of the forces are: 2 ft FX D 0 (both systems) FY1 D 10j 20j D 10j (lb) System 2 y FY2 D 20j C 10j D 10j (lb) Thus the sums of the forces are equal. The sums of the moments about the left end are: 2 ft 20 lb 30 ft-lb 10 lb x M1 D 204k C 50k D 30k (ft lb) 2 ft 2 ft M2 D C102k 30k D 10k (ft lb) The sums of the moments are not equal, hence the systems are not equivalent. No Problem 4.130 Four systems of forces and moments act on an 8-m beam. Which systems are equivalent? System 1 the moments about some point (the left end will be used) must be the same. System 2 10 kN 8m Solution: For equivalence, the sum of the forces and the sum of F kN ML kN-m 10 kN 80 kN-m 8m System 1 10j 80k System 2 10j 80k Systems 1, 2, and 4 are equivalent. System 3 System 4 20 kN 10 kN 8m System 3 10j 160k System 4 10j 80k y x 20 kN 10 kN 80 kN-m 4m 4m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 229 Problem 4.131 The four systems shown in Problem 4.130 can be made equivalent by adding a couple to one of the systems. Which system is it, and what couple must be added? Solution: From the solution to 4.130, all systems have F D 10j kN and systems 1, 2, and 4 have ML D 80k kN-m system 3 has ML D 160k kN-m. Thus, we need to add a couple M D 80k kN-m to system 3 (clockwise moment). Problem 4.132 System 1 is a force F acting at a point O. System 2 is the force F acting at a different point O0 along the same line of action. Explain why these systems are equivalent. (This simple result is called the principle of transmissibility.) System 2 System 1 F F O' O O Solution: The sum of forces is obviously equal for both systems. Let P be any point on the dashed line. The moment about P is the cross product of the distance from P to the line of action of a force times the force, that is, M D rPL ð F, where rPL is the distance from P to the line of action of F. Since both systems have the same line of action, and the forces are equal, the systems are equivalent. 230 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.133 The vector sum of the forces exerted on the log by the cables is the same in the two cases. Show that the systems of forces exerted on the log are equivalent. A 12 m B 16 m C 12 m D 6m Solution: The angle formed by the single cable with the positive E 20 m Solve: x axis is D 180° tan1 12 16 jTL j D 0.3353jT1 j, and D 143.13° . jTR j D 0.7160jT1 j. The single cable tension is T1 D jTji cos 143.13° C j sin 143.13° D jTj0.8i C 0.6j. The position vector to the center of the log from the left end is rc D 10i. The moment about the end of the log is i M D r ð T1 D jT1 j 10 0.8 j k 0 0 D jTj6k (N-m). 0.6 0 The tension in the right hand cable is TR D jT1 j0.71600.9079i C 0.4191j D jT1 j0.6500i C 0.3000. The position vector of the right end of the log is rR D 20i m relative to the left end. The moments about the left end of the log for the second system are i M2 D rR ð TR D jT1 j 20 0.6500 j k 0 0 D jT1 j6k (N-m). 0.3000 0 This is equal to the moment about the left end of the log for System 1, hence the systems are equivalent. For the two cables, the angles relative to the positive x axis are 1 D 180° tan1 2 D 180 tan1 12 6 12 26 D 116.56° , and D 155.22° . The two cable vectors are TL D jTL ji cos 116.56° C j sin 116.56° D jTL j0.4472i C 0.8945j, TR D jTR ji cos 155.22° C j sin 155.22° D jTR j0.9079i C 0.4191j. Since the vector sum of the forces in the two systems is equal, two simultaneous equations are obtained: 0.4472jTL j C 0.9079jTR j D 0.8jT1 j, and 0.8945jTL j C 0.4191jTR j D 0.6jT1 j c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 231 Problem 4.134 Systems 1 and 2 each consist of a couple. If they are equivalent, what is F? System 1 y System 2 y 200 N F 20° 30° Solution: For couples, the sum of the forces vanish for both sys- (5, 4, 0) m 5m 200 N tems. For System 1, the two forces are located at r11 D 4i, and r12 D C5j. The forces are F1 D 200i cos 30° C j sin 30° D 173.21i C 100j. The moment due to the couple in System 1 is i M1 D r11 r12 ð F1 D 4 173.21 30° 2m x j k 5 0 D 1266.05k (N-m). 100 0 x 20° F 4m For System 2, the positions of the forces are r21 D 2i, and r22 D 5i C 4j. The forces are F2 D Fi cos20° C j sin20° D F0.9397i 0.3420j. The moment of the couple in System 2 is i M2 D r21 r22 ð F2 D F 3 0.9397 j k 4 0 D 4.7848Fk, 0.3420 0 from which, if the systems are to be equivalent, FD 1266 D 264.6 N 4.7848 Problem 4.135 Two equivalent systems of forces and moments act on the L-shaped bar. Determine the forces FA and FB and the couple M. System 2 System 1 120 N-m FB 40 N 60 N 3m FA M 3m 50 N 3m Solution: The sums of the forces for System 1 are FX D 50, and The sums of the forces for System 2 are 6m The sum of the moments about the left end for System 2 is FY D FA C 60. 3m M2 D 3FB C M D 150 C M N-m. Equating the sums of the moments, M D 150 180 D 30 N-m FX D FB , and FY D 40. For equivalent systems: FB D 50 N, and FA D 60 40 D 20 N. The sum of the moments about the left end for System 1 is 232 M1 D 3FA 120 D 180 N-m. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.136 Two equivalent systems of forces and moments act on the plate. Determine the force F and the couple M. System 1 30 lb System 2 10 lb 30 lb 100 in-lb 5 in 5 in 8 in 8 in M 50 lb 30 lb F Solution: The sums of the forces for System 1 are FX D 30 lb, FY D 50 10 D 40 lb. The sums of the forces for System 2 are FX D 30 lb, FY D F 30 lb. For equivalent forces, F D 30 C 40 D 70 lb. The sum of the moments about the lower left corner for System 1 is M1 D 530 810 C M D 230 C M in lb. The sum of the moments about the lower left corner for System 2 is M2 D 100 in lb. Equating the sum of moments, M D 230 100 D 130 in lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 233 System 1 Problem 4.137 In Example 4.13, suppose that the 30kN vertical force in system 1 is replaced by a 230-kN vertical force. Draw a sketch of the new system 1. If you represent system 1 by a single force F as in system 3, at what position D on the x axis must the force be placed? y 30j (kN) 20i ⫹ 20j (kN) x O 3m 2m 210 kN-m Solution: The first step is to represent system 1 by a single force F acting at the origin and a couple M (system 2). The force F must equal the sum of the forces in system 1: F2 D F1 F D 230 kN j C 20i C 20j kN F D 20i C 250j kN The moment about the origin in system 2 is M. Therefore M must equal the sum of the moments about the origin due to the forces and moments in system 1: M2 D M1 M D 230 kN3 m C 20 kN5 m C 210 kN-m D 1000 kN-m The next step is to represent system 2 by system 3. The sums of the forces in the two systems are equal. The sums of the moments about the origin must be equal. The j component of F is 250 kN, so M3 D M2 1000 kN-m D 250 kND DD 1000 kN-m D 4m 250 kN DD4m 234 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.138 Three forces and a couple are applied to a beam (system 1). (a) If you represent system 1 by a force applied at A and a couple (system 2), what are F and M? If you represent system 1 by the force F (system 3), what is the distance D? (b) System 1 y 30 lb 40 lb 20 lb 30 ft-lb x A 2 ft 2 ft System 2 y F M x A System 3 y F x A D Solution: The sum of the forces in System 1 is FX D 0i, FY D 20 C 40 30j D 10j lb. The sum of the moments about the left end for System 1 is M1 D 240 430 C 30k D 10k ft lb. (a) For System 2, the force at A is F D 10j lb The moment at A is M2 D 10k ft lb (b) For System 3 the force at D is F D 10j lb. The distance D is the ratio of the magnitude of the moment to the magnitude of the force, where the magnitudes are those in System 1: DD 10 D 1 ft 10 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 235 y Problem 4.139 Represent the two forces and couple acting on the beam by a force F. Determine F and determine where its line of action intersects the x axis. 60i + 60j (N) 280 N-m x – 40 j (N) 3m 3m y Solution: We first represent the system by an equivalent system consisting of a force F at the origin and a couple M: F This system is equivalent if M F D 40j C 60i C 60j x D 60i C 20j N, M D 280 C 660 y D 80 N-m. F We then represent this system by an equivalent system consisting of F alone: x For equivalence, M D dFy , so dD d M 80 D D 4 m. Fy 20 y Problem 4.140 The bracket is subjected to three forces and a couple. If you represent this system by a force F, what is F, and where does its line of action intersect the x axis? 400 N 180 N 0.4 m 140 N-m 200 N 0.2 m x 0.65 m Solution: We locate a single equivalent force along the x axis a distance d to the right of the origin. We must satisfy the following three equations: Fx D 400 N 200 N D Rx Fy D 180 N D Ry MO D 400 N0.6 m C 200 N0.2 m C 180 N0.65 m C 140 Nm D Ry d Solving we find Rx D 200 N, Ry D 180 N, d D 0.317 m 236 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.141 The vector sum of the forces acting on the beam is zero, and the sum of the moments about the left end of the beam is zero. Solution: (a) The sum of the forces is (a) (b) (c) Determine the forces Ax and Ay , and the couple MA . Determine the sum of the moments about the right end of the beam. If you represent the 600-N force, the 200-N force, and the 30 N-m couple by a force F acting at the left end of the beam and a couple M, what are F and M? FY D AY 600 C 200j D 0, from which AY D 400 N. The sum of the moments is ML D MA 0.38600 30 C 0.560200k D 0, from which MA D 146 N-m. (b) The sum of the moments about the right end of the beam is y FX D AX i D 0 and MR D 0.18600 30 C 146 0.56400 D 0. 600 N MA (c) The sum of the forces for the new system is x Ax 30 N-m Ay 200 N 380 mm FY D AY C Fj D 0, from F D AY D 400 N, or F D 400j N. The sum of the moments for the new system is 180 mm M D MA C M D 0, from which M D MA D 146 N-m Problem 4.142 The vector sum of the forces acting on the truss is zero, and the sum of the moments about the origin O is zero. Solution: (a) The sum of the forces is (a) (b) from which AX D 12 kip (c) Determine the forces Ax , Ay , and B. If you represent the 2-kip, 4-kip, and 6-kip forces by a force F, what is F, and where does its line of action intersect the y axis? If you replace the 2-kip, 4-kip, and 6-kip forces by the force you determined in (b), what are the vector sum of the forces acting on the truss and the sum of the moments about O? FX D AX 2 4 6i D 0, FY D AY C Bj D 0. The sum of the moments about the origin is MO D 36 C 64 C 92 C 6B D 0, from which B D 10j kip. (b) Substitute into the force balance equation to obtain AY D B D 10 kip. (b) The force in the new system will replace the 2, 4, and 6 kip forces, F D 2 4 6i D 12i kip. The force must match the moment due to these forces: FD D 36 C 60 D 5 ft, or the action 64 C 92 D 60 kip ft, from which D D 12 line intersects the y axis 5 ft above the origin. (c) The new system is equivalent to the old one, hence the sum of the forces vanish and the sum of the moments about O are zero. 2 kip y 3 ft 4 kip 3 ft 6 kip 3 ft Ax O x B Ay 6 ft c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 237 Problem 4.143 The distributed force exerted on part of a building foundation by the soil is represented by five forces. If you represent them by a force F, what is F, and where does its line of action intersect the x axis? y Solution: The equivalent force must equal the sum of the forces x exerted by the soil: 80 kN 3m F D 80 C 35 C 30 C 40 C 85j D 270j kN 35 kN 30 kN 40 kN 3m 3m 3m 85 kN The sum of the moments about any point must be equal for the two systems. The sum of the moments are M D 335 C 630 C 940 C 1285 D 1665 kN-m. Equating the moments for the two systems FD D 1665 kN-m from which DD 1665 kN-m D 6.167 m. 270 kN Thus the action line intersects the x axis at a distance D D 6.167 m to the right of the origin. Problem 4.144 At a particular instant, aerodynamic forces distributed over the airplane’s surface exert the 88-kN and 16-kN vertical forces and the 22 kN-m counterclockwise couple shown. If you represent these forces and couple by a system consisting of a force F acting at the center of mass G and a couple M, what are F and M? Solution: y 88 kN 16 kN x G 5m 22 kN-m 5.7 m 9m Fy D 88 kN C 16 kN D Ry MG D 88 kN0.7 m C 16 kN3.3 m C 22 kN-m D M Solving we find Ry D 104 kN, M D 13.2 kN-m Problem 4.145 If you represent the two forces and couple acting on the airplane in Problem 4.144 by a force F, what is F, and where does its line of action intersect the x axis? Solution: Fy D 88 kN C 16 kN D Ry MOrigin D 88 kN5 m C 16 kN9 m C 22 kN-m D Ry x Solving we find F D Ry j D 104 kNj, x D 5.83 m 238 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.146 The system is in equilibrium. If you represent the forces FAB and FAC by a force F acting at A and a couple M, what are F and M? y 60° B 40° FAC FAB C A A 100 lb 100 lb x Solution: The sum of the forces acting at A is in opposition to the weight, or F D jWjj D 100j lb. The moment about point A is zero. Problem 4.147 Three forces act on a beam. (a) (b) y Represent the system by a force F acting at the origin O and a couple M. Represent the system by a single force. Where does the line of action of the force intersect the x axis? 30 N 5m x O Solution: (a) The sum of the forces is 30 N 6m 4m 50 N FX D 30i N, and FY D 30 C 50j D 80j N. The equivalent at O is F D 30i C 80j (N). The sum of the moments about O: M D 530 C 1050 D 350 N-m (b) The solution of Part (a) is the single force. The intersection is the 350 D 4.375 m moment divided by the y-component of force: D D 80 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 239 y Problem 4.148 The tension in cable AB is 400 N, and the tension in cable CD is 600 N. (a) If you represent the forces exerted on the left post by the cables by a force F acting at the origin O and a couple M, what are F and M? (b) If you represent the forces exerted on the left post by the cables by the force F alone, where does its line of action intersect the y axis? A 400 mm B C 300 mm D O Solution: From the right triangle, the angle between the positive x axis and the cable AB is D tan1 400 800 D 26.6° . x 800 mm Check. (b) The equivalent single force retains the same scalar components, but must act at a point that duplicates the sum of the moments. The distance on the y axis is the ratio of the sum of the moments to the x-component of the equivalent force. Thus 419 D 0.456 m 919.6 The tension in AB is DD TAB D 400i cos26.6° Cj sin26.6° D 357.77i 178.89j (N). Check: The moment is The angle between the positive x axis and the cable CD is i M D rF ð F D 0 919.6 ˛ D tan1 300 800 D 20.6° . from which D D The tension in CD is 300 mm j k D 0 D 919.6Dk D 419k, 389.6 0 419 D 0.456 m, Check. 919.6 TCD D 600i cos20.6° C j sin20.6° D 561.8i 210.67j. The equivalent force acting at the origin O is the sum of the forces acting on the left post: F D 357.77 C 561.8i C 178.89 210.67j D 919.6i 389.6j (N). The sum of the moments acting on the left post is the product of the moment arm and the x-component of the tensions: M D 0.7357.77k 0.3561.8k D 419k N-m Check: The position vectors at the point of application are rAB D 0.7j, and rCD D 0.3j. The sum of the moments is M D rAB ð TAB C rCD ð TCD i D 0 357.77 j k i 0.7 0 C 0 178.89 0 561.8 j k 0.3 0 210.67 0 D 0.7357.77k 0.3561.8k D 419k 240 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.149 Consider the system shown in Problem 4.148. The tension in each of the cables AB and CD is 400 N. If you represent the forces exerted on the right post by the cables by a force F, what is F, and where does its line of action intersect the y axis? Solution: From the solution of Problem 4.148, the tensions are TAB D 400i cos26.6° Cj sin26.6° D 357.77i C 178.89j, and TCD D 400i cos20.6° Cj sin20.6° D 374.42i C 140.74j. The equivalent force is equal to the sum of these forces: F D 357.77 374.42i C 178.77 C 140.74j D 732.19i C 319.5j (N). The sum of the moments about O is M D 0.3357.77 C 0.8140.74 C 178.89k D 363k (N-m). The intersection is D D 363 D 0.496 m on the positive y axis. 732.19 Problem 4.150 If you represent the three forces acting on the beam cross section by a force F, what is F, and where does its line of action intersect the x axis? y 500 lb 800 lb 6 in Solution: The sum of the forces is FX D 500 500i D 0. x 6 in z FY D 800j. Thus a force and a couple with moment M D 500k ft lb act on the cross section. The equivalent force is F D 800j which acts at a positive 500 D 0.625 ft D 7.5 in to the right of the x axis location of D D 800 origin. 500 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 241 Problem 4.151 In Active Example 4.12, suppose that the force FB is changed to FB D 20i 15j C 30k (kN), and you want to represent system 1 by an equivalent system consisting of a force F acting at the point P with coordinates (4, 3, 2) m and a couple M (system 2). Determine F and M. System 1 y (4, 3, ⫺2) m P FB FA Solution: From Active Example 4.12 we know that x (6, 0, 0) m FA D 10i C 10j 15k kN MC D 90i C 150j C 60k kN-m z MC The force F must equal the sum of the forces in system 1: F2 D F1 : F D FA C FB D 10i 5j C 15k kN In system 2, the sum of the moments about P is M. Therefore equivalence requires that M be equal to the sum of the moments about point P due to the forces and moments in system 1: MP 2 D MP 1 : i M D 4 10 j 3 10 k i 2 C 2 15 20 j 3 15 k 2 C 90i C 150j C 60k kN-m 30 M D 125i C 50j C 20k kN-m Thus F D 10i 5j C 15k kN, M D 125i C 50j C 20k kN-m Problem 4.152 The wall bracket is subjected to the force shown. Determine the moment exerted by the force about the z axis. (b) Determine the moment exerted by the force about the y axis. (c) If you represent the force by a force F acting at O and a couple M, what are F and M? y O (a) z 10i – 30j + 3k (lb) 12 in x Solution: (a) The moment about the z axis is negative, MZ D 130 D 30 ft lb, (b) The moment about the y axis is negative, MY D 13 D 3 ft lb (c) The equivalent force at O must be equal to the force at x D 12 in, thus FEQ D 10i 30j C 3k (lb) The couple moment must equal the moment exerted by the force at x D 12 in. This moment is the product of the moment arm and the y- and zcomponents of the force: M D 130k 13j D 3j 30k (ft lb). 242 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.153 A basketball player executes a “slam dunk” shot, then hangs momentarily on the rim, exerting the two 100-lb forces shown. The dimensions are h D 14 12 in, and r D 9 21 in, and the angle ˛ D 120° . the forces applied: FEQ D 200j. The position vectors of the points of application of the forces are r1 D h C ri, and r2 D ih C r cos ˛ kr sin ˛. The moments about the origin are (a) M D r1 ð F1 C r2 ð F2 D r1 C r2 ð F (b) If you represent the forces he exerts by a force F acting at O and a couple M, what are F and M? The glass backboard will shatter if jMj > 4000 inlb. Does it break? Solution: The equivalent force at the origin must equal the sum of i j D 2h C r1 C cos ˛ 0 0 100 k r sin ˛ 0 D 100r sin ˛i 1002h C r1 C cos ˛k. y –100j (lb) For the values of h, r, and ˛ given, the moment is M D 822.72i 3375k in lb. This is the p couple moment required. (b) The magnitude of the moment is jMj D 822.722 C 33752 D 3473.8 in lb. The backboard does not break. O α r –100j (lb) h x z Problem 4.154 In Example 4.14, suppose that the 30lb upward force in system 1 is changed to a 25-lb upward force. If you want to represent system 1 by a single force F (system 2), where does the line of action of F intersect the xz plane? System 1 y 20j (lb) 30j (lb) (⫺3, 0, –2) ft (6, 0, 2) ft (2, 0, 4) ft of the forces in system 1: z F2 D F1 x O Solution: The sum of the forces in system 2 must equal the sum ⫺10j (lb) System 2 y F D 20 C 25 10j lb F D 35j lb The sum of the moments about a point in system 2 must equal the sum of the moments about the same point is system 1. We sum moments about the origin. F M2 D M1 i x 0 j k i y z D 6 35 0 0 j k i 0 2 C 2 25 0 0 j j k i 0 4 C 3 0 10 0 0 20 k 2 0 x O P z Expanding the determinants results in the equations 35z D 50 C 40 C 40 35x D 150 20 60 Solving yields x D 2.00 ft, z D 0.857 ft c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 243 Problem 4.155 The normal forces exerted on the car’s tires by the road are C A 0.8 m NA D 5104j (N), x NB D 5027j (N), 0.8 m NC D 3613j (N), D 1.4 m 1.4 m B z ND D 3559j (N). y If you represent these forces by a single equivalent force N, what is N, and where does its line of action intersect the xz plane? Solution: We must satisfy the following three equations Fy :5104 N C 5027 N C 3613 N C 3559 N D Ry x Mx :5104 N C 3613 N0.8 m 5027 N C 3559 N0.8 m D Ry z Mz :5104 N C 5027 N1.4 m 3613 N C 3559 N1.4 m D Ry x Solving we find Ry D 17303 N, x D 0.239 m, z D 0.00606 m Problem 4.156 Two forces act on the beam. If you represent them by a force F acting at C and a couple M, what are F and M? Solution: The equivalent force must equal the sum of forces: F D 100j C 80k. The equivalent couple is equal to the moment about C: M D 380j 3100k D 240j 300k y 100 N 80 N z C x 3m 244 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.157 An axial force of magnitude P acts on the beam. If you represent it by a force F acting at the origin O and a couple M, what are F and M? b Pi z Solution: The equivalent force at the origin is equal to the applied force F D Pi. The position vector of the applied force is r D hj C bk. The moment is i M D r ð P D 0 P j h 0 h O k Cb D bPj C hPk. 0 x This is the couple at the origin. y (Note that in the sketch the axis system has been rotated 180 about the x axis; so that up is negative and right is positive for y and z.) Problem 4.158 The brace is being used to remove a screw. (a) (b) If you represent the forces acting on the brace by a force F acting at the origin O and a couple M, what are F and M? If you represent the forces acting on the brace by a force F0 acting at a point P with coordinates xP , yP , zP and a couple M0 , what are F0 and M0 ? Solution: (a) Equivalent force at the origin O has the same value as the sum of forces, y h r h B z O 1 A 2 A B x 1 A 2 FX D B Bi D 0, FY D A C 12 A C 12 A j D 0, thus F D 0. The equivalent couple moment has the same value as the moment exerted on the brace by the forces, MO D rAi. Thus the couple at O has the moment M D rAi. (b) The equivalent force at xP , yP , zP has the same value as the sum of forces on the brace, and the equivalent couple at xP , yP , zP has the same moment as the moment exerted on the brace by the forces: F D 0, M D rAi. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 245 y Problem 4.159 Two forces and a couple act on the cube. If you represent them by a force F acting at point P and a couple M, what are F and M? P FB = 2i – j (kN) Solution: The equivalent force at P has the value of the sum of forces, F = (2 − 1)i + (1 − 1)j + k, FP = i + k (kN). FA = – i + j + k (kN) x MC = 4i – 4j + 4k (kN-m) The equivalentcouple at P has the moment exerted by the forces and moment about P. The position vectors of the forces relative to P are: 1m z rA D i j C k, and rB D Ck. The moment of the couple: M D rA ð FA C rB ð FB C MC i D 1 1 j k i 1 1 C 0 1 1 2 j k 0 1 C MC 1 0 D 3i 2j C 2k (kN-m). Problem 4.160 The two shafts are subjected to the torques (couples) shown. If you represent the two couples by a force F acting at the origin O and a couple M, what are F and M? (b) What is the magnitude of the total moment exerted by the two couples? y (a) 6 kN-m 4 kN-m 40° Solution: The equivalent force at the origin is zero, F D 0 since there is no resultant force on the system. Represent the couples of 4 kN-m and 6 kN-m magnitudes by the vectors M1 and M2 . The couple at the origin must equal the sum: 30° x z M D M1 C M2 . The sense of M1 is (see sketch) negative with respect to both y and z, and the sense of M2 is positive with respect to both x and y. M1 D 4j sin 30° k cos 30° D 2j 3.464k, M2 D 6i cos 40° C j sin 40° D 4.5963i C 3.8567j. Thus the couple at the origin is MO D 4.6i C 1.86j 3.46k (kN-m) (b) The magnitude of the total moment exerted by the two couples is p jMO j D 4.62 C 1.862 C 3.462 D 6.05 (kN-m) 246 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.161 The two systems of forces and moments acting on the bar are equivalent. If y FA D 30i C 30j 20k (kN), FA FB D 40i 20j C 25k (kN), z A 2m MB D 10i C 40j 10k (kN-m), MB 2m B what are F and M? x FB System 1 y Solution: F F D FA C FB D 70i C 10j C 5k kN z M D 2 mi ð FA C 4 mi ð FB C MB x M D 10i 20j 30k kNm System 2 Problem 4.162 Point G is at the center of the block. The forces are FA D 20i C 10j C 20k (lb), Solution: The equivalent force is the sum of the forces: F D 20i C 10 C 10j C 20 10k D 20i C 20j C 10k (lb). FB D 10j 10k (lb). The equivalent couple is the sum of the moments about G. The position vectors are: If you represent the two forces by a force F acting at G and a couple M, what are F and M? rA D 15i C 5j C 10k (in), rB D 15i C 5j 10k. y The sum of the moments: MG D rA ð FA C rB ð FB FB FA 10 in x G i D 15 20 j k i j 5 10 C 15 5 10 20 0 10 k 10 10 D 50i C 250j C 100k (in lb) 20 in z 30 in c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 247 Problem 4.163 The engine above the airplane’s fuselage exerts a thrust T0 D 16 kip, and each of the engines under the wings exerts a thrust TU D 12 kip. The dimensions are h D 8 ft, c D 12 ft, and b D 16 ft. If you represent the three thrust forces by a force F acting at the origin O and a couple M, what are F and M? y T0 c O z h 2 TU y Solution: The equivalent thrust at the point G is equal to the sum of the thrusts: T D 16 C 12 C 12 D 40 kip b The sum of the moments about the point G is x O b M D r1U ð TU C r2U ð TU C rO ð TO D r1U C r2U ð TU C rO ð TO . The position vectors are r1U D Cbi hj, r2U D bi hj, and rO D Ccj. For h D 8 ft, c D 12 ft, and b D 16 ft, the sum of the moments is i M D 0 0 j k i 16 0 C 0 0 12 0 j 12 0 k 0 D 192 C 192i D 0. 16 Thus the equivalent couple is M D 0 Problem 4.164 Consider the airplane described in Problem 4.163 and suppose that the engine under the wing to the pilot’s right loses thrust. Solution: The sum of the forces is now (a) The sum of the moments is now: If you represent the two remaining thrust forces by a force F acting at the origin O and a couple M, what are F and M? (b) If you represent the two remaining thrust forces by the force F alone, where does its line of action intersect the xy plane? F D 12 C 16 D 28k (kip). M D r2U ð TU C rO ð TO . For h D 8 ft, c D 12 ft, and b D 16 ft, using the position vectors for the engines given in Problem 4.163, the equivalent couple is i M D 16 0 j k i 8 0 C 0 0 12 0 j k 12 0 D 96i 192j (ft kip) 0 16 (b) The moment of the single force is i j M D x y 0 0 k z D 28yi 28xj D 96i 192j. 28 From which xD 96 192 D 6.86 ft, and y D D 3.43 ft. 28 28 As to be expected, z can have any value, corresponding to any point on the line of action. Arbitrarily choose z D 0, so that the coordinates of the point of action are (6.86, 3.43, 0). 248 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.165 The tension in cable AB is 100 lb, and the tension in cable CD is 60 lb. Suppose that you want to replace these two cables by a single cable EF so that the force exerted on the wall at E is equivalent to the two forces exerted by cables AB and CD on the walls at A and C. What is the tension in cable EF, and what are the coordinates of points E and F? y y C (4, 6, 0) ft (0, 6, 6) ft E x A x D (7, 0, 2) ft B F (3, 0, 8) ft z z Solution: The position vectors of the points A, B, C, and D are For the systems to be equivalent, the moments about the origin must be the same. The moments about the origin are rA D 0i C 6j C 6k, rB D 3i C 0j C 8k, MO D rA ð FA C rC ð FC i D 0 42.86 rC D 4i C 6j C 0k, and j 6 85.71 k i 6 C 4 28.57 25.72 j 6 51.43 k 0 17.14 rD D 7i C 0j C 2k. D 788.57i C 188.57j 617.14k. The unit vectors parallel to the cables are obtained as follows: rAB D rB rA D 3i 6j C 2k, jrAB j D p 32 C 62 C 22 D 7, This result is used to establish the coordinates of the point E. For the one cable system, the end E is located at x D 0. The moment is M1 D r ð FEF from which i D 0 68.58 j y 137.14 k z 45.71 eAB D 0.4286i 0.8571j C 0.2857k. D 45.71y C 137.14zi C 68.58zj 68.58yk rCD D rD rC D 3i 6j C 2k, D 788.57i C 188.57j 617.14k, jrCD j D p 32 C 62 C 22 D 7, from which eCD D 0.4286i 0.8571j C 0.2857k. Since eAB D eCD , the cables are parallel . To duplicate the force, the single cable EF must have the same unit vector. The force on the wall at point A is FA D 100eAB D 42.86i 85.71j C 28.57k (lb). The force on the wall at point C is from above. From which yD 617.14 D 8.999 . . . D 9 ft 68.58 zD 188.57 D 2.75 ft. 68.58 Thus the coordinates of point E are E (0, 9, 2.75) ft. The coordinates of the point F are found as follows: Let L be the length of cable EF. Thus, from the definition of the unit vector, yF yE D Ley with 9 D 10.5 ft. The other coordithe condition that yF D 0, L D 0.8571 nates are xF xE D LeX , from which xF D 0 C 10.50.4286 D 4.5 ft zF zE D LeZ , from which zF D 2.75 C 10.50.2857 D 5.75 ft The coordinates of F are F (4.5, 0, 5.75) ft FC D 60eCD D 25.72i 51.43j C 17.14k (lb). The total force is FEF D 68.58i 137.14j C 45.71k (lb), jFEF j D 160 lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 249 y Problem 4.166 The distance s D 4 m. If you represent the force and the 200-N-m couple by a force F acting at origin O and a couple M, what are F and M? (2, 6, 0) m s Solution: The equivalent force at the origin is F D 100i C 20j 20k. 100i ⫹ 20j ⫺ 20k (N) O The strategy is to establish the position vector of the action point of the force relative to the origin O for the purpose of determining the moment exerted by the force about the origin. The position of the top of the bar is 200 N-m x (4, 0, 3) m z rT D 2i C 6j C 0k. The vector parallel to the bar, pointing toward the base, is rTB D 2i 6j C 3k, with a magnitude of jrTB j D 7. The unit vector parallel to the bar is eTB D 0.2857i 0.8571j C 0.4286k. The vector from the top of the bar to the action point of the force is rTF D seTB D 4eTB D 1.1429i 3.4286j C 1.7143k. The position vector of the action point from the origin is rF D rT C rTF D 3.1429i C 2.5714j C 1.7143k. The moment of the force about the origin is i MF D r ð F D 3.1429 100 j 2.5714 20 k 1.7143 20 D 85.71i C 234.20j 194.3k. The couple is obtained from the unit vector and the magnitude. The sense of the moment is directed positively toward the top of the bar. MC D 200eTB D 57.14i C 171.42j 85.72k. The sum of the moments is M D MF C MC D 142.86i C 405.72j 280k. This is the moment of the equivalent couple at the origin. 250 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.167 The force F and couple M in system 1 are System 1 System 2 y F D 12i C 4j 3k (lb), y M M D 4i C 7j C 4k (ft-lb). F Suppose you want to represent system 1 by a wrench (system 2). Determine the couple Mp and the coordinates x and z where the line of action of the force intersects the xz plane. O z O x z Mp F x (x, 0, z) Solution: The component of M that is parallel to F is found as follows: The unit vector parallel to F is eF D F D 0.9231i C 0.3077j 0.2308k. jFj The component of M parallel to F is MP D eF Ð MeF D 4.5444i C 1.5148j 1.1361k (ft-lb). The component of M normal to F is MN D M MP D 0.5444i C 5.4858j C 5.1361k (ft-lb). The moment of F must produce a moment equal to the normal component of M. The moment is i MF D r ð F D x 12 k z D 4zi C 3x C 12zj C 4xk, 3 j 0 4 from which zD 0.5444 D 0.1361 ft 4 xD 5.1362 D 1.2840 ft 4 Problem 4.168 A system consists of a force F acting at the origin O and a couple M, where F D 10i (lb), M D 20j (ft-lb). If you represent the system by a wrench consisting of the force F and a parallel couple Mp , what is Mp , and where does the line of action F intersect the yz plane? Solution: The component of M parallel to F is zero, since MP D eF Ð MeF D 0. The normal component is equal to M. The equivalent force must produce the same moment as the normal component i j M D r ð F D 0 y 10 0 from which z D k z D 10zj 10yk D 20j, 0 20 D 2 ft and y D 0 10 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 251 Problem 4.169 A system consists of a force F acting at the origin O and a couple M, where F D i C 2j C 5k (N), M D 10i C 8j 4k (N-m). If you represent it by a wrench consisting of the force F and a parallel couple Mp , (a) determine Mp , and determine where the line of action of F intersects (b) the xz plane, (c) the yz plane. Solution: The unit vector parallel to F is eF D F D 0.1826i C 0.3651j C 0.9129k. jFj from which zD 9.8 5 D 4.9 m, and x D D 2.5 m 2 2 (a) The component of M parallel to F is (c) The intersection with the yz plane is MP D eF Ð MeF D 0.2i C 0.4j C 1.0k (N-m). i MN D r ð F D 0 1 The normal component is j y 2 k z D 5y 2zi C zj yk 5 D 9.8i C 7.6j 5k, MN D M MP D 9.8i C 7.6j 5k. The moment of the force about the origin must be equal to the normal component of the moment. (b) The intersection with the xz plane: i j MN D r ð F D x 0 1 2 k z D 2zi 5x zj C 2xk 5 from which y D 5 m and z D 7.6 m D 9.8i C 7.6j 5k, Problem 4.170 Consider the force F acting at the origin O and the couple M given in Example 4.15. If you represent this system by a wrench, where does the line of action of the force intersect the xy plane? 6j C 2k (N), and M D 12i C 4j C 6k (N-m). The normal component of the moment is MN D 7.592i 4.816j C 3.061k (N-m). y The moment produced by the force must equal the normal component: F i j MN D r ð F D x y 3 6 M O Solution: From Example 4.15 the force and moment are F D 3i C k 0 2 x D 2yi 2xj C 6x 3yk D 7.592i 4.816j C 3.061k, z from which xD 252 4.816 7.592 D 2.408 m and y D D 3.796 m 2 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.171 Consider the force F acting at the origin O and the couple M given in Example 4.15. If you represent this system by a wrench, where does the line of action of the force intersect the plane y D 3 m? Solution: From Example 4.15 (see also Problem 4.170) the force is F D 3i C 6j C 2k, and the normal component of the moment is MN D 7.592i 4.816j C 3.061k. The moment produced by the force must be equal to the normal component: i MN D r ð F D x 3 j k 3 z D 6 6zi 2x 3zj C 6x 9k 6 2 D 7.592i 4.816j C 3.061k, from which xD Problem 4.172 A wrench consists of a force of magnitude 100 N acting at the origin O and a couple of magnitude 60 N-m. The force and couple point in the direction from O to the point (1, 1, 2) m. If you represent the wrench by a force F acting at point (5, 3, 1) m and a couple M, what are F and M? 9 C 3.061 6 7.592 D 2.01 m and z D D 0.2653 m 6 6 Solution: The vector parallel to the force is rF D i C j C 2k, from which the unit vector parallel to the force is eF D 0.4082i C 0.4082j C 0.8165k. The force and moment at the origin are F D jFjeOF D 40.82i C 40.82j C 81.65k (N), and M D 24.492i C 24.492j C 48.99k (N-m). The force and moment are parallel. At the point (5, 3, 1) m the equivalent force is equal to the force at the origin, given above. The moment of this force about the origin is i MF D r ð F D 5 40.82 j 3 40.82 k 1 81.65 D 204.13i 367.43j C 81.64k. For the moments to be equal in the two systems, the added equivalent couple must be MC D M MF D 176.94i C 391.92j 32.65k (N-m) Problem 4.173 System 1 consists of two forces and a couple. Suppose that you want to represent it by a wrench (system 2). Determine the force F, the couple Mp , and the coordinates x and z where the line of action of F intersects the xz plane. Solution: The sum of the forces in System 1 is F D 300j C 600k (N). The equivalent force in System 2 must have this value. The unit vector parallel to the force is eF D 0.4472j C 0.8944k. The sum of the moments in System 1 is M D 6003i C 3004k C 1000i C 600j System 1 y System 2 D 2800i C 600j C 1200k (kN m). y 1000i + 600j (kN-m) The component parallel to the force is MP D 599.963j C 1199.93k (kN-m) D 600j C 1200k (kN-m). 600k (kN) 3m 300j (kN) Mp x x 4m z F z (x, 0, z) The normal component is MN D M MP D 2800i. The moment of the force i MN D x 0 j 0 300 k z D 300zi 600xj C 300xk D 2800i, 600 from which x D 0, z D 2800 D 9.333 m 300 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 253 y Problem 4.174 A plumber exerts the two forces shown to loosen a pipe. What total moment does he exert about the axis of the pipe? (b) If you represent the two forces by a force F acting at O and a couple M, what are F and M? (c) If you represent the two forces by a wrench consisting of the force F and a parallel couple Mp , what is Mp , and where does the line of action of F intersect the xy plane? 12 in 6 in O (a) z x 16 in 16 in 50 k (lb) –70 k (lb) Solution: The sum of the forces is (a) F D 50k 70k D 20k (lb). The total moment exerted on the pipe is M D 1620i D 320i (ft lb). (b) The equivalent force at O is F D 20k. The sum of the moments about O is MO D r1 ð F1 C r2 ð F2 i D 12 0 j 16 0 k i 0 C 18 50 0 j k 16 0 0 70 D 320i C 660j. (c) The unit vector parallel to the force is eF D k, hence the moment parallel to the force is MP D eF Ð MeF D 0, and the moment normal to the force is MN D M MP D 320i C 660j. The force at the location of the wrench must produce this moment for the wrench to be equivalent. i j MN D x y 0 0 k 0 D 20yi C 20xj D 320i C 660j, 20 from which x D 254 660 320 D 33 in, y D D 16 in 20 20 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.175 The Leaning Tower of Pisa is approximately 55 m tall and 7 m in diamteter. The horizontal displacement of the top of the tower from the vertical is approximately 5 m. Its mass is approximately 3.2 ð 106 kg. If you model the tower as a cylinder and assume that its weight acts at the center, what is the magnitude of the moment exerted by the weight about the point at the center of the tower’s base? 5m Solution: The unstretched length of spring is 1 m and the spring N . Assume that the bar is a quarter circle, with m a radius of 4 m. The stretched length of the spring is found from the Pythagorean Theorem: The height of the attachment point is h D 4 sin ˛ m, and the distance from the center is 4 cos ˛. The stretched length of the spring is constant is k D 20 LD 4m B k 3m α 3 h2 C 4 cos ˛2 m. A The spring force is F D 20L 1 N. The angle that the spring makes with a vertical line parallel to A is ˇ D tan1 3h 4 cos ˛ Moment at B as function of alpha 120 . The horizontal component of the spring force is FX D F cos ˇ N. The vertical component of the force is FY D F sin ˇ N. The displacement of the attachment point to the left of point A is d D 41 cos ˛ m, hence the action of the vertical component is negative, and the action of the horizontal component is positive. The moment about A is MA D dFY C hFX . M o m e n t , 100 80 60 40 Collecting terms and equations, N m 20 h D 4 sin ˛ m, 0 FY D F sin ˇ N, 0 10 20 30 40 50 60 70 80 90 Alpha, deg FX D F cos ˇ N, F D 20L 1 N, LD 3 h2 C 4 cos ˛2 m, ˇ D tan1 3h 4 cos ˛ . A programmable calculator or a commercial package such as TK Solver or Mathcad is almost essential to the solution of this and the following problems. The commercial package TK Solver PLUS was used here to plot the graph of M against ˛. Using the graph as a guide, the following tabular values were taken about the maximum: ˛, deg Moment, N-m 41.5 42.0 42.5 101.463 101.483 101.472 The maximum value of the moment is estimated at MB D 101.49 N-m, which occurs at approximately ˛ D 42.2° c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 255 y Problem 4.176 The cable AB exerts a 300-N force on the support A that points from A toward B. Determine the magnitude of the moment the force exerts about point P. B (0.3, 0.6) m A (⫺0.4, 0.3) m Solution: F D 300 N x P (0.5, ⫺0.2) m 0.7i C 0.3j p , 0.58 rPA D 0.9i C 0.5j m MP D rPA ð F D 244 Nmk ) MP D 244 Nm Problem 4.177 Three forces act on the structure. The sum of the moments due to the forces about A is zero. Determine the magnitude of the force F. 30⬚ 45⬚ 2 kN 4 kN Solution: b p MA D 4 kN 2b 2 kN cos 30° 3b A F C 2 kN sin 30° b C F4b D 0 2b b b Solving we find F D 2.463 kN Problem 4.178 Determine the moment of the 400-N force (a) about A, (b) about B. 30° 400 N 220 mm A Solution: Use the two dimensional description of the moment. The vertical and horizontal components of the 200 N force are 260 mm FY D 400 sin 30° D 200 N, FX D C400 cos 30° D 346.41 N. (a) B 500 mm The moment arm from A to the line of action of the horizontal component is 0.22 m. The moment arm from A to the vertical component is zero. The moment about A is negative, MA D 0.22346.41 D 76.21 N-m (b) 256 The perpendicular distances to the lines of action of the vertical and horizontal components of the force from B are d1 D 0.5 m, and d2 D 0.48 m. The action of the vertical component is positive, and the action of the horizontal component is negative. The sum of the moments: MB D C0.5200 0.48346.41 D 66.28 N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.179 Determine the sum of the moments exerted about A by the three forces and the couple. A 5 ft 300 lb 800 ft-lb Solution: Establish coordinates with origin at A, x horizontal, and y vertical with respect to the page. The moment exerted by the couple is the same about any point. The moment of the 300 lb force about A is M300 D 6i 5j ð 300j D 1800k ft-lb. 200 lb 200 lb 6 ft The moment of the downward 200 lb force about A is zero since the line of action of the force passes through A. The moment of the 200 lb force which pulls to the right is 3 ft M200 D 3i 5j ð 200i D 1000k (ft-lb). The moment of the couple is MC D 800k (ft-lb). Summing the four moments, we get MA D 1800 C 0 C 1000 800k D 1600k (ft-lb) Problem 4.180 In Problem 4.179, if you represent the three forces and the couple by an equivalent system consisting of a force F acting at A and a couple M, what are the magnitudes of F and M? Solution: The equivalent force will be equal to the sum of the forces and the equivalent couple will be equal to the sum of the moments about A. From the solution to Problem 4.189, the equivalent couple will be C D MA D 1600k (ft-lb). The equivalent force will be FEQUIV. D 200i 200j C 300j D 200i C 100j (lb) Problem 4.181 The vector sum of the forces acting on the beam is zero, and the sum of the moments about A is zero. (a) (b) 30° 220 mm What are the forces Ax , Ay , and B? What is the sum of the moments about B? 400 N Ay Ax 260 mm Solution: The vertical and horizontal components of the 400 N force are: FX D 400 cos 30° D 346.41 N, 500 mm B FY D 400 sin 30° D 200 N. The sum of the forces is FX D AX C 346.41 D 0, from which AX D 346.41 N FY D AY C B 200 D 0. The sum of the moments about A is MA D 0.5B 0.22346.41 D 0, from which B D 152.42 N. Substitute into the force equation to get AY D 200 B D 47.58 N (b) The moments about B are MB D 0.5AY 0.48346.41 0.26AX C 0.5200 D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 257 Problem 4.182 The hydraulic piston BC exerts a 970lb force on the boom at C in the direction parallel to the piston. The angle ˛ D 40° . The sum of the moment about A due to the force exerted on the boom by the piston and the weight of the suspended load is zero. What is the weight of the suspended load? t 6f t 9f Solution: The horizontal (x) and vertical (y) coordinates of point C relative to point B are C a A x D 9 ft cos ˛ 6 ft D 0.894 ft B 6 ft y D 9 ft sin ˛ D 5.79 ft The angle between the piston BC and the horizontal is ˇ D tan1 y/x D 81.2° The horizontal and vertical components of the force exerted by the piston at C are Cx D 970 lb cos ˇ D 148 lb Cy D 970 lb sin ˇ D 959 lb The sum of the moments about A due to the pistion force and the suspended weight W is MA D W15 ft cos ˛ C Cy 9 ft cos a Cx 9 ft sin ˛ D 0 Solving, yields W D 501 lb y Problem 4.183 The force F D 60i C 60j (lb). F (a) Determine the moment of F about point A. (b) What is the perpendicular distance from point A to the line of action of F? (4, – 4, 2) ft x A (8, 2, 12) ft z Solution: The position vector of A and the point of action are (b) The magnitude of the moment is p 6002 C 6002 C 6002 D 1039.3 ft lb. rA D 8i C 2j C 12k (ft), and rF D 4i 4j C 2k. jMA j D The vector from A to F is p The magnitude of the force is jFj D 602 C 602 D 84.8528 lb. The perpendicular distance from A to the line of action is rAF D rF rA D 4 8i C 4 2j C 2 12k DD D 4i 6j 10k. (a) 1039.3 D 12.25 ft 84.8528 The moment about A is i MA D rAF ð F D 4 60 j k 6 10 60 0 D 600i C 600j 600k (ft lb) 258 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 4.184 The 20-kg mass is suspended by cables attached to three vertical 2-m posts. Point A is at (0, 1.2, 0) m. Determine the moment about the base E due to the force exerted on the post BE by the cable AB. C B D A 1m 1m E 2m 0.3 m x z Solution: The strategy is to develop the simultaneous equations in the unknown tensions in the cables, and use the tension in AB to find the moment about E. This strategy requires the unit vectors parallel to the cables. The position vectors of the points are: The equilibrium conditions are TAB C TAC C TAD D W. Collect like terms in i, j, k: rOA D 1.2j, FX D 0.2281TAB C 0TAC C 0.9284TAD i D 0 FY D C0.6082 Ð TAB C 0.6247 Ð TAC rOB D 0.3i C 2j C 1k, rOC D 2j 1k, rOD D 2i C 2j, rOE D 0.3i C 1k. C 0.3714 Ð TAD 196.2j D 0 FZ D C0.7603 Ð TAB 0.7809 Ð TAC C 0 Ð TAD k D 0 Solve: TAB D 150.04 N, The vectors parallel to the cables are: TAC D 146.08 N, rAB D rOB rOA D 0.3i C 0.8j C 1k, TAD D 36.86 N. rAC D rOC rOA D C0.8j 1k, rAD D rOD rOA D C2i C 0.8j. The unit vectors parallel to the cables are: eAB D rAB D 0.2281i C 0.6082j C 0.7603k : jrAB j eAC D 0i C 0.6247j 0.7809k, The moment about E is ME D rEB ð TAB eAB D TAB rEB ð eAB i D 150 0 0.2281 j 2 C0.6082 k 0 C0.7603 D 228i 68.43k (N-m) eAD D C0.9284i C 0.3714j C 0k. The tensions in the cables are TAB D TAB eAB , TAC D TAC eAC , and TAD D TAD eAD . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 259 Problem 4.185 What is the total moment due to the two couples? (a) Express the answer by giving the magnitude and stating whether the moment is clockwise or counterclockwise. (b) Express the answer as a vector. y 100 N 4m 100 N 2m x 2m 100 N 100 N Solution: (a) (b) 4m The couple in which the forces are 4 m apart exerts a counterclockwise moment of magnitude 100 N4 m D 400 N-m. The couple in which the forces are 8 m apart exerts a clockwise moment of magnitude 100 N8 m D 800 N-m. The sum of their moments is a clockwise moment of 400 N-m. The vector representation of the clockwise moment of 400 N-m magnitude is 400k (N-m). This expression can also be obtained by calculating the sum of the moments of the four forces about any point. The sum of the moments about the origin is M D 2 mi ð 100 Nj C 2 mi ð 100 Nj C 4 mj ð 100 Ni C 4 mj ð 100 Ni D 400 N-mk (a) 400 N-m clockwise (b) 400k N-m Problem 4.186 The bar AB supporting the lid of the grand piano exerts a force F D 6i C 35j 12k (lb) at B. The coordinates of B are (3, 4, 3) ft. What is the moment of the force about the hinge line of the lid (the x axis)? y Solution: The position vector of point B is rOB D 3i C 4j C 3k. The moment about the x axis due to the force is MX D eX Ð rOB ð F D i Ð rOB ð F 1 0 MX D 3 4 6 35 0 3 D 153 ft lb 12 B x A z 260 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.187 Determine the moment of the vertical 800-lb force about point C. y 800 lb A (4, 3, 4) ft B D (6, 0, 0) ft Solution: The force vector acting at A is F D 800j (lb) and the x position vector from C to A is rCA D xA xC i C yA yC j C zA zC k z C (5, 0, 6) ft D 4 5i C 3 0j C 4 6k D 1i C 3j 2k (ft). The moment about C is i MC D 1 0 j 3 800 k 2 D 1600i C 0j C 800k (ft-lb) 0 Problem 4.188 In Problem 4.187, determine the moment of the vertical 800-lb force about the straight line through points C and D. Solution: In Problem 4.197, we found the moment of the 800 lb force about point C to be given by MC D 1600i C 0j C 800j (ft-lb). The vector from C to D is given by rCD D xD xC i C yD yC j C zD zC k D 6 5i C 0 0j C 0 6k D 1i C 0j 6j (ft), and its magnitude is jrCD j D p p 12 C 62 D 37 (ft). The unit vector from C to D is given by 6 1 eCD D p i p k. 37 37 The moment of the 800 lb vertical force about line CD is given by MCD D D 6 1 p i p k Ð 1600i C 0j C 800j (ft-lb) 37 37 1600 4800 p 37 (ft-lb). Carrying out the calculations, we get MCD D 1052 (ft-lb) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 261 Problem 4.189 The system of cables and pulleys supports the 300-lb weight of the work platform. If you represent the upward force exerted at E by cable EF and the upward force exerted at G by cable GH by a single equivalent force F, what is F, and where does its line of action intersect the x axis? Solution: The cable-pulley combination does not produce a moment. Hence the equivalent force does not. The equivalent force is 600 j D 300j (lb). The equal to the total supported weight, or F D C 2 8 force occurs at midpoint of the platform width, x D D 4 ft 2 H F E B y A G 60° D 60° C x 8 ft Problem 4.190 Consider the system in Problem 4.189. Solution: The vertical component of the tension is each cable must equal half the weight supported. (a) What are the tensions in cables AB and CD? (b) If you represent the forces exerted by the cables at A and C by a single equivalent force F, what is F, and where does its line of action intersect the x axis? 262 TAB sin 60° D 150 lb, from which TAB D symmetry, the tension TCD D 173.2 lb. 150 D 173.2 lb. By sin 60° The single force must equal the sum of the vertical components; since there is no resultant moment produced by the cables, the force is F D 300j lb and it acts at the platform width midpoint x D 4 ft. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.191 The two systems are equivalent. Determine the forces Ax and Ay , and the couple MA . System 1 y 20 N 400 mm Solution: The sum of the forces for System 1 is Ax FX D AX C 20i, x 30 N Ay FY D AY C 30j. 600 mm 400 mm The sum of forces for System 2 is System 2 y FX D 20i and 8 N-m FY D 80 10j. Equating the two systems: 400 mm AX C 20 D 20 from which AX D 40 N 20 N MA 10 N x AY C 30 D 80 10 from which AY D 40 N 80 N The sum of the moments about the left end for System 1 is 600 mm 400 mm M1 D 0.420 C 301 D 22 N-m. The sum of moments about the left end for System 2 is M2 D MA 101 8 D MA 18. Equating the moments for the two systems: MA D 18 C 22 D 40 N-m Problem 4.192 If you represent the equivalent systems in Problem 4.191 by a force F acting at the origin and a couple M, what are F and M? Solution: Summing the forces in System 1, F D AX C 20i C AY C 30j. Substituting from the solution in Problem 4.201, F D 20i C 70j. The moment is M D 200.4k C 30k D 22k (N-m) Problem 4.193 If you represent the equivalent systems in Problem 4.191 by a force F, what is F, and where does its line of action intersect the x axis? Solution: The force is F D 20i C 70j. The moment to be represented is i M D r ð F D 22k D x 20 from which x D j k 0 0 D 70xk, 70 0 22 D 0.3143 m 70 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 263 Problem 4.194 The two systems are equivalent. If Solution: The sum of forces in the two systems must be equal, thus F0 D F D 100i C 40j C 30k (lb). F D 100i C 40j C 30k (lb), The moment for the unprimed system is MT D r ð F C M. 0 M D 80i C 120j C 40k (in-lb), The moment for the primed system is M0T D r0 ð F C M0 . determine F0 and M. The position vectors are r D 0i C 6j C 6k, and r0 D 4i C 6j C 6k. Equating the moments and solving for the unknown moment System 1 y System 2 y M MD F 4 in M0 C r0 i r ð F D 80i C 120j C 40k C 4 100 j 0 40 k 0 30 4 in F' D 80i C 200k (in-lb) M' 6 in D 80i C 120j C 40k 120j C 160k 6 in x x 6 in 6 in z z Problem 4.195 The tugboats A and B exert forces FA D 1 kN and FB D 1.2 kN on the ship. The angle D 30° . If you represent the two forces by a force F acting at the origin O and a couple M, what are F and M? y A FA Solution: The sums of the forces are: FX D 1 C 1.2 cos 30° i D 2.0392i (kN) 60 m O x FY D 1.2 sin 30° j D 0.6j (kN). 60 m FB The equivalent force at the origin is FEQ D 2.04i C 0.6j θ B The moment about O is MO D rA ð FA C rB ð FB . The vector positions are 25 m rA D 25i C 60j (m), and rB D 25i 60j (m). The moment: i MO D 25 1 j k i 60 0 C 25 0 0 1.0392 j k 60 0 0.6 0 D 12.648k D 12.6k (kN-m) Check: Use a two dimensional description: The moment is MO D 25FB sin 30° C 60FB cos 30° 60FA D 39.46FB 60FA D 12.6 kN-m 264 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.196 The tugboats A and B in Problem 4.195 exert forces FA D 600 N and FB D 800 N on the ship. The angle D 45° . If you represent the two forces by a force F, what is F, and where does its line of action intersect the y axis? Solution: The equivalent force is Check: Use a two dimensional description: F D 0.6 C 0.8 cos 45° i C 0.8 sin 45° j D 1.1656i C 0.5656j (kN). MO D 25FB sin 45° C 60FB cos 45° 60FA The moment produced by the two forces is D 24.75FB 60FA D 16.20 kN-m. The single force must produce this moment. MO D rA ð FA C rB ð FB . rA D 25i C 60j (m), and rB D 25i 60j (m). i MO D 0 1.1656 The moment: from which The vector positions are j k i 60 0 C 25 0 0 0.5656 i MO D 25 0.6 j k 60 0 D 16.20k (kN-m) 0.5656 0 yD j k y 0 D 1.1656yk D 16.20k, 0.5656 0 16.20 D 13.90 m 1.1656 Problem 4.197 The tugboats A and B in Problem 4.195 want to exert two forces on the ship that are equivalent to a force F acting at the origin O of 2-kN magnitude. If FA D 800 N, determine the necessary values of FB and angle . Solution: The equivalent force at the origin is FA C FB cos 2 C sin 2 D FB must be zero: 20002 . The moment about the origin due to FA and FB MO D 60FA C 60FB cos 25FB sin D 0. These are two equations in two unknowns FB sin and FB cos . For brevity write x D FB cos , y D FB sin , so that the two equations become x2 C 2FA x C F2A C y 2 D 20002 and 60x 25y 60FA D 0. Eliminate y by solving each equation for y2 and equating the results: y 2 D 20002 x 2 2FA x F2A D 60 60 FA C x 25 25 Reduce to obtain the quadratic in x: 1C 60 25 2 C 1C x 2 C 2FA 1 60 25 60 25 Substitute FA D 800 N to obtain 6.76x2 7616x C 326400 D 0. In canonical form: x2 C 2bx C c D 0, where p b D 563.31, and c D 48284.0, with the solutions x D b š b2 c D 1082.0, D 44.62. From the second equation, y D 1812.9, D 676.81. The force FB has two solutions: Solve for FB and : (1) FB D p 44.62 C 1812.92 D 1813.4 N at the angle D tan1 2 . FB D 1812.9 44.6 D 88.6° , and 2 p 676.82 C 1082.02 D 1276.2 N, at the angle 2 x D tan1 676.8 1082.0 D 32.0° 2 F2A 20002 D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 265 y Problem 4.198 If you represent the forces exerted by the floor on the table legs by a force F acting at the origin O and a couple M, what are F and M? 2m 1m Solution: The sum of the forces is the equivalent force at the origin. F D 50 C 48 C 50 C 42j D 190j (N). The position vectors of the legs are, numbering the legs counterclockwise from the lower left in the sketch: 50 N x r1 D C1k, z 42 N 48 N r2 D 2i C 1k, 50 N r3 D 2i, r4 D 0. The sum of the moments about the origin is i MO D 0 0 j 0 48 k i 1 C 2 0 0 j 0 50 k i 1 C 2 0 0 j k 0 0 42 0 D 98i C 184k (N-m). This is the couple that acts at the origin. Problem 4.199 If you represent the forces exerted by the floor on the table legs in Problem 4.198 by a force F, what is F, and where does its line of action intersect the xz plane? Solution: From the solution to Problem 4.198 the equivalent force is F D 190j. This force must produce the moment M D 98i C 184k obtained in Problem 4.198. i M D x 0 j k 0 z D 190zi C 190xk D 98i C 184k, 190 0 from which 266 xD 184 D 0.9684 m and 190 zD 98 D 0.5158 m 190 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 4.200 Two forces are exerted on the crankshaft by the connecting rods. The direction cosines of FA are cos x D 0.182, cos y D 0.818, and cos z D 0.545, and its magnitude is 4 kN. The direction cosines of FB are cos x D 0.182, cos y D 0.818, and cos z D 0.545, and its magnitude is 2 kN. If you represent the two forces by a force F acting at the origin O and a couple M, what are F and M? Solution: The equivalent force is the sum of the forces: FA D 40.182i C 0.818j C 0.545k D 0.728i C 3.272j C 2.18k (kN) FB D 20.182i C 0.818j 0.545k D 0.364iC1.636j1.09k (kN). The sum: FA C FB D 0.364i C 4.908j C 1.09k (kN) y FB The equivalent couple is the sum of the moments. M D rA ð FA C rB ð FB . The position vectors are: FA rA D 0.16i C 0.08k, 360 mm O rB D 0.36i 0.08k. The sum of the moments: z 160 mm 80 mm 80 mm x i M D 0.16 0.728 j 0 3.272 k i 0.08 C 0.36 2.180 0.364 j 0 1.636 k 0.08 1.090 M D 0.1309i 0.0438j C 1.1125k (kN-m) Problem 4.201 If you represent the two forces exerted on the crankshaft in Problem 4.200 by a wrench consisting of a force F and a parallel couple Mp , what are F and Mp , and where does the line of action of F intersect the xz plane? Solution: From the solution to Problem 4.200, F D 0.364i C 4.908j C 1.09k (kN) and M D 0.1309i 0.0438j C 1.1125k (kN-m). The unit vector parallel to F is eF D F D 0.0722i C 0.9737j C 0.2162k. jFj The moment parallel to the force is MP D eF Ð MeF . Carrying out the operations: MP D 0.2073eF D 0.01497i C 0.2019j C 0.0448k (kN-m). This is the equivalent couple parallel to F. The component of the moment perpendicular to F is MN D M MP D 0.1159i 0.2457j C 1.0688k. The force exerts this moment about the origin. i MN D x 0.364 j 0 4.908 k z 1.09 D 4.908zi 1.09x C 0.364zj C 4.908xk D 0.1159i 0.2457j C 1.06884k. From which xD 1.0688 D 0.2178 m, 4.908 zD 0.1159 D 0.0236 m 4.908 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 267 Problem 5.1 In Active Example 5.1, suppose that the beam is subjected to a 6kN-m counterclockwise couple at the right end in addition to the 4-kN downward force. Draw a sketch of the beam showing its new loading. Draw the free-body diagram of the beam and apply the equilibrium equations to determine the reactions at A. 4 kN A 2m Solution: The equilibrium equations are Fx : Ax D 0 Fy : Ay 4 kN D 0 MA : MA 4 kN2 m C 6 kN-m D 0 Solving yields Ax D 0 Ay D 4 kN MA D 2 kN-m Problem 5.2 The beam has a fixed support at A and is loaded by two forces and a couple. Draw the free-body diagram of the beam and apply equilibrium to determine the reactions at A. 4 kN A 2 kN 6 kN-m 60⬚ 1m 1.5 m 1.5 m Solution: The free-body diagram is drawn. The equilibrium equations are Fx : Ax C 2 kN cos 60° D 0 Fy : Ay C 4 kN C 2 kN sin 60° D 0 MA : MA C 6 kN-m C 4 kN2.5 m C 2 kN sin 60° 4 m D 0 We obtain: Ax D 1 kN, Ay D 5.73 kN, MA D 22.9 kN-m 268 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.3 The beam is subjected to a load F D 400 N and is supported by the rope and the smooth surfaces at A and B. (a) (b) Draw the free-body diagram of the beam. What are the magnitudes of the reactions at A and B? F A B 30° 45° 1.2 m y Solution: C FX D 0: A cos 45° B sin 30° D 0 FY D 0: A sin 45° C B cos 30° T 400 N D 0 MA D 0: 1.2T 2.7400 C 3.7B cos 30° D 0 1.5 m 1m A F 45° x B 1.5 m 1.2 m 1m 30° T Solving, we get A D 271 N B D 383 N T D 124 N Problem 5.4 (a) Draw the free-body diagram of the beam. (b) Determine the tension in the rope and the reactions at B. 30⬚ 30⬚ 600 lb B A 5 ft 9 ft Solution: Let T be the tension in the rope. The equilibrium equations are: Fx : T sin 30° 600 lb sin 30° C Bx D 0 Fy : T cos 30° 600 lb cos 30° C By D 0 MB : 600 lb cos 30° 9 ft T cos 30° 14 ft D 0 Solving yields T D 368 lb, Bx D 493 lb, By D 186 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 269 Problem 5.5 (a) Draw the free-body diagram of the 60-lb drill press, assuming that the surfaces at A and B are smooth. (b) Determine the reactions at A and B. Solution: The system is in equilibrium. (a) (b) The free body diagram is shown. The sum of the forces: FX D 0, FY D FA C FB 60 D 0 The sum of the moments about point A: MA D 1060 C 24FB D 0, from which FB D 60 lb 600 D 25 lb 24 Substitute into the force balance equation: FA D 60 FB D 35 lb A B 10 in 14 in 60 lb A B 10 in FA 270 14 in FB c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.6 The masses of the person and the diving board are 54 kg and 36 kg, respectively. Assume that they are in equilibrium. (a) (b) Draw the free-body diagram of the diving board. Determine the reactions at the supports A and B. Solution: (a) (b) A FX D 0: AX D 0 FY D 0: AY C BY 549.81 369.81 D 0 MA D 0: 1.2BY 2.4369.81 B 4.6549.81 D 0 WD WP AX D 0 N Solving: 1.2 m AY D 1.85 kN 2.4 m BY D 2.74 kN 4.6 m 4.6 m 2.4 m 1.2 m AX Draw the free-body diagram of the ironing board. Determine the reactions at A and B. A B x 3 lb 10 lb 12 in 10 in 20 in Solution: The system is in equilibrium. Substitute into the force balance equation: (a) (b) FA D 13 FB D C15.833 lb The free-body diagram is shown. The sums of the forces are: FX D 0, WP y Problem 5.7 The ironing board has supports at A and B that can be modeled as roller supports. (a) (b) WD BY AY y A B x FY D FA C FB 10 3 D 0. 12 in 10 in 10 lb 20 in 3 lb FB 10 lb 3 lb The sum of the moments about A is MA D 12FB 2210 423 D 0, from which FB D FA 346 D 28.833 in. 12 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 271 Problem 5.8 The distance x D 9 m. (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports. 10 kN A B 6m x Solution: 10 kN (a) The FBD (b) The equilibrium equations x= 9 m Ax Fx : Ax D 0 6m Fy : Ay C By 10 kN D 0 Ay By MA : By 6 m 10 kN9 m D 0 Solving we find Ax D 0, Ay D 5 kN, By D 15 kN Problem 5.9 In Example 5.2, suppose that the 200-lb downward force and the 300 ft-lb counterclockwise couple change places; the 200-lb downward force acts at the right end of the horizontal bar, and the 300 ft-lb counterclockwise couple acts on the horizontal bar 2 ft to the right of the support A. Draw a sketch of the object showing the new loading. Draw the free-body diagram of the object and apply the equilibrium equations to determine the reactions at A. 100 lb 30 2 ft 200 lb A 300 ft-lb 2 ft 2 ft 2 ft Solution: The sketch and free-body diagram are shown. The equilibrium equations are Fx : Ax C 100 lb cos 30° D 0 Fy : Ay C 100 lb sin 30° 200 lb D 0 MA : MA C 300 ft-lb C 100 lb sin 30° 4 ft 100 lb cos 30° 2 ft 200 lb6 ft D 0 We obtain Ax D 86.6 lb Ay D 150 lb MA D 873 ft-lb 272 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.10 (a) Draw the free-body diagram of the beam. (b) Determine the reactions at the supports. 100 lb 400 lb 900 ft-lb A B 3 ft Solution: (a) Both supports are roller supports. The free body diagram is shown. (b) The sum of the forces: and 4 ft 3 ft 100 lb 4 ft 400 lb 3 ft 4 ft 3 ft 4 ft FX D 0, A FY D FA C FB C 100 400 D 0. 900 ft lb 100 lb 3 ft 4 ft 3 ft B 400 lb 4 ft The sum of the moments about A is 900 ft lb MA D 3100 C 900 7400 C 11FB D 0. From which FB D FA FB 2200 D 200 lb 11 Substitute into the force balance equation to obtain FA D 300 FB D 100 lb Problem 5.11 The person exerts 20-N forces on the pliers. The free-body diagram of one part of the pliers is shown. Notice that the pin at C connecting the two parts of the pliers behaves like a pin support. Determine the reactions at C and the force B exerted on the pliers by the bolt. Solution: The equilibrium equations MC :B25 mm 20 N cos 45° 80 mm 20 N sin 45° 50 mm D 0 25 mm B C Fx :Cx 20 N sin 45° D 0 Cx Fy :Cy B 20 N cos 45° 80 mm D0 Solving: Cy 50 mm 45⬚ 20 N C B D 73.5 N, Cx D 14.14 N, Cy D 87.7 N 20 N 20 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 273 Problem 5.12 (a) Draw the free-body diagram of the beam. 8 kN 8 kN 2 kN-m A B 30⬚ (b) Determine the reactions at the pin support A. 600 mm Solution: 500 mm 8 kN 600 mm 8 kN (a) The FBD (b) The equilibrium equations MA : 8 kN0.6 m C 8 kN1.1 m 2 kNm 600 mm 30° 2 kN-m B Ax B cos 30° 2.3 m D 0 Ay Fx :Ax B sin 30° D 0 Fy :Ay 8 kN C 8 kN B cos 30° D 0 Solving Ax D 0.502 kN, Ay D 0.870 kN, B D 1.004 kN Problem 5.13 (a) Draw the free-body diagram of the beam. y A (b) Determine the reactions at the supports. 6m 40 kN B x 8m 12 m Solution: A (a) (b) The FBD The equilibrium equations MB : 40 kN4 m C A6 m D 0 40 kN Bx Fx : A C Bx D 0 Fy : 40 kN C By D 0 Solving we find By A D Bx D 26.7 kN, By D 40 kN 274 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.14 (a) Draw the free-body diagram of the beam. A (b) If F D 4 kN, what are the reactions at A and B? 2 kN-m F 0.2 m 0.3 m 0.2 m 0.3 m 0.4 m B 2 kN-m Solution: Ax (a) (b) The free-body diagram The equilibrium equations F = 4 kN MA : 2 kN-m 4 kN0.2 m C B1.0 m D 0 Ay Fx : Ax 4 kN D 0 Fy : Ay C B D 0 Solving: B Ax D 4 kN, Ay D 2.8 kN, B D 2.8 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 275 Problem 5.15 In Example 5.3, suppose that the attachment point for the suspended mass is moved toward point B such that the horizontal distance from A to the attachment point increases from 2 m to 3 m. Draw a sketch of the beam AB showing the new geometry. Draw the free-body diagram of the beam and apply the equilibrium equations to determine the reactions at A to B. 2m 2m B 3m A Solution: From Example 5.3, we know that the mass of the suspended object is 2-Mg. The sketch and free-body diagram are shown. The equilibrium equations are Fx : Ax C Bx D 0 Fy : By 2000 kg9.81 m/s2 D 0 MB : Ax 3 m C 2000 kg9.81 m/s2 1 m D 0 We obtain Ax D 6.54 kN Bx D 6.54 kN By D 19.6 kN 276 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.16 A person doing push-ups pauses in the position shown. His 180-lb weight W acts at the point shown. The dimensions a = 15 in, b = 42 in, and c = 16 in. Determine the normal force exerted by the floor on each of his hands and on each of his feet. c W a b Solution: The free-body diagram is shown. The equilibrium equations are Fy : 2H C 2F 180 lb D 0 MH : Wa C 2Fa C b 0 We find that H D 66.3 lb, F D 23.7 lb Thus 66.3 lb on each hand 23.7 lb on each foot Problem 5.17 The hydraulic piston AB exerts a 400-lb force on the ladder at B in the direction parallel to the piston. Determine the weight of the ladder and the reactions at C. 6 ft W 3 ft A B C 6 ft 3 ft Solution: The free-body diagram of the ladder is shown. The angle between the piston AB and the horizontal is ˛ D tan1 3/6 D 26.6° The equilibrium equations are Fx : Cx C 400 lb cos ˛ D 0 Fy : Cy C 400 lb sin ˛ W D 0 MC : W6 ft 400 lb cos ˛3 ft 400 lb sin ˛3 ft D 0 Solving yields Cx D 358 lb, Cy D 89.4 lb, W D 268 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 277 Problem 5.18 Draw the free-body diagram of the structure by isolating it from its supports at A and E. Determine the reactions at A and E. D 400 lb 2 ft 200 ft-lb A 1 ft B C 100 lb 1 ft E 2 ft 2 ft 2 ft Solution: The free-body diagram is shown. The equilibrium equations are Fx : Ax C 100 lb D 0 Fy : Ay 400 lb C Ey D 0 MA : 100 lb1 ft 400 lb6 ft 200 ft-lb C Ey 4 ft D 0 Solving yields Ax D 100 lb Ay D 225 lb Ey D 625 lb Problem 5.19 (a) Draw the free-body diagram of the beam. (b) Determine the tension in the cable and the reactions at A. A B 30° 30 in 30 in Solution: 800 lb 30 in T (a) The FBD (b) The equilibrium equations C T 30° Ax MA : 800 lb60 in C T30 in C T sin 30° 90 in D 0 Fx :Ax T cos 30° D 0 Ay 800 lb Fy :Ay C T C T sin 30° 800 lb D 0 Solving: Ax D 554 lb, Ay D 160 lb, T D 640 lb 278 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.20 The unstretched length of the spring CD is 350 mm. Suppose that you want the lever ABC to exert a 120-N normal force on the smooth surface at A. Determine the necessary value of the spring constant k and the resulting reactions at B. C k 230 mm D 450 mm 20⬚ 180 mm B A Solution: We have F D k 0.23 m2 C 0.3 m2 0.35 330 mm m 300 mm A D 120 N 30 F0.45 m C A cos 20° 0.18 m MB : p 1429 F 23 30 C A sin 20° 0.33 m D 0 Fx :A cos 20° C Bx C p 30 1429 Fy : A sin 20° C By p FD0 23 1429 Bx FD0 A By Solving we find: 20° k D 3380 N/m, Bx D 188 N, By D 98.7 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 279 Problem 5.21 The mobile is in equilibrium. The fish B weighs 27 oz. Determine the weights of the fish A, C, and D. (The weights of the crossbars are negligible.) 12 in 3 in A 6 in 2 in B 7 in 2 in C D Solution: Denote the reactions at the supports by FAB , FCD , and FBCD as shown. Start with the crossbar supporting the weights C and D. The sum of the forces is FCD D C 7 in 2 in FY D C D C FCD D 0, FBCD from which FCD D C C D. 6 in For the cross bar supporting the weight B, the sum of the forces is FY D B C FBCD FCD D 0, from which, substituting, FBCD D B C C C D. B FCD 2 in FAB FBCD A 12 in 3 in For the crossbar supporting C and D, the sum of the moments about the support is MCD D 7D C 2C D 0, from which D D 2C . 7 For the crossbar supporting B, the sum of the moments is MBCD D 6FCD 2B D 0, from which, substituting from above FCD D 2C 9C 2B DCCD DCC D , 6 7 7 or C D 7B/27 D 7 oz, and D D 2C/7 D 2 oz. The sum of the moments about the crossbar supporting A is MAB D 12A 3FBCD D 0, from which, substituting from above, AD 280 27 C 7 C 2 3B C C C D D D 9 oz 12 4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.22 The car’s wheelbase (the distance between the wheels) is 2.82 m. The mass of the car is 1760 kg and its weight acts at the point x D 2.00 m, y D 0.68 m. If the angle ˛ D 15° , what is the total normal force exerted on the two rear tires by the sloped ramp? x α W Solution: Split W into components: y x α W cos ˛ acts ? to the incline W sin ˛ acts parallel to the incline R FX : f W sin ˛ D 0 FY : NR C NF W cos ˛ D 0 0.68 NF α = 15° W = (1760X9.81) N α m f MR : 2m m 2.82 NR 2W cos ˛ C 0.68W sin ˛ C 2.82NF D 0 Solving: NR D 5930 N, NF D 10750 N Problem 5.23 The link AB exerts a force on the bucket of the excavator at A that is parallel to the link. The weight W = 1500 lb. Draw the free-body diagram of the bucket and determine the reactions at C. (The connection at C is equivalent to a pin support of the bucket.) 14 in 16 in B A 4 in C D W 8 in 8 in Solution: The free-body diagram is shown. The angle between the link AB and the horizontal is ˛ D tan1 12/14 D 40.6° The equilibrium equations are Fx : Cx C TAB cos ˛ D 0 Fy : Cy C TAB sin ˛ 1500 lb D 0 MC : 1500 lb8 in TAB cos ˛4 in TAB sin ˛16 in D 0 Solving yields TAB D 892 lb, Cx D 677 lb, Cy D 919 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 281 Problem 5.24 The 14.5-lb chain saw is subjected to the loads at A by the log it cuts. Determine the reactions R, Bx , and By that must be applied by the person using the saw to hold it in equilibrium. Solution: The sum of the forces are y FX D 5 C BX R cos 60° D 0. FY D 10 14.5 C BY R sin 60° D 0. R The sum of the moments about the origin is 60° By 1.5 in 7 in x A Bx 5 lb MO D 7R cos 60° C 8BY 214.5 1310 51.5 D 0. From which 7R cos 60° C 8BY 166.5 D 0. Collecting equations and reducing to 3 equations in 3 unknowns: 14.5 lb 10 lb BX C 0BY 0.5R D 5 13 in 6 in 2 in 0BX C BY 0.866R D 4.5 0BX C 8BY C 3.5R D 166.5. Solving: BX D 11.257 lb, BY D 15.337 lb, and R D 12.514 lb Problem 5.25 The mass of the trailer is 2.2 Mg (megagrams). The distances a D 2.5 m and b D 5.5 m. The truck is stationary, and the wheels of the trailer can turn freely, which means that the road exerts no horizontal force on them. The hitch at B can be modeled as a pin support. (a) Draw the free-body diagram of the trailer. (b) Determine the total normal force exerted on the rear tires at A and the reactions exerted on the trailer at the pin support B. B Solution: (a) (b) The free body diagram is shown. The sum of forces: FX D BX D 0. FY D FA W C FB D 0. The sum of the moments about A: MA D aW C a C bFB D 0, from which FB D W A 2.52.2 ð 103 9.81 aW D D 6.744 kN aCb 2.5 C 5.5 Substitute into the force equation: a b FA D W FB D 14.838 kN B BX FB W A FA 282 a b c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.26 The total weight of the wheelbarrow and its load is W = 100 lb. (a) What is the magnitude of the upward force F necessary to lift the support at A off the ground? (b) What is the magnitude of the downward force necessary to raise the wheel off the ground? F W B A 40 in 12 in 14 in Solution: The free-body diagram is shown. The equilibrium equations are Fy : A C B C F W D 0 MA : B26 in W12 in F40 in D 0 (a) Set A D 0 and solve. We find that F D 21.2 lb (b) Set B D 0 and solve. We find that F D 30 lb So we have a 21.2 lb, b 30 lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 283 Problem 5.27 The airplane’s weight is W D 2400 lb. Its brakes keep the rear wheels locked. The front (nose) wheel can turn freely, and so the ground exerts no horizontal force on it. The force T exerted by the airplane’s propeller is horizontal. (a) Draw the free-body diagram of the airplane. Determine the reaction exerted on the nose wheel and the total normal reaction on the rear wheels (b) when T D 0, (c) when T D 250 lb. T 4 ft W A 5 ft B 2 ft Solution: (a) The free body diagram is shown. (b) The sum of the forces: FX D BX D 0 FY D AY W C BY D 0. The sum of the moments about A is MA D 5W C 7BY D 0, from which BY D 5W D 1714.3 lb 7 Substitute from the force balance equation: AY D W BY D 685.7 lb (c) The sum of the forces: FX D 250 C BX D 0, from which BX D 250 lb FY D AY W C BY D 0. The sum of the moments about A: MA D 2504 5W C 7BY D 0, from which BY D 1571.4 lb. Substitute into the force balance equation to obtain: AY D 828.6 lb 4 ft AY 284 W A 5 ft B 2 ft BX BY c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.28 A safety engineer establishing limits on the load that can be carried by a forklift analyzes the situation shown. The dimensions are a = 32 in, b = 30 in, and c = 26 in. The combined weight of the forklift and operator is WF = 1200 lb. As the weight WL supported by the forklift increases, the normal force exerted on the floor by the rear wheels at B decreases. The forklift is on the verge of tipping forward when the normal force at B is zero. Determine the value of WL that will cause this condition. WL WF A a B b c Solution: The equilibrium equations and the special condition for this problem are Fy : A C B WL 1200 lb D 0 MA : WL a WF b C Bc D 0 BD0 We obtain WL D 1125 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 285 Problem 5.29 Paleontologists speculate that the stegosaur could stand on its hind limbs for short periods to feed. Based on the free-body diagram shown and assuming that m D 2000 kg, determine the magnitudes of the forces B and C exerted by the ligament–muscle brace and vertebral column, and determine the angle ˛. 580 mm 160 mm mg C B 22° Solution: Take the origin to be at the point of application of the force C. The position vectors of the points of application of the forces B and W are: α 415 mm 790 mm rB D 415i C 160j (mm), rW D 790i C 580j (mm). The forces are C D Ci cos90° ˛ C j sin90° ˛ D Ci sin ˛ C j cos ˛. B D Bi cos270° 22° C j sin270° 22° D B0.3746i 0.9272j. W D 29.81j D 19.62j (kN). The moments about C, i MC D 415 0.3746B i C 790 0 k 160 0 0.9272B 0 j k 580 0 D 0 19.62 0 j D 444.72B 15499.8 D 0, from which BD 15499.8 D 34.85 kN. 444.72 The sums of the forces: FX D C sin ˛ 0.3746Bi D 0, from which C sin ˛ D 13.06 kN. FY D C cos ˛ 0.9272B 19.62j D 0, from which C cos ˛ D 51.93 kN. The angle ˛ is ˛ D tan1 13.06 51.93 D 14.1° . The magnitude of C, CD 286 p 13.062 C 51.932 D 53.55 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.30 The weight of the fan is W D 20 lb. Its base has four equally spaced legs of length b D 12 in. Each leg has a pad near the end that contacts the floor and supports the fan. The height h D 32 in. If the fan’s blade exerts a thrust T D 2 lb, what total normal force is exerted on the two legs at A? T b W h T A B Side View Top View Solution: The free-body diagram is shown. The equilibrium equations are Fy : A C B W D 0 b 2b MB : W p A p Th D 0 2 2 We obtain A D 6.23 lb Problem 5.31 The weight of the fan is W D 20 lb. Its base has four equally spaced legs of length b D 12 in. Each leg has a pad near the end that contacts the floor and supports the fan. The height h D 32 in. As the thrust T of the fan increases, the normal force supported by the two legs at A decreases. When the normal force at A is zero, the fan is on the verge of tipping over. Determine the value of T that will cause this condition. T b W h T A B Side View Top View Solution: The free-body diagram is shown. The equilibrium equations are Fy : A C B W D 0 2b b MB : W p A p Th D 0 2 2 We set A D 0 and solve to obtain T D 5.30 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 287 Problem 5.32 In a measure to decrease costs, the manufacturer of the fan described in Problem 5.31 proposes to support the fan with three equally spaced legs instead of four. An engineer is assigned to analyze the safety implications of the change. The weight of the fan decreases to W D 19.6 lb. The dimensions b and h are unchanged. What thrust T will cause the fan to be on the verge of tipping over in this case? Compare your answer to the answer to Problem 5.31. b T Solution: The free-body diagram is shown. The equilibrium equations are Fy : A C B W D 0 MB : Wb cos 60° Ab C b cos 60° Th D 0 We set A D 0 and solve to obtain T D 3.68 lb This configuration is less stable than the one in Problem 5.31 using four legs. Problem 5.33 A force F D 400 N acts on the bracket. What are the reactions at A and B? F A 80 mm B 320 mm Solution: The joint A is a pinned joint; B is a roller joint. The pinned joint has two reaction forces AX , AY . The roller joint has one reaction force BX . The sum of the forces is FX D AX C BX D 0, F AY AX 80 mm BX 320 mm FY D AY F D 0, from which AY D F D 400 N. The sum of the moments about A is MA D 0.08BX 0.320F D 0, from which BX D 0.320400 D 1600 N. 0.08 Substitute into the sum of forces equation to obtain: AX D BX D 1600 N 288 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.34 The sign’s weight WS D 32 lb acts at the point shown. The 10-lb weight of bar AD acts at the midpoint of the bar. Determine the tension in the cable AE and the reactions at D. 11 in 30 in 11 in E 30⬚ 15⬚ A B C D Ws 33 in Solution: Treat the bar AD and sign as one single object. Let TAE be the tension in the cable. The equilibium equations are Fx : TAE cos 15° C Dx D 0 Fy : TAE sin 15° C Dy Ws 10 lb D 0 MD : TAE cos 15° 52 in tan 30° TAE sin 15° 52 in C 32 lb33 in C 10 lb26 in D 0 Solving yields TAE D 31.0 lb Dx D 29.9 lb Dy D 34.0 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 289 Problem 5.35 The device shown, called a swape or shadoof, helps a person lift a heavy load. (Devices of this kind were used in Egypt at least as early as 1550 B.C. and are still in use in various parts of the world.) The dimensions a D 3.6 m and b D 1.2 m. The mass of the bar and counterweight is 90 kg, and their weight W acts at the point shown. The mass of the load being lifted is 45 kg. Determine the vertical force the person must exert to support the stationary load (a) when the load is just above the ground (the position shown); (b) when the load is 1 m above the ground. Assume that the rope remains vertical. Solution: MO : 441 N F3.6 m cos 883 N1.2 m cos D 0 Solving we find F D 147.2 N Notice that the angle is not a part of this answer therefore (a) F D 147.2 N (b) F D 147.2 N a Oy b θ 25° Ox F W 883 N 441 N Problem 5.36 This structure, called a truss, has a pin support at A and a roller support at B and is loaded by two forces. Determine the reactions at the supports. Strategy: Draw a free-body diagram, treating the entire truss as a single object. 30° 2 kN 45° 4 kN Solution: p MA : 4 kN 2b 2 kN cos 30° 3 b C 2 kN sin 30° b C B4 b D 0 Fx : Ax C 4 kN sin 45° 2 kN sin 30° D 0 Fy : Ay 4 kN cos 45° 2 kN cos 30° C B D 0 Solving: b Ax D 1.828 kN, Ay D 2.10 kN, B D 2.46 kN B A b b b b 45° 4 kN 30° 2 kN Ax Ay 290 B c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.37 An Olympic gymnast is stationary in the “iron cross” position. The weight of his left arm and the weight of his body not including his arms are shown. The distances are a D b D 9 in and c D 13 in. Treat his shoulder S as a fixed support, and determine the magnitudes of the reactions at his shoulder. That is, determine the force and couple his shoulder must support. S 8 lb 144 lb a b c Solution: The shoulder as a built-in joint has two-force and couple reactions. The left hand must support the weight of the left arm and half the weight of the body: FH D FH 8 lb 144 C 8 D 80 lb. 2 The sum of the forces on the left arm is the weight of his left arm and the vertical reaction at the shoulder and hand: FH 8 lb 144 lb FH SX M FX D SX D 0. 8 lb SY b FY D FH SY 8 D 0, c from which SY D FH 8 D 72 lb. The sum of the moments about the shoulder is MS D M C b C cFH b8 D 0, where M is the couple reaction at the shoulder. Thus M D b8 b C cFH D 1688 in lb D 1688 (in lb) 1 ft 12 in D 140.67 ft lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 291 Problem 5.38 Determine the reactions at A. A 5 ft 300 lb 800 ft-lb 200 lb 200 lb 6 ft 3 ft Solution: The built-in support at A is a two-force and couple reaction support. The sum of the forces for the system is FX D AX C 200 D 0, from which AX D 200 lb FY D AY C 300 200 D 0, from which AY D 100 lb The sum of the moments about A: M D 6300 C 5200 800 C MA D 0, from which MA D 1600 ft lb which is the couple at A. AY MA 300 lb AX 5 ft 800 ft-lb 200 lb 200 lb 6 ft 292 3 ft c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.39 The car’s brakes keep the rear wheels locked, and the front wheels are free to turn. Determine the forces exerted on the front and rear wheels by the road when the car is parked (a) on an up slope with ˛ D 15° ; (b) on a down slope with ˛ D 15° . n 70 i n 36 i n 20 i y α 3300 lb Solution: The rear wheels are two force reaction support, and the front wheels are a one force reaction support. Denote the rear wheels by A and the front wheels by B, and define the reactions as being parallel to and normal to the road. The sum of forces: AX D 854.1 lb. 20 in. α FX D AX 3300 sin 15° D 0, from which x BY 3300 lb AX AY 36 in. 70 in. FY D AY 3300 cos 15° C BY D 0. Since the mass center of the vehicle is displaced above the point A, a component of the weight (20W sin ˛) produces a positive moment about A, whereas the other component (36W cos ˛) produces a negative moment about A. The sum of the moments about A: MA D 363300 cos 15° C 203300 sin 15° C BY 106 D 0, from which BY D C97669 D 921.4 lb. 106 Substitute into the sum of forces equation to obtain AY D 2266.1 lb (b) For the car parked down-slope the sum of the forces is FX D AX C 3300 sin 15° D 0, from which AX D 854 lb FY D AY 3300 cos 15° C BY D 0. The component (20W sin ˛) now produces a negative moment about A. The sum of the moments about A is MA D 330036 cos 15° 330020 sin 15° C 106BY D 0, from which BY D 131834 D 1243.7 lb. 106 Substitute into the sum of forces equation to obtain AY D 1943.8 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 293 Problem 5.40 The length of the bar is L D 4 ft. Its weight W D 6 lb acts at the midpoint of the bar. The floor and wall are smooth. The spring is unstretched when the angle ˛ = 0. If the bar is in equilibrium when ˛ D 40° , what is the spring constant k? k α L Solution: The free-body diagram is shown. The stretch in the spring is L L cos ˛, so the upward force exerted on the bar by the spring is F D kL 1 cos ˛. Let N and R be the normal forces exerted by the floor and the wall, respectively. The equilibrium equations for the bar are Fx : R D 0 Fy : F C N W D 0 Mbottom : W L sin ˛ RL cos ˛ 2 FL sin ˛ D 0 Because R D 0, the moment equation can be solved for the force exerted by the spring. F D 0.5W D 3 lb D kL 1 cos ˛ Solving yields k D 3.21 lb/ft 294 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.41 The weight W of the bar acts at its midpoint. The floor and wall are smooth. The spring is unstretched when the angle ˛ D 0. Determine the angle ˛ at which the bar is in equilibrium in terms of W, k, and L. k α L Solution: The free-body diagram is shown. The stretch in the spring is L L cos ˛, so the upward force exerted on the bar by the spring is F D kL1 cos ˛. Let N and R be the normal forces exerted by the floor and the wall, respectively. The equilibrium equations for the bar are Fx : R D 0 Fy : F C N W D 0 Mbottom : W L sin ˛ RL cos ˛ 2 FL sin ˛ D 0 Because R D 0, the moment equation can be solved for the force exerted by the spring. FD W D kL1 cos ˛ 2 W Solving yields ˛ D cos1 1 2L c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 295 Problem 5.42 The plate is supported by a pin in a smooth slot at B. What are the reactions at the supports? 2 kN-m 6 kN-m A B 60° 2m Solution: The pinned support is a two force reaction support. The smooth pin is a roller support, with a one force reaction. The reaction at B forms an angle of 90° C 60° D 150° with the positive x axis. The sum of the forces: 6 kN-m 2 kN-m FX D AX C B cos 150° D 0 FY D AY C B sin 150° D 0 A 60° B The sum of the moments about B is 2m MB D 2AY C 2 6 D 0, from which AY D 4 D 2 kN. 2 2 kN-m 6 kN-m Substitute into the force equations to obtain BD AY D 4 kN, sin 150° and AX D B cos 150° D 3.464 kN. AX 150° AY B 2m The horizontal and vertical reactions at B are BX D 4 cos 150° D 3.464 kN, and BY D 4 sin 150° D 2 kN. 296 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.43 Determine the reactions at the fixed support A. Solution: The free-body diagram is shown. The equilibrium equations are y 30 lb 40 lb 150 ft-lb 45⬚ A x Fx : Ax C 40 lb cos 45° D 0 3 ft 3 ft 6 ft Fy : Ay C 30 lb C 40 lb sin 45° D 0 MA : MA C 30 lb3 ft C 40 lb sin 45° 6 ft C 150 ft-lb D 0 Solving yields Ax D 28.3 lb, Ay D 58.3 lb, MA D 410 ft-lb. Problem 5.44 Suppose the you want to represent the two forces and couple acting on the beam in Problem 5.43 by an equivalent force F as shown. (a) Determine F and the distance D at which its line of action crosses the x axis. (b) Assume that F is the only load acting on the beam and determine the reactions at the fixed support A. Compare your answers to answers to Problem 5.43. y F A x D Solution: The free-body diagram is shown. (a) To be equivalnet, F must equal the sum of the two forces: F D 30 lbj C 40 lbcos 45° i C sin 45° j F D 28.3i C 58.3j lb The force F must be placed so that the moment about a point due to F is equal to the moment about the same point due to the two forces and couple. Evaluating the moments about the origin, 58.3 lbD D 30 lb3 ft C 40 lb sin 45° 6 ft C 150 ft-lb The distance D D 7.03 ft (b) The equilibrium equations are Fx : Ax C 28.3 lb D 0 Fy : Ay C 58.3 lb D 0 MA : MA C 58.3 lb7.03 ft D 0 Solving yields Ax D 28.3 lb, Ay D 58.3 lb, MA D 410 ft-lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 297 Problem 5.45 The bicycle brake on the right is pinned to the bicycle’s frame at A. Determine the force exerted by the brake pad on the wheel rim at B in terms of the cable tension T. T 35° 40 mm B Brake pad Wheel rim 45 mm A 40 mm Solution: From the force balance equation for the cables: the force TB on the brake mechanism TB in terms of the cable tension T is 35° T 2TB sin 35° D 0, from which TB D T D 0.8717T. 2 sin 35° 40 mm B Take the origin of the system to be at A. The position vector of the point of attachment of B is rB D 45j (mm). The position vector of the point of attachment of the cable is rC D 40i C 85j (mm). The force exerted by the brake pad is B D Bi. The force vector due the cable tension is AY 45 mm AX 40 mm TB D TB i cos 145° C j sin 145° D TB 0.8192i C 0.5736j. The moment about A is MA D rB ð B C rC ð TB D 0 i MA D 0 B j 45 0 i k 45 C 40 0 0.8192 k 85 85 TB D 0 0.5736 0 j MA D 45B C 92.576TB k D 0, from which B D 92.576TB D 2.057TB . 45 Substitute the expression for the cable tension: B D 2.0570.8717T D 1.793T 298 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.46 The mass of each of the suspended weights is 80 kg. Determine the reactions at the supports at A and E. A B C 300 mm D E 200 mm Solution: From the free body diagram, the equations of equilibrium for the rigid body are and Fx D AX C EX D 0, 200 mm y AY 0.2 m 0.2 m AX A x mg 0.3 m Fy D AY 2809.81 D 0, mg E EX MA D 0.3EX 0.2809.81 0.4809.81 D 0. We have three equations in the three components of the support reactions. Solving for the unknowns, we get the values AX D 1570 N, AY D 1570 N, and EX D 1570 N. Problem 5.47 The suspended weights in Problem 5.46 are each of mass m. The supports at A and E will each safely support a force of 6 kN magnitude. Based on this criterion, what is the largest safe value of m? Solution: Written with the mass value of 80 kg replaced by the symbol m, the equations of equilibrium from Problem 5.46 are and Fx D AX C EX D 0, Fy D AY 2 m9.81 D 0, MA D 0.3EX 0.2 m9.81 0.4 m9.81 D 0. We also need the relation jAj D A2X C A2Y D 6000 N. We have four equations in the three components of the support reactions plus the magnitude of A. This is four equations in four unknowns. Solving for the unknowns, we get the values AX D 4243 N, AY D 4243 N, EX D 4243 N, and m D 216.5 kg. Note: We could have gotten this result by a linear scaling of all of the numbers in Problem 5.46. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 299 Problem 5.48 The tension in cable BC is 100 lb. Determine the reactions at the built-in support. C 6 ft A B 300 ft-lb 200 lb 3 ft 3 ft 6 ft Solution: The cable does not exert an external force on the system, and can be ignored in determining reactions. The built-in support is a two-force and couple reaction support. The sum of forces: FX D AX D 0. MA AY 300 ft-lb AX 200 lb 3 ft FY D AY 200 D 0, from which AY D 200 lb. The sum of the moments about A is M D MA 3200 300 D 0, from which MA D 900 ft lb Problem 5.49 The tension in cable AB is 2 kN. What are the reactions at C in the two cases? 60° A B 2m C A 1m 2m (a) Solution: First Case: The sum of the forces: FX D CX T cos 60° D 0, from which CX D 20.5 D 1 kN T T C 1m (b) CY 2m 1m Case (a) MC CX CY from which CY D 1.8662 D 3.732 kN. B 60° FY D CY C T sin 60° C T D 0, The sum of the moments about C is 60° Case (b) MC CX M D MC T sin 60° 3T D 0, from which MC D 3.8662 D 7.732 kN Second Case: The weight of the beam is ignored, hence there are no external forces on the beam, and the reactions at C are zero. 300 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.50 Determine the reactions at the supports. 6 in 5 in 50 lb A 3 in 100 in-lb Solution: The reaction at A is a two-force reaction. The reaction 3 in at B is one-force, normal to the surface. B The sum of the forces: 30° FX D AX B cos 60° 50 D 0. AX FY D AY C B sin 60° D 0. 50 lb AY The sum of the moments about A is 6 in. 100 MA D 100 C 11B sin 60° 6B cos 60° D 0, B 11 in. from which 60° 100 D 15.3 lb. BD 11 sin 60° 6 cos 60° Substitute into the force equations to obtain AY D B sin 60° D 13.3 lb and AX D B cos 60° C 50 D 57.7 lb Problem 5.51 The weight W D 2 kN. Determine the tension in the cable and the reactions at A. 30° A W 0.6 m AY Solution: Equilibrium Eqns: C FX D 0: 0.6 m T AX C T cos 30° D 0 AX FY D 0: AY C T C T sin 30° MA D 0: 0, 6W C 0.6T sin 30° WD0 T 30° 0.6 m 0.6 m W = 2 kN = 2000 N C 1, 2T D 0 Solving, we get AX D 693 N, AY D 800 N, T D 800 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 301 Problem 5.52 The cable shown in Problem 5.51 will safely support a tension of 6 kN. Based on this criterion, what is the largest safe value of the weight W? Solution: The equilibrium equations in the solution of problem are C FX D 0: AX C T cos 30° D 0 FY D 0: AY C T C T sin 30° W D 0 MA D 0: 0, 6W C 0, 6T sin 30° C 1, 2T D 0 We previously had 3 equations in the 3 unknowns AX , AY and T (we knew W). In the current problem, we know T but don’t know W. We again have three equations in three unknowns (AX , AY , and W). Setting T D 6 kN, we solve to get AX D 5.2 kN AY D 6.0 kN W D 15.0 kN Problem 5.53 The blocks being compressed by the clamp exert a 200-N force on the pin at D that points from A toward D. The threaded shaft BE exerts a force on the pin at E that points from B toward E. 50 mm (a) 50 mm Draw a free-body diagram of the arm DCE of the clamp, assuming that the pin at C behaves like a pin support. (b) Determine the reactions at C. E A C 50 mm D FBE (a) The free-body diagram (b) The equilibrium equations 125 mm B Solution: 125 mm 125 mm Cy MC : 200 N0.25 m FBE 0.1 m D 0 200 N Fx : Cx C FBE D 0 Cx Fy : Cy 200 N D 0 Solving Cx D 500 N, Cy D 200 N 302 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.54 Consider the clamp in Problem 5.53. The blocks being compressed by the clamp exert a 200N force on the pin at A that points from D toward A. The threaded shaft BE exerts a force on the pin at B that points from E toward B. (a) Draw a free-body diagram of the arm ABC of the clamp, assuming that the pin at C behaves like a pin support. Determine the reactions at C. (b) FBE Solution: (a) (b) The free-body diagram The equilibrium equations Cx MC : 200 N0.25 m C FBE 0.1 m D 0 Fx : FBE C Cx D 0 Cy 200 N Fy : 200 N C Cy D 0 Solving we find Cx D 500 N, Cy D 200 N Problem 5.55 Suppose that you want to design the safety valve to open when the difference between the pressure p in the circular pipe diameter D 150 mm and the atmospheric pressure is 10 MPa (megapascals; a pascal is 1 N/m2 ). The spring is compressed 20 mm when the valve is closed. What should the value of the spring constant be? 250 mm 150 mm k A p 150 mm Solution: The area of the valve is aD 0.15 2 150 mm 2 250 mm D 17.671 ð 103 m2 . k The force at opening is A F D 10a ð 106 D 1.7671 ð 105 N. 150 mm The force on the spring is found from the sum of the moments about A, MA D 0.15F 0.4kL D 0. k∆L A Solving, F 0.15F 0.151.7671 ð 105 kD D 0.4L 0.40.02 D 3.313 ð 106 0.15 m 0.25 m N m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 303 Problem 5.56 The 10-lb weight of the bar AB acts at the midpoint of the bar. The length of the bar is 3 ft. Determine the tension in the string BC and the reactions at A. C B 3 ft A 30⬚ 1 ft Solution: Geometry: tan D 3 ft 3 ft sin 30° D 0.4169 ) D 22.63° 1 ft C 3 ft cos 30° The equilibrium equations MA : TBC cos 3 ft sin 30° C TBC sin 3 ft cos 30° 10 lb1.5 ft cos 30° D 0 Fx : TBC cos C Ax D 0 Fy : TBC sin 10 lb C Ay D 0 Solving: Ax D 5.03 lb, Ay D 7.90 lb, T D 5.45 lb TBC θ 10 lb Ax Ay 304 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.57 The crane’s arm has a pin support at A. The hydraulic cylinder BC exerts a force on the arm at C in the direction parallel to BC. The crane’s arm has a mass of 200 kg, and its weight can be assumed to act at a point 2 m to the right of A. If the mass of the suspended box is 800 kg and the system is in equilibrium, what is the magnitude of the force exerted by the hydraulic cylinder? C A 2.4 m 1m B 1.8 m 1.2 m 7m Solution: The free-body diagram of the arm is shown. 2.4 D 63.4° 1.2 The equilibrium equations are The angle D tan1 Fx : Ax C FH cos D 0 Fy : Ay C FH sin 200 kg9.81 m/s2 800 kg9.81 m/s2 D 0 MA : FH sin 3 m FH cos 1.4 m 200 kg9.81 m/s2 2 m 800 kg9.81 m/s2 7 m D 0 We obtain Ax D 12.8 kN, Ay D 15.8 kN, FH D 28.6 kN Thus FH D 28.6 kN Problem 5.58 In Problem 5.57, what is the magnitude of the force exerted on the crane’s arm by the pin support at A? Solution: See the solution to Problem 5.57. Ax D 12.8 kN, Ay D 15.8 kN, FH D 28.6 kN jAj D 12.8 kN2 C 15.8 kN2 D 20.3 kN C A 2.4 m 1m B Thus jAj D 20.3 kN 1.8 m 1.2 m 7m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 305 Problem 5.59 A speaker system is suspended by the cables attached at D and E. The mass of the speaker system is 130 kg, and its weight acts at G. Determine the tensions in the cables and the reactions at A and C. 0.5 m 0.5 m 0.5 m 0.5 m 1m E C A 1m B D G Solution: The weight of the speaker is W D mg D 1275 N. The 1m AY equations of equilibrium for the entire assembly are 1.5 m CY E A Fx D CX D 0, Fy D AY C CY mg D 0 C CX B D (where the mass m D 130 kg), and mg MC D 1AY 1.5mg D 0. Solving these equations, we get 1.5 m CX D 0, 1m T2 CY D 3188 N, T1 and AY D 1913 N. From the free body diagram of the speaker alone, we get and Mleft mg Fy D T1 C T2 mg D 0, support D 1mg C 1.5T2 D 0. Solving these equations, we get T1 D 425. N and T2 D 850 N 306 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.60 The weight W1 D 1000 lb. Neglect the weight of the bar AB. The cable goes over a pulley at C. Determine the weight W2 and the reactions at the pin support A. B 50° 35° W1 A C W2 Solution: The strategy is to resolve the tensions at the end of bar AB into x- and y-components, and then set the moment about A to zero. The angle between the cable and the positive x axis is 35° . The tension vector in the cable is T2 D W2 i cos35° C j sin35° . 35° T2 rB T1 50° AY D W2 0.8192i 0.5736jlb. AX Assume a unit length for the bar. The angle between the bar and the positive x axis is 180° 50° D 130° . The position vector of the tip of the bar relative to A is rB D i cos130° C j sin130° , D 0.6428i C 0.7660j. The tension exerted by W1 is T1 D 1000j. The sum of the moments about A is: MA D rB ð T1 C rB ð T2 D rB ð T1 C T2 i D L 0.6428 0.8191W2 j 0.7660 0.5736W2 1000 MA D 0.2587W2 C 642.8k D 0, from which W2 D 2483.5 lb The sum of the forces: FX D AX C W2 0.8192i D 0, from which AX D 2034.4 lb FY D AY W2 0.5736 1000j D 0, from which AY D 2424.5 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 307 Problem 5.61 The dimensions a D 2 m and b D 1 m. The couple M D 2400 N-m. The spring constant is k D 6000 N/m, and the spring would be unstretched if h D 0. The system is in equilibrium when h D 2 m and the beam is horizontal. Determine the force F and the reactions at A. k h A M F a Solution: We need to know the unstretched length of the spring, l0 b Unstretched l0 D a C b D 3 m (a + b) We also need the stretched length l2 D h2 C a C b2 AY θ M l D 3.61 m AX a FS D kl l0 tan D F b h a C b D 33.69° Equilibrium eqns: C FX : AX FS cos D 0 FY : AY C FS sin F D 0 MA : M aF C a C bFS sin D 0 a D 2 m, b D 1 m, M D 2400 N-m, h D 2 m, k D 6000 N/m. Substituting in and solving, we get FS D 6000l l0 D 3633 N and the equilibrium equations yield AX D 3023 N AY D 192 N F D 1823 N 308 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.62 The bar is 1 m long, and its weight W acts at its midpoint. The distance b D 0.75 m, and the angle ˛ D 30° . The spring constant is k D 100 N/m, and the spring is unstretched when the bar is vertical. Determine W and the reactions at A. k α W A b p Solution: The unstretched length of the spring is L D b2 C 12 D 1.25 m. The obtuse angle is 90 C ˛, so the stretched length can be determined from the cosine law: L22 D 12 C 0.752 20.75 cos90 C ˛ D 2.3125 β m2 α W from which L2 D 1.5207 m. The force exerted by the spring is A T D kL D 1001.5207 1.25 D 27.1 N. b The angle between the spring and the bar can be determined from the sine law: 1.5207 b D , sin ˇ sin90 C ˛ T β from which sin ˇ D 0.4271, α ˇ D 25.28° . W The angle the spring makes with the horizontal is 180 25.28 90 ˛ D 34.72° . The sum of the forces: AX FX D AX T cos 34.72° D 0, AY from which AX D 22.25 N. FY D AY W T sin 34.72° D 0. The sum of the moments about A is MA D T sin 25.28° W 2 sin ˛ D 0, from which WD 2T sin 25.28° D 46.25 N. sin ˛ Substitute into the force equation to obtain: AY D W C T sin 34.72° D 61.66 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 309 B Problem 5.63 The boom derrick supports a suspended 15-kip load. The booms BC and DE are each 20 ft long. The distances are a D 15 ft and b D 2 ft, and the angle D 30° . Determine the tension in cable AB and the reactions at the pin supports C and D. E θ Solution: Choose a coordinate system with origin at point C, with the y axis parallel to CB. The position vectors of the labeled points are: C A D rD D 2i a b rE D rD C 20i sin 30° C j cos 30° D 12i C 17.3j, The components: rB D 20j, Dx D 0.6jDj D 7.67 kip, rA D 15i. The unit vectors are: eDE D rE r D D 0.5i C 0.866j, jrE rD j eEB D rB r E D 0.976i C 0.2179j. jrB rE j eCB D rB r C D 1j, jrB rC j eAB D rA r B D 0.6i 0.8j. jrA rB j Dy D 0.866jDj D 13.287 kip, and Cy D 1jCj D 11.94 kip B E θ A C Fx D 0.5jDj 0.976jTEB j D 0, TAB Fy D 0.866jDj C 0.2179jTEB j 15 D 0, from which b a Isolate the juncture at E: The equilibrium conditions are D TEB TEB 15 kip C D Juncture B Juncture E jDj D 15.34 kip and jTEB j D 7.86 kip. Isolate the juncture at B: The equilibrium conditions are: and Fx D 0jCj 0.6jTAB j C 0.976jTEB j, Fy D 1jCj 0.6jTAB j 0.2179jTEB j D 0, from which jTAB j D 12.79 kip, and jCj D 11.94 kip. 310 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.64 The arrangement shown controls the elevators of an airplane. (The elevators are the horizontal control surfaces in the airplane’s tail.) The elevators are attached to member EDG. Aerodynamic pressures on the elevators exert a clockwise couple of 120 in-lb. Cable BG is slack, and its tension can be neglected. Determine the force F and the reactions at pin support A. A D 2.5 in C F 3.5 in Elevator E B 6 in 120 in-lb G 2 in 2.5 in 2.5 in 1.5 in 120 in (Not to scale) Solution: Begin at the elevator. The moment arms at E and G are 6 in. The angle of the cable EC with the horizontal is ˛ D tan1 12 D 5.734° . 119.5 Denote the horizontal and vertical components of the force on point E by FX and FY . The sum of the moments about the pinned support on the member EG is 2 in FX α C A TEC 2.5 in F 3.5 in E FY D TEC α C 6 in 120 in-lb 2.5 in MEG D 2.5FY C 6FX 120 D 0. This is the tension in the cable EC. Noting that FX D TEC cos ˛, and FY D TEC sin ˛, then TEC D 120 . 2.5 sin ˛ C 6 cos ˛ The sum of the moments about the pinned support BC is MBC D 2TEC sin ˛ C 6TEC cos ˛ 2.5F D 0. Substituting: FD 120 2.5 6 cos ˛ 2 sin ˛ 6 cos ˛ C 2.5 sin ˛ D 480.9277 D 44.53 lb. The sum of the forces about the pinned joint A: Fx D Ax F C TEC cos ˛ D 0 from which Ax D 25.33 lb, Fy D Ay C TEC sin ˛ D 0 from which Ay D 1.93 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 311 Problem 5.65 In Example 5.4 suppose that ˛ D 40° , d D 1 m, a D 200 mm, b D 500 mm, R D 75 mm, and the mass of the luggage is 40 kg. Determine F and N. Solution: (See Example 5.4.) The sum of the moments about the center of the wheel: MC D dF cos ˛ C aW sin ˛ bW cos ˛ D 0, from which F D b a tan ˛W D 130.35 N. d The sum of the forces: FY D N W C F D 0, F from which N D 262.1 N d A d b h a a W α R C W b F R h a C N Problem 5.66 In Example 5.4 suppose that ˛ D 35° , d D 46 in, a D 10 in, b D 14 in, R D 3 in, and you don’t want the user to have to exert a force F larger than 20 lb. What is the largest luggage weight that can be placed on the carrier? N Solution: (See Example 5.4.) From the solution to Problem 5.65, the force is FD b a tan ˛W . d Solve for W: WD Fd . b a tan ˛ For F D 20 lb, W D 131.47 D 131.5 lb Problem 5.67 One of the difficulties in making design decisions is that you don’t know how the user will place the luggage on the carrier in Example 5.4. Suppose you assume that the point where the weight acts may be anywhere within the “envelope” R a 0.75c and 0 b 0.75d. If ˛ D 30° , c D 14 in, d D 48 in, R D 3 in, and W D 80 lb, what is the largest force F the user will have to exert for any luggage placement? Solution: (See Example 5.4.) From the solution to Problem 5.65, the force is FD b a tan ˛W . d The force is maximized as b ! 0.75d, and a ! R. Thus FMAX D 312 0.75d R tan ˛W D 57.11 lb d c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.68 In our design of the luggage carrier in Example 5.4, we assumed a user that would hold the carrier’s handle at h D 36 in above the floor. We assumed that R D 3 in, a D 6 in, and b D 12 in, and we chose the dimension d D 4 ft. The resulting ratio of the force the user must exert to the weight of the luggage is F/W D 0.132. Suppose that people with a range of heights use this carrier. Obtain a graph of F/W as a function of h for 24 h 36 in. Solution: (See Example 5.4.) From the solution to Problem 5.67, the force that must be exerted is FD b a tan ˛W , d b a tan ˛ F D . W d from which The angle a is given by ˛ D sin1 hR d . F / W , d i m e n s i o n l e x e F/W versus height .2 .19 .18 .17 .16 .15 .14 .13 24 26 28 30 32 height h, in 34 36 The commercial package TK Solver Plus was used to plot a graph of F as a function of h. W Problem 5.69 (a) Draw the free-body diagram of the beam and show that it is statically indeterminate. (See Active Example 5.5.) (b) Determine as many of the reactions as possible. 20 N-m A 800 mm Solution: (a) The free body diagram shows that there are four unknowns, whereas only three equilibrium equations can be written. (b) The sum of moments about A is MA D M C 1.1BY D 0, from which BY D 20 D 18.18 N. 1.1 The sum of forces in the vertical direction is 300 mm 20 N-m A 800 mm B B 300 mm BX AX AY 800 mm 300 mm BY FY D AY C BY D 0, from which AY D BY D 18.18 N. The sum of forces in the horizontal direction is FX D AX C BX D 0, from which the values of AX and BX are indeterminate. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 313 Problem 5.70 Consider the beam in Problem 5.69. Choose supports at A and B so that it is not statically indeterminate. Determine the reactions at the supports. Solution: One possibility is shown: the pinned support at B is replaced by a roller support. The equilibrium conditions are: 20 N-m A B FX D AX D 0. 800 mm The sum of moments about A is AX MA D M C 1.1BY D 0, from which BY D 300 mm 20 N-m AY 800 mm 300 mm BY 20 D 18.18 N. 1.1 The sum of forces in the vertical direction is FY D AY C BY D 0, from which AY D BY D 18.18 N. Problem 5.71 (a) Draw the free-body diagram of the beam and show that it is statically indeterminate. (The external couple M0 is known.) M0 A B (b) By an analysis of the beam’s deflection, it is determined that the vertical reaction B exerted by the roller support is related to the couple M0 by B D 2M0 /L. What are the reactions at A? Solution: (a) C L Eqn (3) and Eqn (4) yield FX : AX D 0 (1) MA D MO 2MO FY : AY C B D 0 (2) MA D MO MA : MA MO C BL D 0 (3) MA was assumed counterclockwise Unknowns: MA , AX , AY , B. 3 Eqns in 4 unknowns MA D jMO j clockwise AX D 0 AY D 2MO /L ∴ Statistically indeterminate (b) Given B D 2MO /L (4) We now have 4 eqns in 4 unknowns and can solve. Eqn (1) yields AX D 0 AY MO MA AX L B Eqn (2) and Eqn (4) yield AY D 2MO /L 314 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.72 Consider the beam in Problem 5.71. Choose supports at A and B so that it is not statically indeterminate. Determine the reactions at the supports. Solution: This result is not unique. There are several possible answers FX : A MO L AX D 0 FY : AY C BY D 0 MA : Mo C BL D 0 O MO AX AX D 0 L B AY B D MO /L AY D MO /L Problem 5.73 Draw the free-body diagram of the L-shaped pipe assembly and show that it is statically indeterminate. Determine as many of the reactions as possible. B 80 N A Strategy: Place the coordinate system so that the x axis passes through points A and B. 300 mm Solution: The free body diagram shows that there are four reactions, hence the system is statically indeterminate. The sum of the forces: and FX D AX C BX D 0, FY D AY C BY C F D 0. 300 mm 100 N-m 700 mm The moment about the point A is MA D LBN 0.3F C M D 0, from which BN D 76 M C 0.3F D D 99.79 N, L 0.76157 from which A strategy for solving some statically indeterminate problems is to select a coordinate system such that the indeterminate reactions vanish from the sum of the moment equations. The choice here is to locate the x axis on a line passing through both A and B, with the origin at A. Denote the reactions at A and B by AN , AP , BN , and BP , where the subscripts indicate the reactions are normal to and parallel to the new x axis. Denote The sum of the forces normal to the new axis is F D 80 N, The reactions parallel to the new axis are indeterminate. FN D AN C BN C F cos D 0, from which AN D BN F cos D 26.26 lb M D 100 N-m. The length from A to B is LD p 0.32 C 0.72 D 0.76157 m. The angle between the new axis and the horizontal is D tan1 0.3 0.7 BN 80 N AN D 23.2° . 300 mm BP 300 mm AP 100 N-m 700 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 315 Problem 5.74 Consider the pipe assembly in Problem 5.73. Choose supports at A and B so that it is not statically indeterminate. Determine the reactions at the supports. Solution: This problem has no unique solution. Problem 5.75 State whether each of the L-shaped bars shown is properly or improperly supported. If a bar is properly supported, determine the reactions at its supports. (See Active Example 5.6.) F C –12 L F L –1 L 2 Solution: (1) is properly constrained. The sum of the forces FX D F C BX D 0, A B L (1) (2) FY D BY C Ay D 0, –12 L F –12 L A MB D LAY C LF D 0, B 45° from which AY D F, and By D F (2) is improperly constrained. The reactions intersect at B, while the force produces a moment about B. (3) is properly constrained. The forces are neither concurrent nor parallel. The sum of the forces: 45° C 45° from which By D Ay . The sum of the moments about B: B L from which BX D F. A L (3) FX D C cos 45° B A cos 45° C F D 0. FY D C sin 45° A sin 45° D 0 from which A D C. The sum of the moments about A: MA D 21 LF C LC cos 45° C LC sin 45° D 0, F F from which C D p . Substituting and combining: A D p , 2 2 2 2 F BD 2 316 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.76 State whether each of the L-shaped bars shown is properly or improperly supported. If a bar is properly supported, determine the reactions at its supports. (See Active Example 5.6.) C C –12 L F F –12 L –12 L Solution: (1) (2) (3) is improperly constrained. The reactions intersect at a point P, and the force exerts a moment about that point. is improperly constrained. The reactions intersect at a point P and the force exerts a moment about that point. is properly constrained. The sum of the forces: A L 45° (2) (1) C FX D C F D 0, –12 L F –12 L from which C D F. A B FY D A C B D 0, from which A D B. The sum of the moments about B: LA C L 1 1 F LC D 0, from which A D F, and B D F 2 2 2 B A B L –12 L L (3) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 317 Problem 5.77 The bar AB has a built-in support at A and is loaded by the forces y A FB D 2i C 6j C 3k (kN), FB z FC D i 2j C 2k (kN). B 1m (a) Draw the free-body diagram of the bar. (b) Determine the reactions at A. C 1m Strategy: (a) Draw a diagram of the bar isolated from its supports. Complete the free-body diagram of the bar by adding the two external forces and the reactions due to the built-in support (see Table 5.2). (b) Use the scalar equilibrium equations (5.16)–(5.21) to determine the reactions. FC Solution: x AY MA = MAX i + MAY j + MAZ K AX MA D MAX i C MAY j C MAZ k FB (b) Equilibrium Eqns (Forces) 1m FX : AX C FBX C FCX D 0 FY : AY C FBY C FCY D 0 FZ : AZ B C 1m x FC AZ C FBZ C FCZ D 0 Equilibrium Equations (Moments) Sum moments about A rAB ð FB D 1i ð 2i C 6j C 3k kN-m rAB ð FB D 3j C 6k (kN-m) rAC ð FC D 2i ð 1i 2j C 2k kN-m rAC ð FC D 4j 4k (kN-m) x: y: z: MA : M AX D 0 MA : M AY 3 4 D 0 MA : M AZ C 6 4 D 0 Solving, we get AX D 3 kN, AY D 4 kN, AZ D 5 kN MAx D 0, MAy D 7 kN-m, MAz D 2 kN-m 318 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.78 The bar AB has a built-in support at A. The tension in cable BC is 8 kN. Determine the reactions at A. A z C (3,0.5,–0.5)m 2m B x Solution: AY MA = MAX i + MAY j + MAZ K AX MA D MAx i C MAy j C MAz k AZ We need the unit vector eBC eBC 2m xC xB i C yC yB j C zC zB k D xC xB 2 C yC yB 2 C zC zB 2 TBC C (3, 0.5, −0.5) B (2, 0, 0) x eBC D 0.816i C 0.408j 0.408k TBC D 8 kNeBC TBC D 6.53i C 3.27j 3.27k (kN) The moment of TBC about A is MBC D rAB ð TBC i D 2 6.53 j 0 3.27 k 0 3.27 MBC D rAB ð TBC D 0i C 6.53j C 6.53k (kN-m) Equilibrium Eqns. FX : AX C TBCX D 0 FY : AY C TBCY D 0 FZ : AZ C TBCZ D 0 MX : MAX C MBCX D 0 MY : MAY C MBCY D 0 MZ : MAZ C MBCZ D 0 Solving, we get AX D 6.53 (kN), AY D 3.27 (kN), AZ D 3.27 (kN) MAx D 0, MAy D 6.53 (kN-m), MAz D 6.53 (kN-m) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 319 y Problem 5.79 The bar AB has a fixed support at A. The collar at B is fixed to the bar. The tension in the rope BC is 300 lb. (a) Draw the free-body diagram of the bar. (b) Determine the reactions at A. B (6, 6, 2) ft A x C (8, 0, 3) ft z Solution: (a) The free-body diagram is shown. (b) We need to express the force exerted by the rope in terms of its components. The vector from B to C is rBC D [8 6i C 0 6j C 3 2k] ft D 2i 6j C k ft The force in the rope can now be written T D TBC rBC D TBC 0.312i 0.937j C 0.156k jrBC j The equilibrium equations for the bar are Fx : Ax C 0.312TBC D 0 Fy : Ay 0.937TBC D 0 Fz : Az C 0.156TBC D 0 i MA : MAx i C MAy j C MAz k C 6 ft 0.312TBC j 6 ft 0.937TBC k 2 ft D 0 0.156TBC These last equations can be written as MAx D 0.218 ftTBC , MAy D 0.312 ftTBC , MAz D 7.50 ftTBC Setting TBC D 300 lb, expanding and solving we have Ax D 93.7 lb, Ay D 281 lb, Az D 46.9 lb MAx D 843 ft-lb, MAy D 93.7 ft-lb, MAz D 2250 ft-lb 320 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.80 The bar AB has a fixed support at A. The collar at B is fixed to the bar. Suppose that you don’t want the support at A to be subjected to a couple of magnitude greater than 3000 ft-lb. What is the largest allowable tension in the rope BC? B (6, 6, 2) ft Solution: See the solution to Problem 5.79. The magnitude of the couple at A can be expressed in terms of the tension in the rope as A 2 MAx C MAy C MAz jMA j D D 2 2 x 2.81 ft2 C 0.312 ft2 C 7.50 ft2 TBC Setting jMA j D 3000 ft-lb and solving for TBC yields TBC D 374 lb C (8, 0, 3) ft z Problem 5.81 The total force exerted on the highway sign by its weight and the most severe anticipated winds is F D 2.8i 1.8j (kN). Determine the reactions at the fixed support. y F 8m Solution: The applied load is F D 2.8i 1.8j kN applied at r D 8j C 8k m The force reaction at the base is 8m O x R D Ox i C Oy j C Oz k The moment reaction at the base is MO D MOx i C MOy j C MOz k z For equilibrium we need Fx : 2.8 kN C Ox D 0 FDFCRD0) Fy : 1.8 kN C Oy D 0 Fz : 0 C Oz D 0 Ox D 2.8 kN ) Oy D 1.8 kN Oz D 0 Mx : 14.4 kN-m C MOx D 0 M D r ð F C MO D 0 ) My : 22.4 kN-m C MOy D 0 Mz : 22.4 kN-m C MOz D 0 MOx D 14.4 kN-m ) MOy D 22.4 kN-m MOz D 22.4 kN-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 321 Problem 5.82 The tension in cable AB is 800 lb. Determine the reactions at the fixed support C. y 4 ft Solution: The force in the cable is F D 800 lb 2i 4j k p 21 C 5 ft We also have the position vector 4 ft rCA D 4i C 5k ft A x The force reaction at the base is R D Cx i C Cy j C Cz k B The moment reaction at the base is z (6, 0, 4) ft MC D MCx i C MCy j C MCz k For equilibrium we need Fx : Cx C 349 lb D 0 FDFCRD0) Fy : Cy 698 lb D 0 Fz : Cz 175 lb D 0 Cx D 349 lb ) Cy D 698 lb Cz D 175 lb Mx : MCx C 3490 ft-lb D 0 MDrCFCRD0) My : MCy C 2440 ft-lb D 0 Mz : MCz 2790 ft-lb D 0 MCx D 3490 ft-lb ) MCy D 2440 ft-lb MCz D 2790 ft-lb 322 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.83 The tension in cable AB is 24 kN. Determine the reactions in the built-in support D. 2m C A 2m Solution: The force acting on the device is D B F D FX i C FY j C FZ k D 24 kNeAB , 3m and the unit vector from A toward B is given by eAB D 1m 1i 2j C 1k p . 6 The force, then, is given by F D 9.80i 19.60j C 9.80k kN. The position from D to A is r D 2i C 2j C 0k m. The force equations of equilibrium are DX C FX D 0, DY C FY D 0, and DZ C FZ D 0. The moment equation, in vector form, is M D MD C r ð F. Expanded, we get i M D MDX i C MDY j C MDZ k C 2 9.80 j 2 19.60 k 0 D 0. 9.80 The corresponding scalar equations are MDX C 29.80 D 0, MDY 29.80 D 0, and MDZ C 219.60 29.80 D 0. Solving for the support reactions, we get DX D 9.80 kN, OY D 19.60 kN, OZ D 9.80 kN. MDX D 19.6 kN-m, MDY D 19.6 kN-m, and MDZ D 58.8 kN-m. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 323 Problem 5.84 The robotic manipulator is stationary and the y axis is vertical. The weights of the arms AB and BC act at their midpoints. The direction cosines of the centerline of arm AB are cos x D 0.174, cos y D 0.985, cos z D 0, and the direction cosines of the centerline of arm BC are cos x D 0.743, cos y D 0.557, cos z D 0.371. The support at A behaves like a built-in support. y 600 mm C 160 N B (a) What is the sum of the moments about A due to the weights of the two arms? (b) What are the reactions at A? 600 mm 200 N A z x Solution: Denote the center of mass of arm AB as D1 and that of BC as D2 . We need rAD , (a) rAB , and rBD2 . We now have the geometry determined and are ready to determine the moments of the weights about A. MW D rAD1 ð W1 C rAD2 ð W2 where i rAD1 ð W1 D 0.0522 0 To get these, use the direction cosines to get the unit vectors eAB and eBC . Use the relation e D cos X i C cos Y j C cos Z k j k 0.2955 0 200 0 rAD1 ð W1 D 10.44k N-m eAB D 0.174i C 0.985j C 0k and i rAD2 ð W2 D 0.3273 0 eBC D 0.743i C 0.557j 0.371k rAD1 D 0.3eAB m j 0.7581 160 k 0.1113 0 rAD2 ð W2 D 17.81i 52.37k rAB D 0.6eAB m Thus, MW D 17.81i 62.81k (N-m) rBC D 0.6eBC m rBD2 D 0.3eBC m (b) WAB D 200j N Equilibrium Eqns FX : AX D 0 WBC D 160j N Thus rAD1 D 0.0522i C 0.2955j m rAB D 0.1044i C 0.5910j m rBD2 D 0.2229i C 0.1671j 0.1113k m rBC D 0.4458i C 0.3342j 0.2226k m and rAD2 D rAB C rBD2 rAD2 D 0.3273i C 0.7581j 0.1113k m 324 FY : AY W1 W2 D 0 FZ : AZ D 0 Sum Moments about A : MA C MW D 0 MX : MAx 17.81 D 0 (N-m) MY : MAy C 0 D 0 MZ : MZ 62.81 D 0 (N-m) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 5.84 (Continued ) Thus: AX D 0, AY D 360 (N), AZ D 0, MAx D 17.81 (N-m), MAy D 0, MAz D 62.81 (N-m) C W2 W1 D2 B D1 MA MA = MAXi + MAYj + MAZk W1 = 200 N W2 = 160 N AX AZ AY c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 325 Problem 5.85 The force exerted on the grip of the exercise machine is F D 260i 130j (N). What are the reactions at the built-in support at O? 150 mm y F O 200 mm z 250 mm Solution: MO D MOx i C MOy j C MOz k OX rOP D 0.25i C 0.2j 0.15k z Equilibrium (Forces) 0.15 P m y MO OZ x OY 0.2 F = 260 i – 130 j (N) 5m 0.2 m x FX : OX C FX D OX C 260 D 0 (N) FY : OY C FY D OY 130 D 0 (N) FZ : OZ C FZ D OZ D 0 (N) Thus, OX D 260 N, OY D 130 N, OZ D 0 Summing Moments about O MX : MOX C MFX D 0 MY : MOY C MFY D 0 MZ : MOZ C MFZ D 0 where i MF D rOP ð F D 0.25 260 j 0.2 130 k 0.15 0 MF D 19.5i 39j 84.5k (N-m) and from the moment equilibrium eqns, MOX D 19.5 (N-m) MOY D 39.0 (N-m) MOZ D 84.5 (N-m) 326 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.86 In Active Example 5.7, suppose that cable BD is lengthened and the attachment point D moved form (0, 600, 400) mm to (0, 600, 600) mm. (The end B of bar AB remains where it is.) Draw a sketch of the bar and its supports showing cable BD in its new position. Draw the free-body diagram of the bar and apply equilibrium to determine the tensions in the cables and the reactions at A. 400 mm 1000 mm C B D 600 mm 600 mm A x ⫺200j (N) z Solution: The sketch and free-body diagram are shown. We must express the force exerted on the bar by cable BD in terms of its components. The vector from B to D is rBD D [0 1000i C 600 600j C 600 400k] mm D 1000i C 200j mm The force exerted by cable BD can be expressed as TBD rBD D TBD 0.981i C 0.196k jrBD j The equilibrium equations are Fx : Ax 0.981TBD D 0 Fy : Ay 200 N D 0 Fz : Az C 0.196TBD TBC D 0 i 1 MA : 0.981TBD i j k C 0.5 0.6 0.4 0 0.196TBD TBC 0 j 0.3 200 k 0.2 D 0 0 Expanding and solving these equations, we find Ax D 166.7 N, Ay D 200 N, Az D 66.7 N, TBC D 100 N, TBD D 170 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 327 Problem 5.87 The force F acting on the boom ABC at C points in the direction of the unit vector 0.512i 0.384j C 0.768k and its magnitude is 8 kN. The boom is supported by a ball and socket at A and the cables BD and BE. The collar at B is fixed to the boom. y 1.5 m 2m D E (a) Draw the free-body diagram of the boom. (b) Determine the tensions in the cables and the reactions at A. 1m 2m A B z 2m C 2m x Solution: F (a) The free-body diagram (b) We identify the following forces, position vectors, and reactions rAC D 4 mi, F D 8 kN0.512i 0.384j C 0.768k 2i C 2j C 1.5k p T D T BD BD 10.25 D 2 mi, 2i C j 2k TBE D TBE 3 rAB Az Ax TBE B Ay TBD C R D Ax i C Ay j C Az k Force equilibrium requires: F D R C TBD C TBE C F D 0. F In component form we have 2 2 TBD TBE D 0 Fx : Ax C 8 kN0.512 p 3 10.25 2 1 TBD C TBE D 0 Fy : Ay 8 kN0.384 C p 3 10.25 1.5 2 TBD TBE D 0 Fz : Az C 8 kN0.768 C p 3 10.25 Moment equilibrium requires: MA D rAB ð TBD C TBE C rAC ð F D 0. In components: Mx : 0 D 0 My : 8 kN0.7684 m p C TBD 2 m 2 TBE 2 m D 0 3 Mz : 8 kN0.3844 m C p C 1.5 10.25 2 10.25 TBD 2 m 1 TBE 2 m D 0 3 Solving five equations for the five unknowns we find Ax D 8.19 kN, Ay D 3.07 kN, Az D 6.14 kN, TBD D 0, TBE D 18.43 kN 328 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.88 The cables BD and BE in Problem 5.87 will each safely support a tension of 25 kN. Based on this criterion, what is the largest acceptable magnitude of the force F? Solution: We have the force and distances: rAC D 4 mi, F D F0.512i 0.384j C 0.768k rAB 2i C 2j C 1.5k p TBD D TBD 10.25 D 2 mi, 2i C j 2k TBE D TBE 3 The moment equations are 1.5 2 TBD 2 m C TBE 2 m D 0 My : F0.7684 m p 3 10.25 2 1 TBD 2 m C TBE 2 m D 0 Mz : F0.3844 m C p 3 10.25 Solving we find TBE D 2.304F, TBD D 0 Thus: 25 kN D 2.304F ) F D 10.85 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 329 Problem 5.89 The suspended load exerts a force F D 600 lb at A, and the weight of the bar OA is negligible. Determine the tensions in the cables and the reactions at the ball and socket support O. y C (0, 6, –10) ft A (8, 6, 0) ft B (0, 10, 4) ft –Fj x O Solution: From the diagram, the important points in this problem z are A (8, 6, 0), B (0, 10, 4), C (0, 6, 10), and the origin O (0, 0, 0) with all dimensions in ft. We need unit vectors in the directions A to B and A to C. Both vectors are of the form eAP D xP xA i C yP yA j C zP zA k, If we carry through these operations in the sequence described, we get the following vectors: where P can be either A or B. The forces in cables AB and AC are TAB D TAB eAB D TABX i C TABY j C TABZ k, and TAC D TAC eAB D TACX i C TACY j C TACZ k. eAB D 0.816i C 0.408j C 0.408k, eAC D 0.625i C 0j 0.781k, TAB D 387.1i C 193.5j C 193.5k lb, The weight force is jTAB j D 474.1 lb, F D 0i 600j C 0k, and the support force at the ball joint is TAC D 154.8i C 0j 193.5k lb, jTAC j D 247.9 lb, S D SX i C SY j C SZ k. The vector form of the force equilibrium equation (which gives three scalar equations) for the bar is MAB D rOA ð TAB D 1161i 1548j C 3871k ft-lb, MAC D rOA ð TAC D 1161i C 1548j C 929k ft-lb, TAB C TAC C F C S D 0. Let us take moments about the origin. The moment equation, in vector form, is given by and S D 541.9i C 406.5j C 0k lb MO D rOA ð TAB C rOA ð TAC C rOA ð F D 0, where rOA D 8i C 6j C 0k. The cross products are evaluated using the form i M D r ð H D 8 HX j 6 HY k 0 , HZ where H can be any of the three forces acting at point A. The vector moment equation provides another three equations of equilibrium. Once we have evaluated and applied the unit vectors, we have six vector equations of equilibrium in the five unknowns TAB , TAC , SX , SY , and SZ (there is one redundant equation since all forces pass through the line OA). Solving these equations yields the required values for the support reactions at the origin. 330 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.90 In Problem 5.89, suppose that the suspended load exerts a force F D 600 lb at A and bar OA weighs 200 lb. Assume that the bar’s weight acts at its midpoint. Determine the tensions in the cables and the reactions at the ball and socket support O. Solution: Point G is located at (4, 3, 0) and the position vector of G with respect to the origin is rOG D 4i C 3j C 0k ft. The weight of the bar is WB D 0i 200j C 0k lb, and its moment around the origin is MWB D 0i C 0j 800k ft-lb. The mathematical representation for all other forces and moments from Problem 5.89 remain the same (the numbers change!). Each equation of equilibrium has a new term reflecting the addition of the weight of the bar. The new force equilibrium equation is TAB C TAC C F C S C WB D 0. The new moment equilibrium equation is MO D rOA ð TAB C rOA ð TAC C rOA ð F C rOG ð WB D 0. As in Problem 5.89, the vector equilibrium conditions can be reduced to six scalar equations of equilibrium. Once we have evaluated and applied the unit vectors, we have six vector equations of equilibrium in the five unknowns TAB , TAC , SX , SY , and SZ (As before, there is one redundant equation since all forces pass through the line OA). Solving these equations yields the required values for the support reactions at the origin. If we carry through these operations in the sequence described, we get the following vectors: eAB D 0.816i C 0.408j C 0.408k, eAC D 0.625i C 0j 0.781k, TAB D 451.6i C 225.8j C 225.8k lb, jTAB j D 553.1 lb, TAC D 180.6i C 0j 225.8k lb, jTAC j D 289.2 lb, MAB D rOA ð TAB D 1355i 1806j C 4516k ft-lb, MAC D rOA ð TAC D 1354i C 1806j C 1084k ft-lb, and S D 632.3i C 574.2j C 0k lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 331 Problem 5.91 The 158,000-kg airplane is at rest on the ground (z D 0 is ground level). The landing gear carriages are at A, B, and C. The coordinates of the point G at which the weight of the plane acts are (3, 0.5, 5) m. What are the magnitudes of the normal reactions exerted on the landing gear by the ground? 21 m 6m B G A x C 6m y Solution: mg 3m FY D NL C NR C NF W D 0 MR D 3 mg C 21NF D 0 21 m Side View x R Solving, NF D 221.4 kN F (NL + NR) (1) NF Z NL C NR D 1328.6 kN (2) 0.5 m W FY D NR C NL C NF W D 0 (same equation as before) C MO D 0.5 W 6NR C 6NL D 0 (3) Front View y Solving (1), (2), and (3), we get 6 6 NF D 221.4 kN NR D 728.9 kN NF NR NL z NL D 599.7 kN 332 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.92 The horizontal triangular plate is suspended by the three vertical cables A, B, and C. The tension in each cable is 80 N. Determine the x and z coordinates of the point where the plate’s weight effectively acts. y A B C 0.3 m 0.4 m (x, 0, z) x z Solution: 80 N 80 N Mx : 240 Nz 80 N0.4 m D 0 Mz : 80 N0.3 m 240 Nx D 0 z 80 N x X Solving x D 0.1 m, z D 0.1333 m 240 N Z Problem 5.93 The 800-kg horizontal wall section is supported by the three vertical cables, A, B, and C. What are the tensions in the cables? B 7m Solution: All dimensions are in m and all forces are in N. Forces A, B, C, and W act on the wall at (0, 0, 0), (5, 14, 0), (12, 7, 0), and (4, 6, 0), respectively. All forces are in the z direction. The force equilibrium equation in the z direction is A C B C C W D 0. The moments are calculated from MB D rOB ð Bk, C A 7m 7m 6m 4m 8m mg MC D rOC ð Ck, and MG D rOG ð Wk. The moment equilibrium equation is MO D MB C MC C MG D 0. Carrying out these operations, we get A D 3717 N, B D 2596 N, C D 1534 N, and W D 7848 N. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 333 Problem 5.94 The bar AC is supported by the cable BD and a bearing at A that can rotate about the z axis. The person exerts a force F D 10j (lb) at C. Determine the tension in the cable and the reactions at A. y A x B C 8 in 14 in z (18, ⫺8, 7) in D Solution: The force in the cable is TBD D TBD 10i 8j C 7k p 213 Ax, MAx Az F = 10 lbj We have the following six equilibrium equations 10 TBD D 0 Fx : Ax C p 213 8 TBD C 10 lb D 0 Fy : Ay p 213 Ay MAy TBD 7 TBD D 0 Fz : Az C p 213 Mx : MAx D 0 My : MAy p 7 213 TBD 8 in D 0 8 TBD 8 in C 10 lb22 in D 0 Mz : p 213 Solving we find Ax D 34.4 lb, Ay D 17.5 lb, Az D 24.1 lb MAx D 0, MAy D 192.5 lb in, TBD D 50.2 lb 334 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.95 The L-shaped bar is supported by a bearing at A and rests on a smooth horizontal surface at B. The vertical force F D 4 kN and the distance b D 0.15 m. Determine the reactions at A and B. F b A x Solution: Equilibrium Eqns: B FX : 0.2 m OD0 0.3 m z FY : AY C B F D 0 FZ : AZ D 0 y Sum moments around A F b x: Fb 0.3B D 40.15 0.3B D 0 y: M AY D 0 z: MAZ C 0.2F 0.2B D 0 MZ 0.2 m A B 3 0. m x AY b = 0.15 m F = 4 kN AZ B z (MAX ≡ 0) AX ≡ 0 MY Solving, AX D 0, AY D 2 (kN), AZ D 0 MAX D 0, MAY D 0, MAZ D 0.4 (kN-m) Problem 5.96 In Problem 5.95, the vertical force F D 4 kN and the distance b D 0.15 m. If you represent the reactions at A and B by an equivalent system consisting of a single force, what is the force and where does its line of action intersect the xz plane? Solution: We want to represent the forces at A & B by a single zR 4 D C0.32 force. From Prob. 5.95 zR D C0.15 m A D C2j (kN), xR 4 D 0.22 B D C2j (kN) xR D 0.1 m MA D 0.4k (kN-m) y We want a single equivalent force, R that has the same resultant force and moment about A as does the set A, B, and MA . R F b A R D A C B D 4j (kN) x Let R pierce the xz plane at xR , zR B MX : zR R D 0.3B z MZ : 0.2 m 0.3 m xR R D 0.2AY c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 335 Problem 5.97 In Problem 5.95, the vertical force F D 4 kN. The bearing at A will safely support a force of 2.5-kN magnitude and a couple of 0.5 kN-m magnitude. Based on these criteria, what is the allowable range of the distance b? Solution: The solution to Prob. 5.95 produced the relations AY C B F D 0 F D 4 kN Fb 0.3B D 0 MAZ C 0.2F 0.2B D 0 AX D AZ D MAX D MAY D 0 Set the force at A to its limit of 2.5 kN and solve for b. In this case, MAZ D 0.5 (kN-m) which is at the moment limit. The value for b is b D 0.1125 m We make AY unknown, b unknown, and B unknown F D 4 kN, MAY D C0.5 (kN-m), and solve we get AY D 2.5 at b D 0.4875 m However, 0.3 is the physical limit of the device. Thus, 0.1125 m b 0.3 m y F b A x B z 336 0.2 m 0.3 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.98 The 1.1-m bar is supported by a ball and socket support at A and the two smooth walls. The tension in the vertical cable CD is 1 kN. (a) (b) y B Draw the free-body diagram of the bar. Determine the reactions at A and B. 400 mm D x C (a) (b) 600 mm 700 mm z Solution: A From which, The ball and socket cannot support a couple reaction, but can support a three force reaction. The smooth surface supports oneforce normal to the surface. The cable supports one force parallel to the cable. The strategy is to determine the moments about A, which will contain only the unknown reaction at B. This will require the position vectors of B and D relative to A, which in turn will require the unit vector parallel to the rod. The angle formed by the bar with the horizontal is required to determine the coordinates of B: ˛ D cos1 p 0.62 C 0.72 1.1 BZ D 0.3819 D 0.6365 kN, 0.6 BX D 0.4455 D 0.7425 kN. 0.6 The reactions at A are determined from the sums of the forces: D 33.1° . FX D BX C AX i D 0, from which AX D 0.7425 kN. FY D AY 1j D 0, from which AY D 1 kN. FZ D BZ C AZ k D 0, from which AZ D 0.6365 kN The coordinates of the points are: A (0.7, 0, 0.6), B 0, 1.1 sin 33.1° , 0 D 0, 0.6, 0, from which the vector parallel to the bar is FB rAB D rB rA D 0.7i C 0.6j 0.6k (m). The unit vector parallel to the bar is eAB T FY rAB D 0.6364i C 0.5455j 0.5455k. D 1.1 FZ The vector location of the point D relative to A is FX rAD D 1.1 0.4eAB D 0.7eAB D 0.4455i C 0.3819j 0.3819k. The reaction at B is horizontal, with unknown x-component and z-components. The sum of the moments about A is i MA D rAB ð B C rAD ð D D 0 D 0.7 BX i C 0.4455 0 j 0.3819 1 j 0.6 0 k 0.6 BZ k 0.3819 D 0 0 Expand and collect like terms: MA D 0.6BZ 0.3819i 0.6BX 0.7BZ j C0.6BX C 0.4455k D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 337 Problem 5.99 The 8-ft bar is supported by a ball and socket at A, the cable BD, and a roller support at C. The collar at B is fixed to the bar at its midpoint. The force F D 50k (lb). Determine the tension in the cable BD and the reactions at A and C. y A 3 ft Solution: The strategy is to determine the sum of the moments B F about A, which will involve the unknown reactions at B and C. This will require the unit vectors parallel to the rod and parallel to the cable. z The angle formed by the rod is 4 ft D 2 ft 3 D 22° . ˛ D sin1 8 C x The vector positions are: rA D 3j, from which jTj D rD D 4i C 2k CY D and rC D 8 cos 22° i D 7.4162i. 75 D 62.92 lb 1.192 2.036jTj D 17.27 lb. 7.4162 The reaction at A is determined from the sums of forces: The vector parallel to the rod is FX D AX C 0.1160jTji D 0, rAC D rC rA D 7.4162i 3j. from which AX D 7.29 lb, The unit vector parallel to the rod is eAC D 0.9270i 0.375j. FY D AY 0.5960jTj C CY j D 0, from which AY D 20.23 lb The location of B is rAB D 4eAC D 3.7081i 1.5j. FZ D AZ C 0.7946jTj 50k D 0, from which AZ D 0 lb The vector parallel to the cable is rBD D rD rA C rAB D 0.2919i 1.5j C 2k. y The unit vector parallel to the cable is A eBD D 0.1160i 0.5960j C 0.7946k. F B 3 ft The tension in the cable is T D jTjeBD . The reaction at the roller support C is normal to the xz plane. The sum of the moments about A C D z 4 ft MA D rAB ð F C rAB ð T C rAC ð C D 0 i D 3.7081 0 j k 1.5 0 0 50 i C jTj 3.7081 0.1160 i C 7.4162 0 2 ft j 1.5 0.5960 x y k 0 0.7946 j k 3 0 D 0 CY 0 AY AZ AX T F z D D 75i C 185.4j C jTj1.192i 2.9466j 2.036k CY x C 7.4162CY k D 0, 338 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.100 Consider the 8-ft bar in Problem 5.99. The force F D Fy j 50k (lb). What is the largest value of Fy for which the roller support at C will remain on the floor? Solution: From the solution to Problem 5.99, the sum of the moments about A is j k 1.5 0 FY 50 i MA D 3.7081 0 i C jTj 3.7081 0.1160 i C 7.4162 0 j 1.5 0.5960 k 0 0.7946 j k 3 0 D 0 CY 0 D 75i C 185.4j C 3.7081FY k C jTj1.192i 2.9466j 2.036k C 7.4162CY k D 0, from which, jTj D 75 D 62.92 lb. 1.192 Collecting terms in k, 3.7081FY C 2.384jTj 7.4162CY D 0. For CY D 0, FY D 128.11 D 34.54 lb 3.708 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 339 Problem 5.101 The tower is 70 m tall. The tension in each cable is 2 kN. Treat the base of the tower A as a built-in support. What are the reactions at A? y B Solution: The strategy is to determine moments about A due to the cables. This requires the unit vectors parallel to the cables. C 40 m The coordinates of the points are: A0, 0, 0, B0, 70, 0, C50, 0, 0, 50 m E A D20, 0, 50, E40, 0, 40. 40 m D The unit vectors parallel to the cables, directed from B to the points E, D, and C 50 m z 20 m x rBE D 40i 70j 40k, The force reactions at A are determined from the sums of forces. (Note that the sums of the cable forces have already been calculated and used above.) rBD D 20i 70j C 50k, rBC D 50i 70j. FX D AX C 0.17932i D 0, The unit vectors parallel to the cables, pointing from B, are: from which AX D 0.179 kN, eBE D 0.4444i 0.7778j 0.4444k, FY D AY 4.7682j D 0, eBD D 0.2265i 0.7926j C 0.5661k, from which AY D 4.768 kN, eBC D 0.5812i 0.8137j C 0k. The tensions in the cables are: TBD D 2eBD D 0.4529i 1.5852j C 1.1323k (kN), FZ D AZ C 0.2434k D 0, from which AZ D 0.2434 kN TBE D 2eBE D 0.8889i 1.5556j 0.8889k (kN), y B TBC D 2eBC D 1.1625i 1.6275j 0k. TBC The sum of the moments about A is TBD MA D MA C rAB ð TBE AY , C A MY rAB ð TBD C rAB ð TBC D 0 C D MA C rAB ð TBE C TBC C TBD i A MA D M C 0 0.1793 TBE j 70 4.7682 k 0 D 0 0.2434 AZ ,MA EA AX , MA Z z D X x D MAX C 17.038i C MAY C 0j C MAZ 12.551k D 0 from which MAX D 17.038 kN-m, MAY D 0, MAZ D 12.551 kN-m. 340 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.102 Consider the tower in Problem 5.101. If the tension in cable BC is 2 kN, what must the tensions in cables BD and BE be if you want the couple exerted on the tower by the built-in support at A to be zero? What are the resulting reactions at A? Solution: From the solution to Problem 5.101, the sum of the moments about A is given by MA D MA C rAB ð TBE C TBC C TBD D 0. If the couple MA D 0, then the cross product is zero, which is possible only if the vector sum of the cable tensions is zero in the x and z directions. Thus, from Problem 5.101, ex Ð TBC C jTBE jeBE C jTBD jeBD D 0, and ez Ð TBC C jTBE jeBE C jTBD jeBD D 0. Two simultaneous equations in two unknowns result; 0.4444jTBE j C 0.2265jTBD j D 1.1625 0.4444jTBE j C 0.5661jTBD j D 0. Solve: jTBE j D 1.868 kN, jTBD j D 1.467 kN. The reactions at A oppose the sum of the cable tensions in the x-, y-, and z-directions. AX D 0, AY D 4.243 kN, AZ D 0. (These results are to be expected if there is no moment about A.) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 341 Problem 5.103 The space truss has roller supports at B, C, and D and is subjected to a vertical force F D 20 kN at A. What are the reactions at the roller supports? y F A (4, 3, 4) m B Solution: The key to this solution is expressing the forces in terms D (6, 0, 0) m of unit vectors and magnitudes-then using the method of joints in three dimensions. The points A, B, C, and D are located at x A4, 3, 4 m, B0, 0, 0 m, C5, 0, 6 m, D6, 0, 0 m z C (5, 0, 6) m we need eAB , eAC , eAD , eBC , eBD , and eCD . Use the form ePQ D xQ xP i C yQ yP j C zQ zP k [xQ xP 2 C yQ yP 2 C zQ zP 2 ]1/2 F Joint A : eAB D 0.625i 0.469j 0.625k TAD TAB eAC D 0.267i 0.802j C 0.535k eAD D 0.371i 0.557j 0.743k TAC D B eBC D 0.640i C 0j C 0.768k C eBD D 1i C 0j C 0k eCD D 0.164i C 0j 0.986k Joint B : –TAB We will write each force as a magnitude times the appropriate unit vector. TAB D TAB eAB , TAC D TAC eAC TBD NBJ TBC TAD D TAD eAD , TBC D TBC eBC Joint C : TBD D TBD eBD , TCD D TCD eCD –TAC Each force will be written in component form, i.e. TABX D TAB eABX TABY D TAB eABY etc. TABZ D TAB eABZ Joint A: TAB C TAC C TAD C F D 0 TABX C TACX C TADX D 0 TABY C TACY C TADY 20 D 0 TCD –TBC NCJ Joint D : –TAD –TBD –TCD NDJ TABZ C TACZ C TADZ D 0 Joint B: TAB C TBC C TBD C NB j D 0 Joint C: TAC TBC C TCD C NC j D 0 Joint D: TAD TBD TCD C ND j D 0 Solving for all the unknowns, we get NB D 4.44 kN NC D 2.22 kN ND D 13.33 kN Also, TAB D 9.49 kN, TAC D 16.63 kN TAD D 3.99 kN, TBC D 7.71 kN TBD D 0.99 kN, TCD D 3.00 kN 342 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.104 In Example 5.8, suppose that the cable BD is lengthened and the attachment point B is moved to the end of the bar at C. The positions of the attachment point D and the bar are unchanged. Draw a sketch of the bar showing cable BD in its new position. Draw the free-body diagram of the bar and apply equilibrium to determine the tension in the cable and the reactions at A. D (2, 2, ⫺1) ft x A 30⬚ B z C ⫺100j (lb) Solution: The sketch and free-body diagram are shown. We must express the force exerted on the bar by cable BD in terms of its components. The bar AC is 4 ft long. The vector from C to D is rCD D [2 4 cos 30° i C 2 f4 sin 30° gj C 1 0j] ft rCD D 1.46i C 4j k ft The force exerted by the cable CD can be expressed T rCD D T0.335i C 0.914j 0.229k jrCD j The equilibrium equations are Fx : Ax 0.335T D 0 Fy : Ay C 0.914T 100 lb D 0 Fz : Az 0.229T D 0 MA : MAx i C MAy j i C 3.464 0.335T j 2 0.914T 100 k D0 0 0.229T Expanding the determinant and solving the six equations, we obtain T D 139 lb, Ax D 46.4 lb, Ay D 26.8 lb, Az D 31.7 lb MAx D 63.4 ft-lb, MAy D 110 ft-lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 343 Problem 5.105 The 40-lb door is supported by hinges at A and B. The y axis is vertical. The hinges do not exert couples on the door, and the hinge at B does not exert a force parallel to the hinge axis. The weight of the door acts at its midpoint. What are the reactions at A and B? y 4 ft 1 ft B Solution: The position vector of the midpoint of the door: 5 ft rCM D 2 cos 50° i C 3.5j C 2 cos 40° k A D 1.2856i C 3.5j C 1.532k. 1 ft x The position vectors of the hinges: 40 rA D j, rB D 6j. z The forces are: W D 40j, y A D AX i C AY j C AZ k, BX B D BX i C BZ k. BZ The position vectors relative to A are AY rACM D rCM rA D 1.2856i C 2.5j C 1.532k, rAB D rB rA D 5j. AZ The sum of the moments about A AX x z MA D rACM ð W C rAB ð B i D 1.2856 0 W j 2.5 40 k i 1.532 C 0 0 BX j k 5 0 D 0 0 BZ MA D 5BZ C 401.532i C 5BX 401.285k D 0, from which BZ D 401.532 D 12.256 lb 5 and 401.285 D 10.28 lb. 5 BX D The reactions at A are determined from the sums of forces: FX D AX C BX i D 0, from which AX D 10.28 lb, FY D AY 40j D 0, from which AY D 40 lb, FZ D AZ C BZ k D 0, from which AZ D 12.256 lb 344 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.106 The vertical cable is attached at A. Determine the tension in the cable and the reactions at the bearing B due to the force F D 10i 30j 10k (N). y 200 mm 100 mm 100 mm Solution: The position vector of the point of application of the force is B 200 mm rF D 0.2i 0.2k. F z The position vector of the bearing is x A rB D 0.1i. The position vector of the cable attachment to the wheel is rC D 0.1k. he position vectors relative to B are: rBC D rC rB D 0.1i C 0.1k, rBF D rF rB D 0.1i 0.2k. The sum of the moments about the bearing B is or MB D MB C rBF ð F C rBC ð C D 0, i MB D MB C 0.1 10 j j k i 0 0.2 C 0.1 0 T 30 10 0 k 0.1 0 D 6 C 0.1Ti C MBY 1j C MBZ 3 C 0.1Tk D 0, from which T D 6 D 60 N, 0.1 MBY D C1 N-m, MBZ D 0.1T C 3 D 3 N-m. The force reactions at the bearing are determined from the sums of forces: FX D BX C 10i D 0, from which BX D 10 N. FY D BY 30 60j D 0, from which BY D 90 N. FZ D BZ 10j D 0, from which BZ D 10 N. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 345 Problem 5.107 In Problem 5.106, suppose that the z component of the force F is zero, but otherwise F is unknown. If the couple exerted on the shaft by the bearing at B is MB D 6j 6k N-m, what are the force F and the tension in the cable? Solution: From the diagram of Problem 5.106, the force equilibrium equation components are and Fx D BX C FX D 0, Fy D BY C FY D 0, Fz D BZ C FZ D 0, where FZ D 0 is given in the problem statement. The moment equations can be developed by inspection of the figure also. They are and Mx D MBX C MAX C MFX D 0, MY D MBY C MAY C MFY D 0, MZ D MBZ C MAZ C MFZ D 0, where MB D 6j 6k N-m. Note that MBX D 0 can be inferred. The moments which need to be substituted into the moment equations are MA D 0.1Ai C 0j C 0.1Ak N-m, and MF D 0.2FY i 0.2FX j C 0.1FY k N-m. Substituting these values into the equilibrium equations, we get F D 30i 60j C 0k N, and A D 120 N. 346 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.108 The device in Problem 5.106 is badly designed because of the couples that must be supported by the bearing at B, which would cause the bearing to “bind”. (Imagine trying to open a door supported by only one hinge.) In this improved design, the bearings at B and C support no couples, and the bearing at C does not exert a force in the x direction. If the force F D 10i 30j 10k (N), what are the tension in the vertical cable and the reactions at the bearings B and C? 200 mm 50 mm 100 mm 50 mm 200 mm B C F z x A Solution: The position vectors relative to the bearing B are: the position vector of the cable attachment to the wheel is y rBT D 0.05i C 0.1k. BY The position vector of the bearing C is: BX CY BZ F CZ z rBC D 0.1i. T x The position vector of the point of application of the force is: rBF D 0.15i 0.2k. The sum of the moments about B is MB D rBT ð T C rBC ð C C rBF ð F D 0 i j MB D 0.05 0 0 T i C 0.15 10 j 0 30 j k i 0.1 C 0.1 0 0 0 CY k 0 CZ k 0.2 D 0 10 MB D 0.1T 6i C 0.1CZ C 1.5 2j C 0.05T C 0.1CY 4.5k D 0. From which: T D 60 N, CZ D 0.5 D 5 N, 0.1 CY D 4.5 0.05T D 15 N. 0.1 The reactions at B are found from the sums of forces: FX D BX C 10i D 0, from which BX D 10 N. FY D BY C CY T 30j D 0, from which BY D 75 N. FZ D BZ C CZ 10k D 0, from which BZ D 15 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 347 Problem 5.109 The rocket launcher is supported by the hydraulic jack DE and the bearings A and B. The bearings lie on the x axis and support shafts parallel to the x axis. The hydraulic cylinder DE exerts a force on the launcher that points along the line from D to E. The coordinates of D are (7, 0, 7) ft, and the coordinates of E are (9, 6, 4) ft. The weight W D 30 kip acts at (4.5, 5, 2) ft. What is the magnitude of the reaction on the launcher at E? y W E A B x D 3 ft 3 ft Solution: The position vectors of the points D, E and W are rD D 7i C 7k, rE D 9i C 6j C 4k (ft), rW D 4.5i C 5j C 2k (ft). The vector parallel to DE is rDE D rE rD D 2i C 6j 3k. The unit vector parallel to DE is eDE D 0.2857i C 0.8571j 0.4286k. Since the bearings cannot exert a moment about the x axis, the sum of the moments due to the weight and the jack force must be zero about the x axis. The sum of the moments about the x axis is: 1 1 0 0 5 2 CjFDE j 9 MX D 4.5 0 30 0 0.2857 0 0 D0 6 4 0.8571 0.4286 D 60 6jFDE j D 0. From which jFDE j D 60 D 10 kip 6 y 30 kip BY AY AX AZ BZ x FDE z 348 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.110 Consider the rocket launcher described in Problem 5.109. The bearings at A and B do not exert couples, and the bearing B does not exert a force in the x direction. Determine the reactions at A and B. Solution: See the solution of Problem 5.109. The force FDE can be written FDE D FDE 0.2857i C 0.8571j 0.4286k. The equilibrium equations are FX D AX C 0.2857FDE D 0, FY D AY C BY C 0.8571FDE 30 D 0, FZ D AZ C BZ 0.4286FDE D 0, Morigin i D 3 AX j 0 AY k i 0 C 6 AZ 0 i C FDE 7 0.2857 j 0 BY j 0 0.8571 k 0 BZ k 7 0.4286 i j k C 4.5 5 2 D 0 0 30 0 The components of the moment eq. are 5.9997FDE C 60 D 0, 3AZ 6BZ C 5.0001FDE D 0, 3AY C 6BY C 5.9997FDE 135 D 0. Solving, we obtain FDE D 10.00 kip, AX D 2.86 kip, AY D 17.86 kip, AZ D 8.09 kip, BY D 3.57 kip, BZ D 12.38 kip. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 349 Problem 5.111 The crane’s cable CD is attached to a stationary object at D. The crane is supported by the bearings E and F and the horizontal cable AB. The tension in cable AB is 8 kN. Determine the tension in the cable CD. Strategy: Since the reactions exerted on the crane by the bearings do not exert moments about the z axis, the sum of the moments about the z axis due to the forces exerted on the crane by the cables AB and CD equals zero. (See the discussion at the end of Example 5.9.) y Solution: The position vector from C to D is C rCD D 3i 6j 3k (m), A so we can write the force exerted at C by cable CD as B TCD rCD D TCD D TCD 0.408i 0.816j 0.408k. jrCD j The coordinates of pt. B are x D 4 3 D 2 m, y D 4 m. 6 F E z The moment about the origin due to the forces exerted by the two cables is i MO D 2 8 i j k 3 4 0 C 0 0 0.408TCD j 6 0.816TCD 2m k 0 0.408TCD 2m D 3m y x C D 32k 2.448TCD i C 1.224TCD j 4.896TCD k. A The moment about the z axis is B 6m k Ð MO D 32 4.896TCD D 0, so TCD D 6.54 kN. 4m D 3m 350 x 3m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.112 In Example 5.9, suppose that the cable CE is shortened and its attachment point E is moved to the point (0, 80, 0) mm. The plate remains in the same position. Draw a sketch of the plate and its supports showing the new position of cable CE. Draw the freebody diagram of the plate and apply equilibrium to determine the reactions at the hinges and the tension in the cable. 100 mm E 80 mm A B z C 200 mm D 200 mm x ⫺400j (N) Solution: The sketch and free-body diagram are shown. The vector from C to E is rCE D [0 200i C 80 0j C 0 0k] mm D 200i C 80j mm The force exerted by cable CE can be expresses as T rCE D T0.928i C 0.371j jrCE j The equilibrium equations for the plate are Fx : Ax C Bx 0.928T D 0 Fy : Ay C By C 0.371T 400 N D 0 Fz : Az C Bz D 0 i MB : 0.2 0.928T i C 0 Ax j 0 Ay j k 0 0 0.371T 0 j k i 0 0.2 C 0.2 0 0 400 k 0.2 D 0 0 Expanding and solving we find Ax D 0, Ay D 400 N, T D 1080 N Bx D 1000 N, By D 400 N, Bz D 0, c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 351 Problem 5.113 The plate is supported by hinges at A and B and the cable CE, and it is loaded by the force at D. The edge of the plate to which the hinges are attached lies in the yz plane, and the axes of the hinges are parallel to the line through points A and B. The hinges do not exert couples on the plate. What is the tension in cable CE? y 3m 2i – 6j (kN) E A D 2m 1m Solution: B z C 20° 2m F D A C B C FD C TCE D 0 However, we just want tension in CE. This quantity is the only unknown in the moment equation about the line AB. To get this, we need the unit vector along CE. y 3m Point C is at (2, 2 sin 20° , 2 cos 20° ) Point E is at (0, 1, 3) eCE AX A 1m z eCE D 0.703i C 0.592j C 0.394k BZ BY AZ B 20° FD = 2i – 6j AY E rCE D jrCE j We also need the unit vector eAB . A(0, 0, 0), B0, 2 sin 20° , 2 cos 20° x D x TCE 2m BX 2m C eAB D 0i 0.342j C 0.940k The moment of FD about A (a point on AB) is MFD D rAD ð FD1 D 2i ð 2i 6j MFD D 12k The moment of TCE about B (another point on line CE) is MTCE D rBC ð TCE eCE D 2i ð TCE eCE , where eCE is given above. The moment of FD about line AB is MFDAB D MFD Ð eAB MFDAB D 11.27 N-m The moment of TCE about line AB is MCEAB D TCE 2i ð eCE Ð eAB MCEAB D TCE 0.788j C 1.184k Ð eAB MCEAB D 1.382TCE The sum of the moments about line AB is zero. Hence MFDAB C MCEAB D 0 11.27 C 1.382TCE D 0 TCE D 8.15 kN 352 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.114 In Problem 5.113, the hinge at B does not exert a force on the plate in the direction of the hinge axis. What are the magnitudes of the forces exerted on the plate by the hinges at A and B? Solution: From the solution to Problem 5.113, TCE D 8.15 kN y Also, from that solution, F = + 2i – 6j (kN) AY eAB D 0i 0.342j C 0.940k AX We are given that the force at force at hinge B does not exert a force parallel to AB at B. This implies θ z BY B Ð eAB D 0. B Ð eAB D 0.342BY C 0.940BZ D 0 D AZ x TCE 2 BX C (2, –2sinθ , + 2cosθ) (1) BZ We also had, in the solution to Problem 5.113 eCE D 0.703i C 0.592j C 0.394k and TCE D TCE eCE (kN) For Equilibrium, F D A C B C TCE C F D 0 FX : AX C BX C TCE eCEX C 2 D 0 (kN) (2) FY : AY C BY C TCE eCEY 6 D 0 (kN) (3) FZ : AZ C BZ C TCE eCEZ D 0 (kN) (4) Summing Moments about A, we have rAD ð F C rAC ð TCE C rAB ð B D 0 rAD ð F D 2i ð 2i 6j D 12k (kN) rAC ð TCE D 2 sin TZ 2 cos TY i C 2 cos TX 2TZ j C 2TY C 2TX sin k rCE ð B D 2BZ sin 2BY cos i C 2BX cos j C C2BX sin k MA D 0, Hence x: 2 sin TZ 2 cos TY 2BZ sin 2BY cos D 0 (5) y: 2 cos TX 2TZ C 2BX cos D 0 (6) z: 12 C 2TY C 2TX sin C 2BX sin D 0 (7) Solving Eqns (1)–(7), we get jAj D 8.53 (kN), jBj D 10.75 (kN) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 353 Problem 5.115 The bar ABC is supported by ball and socket supports at A and C and the cable BD. The suspended mass is 1800 kg. Determine the tension in the cable. (⫺2, 2, ⫺1) m y D 2m 4m B A C x 4m z Solution: We take moments about the line AC to eliminate the TBD reactions at A and C. We have rAB D 4 mk, TBD D TBD 2i C 2j k 3 Az rAW D 2i 4k m, W D 1800 kg9.81 m/s2 j eCA D 1 6i 4k p D p 3i 2k 52 13 Ax Cx Ay 17.66 kN The one equilibrium equation we need is MAC D eCA Ð rAB ð TAB C rAW ð W D 0 Cz Cy This equation reduces to the scalar equation 1 p 3i 2k Ð 13 8 8 m TBD i C m TBD j [4 m][17.66 kN]i 3 3 [2 m][17.66 kN]k D 0 8 1 p 3 m TBD [4 m][17.66 kN] C 2 f[2 m][17.66 kN]g D 0 3 2 Solving we find TBD D 17.66 kN 354 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.116* In Problem 5.115, assume that the ball and socket support at A is designed so that it exerts no force parallel to the straight line from A to C. Determine the reactions at A and C. Solution: We have TBD D TBD 2i C 2j k 3 There are 7 unknowns. We have the following 6 equilibrium equations Fx : Ax C Cx 2 TBD D 0 3 Fy : Ay C Cy C 2 TBD 17.66 kN D 0 3 Fz : Az C Cz 1 TBD D 0 3 Mx : Cy 4 m D 0 My : Cx 4 m Az 6 m D 0 Mz : Ay 6 m 17.66 kN2 m D 0 The last equation comes from the fact that the ball and socket at A exerts no force in the direction of the line from A to C 6i C 4k 1 p Ax i C Ay j C Az k Ð D p 6Ax C 4Az D 0 52 52 Solving these 7 equations we find Ax D 3.62 kN, Ay D 5.89 kN, Az D 5.43 kN Cx D 8.15 kN, Cy D 0, Cz D 0.453 kN TBD D 17.66 kN TBD Az Ax Cx Ay 17.66 kN Cz Cy c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 355 y Problem 5.117 The bearings at A, B, and C do not exert couples on the bar and do not exert forces in the direction of the axis of the bar. Determine the reactions at the bearings due to the two forces on the bar. 200 i (N) 300 mm x Solution: The strategy is to take the moments about A and solve C 180 mm the resulting simultaneous equations. The position vectors of the bearings relative to A are: B z rAB D 0.15i C 0.15j, A rAC D 0.15i C 0.33j C 0.3k. 100 k (N) Denote the lower force by subscript 1, and the upper by subscript 2: 150 mm rA1 D 0.15i, rA2 D 0.15i C 0.33j. y CY The sum of the moments about A is: i C 0.15 200 k i 0 C 0.15 100 BX j k 0.15 0 0 BZ j k i 0.33 0 C 0.15 0 0 CX j 0.33 CY k 0.3 D 0 0 200 N x MA D rA1 ð F1 C rAB ð B C rA2 ð F2 C rAC ð C D 0 i j MA D 0.15 0 0 0 150 mm z CX BX BZ 100 N AY AZ MA D 0.15BZ 0.3CY i C 15 C 0.15BZ C 0.3CX j C 0.15BX 66 0.15CY 0.33CX k D 0. This results in three equations in four unknowns; an additional equation is provided by the sum of the forces in the x-direction (which cannot have a reaction term due to A) FX D BX C CX C 200i D 0. The four equations in four unknowns: 0BX C 0.15BZ C 0CX 0.3CZ D 0 0BX C 0.15BZ C 0.3CX C 0CY D 15 0.15BX C 0BZ 0.33CX 0.15CY D 66 BX C 0BZ C CX C 0CZ D 200. (The HP-28S hand held calculator was used to solve these equations.) The solution: BX D 750 N, BZ D 1800 N, CX D 950 N, CY D 900 N. The reactions at A are determined by the sums of forces: 356 FY D AY C CY j D 0, from which AY D CY D 900 N FZ D AZ C BZ C 100k D 0, from which AZ D 1900 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.118 The support that attaches the sailboat’s mast to the deck behaves like a ball and socket support. The line that attaches the spinnaker (the sail) to the top of the mast exerts a 200-lb force on the mast. The force is in the horizontal plane at 15° from the centerline of the boat. (See the top view.) The spinnaker pole exerts a 50lb force on the mast at P. The force is in the horizontal plane at 45° from the centerline. (See the top view.) The mast is supported by two cables, the back stay AB and the port shroud ACD. (The forestay AE and the starboard shroud AFG are slack, and their tensions can be neglected.) Determine the tensions in the cables AB and CD and the reactions at the bottom of the mast. A A Spinnaker 50 ft C C F P P 6 ft E x B D Side View G D 15 ft 21 ft Aft View x z (Spinnaker not shown) Top View F 200 lb G 15° A B E P C 50 lb 45° D Solution: Although the dimensions are not given in the sketch, assume that the point C is at the midpoint of the mast (25 ft above the deck). The position vectors for the points A, B, C, D, and P are: (5) The force due to the spinnaker pole: rA D 50j, The sum of the moments about the base of the mast is FP D 500.707i C 0.707k D 35.35i C 35.35k. MQ D rA ð FA C rA ð TAB C rA ð TAC C rC ð TCE rB D 21i, rP D 6j, C rP ð FP D 0 MQ D rA ð FA C TAB C TAC C rC ð TCE C rP ð FP D 0. rC D 25j 7.5k. The vector parallel to the backstay AB is From above, FA C TAB C TAC D FTX i C FTY j C FTZ k rAB D rB rA D 21i 50j. D 193.2 0.3872jTAB ji C 0.922jTAB j The unit vector parallel to backstay AB is 0.9578jTAC jj C 51.76 0.2873jTAC jk eAB D 0.3872i 0.9220j. The vector parallel to AC is i MQ D 0 FTX i j C 0 6 35.35 0 rAC D rC rA D 25j 7.5k. The forces acting on the mast are: (1) The force due to the spinnaker at the top of the mast: j 25 0 k 7.5 0.2873jTAC j k 6 D 0 35.35 C 50FTX C 212.1k D 0. (2) The reaction due to the backstay: (3) The reaction due to the shroud: k i 50 C 0 FTZ 0 D 50FTZ C 250.2873jTAC j C 212.1i FA D 200i cos 15° C k cos 75° D 193.19i C 51.76k. TAB D jTAB jeAB j 50 FTY Substituting and collecting terms: 2800 7.1829jTAC ji C 9447.9 C 19.36jTAB jk D 0, from which TAC D jTAC jeAC (4) The force acting on the cross spar CE: TCE D k Ð TAC k D 0.2873jTAC jk. jTAC j D 2800 D 389.81 lb, 7.1829 jTAB j D 488.0 lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 357 5.118 (Continued ) The tension in cable CD is the vertical component of the tension in AC, jTCD j D jTAC jj Ð eAC D jTAC j0.9578 D 373.37 lb. The reaction at the base is found from the sums of the forces: FX D QX C 193.19 35.35 jTAB j0.3872 D 0, from which QX D 31.11 lb FY D QY 0.922jTAB j 0.9578jTAC jj D 0, from which QY D 823.24 lb FZ D QZ C 51.76 C 0.2873jTAC j 0.2873jTAC j C 35.35k D 0, from which QZ D 87.11 lb Collecting the terms, the reaction is Q D 31.14i C 823.26j 87.12k (lb) y y FA TAC FA TCD TCE TAB z FP QX QY TCD FP QY QB z SIDE VIEW z AFT VIEW FA QB x QX FP TAB TOP VIEW 358 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 5.119* The bar AC is supported by the cable BD and a bearing at A that can rotate about the axis AE. The person exerts a force F D 50j (N) at C. Determine the tension in the cable. (0.3, 0.5, 0) m E Strategy: Use the fact that the sum of the moments about the axis AE due to the forces acting on the freebody diagram of the bar must equal zero. C (0.82, 0.60, 0.40) m (0.3, 0.4, 0.3) m A B (0.46, 0.46, 0.33) m x Solution: We will take moments about the line AE in order to eliminate all of the reactions at the bearing A. We have: eAE D z 0.1j 0.3k p D 0.316j 0.949k 0.1 D (0.7, 0, 0.5) m rAB D 0.16i C 0.06j C 0.03km, TBD D TBD 0.24i 0.46j C 0.17k p 0.2981 rAC D 0.52i C 0.2j C 0.1km, F D 50jN Then the equilibrium equation is MAE D eAE Ð rAB ð TBD C rAC ð F D 0 This reduces to the single scalar equation TBD D 174.5 N Problem 5.120* In Problem 5.119, determine the reactions at the bearing A. Solution: See the previous problem for setup. We add the reactions Strategy: Write the couple exerted on the free-body diagram of the bar by the bearing as MA D MAx i C MAy j C MAz k. Then, in addition to the equilibrium equations, obtain an equation by requiring the component of MA parallel to the axis AE to equal zero. A D Ax i C Ay j C Az k, MA D MAx i C MAy j C MAz k (force, moment) This gives us too many reaction moments. We will add the constrain that MA Ð eAE D 0 We have the following 6 equilibrium equations: Fx : Ax C 0.440TBD D 0 F D A C TBD C F D 0 ) Fy : Ay 0.843TBD C 50 N D 0 Fx : Az C 0.311TBD D 0 MA D MA C rAB ð TBD C rAC ð F D 0 MAx : MAx 5 N-m C 0.0440 mTBD D 0 ) MAy : MAy 0.0366 mTBD D 0 MAz : MAz C 26 N-m 0.161 mTBD D 0 eAE Ð MA D 0 ) 0.316MAy 0.949MAz D 0 Solving these 7 equations we find Ax D 76.7 N, Ay D 97.0 N, Az D 54.3 N MAx D 2.67 N-m, MAy D 6.39 N-m, MAz D 2.13 N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 359 Problem 5.121 In Active Example 5.10, suppose that the support at A is moved so that the angle between the bar AB and the vertical decreases from 45° to 30° . The position of the rectangular plate does not change. Draw the free-body diagram of the plate showing the point P where the lines of action of the three forces acting on the plate intersect. Determine the magnitudes of the reactions on the plate at B and C. 45 A C B 4 ft Solution: The equilibrium equations are Fx : B sin 30° C sin 30° D 0 Fy : B cos 30° C C cos 30° 100 lb D 0 Solving yields B D C D 57.7 lb 360 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.122 The magnitude of the reaction exerted on the L-shaped bar at B is 60 lb. (See Example 5.11.) (a) What is the magnitude of the reaction exerted on the bar by the support at A? (b) What are the x and y components of the reaction exerted on the bar by the support at A? y 14 in B 17 in A x Solution: The angle between the line AB and the x axis is D tan1 17/14 D 50.5° (a) The bar is a two-force member, so the magnitude of the reaction at A is A D 60 lb (b) The reaction at A must be parallel to the line from A to B, but it may point either from A toward B or from B toward A. In the first case, the components are Ax D 60 lb cos , Ay D 60 lb sin In the second case the components are Ax D 60 lb cos , Ay D 60 lb sin Thus Ax D 38.1 lb, Ay D 46.3 lb or Ax D 38.1 lb, Ay D 46.3 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 361 Problem 5.123 The suspended load weighs 1000 lb. The structure is a three-force member if its weight is neglected. Use this fact to determine the magnitudes of the reactions at A and B. A 5 ft B 10 ft Solution: The pin support at A is a two-force reaction, and the roller support at B is a one force reaction. The moment about A is MA D 5B 101000 D 0, from which the magnitude at B is B D 2000 lb. The sums of the forces: A 5 ft FX D AX C B D AX C 2000 D 0, from which AX D 2000 lb. B 10 ft FY D AY 1000 D 0, from which AY D 1000 lb. 1000 lb p The magnitude at A is A D 20002 C 10002 D 2236 lb AX AY 5 ft B 1000 lb 10 ft Problem 5.124 The weight W D 50 lb acts at the center of the disk. Use the fact that the disk is a threeforce member to determine the tension in the cable and the magnitude of the reaction at the pin support. 60° Solution: Denote the magnitude of the reaction at the pinned joint by B. The sums of the forces are: and FX D BX T sin 60° D 0, W FY D BY C T cos 60° W D 0. The perpendicular distance to the action line of the tension from the center of the disk is the radius R. The sum of the moments about the center of the disk is MC D RBY C RT D 0, from which BY D T. Substitute into the sum of the forces to obtain: T C T0.5 W D 0, from which TD 60° W 2 W D 33.33 lb. 3 Substitute into the sum of forces to obtain 60° BX D T sin 60° D 28.86 lb. The magnitude of the reaction at the pinned joint is BD 362 p T R BX W BY 33.332 C 28.862 D 44.1 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.125 The weight W D 40 N acts at the center of the disk. The surfaces are rough. What force F is necessary to lift the disk off the floor? F 150 mm W 50 mm Solution: The reaction at the obstacle acts through the center of the disk (see sketch) Denote the contact point by B. When the moment is zero about the point B, the disk is at the verge of leaving the floor, hence the force at this condition is the force required to lift the disk. The perpendicular distance from B to the action line of the weight is d D R cos ˛, where ˛ is given by (see sketch) ˛ D sin1 Rh R D sin1 150 50 150 F 150 mm 50 mm W D 41.81° . α The perpendicular distance to the action line of the force is F D D 2R h D 300 50 D 250 mm. The sum of the moments about the contact point is W MB D R cos ˛W C 2R hF D 0, from which F D 150 cos 41.81° W D 0.4472W D 17.88 N 250 h b c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 363 Problem 5.126 Use the fact that the horizontal bar is a three-force member to determine the angle ˛ and the magnitudes of the reactions at A and B. Assume that 0 ˛ 90° . 2m Solution: The forces at A and B are parallel to the respective bars since these bars are 2-force members. Since the horizontal bar is a 3-force member, all of the forces must intersect at a point. Thus we have the following picture: a 3 kN 60⬚ B A 1m 30⬚ From geometry we see that d D 1 m cos 30° d sin 30° D e sin ˛ d cos 30° C e cos ˛ D 3 m Solving we find ˛ D 10.89° To find the other forces we look at the force triangle FB D 3 kN cos 40.89° D 2.27 kN FA D 3 kN sin 40.89° D 1.964 kN FA e d α 60° 30° α 30° 2m 3 kN 1m FB 3 kN 40.89° FA FB 364 90° c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.127 The suspended load weighs 600 lb. Use the fact that ABC is a three-force member to determine the magnitudes of the reactions at A and B. 3 ft B 4.5 ft 30⬚ C 45⬚ A Solution: All of the forces must intersect at a point. From geometry tan D 3 ft D 0.435 3 C 4.5 cos 30° ft ) D 23.5° Now using the force triangle we find FB D 600 lb cot D 1379 lb FA D 600 lb csc D 1504 lb FB θ FA 600 lb FB θ 600 lb FA c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 365 2 kN Problem 5.128 (a) Is the L-shaped bar a three-force member? (b) Determine the magnitudes of the reactions at A and B. (c) Are the three forces acting on the L-shaped bar concurrent? 3 kN-m B 300 mm 150 mm 700 mm A 250 mm 500 mm Solution: (a) No. The reaction at B is one-force, and the reaction at A is two-force. The couple keeps the L-shaped bar from being a three force member.(b) The angle of the member at B with the horizontal is ˛ D tan1 150 250 D 30.96° . The sum of the moments about A is MA D 3 0.52 C 0.7B cos ˛ D 0, from which B D 6.6637 kN. The sum of forces: FX D AX C B cos ˛ D 0, from which AX D 5.7143 kN. FY D AY B sin ˛ 2 D 0, from which AY D 5.4281 kN. The magnitude at A: AD p 5.712 C 5.432 D 7.88 kN (c) No, by inspection. 0.5 m 3 kN-m α 2 kN 0.3 m B 0.7 m Ay Ax 366 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.129 The hydraulic piston exerts a horizontal force at B to support the weight W D 1500 lb of the bucket of the excavator. Determine the magnitude of the force the hydraulic piston must exert. (The vector sum of the forces exerted at B by the hydraulic piston, the two-forces member AB, and the two-force member BD must equal zero.) Solution: See the solution to Problem 5.23. 14 in 16 in B A 4 in C The angle between the two-force member AB and the horizontal is D ˛ D tan1 12/14 D 40.6° and the magnitude of the force exerted at B by member AB is TAB D 892 lb W in the direction from B toward A. Let F be the force exerted by the piston, and let TBD be the force exerted at B by member BD in the direction from B toward D. The angle between member BD and the horizontal is ˇ D tan1 16/12 D 53.1° 8 in 8 in The sum of the forces at B is Fx : F C TBD cos ˇ TAB cos ˛ D 0 Fy : TBD sin ˇ TAB sin ˛ D 0 Solving yields TBD D 726 lb, F D 1110 lb Thus F D 1110 lb Problem 5.130 The member ACG of the front-end loader is subjected to a load W D 2 kN and is supported by a pin support at A and the hydraulic cylinder BC. Treat the hydraulic cylinder as a two-force member. A 0.75 m B (a) (b) Draw a free-body diagrams of the hydraulic cylinder and the member ACG. Determine the reactions on the member ACG. C 1m G 0.5 m W 1.5 m Solution: This is a very simple Problem. The free body diagrams are shown at the right. From the free body diagram of the hydraulic cylinder, we get the equation BX C CX D 0. This will enable us to find BX once the loads on member ACG are known. From the diagram of ACG, the equilibrium equations are and CX BX B AX Fx D AX C CX D 0, 0.75 m Fy D AY W D 0, 1m MA D 0.75CX 3W D 0. 1.5 m AY CX 0.5 m 1.5 m 1.5 m W Using the given value for W and solving these equations, we get AX D 8 kN, AY D 2 kN, CX D 8 kN, and BX D 8 kN. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 367 Problem 5.131 In Problem 5.130, determine the reactions of the member ACG by using the fact that it is a three-force member. Solution: The easiest way to do this is take advantage of the fact that for a three force member, the three forces must be concurrent. The fact that the force at C is horizontal and the weight is vertical make it very easy to find the point of concurrency. We then use this point to determine the direction of the force through A. We can even know which direction this force must take along its line — it must have an upward component to support the weight — which is down. From the geometry, we can determine the angle between the force A and the horizontal. y A A 0.75 m 1m θ CX C 1.5 m x 1.5 m G W = 2 kN tan D 0.75/3, or D 14.04° . Using this, we can write force equilibrium equations in the form Fx D A cos C CX D 0, and Fy D A sin W D 0. Solving these equations, we get A D 8.246 kN, and CX D 8 kN. The components of A are as calculated in Problem 5.130. 368 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.132 A rectangular plate is subjected to two forces A and B (Fig. a). In Fig. b, the two forces are resolved into components. By writing equilibrium equations in terms of the components Ax , Ay , Bx , and By , show that the two forces A and B are equal in magnitude, opposite in direction, and directed along the line between their points of application. B B A h A b (a) y By Bx B h Ay Ax A x b (b) Solution: The sum of forces: b B FX D AX C BX D 0, h A from which AX D BX By y FY D AY C BY D 0, Ay from which AY D By . These last two equations show that A and B are equal and opposite in direction, (if the components are equal and opposite, the vectors are equal and opposite). To show that the two vectors act along the line connecting the two points, determine the angle of the vectors relative to the positive x axis. The sum of the moments about A is Fig a Ax Bx Fig b x MA D Bx h bBy D 0, from which the angle of direction of B is tan1 BY BX D tan1 h D ˛B . b or 180 C ˛B . Similarly, by substituting A: tan1 AY AX D tan1 h D ˛A , b or 180 C ˛A . But ˛ D tan1 h b describes direction of the line from A to B. The two vectors are opposite in direction, therefore the angles of direction of the vectors is one of two possibilities: B is directed along the line from A to B, and A is directed along the same line, oppositely to B. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 369 Problem 5.133 An object in equilibrium is subjected to three forces whose points of application lie on a straight line. Prove that the forces are coplanar. F2 F3 Solution: The strategy is to show that for a system in equilibrium under the action of forces alone, any two of the forces must lie in the same plane, hence all three must be in the same plane, since the choice of the two was arbitrary. Let P be a point in a plane containing the straight line and one of the forces, say F2 . Let L also be a line, not parallel to the straight line, lying in the same plane as F2 , passing through P. Let e be a vector parallel to this line L. First we show that the sum of the moments about any point in the plane is equal to the sum of the moments about one of the points of application of the forces. The sum of the moments about the point P: F1 F2 F3 F1 P L M D r1 ð F1 C r2 ð F2 C r3 ð F3 D 0, where the vectors are the position vectors of the points of the application of the forces relative to the point P. (The position vectors lie in the plane.) Define d12 D r2 r1 , and d13 D r3 r1 . Then the sum of the moments can be rewritten, M D r1 ð F1 C F2 C F3 C d12 ð F2 C d13 ð F3 D 0. Since the system is in equilibrium, F1 C F2 C F3 D 0, and the sum of moments reduces to M D d12 ð F2 C d13 ð F3 D 0, which is the moment about the point of application of F1 . (The vectors d12 , d13 are parallel to the line L.) The component of the moment parallel to the line L is e Ð d12 ð F2 e C e Ð d13 ð F3 e D 0, or F2 Ð d12 ð ee C F3 Ð d13 ð ee D 0. But by definition, F2 lies in the same plane as the line L, hence it is normal to the cross product d12 ð e 6D 0, and the term F2 Ð d12 ð e D 0. But this means that F3 Ð d13 ð ee D 0, which implies that F3 also lies in the same plane as F2 , since d13 ð e 6D 0. Thus the two forces lie in the same plane. Since the choice of the point about which to sum the moments was arbitrary, this process can be repeated to show that F1 lies in the same plane as F2 . Thus all forces lie in the same plane. 370 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.134 The suspended cable weighs 12 lb. (a) B Draw the free-body diagram of the cable. (The tensions in the cable at A and B are not equal.) Determine the tensions in the cable at A and B. What is the tension in the cable at its lowest point? (b) (c) 50⬚ A 32⬚ Solution: (a) (b) The FBD The equilibrium equations Fx : TA cos 32° C TB cos 50° D 0 Fy : TA sin 32° C TB sin 50 12 lb D 0 Solving we find TA D 7.79 lb, TB D 10.28 lb (c) Consider the FBD where W represents only a portion of the total weight. We have Fx : TA cos 32° C T D 0 Solving T D 6.61 lb TB 50° TA 32° 12 lb TA 32° T W c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 371 Problem 5.135 Determine the reactions at the fixed support. 4 kN 3m A 20 kN-m 2 kN 5m 3 kN 3m Solution: The equilibrium equations 4 kN Fx : Ax C 4 kN D 0 Fy : Ay 2 kN 3 kN D 0 Ax 20 kN-m MA : MA 2 kN5 m 4 kN3 m 3 kN8 m C 20 kN-m D 0 MA Ay Solving 2 kN 3 kN Ax D 4 kN, Ay D 5 kN, MA D 26 kN-m Problem 5.136 (a) Draw the free-body diagram of the 50-lb plate, and explain why it is statically indeterminate. y (b) Determine as many of the reactions at A and B as possible. A 12 in Solution: (a) (b) The pin supports at A and B are two-force supports, thus there are four unknown reactions AX , AY , BX , and BY , but only three equilibrium equations can be written, two for the forces, and one for the moment. Thus there are four unknowns and only three equations, so the system is indeterminate. B x 20 in 50 lb Sums the forces: A FX D AX C BX D 0, or AX D BX , and 8 in 12 in 8 in B 20 in FY D AY C BY 50 D 0. The sum of the moments about B 50 lb x AY AX MB D 20AX 5020 D 0, from which AX D 50 lb, and from the sum of forces BX D 50 lb. 372 12 in. BY 8 in. BX 50 lb x 20 in. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.137 The mass of the truck is 4 Mg. Its wheels are locked, and the tension in its cable is T D 10 kN. (003) 676-5942 (a) (b) Draw the free-body diagram of the truck. Determine the normal forces exerted on the truck’s wheels by the road. 30° AL's To w i n g 2m T 2.5 m 3m 2.2 m mg Solution: The weight is 40009.81 D 39.24 kN. The sum of the moments about B MB D 3T sin 30° 2.2T cos 30° C 2.5W 4.5AN D 0 from which AN D D 2.5W T3 sin 30° C 2.2 cos 30° 4.5 64.047 D 14.23 N 4.5 The sum of the forces: FY D AN W C BN T cos 30° D 0, from which BN D T cos 30° AN C W D 33.67 N 30° AX A AN W BX T B BN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 373 Problem 5.138 Assume that the force exerted on the head of the nail by the hammer is vertical, and neglect the hammer’s weight. (a) Draw the free-body diagram of the hammer. (b) If F D 10 lb, what are the magnitudes of the forces exerted on the nail by the hammer and the normal and friction forces exerted on the floor by the hammer? F 11 in. Solution: Denote the point of contact with the floor by B. The 65° perpendicular distance from B to the line of action of the force is 11 in. The sum of the moments about B is MB D 11F 2FN D 0, from 11F D 5.5F. The which the force exerted by the nail head is FN D 2 sum of the forces: FX D F cos 25 C Hx D 0, 2 in. from which the friction force exerted on the hammer is HX D 0.9063F. FY D NH FN C F sin 25° D 0, from which the normal force exerted by the floor on the hammer is NH D 5.077F If the force on the handle is F D 10 lb, then FN D 55 lb, HX D 9.063 lb, and NH D 50.77 lb F 11 in. 65° 2 in. F 11 in. 65° HX 374 B NH FN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.139 The spring constant is k D 9600 N/m and the unstretched length of the spring is 30 mm. Treat the bolt at A as a pin support and assume that the surface at C is smooth. Determine the reactions at A and the normal force at C. A 24 mm B 15 mm 30 mm 30° C k Solution: The length of the spring is lD p 302 C 302 mm D p 1800 mm 30 mm AY l D 42.4 mm D 0.0424 m The spring force is kυ where υ D l l0 . l0 is give as 30 mm. (We must be careful because the units for k are given as N/m) We need to use length units as all mm or all meters). k is given as 9600 N/m. Let us use l0 D 0.0300 m and l D 0.0424 m AX 24 mm B kδ Equilibrium equations: FX D 0: AX kl l0 sin 45° NC cos 60° D 0 FY D 0: 50 mm 15 mm 30 mm 45° 60° 50 mm 30 mm 30° C NC AY kl l0 cos 45° Solving, we get C NC sin 60° D 0 MB D 0: 0.024AX C 0.050NC sin 60° 0.015NC cos 60° D 0 AX D 126.7 N AY D 10.5 N NC D 85.1 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 375 Problem 5.140 The engineer designing the release mechanism shown in Problem 5.139 wants the normal force exerted at C to be 120 N. If the unstretched length of the spring is 30 mm, what is the necessary value of the spring constant k? Solution: Refer to the solution of Problem 5.139. The equilibrium equations derived were FX D 0: AX kl l0 sin 45 NC cos 60° D 0 FY D 0: AY kl l0 cos 45 C NC sin 60° D 0 MB D 0: 0.024AX C 0.050NC sin 60° 0.015NC cos 60° D 0 where l D 0.0424 m, l0 D 0.030 m, NC D 120 N, and AX , AY , and k are unknowns. Solving, we get AX D 179.0 N, AY D 15.1 N, k D 13500 N/m 376 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.141 The truss supports a 90-kg suspended object. What are the reactions at the supports A and B? 400 mm 700 mm 300 mm B A Solution: Treat the truss as a single element. The pin support at A is a two force reaction support; the roller support at B is a single force reaction. The sum of the moments about A is MA D B400 W1100 D 0, from which B D 1100W D 2.75W 400 B D 2.75909.81 D 2427.975 D 2.43 kN. The sum of the forces: FX D AX D 0 FY D AY C B W D 0, from which AY D W B D 882.9 2427.975 D 1.545 kN AX W B AY 400 mm 700 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 377 Problem 5.142 The trailer is parked on a 15° slope. Its wheels are free to turn. The hitch H behaves like a pin support. Determine the reactions at A and H. y 1.4 ft H x 870 lb 1.6 ft A 8 ft 2.8 ft 15° Solution: The coordinate system has the x axis parallel to the road. The wheels are a one force reaction normal to the road, the pin H is a two force reaction. The position vectors of the points of the center of mass and H are: rW D 1.4i C 2.8j ft and rH D 8i C 1.6j. The angle of the weight vector realtive to the positive x axis is ˛ D 270° 15° D 255° . The weight has the components W D Wi cos 255° C j sin 255° D 8700.2588i 0.9659j D 225.173i 840.355j (lb). The sum of the moments about H is MH D rW rH ð W C rA rH ð A, j k i 1.2 0 C 8 840.355 0 0 i MH D 6.6 225.355 j k 1.6 0 D 0 AY 0 D 5816.55 8AY D 0, from which AY D 5816.55 D 727.1 lb. 8 The sum of the forces is FX D HX 225.173i D 0, from which HX D 225.2 lb, FY D AY C HY 840.355j D 0, from which HY D 113.3 lb 1.4 ft 1.2 ft W 15° AY 378 HX HY 8 ft c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.143 To determine the location of the point where the weight of a car acts (the center of mass), an engineer places the car on scales and measures the normal reactions at the wheels for two values of ˛, obtaining the following results: y h x B W Ay (kN) ˛ B (kN) 10° 10.134 4.357 20° 10.150 3.677 α Ax Ay b 2.7 m What are the distances b and h? Solution: The position vectors of the cm and the point B are These two simultaneous equations in two unknowns were solved using the HP-28S hand held calculator. rCM D 2.7 bi C hj, b D 1.80 m, rB D 2.7i. h D 0.50 m The angle between the weight and the positive x axis is ˇ D 270 ˛. The weight vector at each of the two angles is W10 D Wi cos 260° C j sin 260° W10 D W0.1736i 0.9848j W20 D Wi cos 250° C j sin 250° or W20 D W0.3420i 0.9397j The weight W is found from the sum of forces: FY D AY C BY C W sin ˇ D 0, from which Wˇ D AY C BY . sin ˇ Taking the values from the table of measurements: W10 D 10.134 C 4.357 D 14.714 kN, sin 260° [check :W20 D 10.150 C 3.677 D 14.714 kN check ] sin 250° The moments about A are MA D rCM ð W C rB ð B D 0. Taking the values at the two angles: M10 A i D 2.7 b 2.5551 j k j k i 0 0 D 0 h 0 C 2.7 14.4910 0 0 4.357 0 D 14.4903b C 2.5551h 27.3618 D 0 M20 A i D 2.7 b 5.0327 j k j k i 0 0 h 0 C 2.7 13.8272 0 0 3.677 0 D 013.8272b C 5.0327h 27.4054 D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 379 Problem 5.144 The bar is attached by pin supports to collars that slide on the two fixed bars. Its mass is 10 kg, it is 1 m in length, and its weight acts at its midpoint. Neglect friction and the masses of the collars. The spring is unstretched when the bar is vertical (˛ D 0), and the spring constant is k D 100 N/m. Determine the values of ˛ in the range 0 ˛ 60° at which the bar is in equilibrium. k α Solution: The force exerted by the spring is given by FS D kL L cos ˛. The equations of equilibrium, from the free body diagram, are and Spring Constant (K) in N/m vs Alpha (deg) 90000 80000 70000 Fx D NB D 0, Fy D FS C NA mg D 0, MB D L sin ˛NA C L sin ˛ mg D 0. 2 60000 K = 50000 N − 40000 m 30000 20000 10000 These equations can be solved directly with most numerical solvers and the required plot can be developed. The plot over the given ˛ range is shown at the left and a zoom-in is given at the right. The solution and the plot were developed with the TK Solver Plus commercial software package. From the plot, the required equilibrium value is ˛ ¾ D 59.4° . 0 0 10 20 30 40 50 60 Alpha (deg) Spring Constant (K) in N/m vs Alpha (deg) 116 114 FS 112 B y NB α mg 110 K = 108 N − 106 m 104 102 A x NA 100 98 55 55.5 56 56.5 57 57.5 58 58.5 59 59.5 60 Alpha (deg) 380 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.145 With each of the devices shown you can support a load R by applying a force F. They are called levers of the first, second, and third class. (a) (b) R F A A The ration R/F is called the mechanical advantage. Determine the mechanical advantage of each lever. Determine the magnitude of the reaction at A for each lever. (Express you answers in terms of F) L L L First-class lever R A Solution: R/F D 1 First Class MA : LF LR D 0 ) F D R R/F D 2 Second Class MA : 2LF LR D 0 ) F D R/2 Third Class MA : LF 2LR D 0 ) F D 2R (b) L Second-class lever F (a) F R R/F D 1/2 First Class Fy : F C A R D 0 ) A D 2F Second Class Fy : A R C F D 0 ) ADF Third Class Fy : A C F R D 0 ) L L Third-class lever A D F/2 Problem 5.146 The force exerted by the weight of the horizontal rectangular plate is 800 N. The weight of the plate acts at its midpoint. If you represent the reactions exerted on the plate by the three cables by a single equivalent force, what is the force, and where does its line of action intersect the plate? A C B 2m 0.5 m 1m Solution: The equivalent force must equal the sum of the reactions: FEQ D TA C TB C TC . FEQ D 300 C 100 C 400 D 800 N. The moment due to the action of the equivalent force must equal the moment due to the reactions: The moment about A is i MA D 2 0 j j k i 0 0 0 C 1.5 100 0 0 400 k i 1 D x 0 0 j k 0 z 800 0 MA D 400i C 800k D 800zi C 800xk, y TA TC 1m z TB 2m 0.5 m 800 N x from which z D 0.5 m, and x D 1 m, which corresponds to the midpoint of the plate. Thus the equivalent force acts upward at the midpoint of the plate. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 381 y Problem 5.147 The 20-kg mass is suspended by cables attached to three vertical 2-m posts. Point A is at (1, 1.2, 0) m. Determine the reactions at the built-in support at E. C B D A Solution: All distances will be in meters, all forces in Newtons, and all moments in Newton-meters. To solve the three dimensional point equilibrium problem at A, we will need unit vectors eAB , eAC , and eAD . To determine these, we need the coordinates of points A, B, C, and D. The rest of the problem will require knowing where points E, G (under C), and H(under D) are located. From the diagram, the required point locations are A (0, 1.2, 0), B (0.3, 2, 1), C (0, 2, 1), D (2, 2, 0), E (0.3, 0, 1), G(0, 0, 1), and H(2, 0, 0). The required unit vectors are calculated from the coordinates of the points of the ends of the lines defining the vector. These are 1m 2m 0.3 m x z The couple ME is the couple exerted on the post by the built in support. Solving these equations, we get E D 34.2i C 91.3j C 114.1k N eAB D 0.228i C 0.608j C 0.760k, eAC D 0i C 0.625j 0.781k, 1m E and ME D 228.1i C 0j C 68.44k N-m. and eAD D 0.928i C 0.371j C 0k. Also, The force TAB in cable AB can be written as Using a procedure identical to that followed for post EB above, we can find the built-in support forces and moments for posts CG and DH. The results for CG are: TAB D TABX i C TABY j C TABZ k, where TABX D jTAB jeABX , etc. Similar equations can be written for the forces in AC and AD. The free body diagram of point A yields the following three equations of equilibrium. and jME j D 238.2 N-m. G D 0i C 91.3j 114.1k N and MG D 228.1i C 0j C 0k N-m. Fx D TABX C TACX C TADX D 0, Also, jMG j D 228.1 N-m. Fy D TABY C TACY C TADY W D 0, The results for DH are: H D 34.2i C 13.7j C 0 kN Fz D TABZ C TACZ C TADZ D 0, and MH D 0i C 0j C 68.4k N-m. where W D mg D 209.81 D 196.2 N. Solving the equations above after making the substitutions related to the force components yields the tensions in the cables. They are jTAB j D 150 N, Also, jMG j D 68.4 N-m B jTAC j D 146 N, and jTAD j D 36.9 N. Now that we know the tensions in the cables, we are ready to tackle the reactions at E (also G and H). The first step is to draw the free body diagram of the post EB and to write the equations of equilibrium for the post. A key point is to note that the force on the post from cable AB is opposite in direction to the force found in the first part of the problem. The equations of equilibrium for post EB are Z −TAD C −TAB MG ZM ME E x EY D −TAC EX H GZ HX G HZ GY GX HY EZ Fx D EX TABX D 0, Fy D EY TABY D 0, Fz D EZ TABZ D 0, and, summing moments around the base point E, 382 M D ME C 2j ð TAB D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.148 In Problem 5.147, the built-in support of each vertical post will safely support a couple of 800 N-m magnitude. Based on this criterion, what is the maximum safe value of the suspended mass? Solution: We have all of the information necessary to solve this problem in the solution to Problem 5.147 above. All of the force and moment equations are linear and we know from the solution that a 20 kg mass produces a couple of magnitude 238.2 N-m at support E and that the magnitudes of the couples at the other two supports are smaller than this. All we need to do is scale the Problem. The scale factor is f D 800/238.2 D 3.358 and the maximum value for the suspended mass is mmax D 20f D 67.16 kg c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 383 y Problem 5.149 The 80-lb bar is supported by a ball and socket support at A, the smooth wall it leans against, and the cable BC. The weight of the bar acts at its midpoint. 5 ft 3 ft (a) Draw the free-body diagram of the bar. (b) Determine the tension in cable BC and the reactions at A. B C 4 ft 3 ft 3 ft x A z Solution: (a) The ball and socket is a three reaction force support; the cable and the smooth wall are each one force reaction supports. (b) The coordinates of the points A, B and C are A (3, 0, 3), B (5, 4, 0), and C(0, 4, 3). The vector parallel to the bar is rAB D rB rA D 2i C 4j 3k. Solve: jTj D 80 D 23.32 lb 3.43 jBj D 120 2.058jTj D 18.00 lb. 4 The reactions at A are found from the sums of forces: The length of the bar is p jrAB j D 22 C 42 C 32 D 5.3852. The unit vector parallel to the bar is eAB D 0.3714i C 0.7428j 0.5571k. FX D AX jTj0.8575 D 0 from which AX D 20 lb FY D AY 80 D 0, from which AY D 80 lb FZ D AZ C jTj0.5145 C jBj D 0, from which AZ D 30 lb The vector parallel to the cable is rBC D rC rB D 5i C 3k. B The unit vector parallel to the cable is eBC D 0.8575i C 0.5145k. The cable tension is T D jTjeBC . The point of application of the weight relative to A is AY AX rAW D 2.6936eAB AZ rAW D 1.000i C 2.000j 1.500k. The reaction at B is B D jBjk, since it is normal to a wall in the yz plane. The sum of the moments about A is MA D rAW ð W C rAB ð B C rAB ð T D 0 i MA D 1 0 j 2 80 k i 1.5 C 2 0 0 i j C 2 4 0.8575 0 j 4 0 k 3 jBj k 3 jTj D 0 0.5145 MA D 120 C 4jBj C 2.058jTji 2jBj 1.544jTjj C 3.43jTj 80k D 0. 384 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 5.150 The horizontal bar of weight W is supported by a roller support at A and the cable BC. Use the fact that the bar is a three-force member to determine the angle ˛, the tension in the cable, and the magnitude of the reaction at A. C A B α W L/2 L/2 Solution: The sum of the moments about B is MB D LAY C L W D 0, 2 from which AY D W . The sum of the forces: 2 FX D T cos ˛ D 0, from which T D 0 or cos ˛ D 0. The choice is made from the sum of forces in the y-direction: FY D AY W C T sin ˛ D 0, W . This equation cannot be satisfied 2 W if T D 0, hence cos ˛ D 0, or ˛ D 90° , and T D 2 from which T sin ˛ D W AY D T α AY W L/ 2 L/ 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 385 Problem 6.1 In Active Example 6.1, suppose that in addition to the 2-kN downward force acting at point D, a 2-kN downward force acts at point C. Draw a sketch of the truss showing the new loading. Determine the axial forces in members AB and AC of the truss. A 3m C 3m B D 5m 5m 2 kN Solution: The new sketch, a free-body diagram of the entire truss and a free-body diagram of the joint at A are shown. The angle ˛ between CD and BD is ˛ D tan1 6/10 D 31.0° Using the entire truss, the equilibrium equations are Fx : Ax C B D 0 Fy : Ay 2 kN 2 kN D 0 MA : 2 kN5 m 2 kN10 m C B6 m D 0 Solving yields Ax D 5 kN, Ay D 4 kN, B D 5 kN Using the free-body diagram of joint A, the equilibrium equations are: Fx : Ax C TAC cos ˛ D 0 Fy : Ay TAB TAC sin ˛ D 0 Solving yields TAB D 1 kN, TAC D 5.83 kN Because both values are positive, we know that both are in tension AB : 1 kN (T), 386 AC : 5.83 kN (T) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.2 Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C). 20⬚ 800 N A 0.4 m C B 0.7 m 0.7 m Solution: We start at joint A Next we move to joint C 7 7 Fx : p FAB C p FAC 800 N sin 20° D 0 65 65 4 4 Fy : p FAB p FAC 800 N cos 20° D 0 65 65 Solving we have 7 Fx : p FAC FBC D 0 ) FBC D 521 N 65 FAC 7 4 FAB D 915 N, FAC D 600 N C 20° 800 N FCB Cy A 4 In summary we have 4 7 7 FAB FAB D 915 NC, FAC D 600 NC, FBC D 521 NT FAC A Problem 6.3 Member AB of the truss is subjected to a 1000-lb tensile force. Determine the weight W and the axial force in member AC. 60 in W B C 60 in 60 in Solution: Using joint A 1 1 2 Fx : p 1000 lb p FAC D 0 5 2 1 1000 lb 1 1 Fy : p 1000 lb p FAC W D 0 5 2 Solving we have FAC D 1265 lb, W D 447 lb In summary we have A 2 1 FAC W W D 447 lb, FAC D 1265 lbC c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 387 Problem 6.4 Determine the axial forces in members BC and CD of the truss. 600 lb E 3 ft C D 3 ft A B 3 ft 3 ft Solution: The free-body diagrams for joints E, D, and C are shown. The angle ˛ is ˛ D tan1 3/4 D 36.9° Using Joint E, we have Fx : 600 lb TCE sin ˛ D 0 Fy : TCE cos ˛ TDE D 0 Using Joint D, we have Fx : TCD TBD sin ˛ D 0 Fy : TDE TBD cos ˛ D 0 Finally, using Joint C, we have Fx : TCD C TCD sin ˛ TAC sin ˛ D 0 Fy : TCE cos ˛ TAC cos ˛ TBC D 0 Solving these six equations yields TCE D 1000 lb, TDE D 800 lb TCD D 600 lb, TAC D 2000 lb TBC D 800 lb, TBD D 1000 lb A positive value means tension and a negative value means compression Thus 388 BC : 800 lb (T), CD : 600 lb (C) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.5 Each suspended weight has mass m D 20 kg. Determine the axial forces in the members of the truss and indicate whether they are in tension (T) or compression (C). A 0.4 m C B D m 0.32 m 0.16 m 0.16 m Solution: Assume all bars are in tension. Start with joint D Finally work with joint A 5 Fy : p TAD 196.2 N D 0 61 6 Fx : p TAD TCD D 0 61 5 5 Fy : p TAB C TAC p TAD D 0 29 61 ) TAB D 423 N T A TAD D 306 N, TCD D 235 N Solving: m 6 2 TAD 5 5 6 5 5 2 D TAD TCD TAB TAC In summary: TAB D 423 NC 196.2 N TAC D 211 NT TAD D 306 NT Now work with joint C TBC D 314 NC 5 Fy : p TAC 196.2 N D 0 29 TCD D 235 NC 2 Fx : p TAC TBC C TCD D 0 29 Solving: TAC D 211 N, TBC D 313 N TAC 5 2 C TBC TCD 196.2 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 389 Problem 6.6 Determine the largest tensile and compressive forces that occur in the members of the truss, and indicate the members in which they occur if (a) (b) the dimension h D 0.1 m; the dimension h D 0.5 m. Observe how a simple change in design affects the maximum axial loads. B A h D 0.7 m 1 kN 0.4 m C 0.6 m Solution: To get the force components we use equations of the y form TPQ D TPQ ePQ D TPQX i C TPQY j where P and Q take on the designations A, B, C, and D as needed. Equilibrium yields At joint A: and Fx D TABX C TACX D 0, h 0.4 m 1.2 m B −TAB TAB −TBD TBD T BC TAC DX D −TBC T −TCD CD −TAC DY C CY 0.6 m A 1 kN 0.7 m x 1.2 m Fy D TABY C TACY 1 kN D 0. At joint B: and Fx D TABX C TBCX C TBDX D 0, Fy D TABY C TBCY C TBDY D 0. At joint C: and and eAB D 0.986i C 0.164j, eAC D 0.864i 0.504j, Fx D TBCX TACX C TCDX D 0, eBC D 0i 1j, Fy D TBCY TACY C TCDY C CY D 0. eBD D 0.768i 0.640j, At joint D: (b) For this part of the problem, we set h D 0.5 m. The unit vectors change because h is involved in the coordinates of point B. The new unit vectors are Fx D TCDX TBDX C DX D 0, and eCD D 0.832i C 0.555j. We get the force components as above, and the equilibrium forces at the joints remain the same. Solving the equilibrium equations simultaneously for this situation yields Fy D TCDY TBDY C DY D 0. TAB D 1.35 kN, Solve simultaneously to get TAC D 1.54 kN, TAB D TBD D 2.43 kN, TBC D 1.33, TAC D 2.78 kN, TBD D 1.74 kN, TBC D 0, TCD D 2.88 kN. and TCD D 1.60 kN. Note that with appropriate changes in the designation of points, the forces here are the same as those in Problem 6.4. This can be explained by noting from the unit vectors that AB and BC are parallel. Also note that in this configuration, BC carries no load. This geometry is the same as in Problem 6.4 except for the joint at B and member BC which carries no load. Remember member BC in this geometry — we will encounter things like it again, will give it a special name, and will learn to recognize it on sight. 390 These numbers differ significantly from (a). Most significantly, member BD is now carrying a compressive load and this has reduced the loads in all members except member BD. “Sharing the load” among more members seems to have worked in this case. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.7 This steel truss bridge is in the Gallatin National Forest south of Bozeman, Montana. Suppose that one of the tandem trusses supporting the bridge is loaded as shown. Determine the axial forces in members AB, BC, BD, and BE. B D F A C E MH : 10 kip17 ft C 10 kip34 ft C 10 kip51 ft G 10 kip 17 ft 17 ft Solution: We start with the entire structure in order to find the reaction at A. We have to assume that either A or H is really a roller instead of a pinned support. 10 kip 17 ft 10 kip 17 ft Finally work with joint B A68 ft D 0 ) A D 15 kip 17 17 FAB C p FBE C FBD D 0 Fx : p 353 353 8 8 FAB p FBE FBC D 0 Fy : p 353 353 Solving we find FBD D 42.5 kip, FBE D 11.74 kip B 17 ft 17 ft 8 8 A 10 kip 10 kip 17 10 kip H FBC 8 FAB C A D 0 ) FAB D 35.2 kip Fy : p 353 17 FBE FAB Now we examine joint A FBD 17 17 ft 17 ft 8 ft H FAB In Summary we have FAB D 35.2 kipC, FBC D 10 kipT, 8 FBD D 42.5 kipC, FBE D 11.74 kipT FAC A Now work with joint C Fy : FBC 10 kip D 0 ) FBC D 10 kip FBC FCE FAC C 10 kip c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 391 Problem 6.8 For the bridge truss in Problem 6.7, determine the largest tensile and compressive forces that occur in the members, and indicate the members in which they occur. Solution: Continuing the solution to Problem 6.7 will show the largest tensile and compressive forces that occur in the structure. Examining joint A we have 17 FAB C FAC D 0 ) FAC D 31.9 kip Fx : p 353 Examining joint C Fx : FAC C FCE D 0 ) FCE D 31.9 kip Examining joint D Fy : FDE D 0 ) FDE D 0 D FBD FDF FDE The forces in the rest of the members are found by symmetry. We have FAB D FFH D 35.2 kipC FAC D FGH D 31.9 kipT FBC D FFG D 10 kipT FBD D FDF D 42.5 kipC FBE D FEF D 11.74 kipT FCE D FEG D 31.9 kipT FDE D 0 The largest tension and compression members are then FAC D FEG D FCE D FGH D 31.9 kipT FBD D FDH D 42.5 kipC 392 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.9 The trusses supporting the bridge in Problems 6.7 and 6.8 are called Pratt trusses. Suppose that the bridge designers had decided to use the truss shown instead, which is called a Howe truss. Determine the largest tensile and compressive forces that occur in the members, and indicate the members in which they occur. Compare your answers to the answers to Problem 6.8. Solution: We start with the entire structure in order to find the reaction at A. We have to assume that either A or H is really a roller instead of a pinned support. MH : 10 kip17 ft C 10 kip34 ft C 10 kip51 ft B D F A H C 17 ft E 8 ft G 10 kip 17 ft 10 kip 17 ft 10 kip 17 ft Next work with joint C A68 ft D 0 ) A D 15 kip Fy : FBC C p 8 353 FCD 10 kip D 0 ) FCD D 11.74 kip 17 FCD FAC D 0 ) FCE D 42.5 kip Fx : FCE C p 353 FBC FCD 17 8 FAC FCE C A 10 kips 10 kips 10 kips H 10 kip Now we examine joint A Finally from joint E we find 8 FAB C A D 0 ) FAB D 35.2 kip Fy : p 353 17 FAB C FAC D 0 ) FAC D 31.9 kip Fx : p 353 FAB 17 Fy : FDE 10 kip D 0 ) FDE D 10 kip FDE FCE FEG E 8 FAC 10 kip The forces in the rest of the members are found by symmetry. We have A FAB D FFH D 35.2 kipC Now work with joint B FAC D FGH D 31.9 kipT FBD D FDF D 31.9 kipC 17 FAB C FBD D 0 ) FBD D 31.9 kip Fx : p 353 8 FAB FBC D 0 ) FBC D 15 kip Fy : p 353 B FBD 17 FBC D FFG D 15 kipT FCD D FDG D 11.74 kipC FCE D FEG D 42.5 kipT 8 FDE D 10 kipT FAB The largest tension and compression members are then FBC FCE D FEG D 42.5 kipT FAB D FFH D 35.2 kipC c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 393 Problem 6.10 Determine the axial forces in members BD, CD, and CE of the truss. G 300 mm E C F Solution: The free-body diagrams of the entire truss and of joints A, B, and C are shown. The angle 300 mm A ˛ D tan1 3/4 D 36.9° From the free-body diagram of the entire truss B 400 mm D 400 mm 6 kN 400 mm Fy : Ay 6 kN D 0 MG : 6 kN400 mm C Ax 600 mm Ay 1200 mm D 0 Solving, Ax D 8 kN, Ay D 6 kN Using joint A, Fx : Ax C TAB C TAC cos ˛ D 0 Fy : Ay C TAC sin ˛ D 0 Solving we find TAB D 0, TAC D 10 kN Because joint B consists of three members, two of which are parallel, and is subjected to no external load, we can recognize that TBD D TAB D 0 and TBD D 0 Finally we examine joint C Fx : TCE C TCD cos ˛ TAC cos ˛ D 0 Fy : TAC sin ˛ TCD sin ˛ TBC D 0 In summary 394 BD : 0, CD : 10 kN (T), ) TCD D 10 kN, TCE D 16 kN CE : 16 kN (C) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.11 The loads F1 D F2 D 8 kN. Determine the axial forces in members BD, BE, and BG. F1 D 3m F2 B E 3m Solution: First find the external support loads and then use the method of joints to solve for the required unknown forces. (Assume all unknown forces in members are tensions). G A C External loads: 4m F1 = 8 kN D y 3m B E C 8m DE D 6 kN C F2 = 8 kN Joint E : y x C AY BD D 10 kN T 3m G A AX Solving, 4m DE GY F2 = 8 kN BE Fx : Ax C F1 C F2 D 0 (kN) x Fy : Ay C Gy D 0 MA : 8Gy 3F2 6F1 D 0 EG Solving for the external loads, we get DE D 6 kN Ax D 16 kN to the left Ay D 9 kN downward Gy D 9 kN upward Now use the method of joints to determine BD, BE, and BG. Start with joint D. Fx D DE EG D 0 Fy D BE C F2 D 0 Solving: EG D 6 kN C BE D 8 kN T Joint D: y D F1 = 8 kN DE θ x BD cos D 0.8 sin D 0.6 D 36.87° Fx : F1 BD cos D 0 Fy : BD sin DE D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 395 6.11 (Continued ) Joint G: Fx : CG BG cos D 0 y EG BG Fy : BG sin C EG C Gy D 0 Solving, we get BG D 5 kN C θ x CG D 4 kN T CG GY Thus, we have BD D 10 kN T BE D 8 kN T BG D 5 kN C EG D 6 kN C Gy D 9 kN Problem 6.12 Determine the largest tensile and compressive forces that occur in the members of the truss, and indicate the members in which they occur if (a) (b) B D h C E the dimension h D 5 in; the dimension h D 10 in. 20 in A 20 in 20 in 30⬚ 800 lb Observe how a simple change in design affects the maximum axial loads. Solution: Starting at joint A Finally joint C 20 FAB FAC C 800 lb sin 30° D 0 Fx : p h2 C 202 h FAB 800 lb cos 30° D 0 Fy : p h2 C 202 20 20 FCD C p FBC FCE C FAC D 0 Fx : p h2 C 202 h2 C 202 h h FCD C p FBC D 0 Fy : p h2 C 202 h2 C 202 FBC FCD FAB 20 h 20 h 20 h A FAC C FCE FAC (a) Using h D 5 in we find: FAB D 2860 lbT, FAC D 2370 lbC, FBD D 5540 lbT 800 lb FBC D 2860 lbC, FCD D 2860 lbT, FCE D 7910 lbC Next joint B 20 20 FBC C p FAB D 0 Fx : FBD p h2 C 202 h2 C 202 FCE D 7910 lbC h h FBC p FAB D 0 Fy : p h2 C 202 h2 C 202 FBD B h 20 FBC 396 FBD D 5540 lbT ) (b) Using h D 10 in we find: FAB D 1549 lbT, FAC D 986 lbC, FBD D 2770 lbT FBC D 1549 lbC, FCD D 1549 lbT, FCE D 3760 lbC h 20 FBD D 2770 lbT FAB ) FCE D 3760 lbC c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.13 The truss supports loads at C and E. If F D 3 kN, what are the axial forces in members BC and BE? 1m 1m A 1m B D 1m G C E F 2F Solution: The moment about A is AY AX MA D 1F 4F C 3G D 0, 1m from which G D 5 F D 5 kN. The sums of forces: 3 F 1m FY D AY 3F C G D 0, 1m 2F 1m DG from which AY D 45° 4 F D 4 kN. 3 BD EG G Joint G DG DE 45° Joint D BE 45° CE DE EG Joint E FX D AX D 0, AY from which AX D 0. The interior angles GDE, EBC are 45° , AC AB 45° 45° BC AC CE F Joint C 1 from which sin ˛ D cos ˛ D p . 2 Joint A Denote the axial force in a member joining I, K by IK. from which (1) 5 BD D F D 5 kN C. 3 Joint G: DG Fy D p C G D 0, 2 from which p p p 5 2 F D 5 2 kN C. DG D 2G D 3 DG Fx D p EG D 0, 2 from which 5 DG EG D p D F D 5kN T. 3 2 (2) G Joint D: (3) Joint E : BE Fy D p 2F C DE D 0, 2 p p from which BE D 2 2F 2DE D p p 2 F D 2 kN T. 3 BE Fx D CE p C EG D 0, 2 from which 4 BE CE D EG p D F D 4 kN T. 3 2 DG Fy D DE p D 0, 2 from which DE D 5 F D 5 kN T. 3 DG Fx D BD C p D 0, 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 397 6.13 (Continued ) (4) Joint A: AC Fy D Ay p D 0, 2 from which AC D p p 4 2 F D 4 2 kN T. 3 AC Fx D AB C p D 0, 2 4 from which AB D F D 4 kN C. 3 (5) Joint C : AC Fy D BC C p F D 0, 2 AC 1 from which BC D F p D F D 1 kN C. 3 2 Problem 6.14 If you don’t want the members of the truss to be subjected to an axial load (tension or compression) greater than 20 kN, what is the largest acceptable magnitude of the downward force F? A 12 m F B Solution: Start with joint A 4m Fx : FAB cos 36.9° FAC sin 30.5° D 0 C D Fy : FAB sin 36.9° FAC cos 30.5° F D 0 3m A Finally examine joint D 36.9° Fy : FBD D 0 30.5° FBD FAB F FAC Dx Now work with joint C Fx : FCD FBC sin 36.9° FAC 36.9° FCD 398 Solving we find C FAC sin 30.5° Fy : FBC cos 36.9° C FAC cos 30.5° D 0 FBC 30.5° FCD D D0 FAB D 1.32F, FAC D 2.08F, FCD D 2.4F, FBC D 2.24F, FBD D 0 The critical member is CD. Thus 2.4F D 20 kN ) F D 8.33 kN C c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.15 The truss is a preliminary design for a structure to attach one end of a stretcher to a rescue helicopter. Based on dynamic simulations, the design engineer estimates that the downward forces the stretcher will exert will be no greater than 1.6 kN at A and at B. What are the resulting axial forces in members CF, DF, and FG? G 300 mm 290 mm 390 mm 150 mm F 480 mm C E D 200 mm B A Solution: Start with joint C 48 FCF 1.6 kN D 0 ) FCF D 2.06 kN Fy : p 3825 FCF 39 48 C FCD 1.6 kN Now use joint F 59 29 39 FFG p FDF C p FCF D 0 Fx : p 3706 3145 3825 15 48 48 FFG p FDF p FCF D 0 Fy : p 3706 3145 3825 Solving we find FDF D 1.286 kN, FCF D 2.03 kN 59 FFG 15 F 39 48 48 29 FDF FCF In Summary FCF D 2.06 kNT, FDF D 1.29 kNC, FCF D 2.03 kNT c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 399 Problem 6.16 Upon learning of an upgrade in the helicopter’s engine, the engineer designing the truss does new simulations and concludes that the downward forces the stretcher will exert at A and at B may be as large as 1.8 kN. What are the resulting axial forces in members DE, DF, and DG? Solution: Assume all bars are in tension. Next work with joint B Start at joint C 16 TCF 1.8 kN D 0 ) TCF D 2.32 kN Fy : p 425 13 TCF TCD D 0 ) TCD D 1.463 kN Fx : p 425 TBE 3 Fx : p TBE D 0 ) TBE D 0 13 2 Fy : p TBE C TBD 1.8 kN D 0 ) TBD D 1.8 kN 13 TBD TCF 2 13 B 3 16 C 1.8 kN TCD Finally work with joint D 1.8 kN Next work with joint F 59 29 13 TFG p TDF C p TCF D 0 Fx : p 3706 3145 425 10 29 TDG C p TDF C TCD D 0 Fx : TDE p 541 3145 21 48 TDG C p TDF TBD D 0 Fy : p 541 3145 Solving: TDG D 6.82 kN, TDE D 7.03 kN TDF TDG 15 48 48 TFG p TDF p TCF D 0 Fy : p 3706 3145 425 21 Solving TDF D 5.09 kN, TFG D 4.23 kN 48 29 10 59 TFG 15 TDE F D TCD 13 48 TBD 16 29 TDF TCF In summary: TDE D 7.03 kNC, TDF D 5.09 kNC, TDG D 6.82 kNT 400 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.17 Determine the axial forces in the members in terms of the weight W. B E 1m A D W 1m C 0.8 m 0.8 m 0.8 m Solution: Denote the axial force in a member joining two points I, K by IK. The angle between member DE and the positive x axis is ˛ D tan1 0.8 D 38.66° . The angle formed by member DB with the positive x axis is 90° C ˛. The angle formed by member AB with the positive x axis is ˛. Joint E : Fy D DE cos ˛ W D 0, from which DE D 1.28W C . Fy D BE DE sin ˛ D 0, from which BE D 0.8W T Joint D: Fx D DE cos ˛ C BD cos ˛ CD cos ˛ D 0, from which BD CD D DE. Fy D BD sin ˛ C DE sin ˛ CD sin ˛ D 0, from which BD C CD D DE. Solving these two equations in two unknowns: CD D DE D 1.28W C , BD D 0 Joint B : Fx D BE AB sin ˛ BD sin ˛ D 0, from which AB D BE D 1.28WT sin ˛ Fy D AB cos ˛ BC D 0, from which BC D AB cos ˛ D WC c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 401 Problem 6.18 The lengths of the members of the truss are shown. The mass of the suspended crate is 900 kg. Determine the axial forces in the members. A 12 m B 13 m 5m C 13 m 12 m D 40⬚ Solution: Start with joint A Finally work with joint B Fx : FAB cos 40° FAC sin 27.4° D 0 Fy : FAB cos 50° FBC sin 50° FBD cos 27.4° D 0 Fy : FAB sin 40° FAC cos 27.4° 900 kg9.81 m/s2 D 0 FAB 50° A 40° T B 50° 27.4° FAB 27.4° FBC FAC 8829 N Solving we find Next work with joint C FBD Fx : FCD cos 40° FBC cos 50° C FAC sin 27.4° D 0 FAB D 10.56 kN D 10.56 kNT FAC D 17.58 kN D 17.58 kNC Fy : FCD sin 40° C FBC sin 50° C FAC cos 27.4° D 0 FAC FBC 27.4° FCD D 16.23 kN D 16.23 kNC FBC D 6.76 kN D 6.76 kNT FBD D 1.807 kN D 1.807 kNT 50° C 40° FCD 402 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.19 The loads F1 D 600 lb and F2 D 300 lb. Determine the axial forces in members AE, BD, and CD. F1 G D F2 B 6 ft C 3 ft E A 4 ft 4 ft Solution: The reaction at E is determined by the sum of the moments about G: F1 GX MG D C6E 4F1 8F2 D 0, F2 GY 6 ft from which ED 4F1 C 8F2 D 800 lb. 6 E EG The interior angle EAG is ˛D tan1 6 D 36.87° . 8 E AC AE Joint E From similar triangles this is also the value of the interior angles ACB, CBD, and CGD. Method of joints: Denote the axial force in a member joining two points I, K by IK. from which BD D Joint E : 4 ft α AE AB 4 ft BD α BC Joint A F2 F1 DG AB Joint B CD α BD Joint D 300 F2 C AB D D 500 lbC . 0.6 0.6 Fx D BC BD cos ˛ D 0, Fy D E C AE D 0, from which BC D BD0.8 D 400 lbT. from which AE D E D 800 lb C . Joint D: Fy D EG D 0, from which EG D 0. Fy D BD sin ˛ CD F1 D 0, from which CD D F1 BD0.6 D 300 lbC Joint A: Fy D AE AC cos ˛ D 0, from which AC D AE D 1000 lbT. 0.8 Fy D AC sin ˛ C AB D 0, from which AB D AC0.6 D 600 lbC. Joint B : Fy D BD sin ˛ AB F1 D 0, c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 403 Problem 6.20 Consider the truss in Problem 6.19. The loads F1 D 450 lb and F2 D 150 lb. Determine the axial forces in members AB, AC, and BC. Solution: From the solution to Problem 6.19 the angle ˛ D 36.87° 4F1 C 8F2 D 500 lb. Denote the axial 6 force in a member joining two points I, K by IK. EG and the reaction at E is E D E Joint E : AE Joint E AC AB α AE Joint A BD α BC F2 AB Joint B Fy D EG D 0. Fx D AE C E D 0, from which AE D E D 500 lbC. Joint A: Fx D AE AC cos ˛ D 0, from which AC D AE D 625 lbT . 0.8 Fy D AC sin ˛ C AB D 0, from which AB D AC0.6 D 375 lbC Joint B: Fy D BD sin ˛ F2 AB D 0, from which BD D F2 C AB D 375 lbC 0.6 Fx D BC BD cos ˛ D 0, from which BC D BD0.8 D 300 lbT 404 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. C Problem 6.21 Determine the axial forces in members BD, CD, and CE of the truss. E 4 ft G B D F 12 kip 4 ft H A 4 ft 4 ft 4 ft Solution: The free-body diagrams for the entire truss as well as for joints A, B and C are shown. From the entire truss: Fx : Ax D 0 FH : 12 kip8 ft Ay 12 ft D 0 Solving, yields Ax D 0, Ay D 8 kip From joint A: Fx : Ax C TAD cos 45° D 0 Fy : Ay C TAB C TAD sin 45° D 0 Solving yields TAB D 8 kip, TAD D 0 From joint B: Fx : TBD C TBC cos 45° D 0 Fy : TBC C sin 45° TAB D 0 Solving yields TBD D 8 kip, TBC D 11.3 kip From joint C: Fx : TCE TBC cos 45° D 0 Fy : TBC sin 45° TCD D 0 Solving yields TCD D 8 kip, TCE D 8 kip Thus we have BC : 11.3 kip (C), CD : 8 kip (T), CE : 8 kip (C) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 405 Problem 6.22 The Warren truss supporting the walkway is designed to support vertical 50-kN loads at B, D, F, and H. If the truss is subjected to these loads, what are the resulting axial forces in members BC, CD, and CE? B D F H 2m A C 6m Solution: Assume vertical loads at A and I Find the external loads at A and I, then use the method of joints to work through the structure to the members needed. 50 kN 50 kN 6m 50 kN 6m 3m AY 3m 6m I 6m D 33.69° 6m 6m G AB D 180.3 kN 50 kN E Fx : BC cos C BD AB cos D 0 Fy : 50 AB sin BC sin D 0 x IY Solving, BC D 90.1 kN T BD D 225 kN C Fy : Ay C Iy 450 D 0 (kN) MA : Solving Joint C : 350 950 1550 2150 C 24 Iy D 0 y Ay D 100 kN BC Iy D 100 kN θ CD θ Joint A: AC y C CE x D 33.69° AB AC D 150 kN T BC D 90.1 kN T θ A x AC AY CE D 300 kN T D 33.69° Fy : CD sin C BC sin D 0 Solving, 2 3 tan D Fx : CE AC C CD cos BC cos D 0 CD D 90.1 kN C Fx : AB cos C AC D 0 Fy : AB sin C Ay D 0 Hence Solving, BC D 90.1 kN T CD D 90.1 kN C CE D 300 kN T AB D 180.3 kN C AC D 150 kN T Joint B : y 50 kN B θ BD x θ BC AB 406 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.23 For the Warren truss in Problem 6.22, determine the axial forces in members DF, EF, and FG. Solution: In the solution to Problem 6.22, we solved for the forces Solving, we get in AB, AC, BC, BD, CD, and CE. Let us continue the process. We ended with Joint C. Let us continue with Joint D. EF D 0 Joint D: EG D 300 kN T y Note: The results are symmetric to this point! 50 kN Joint F : D BD DF y DF DE CD 50 kN x θ θ x θ θ D 33.69° FH F BD D 225 kN C EF CD D 90.1 kN C D 33.69° Fx : DF BD C DE cos CD cos D 0 Fy : 50 CD sin DE sin D 0 Solving, FG DF D 300 kN C EF D 0 DF D 300 kN C DE D 0 At this point, we have solved half of a symmetric truss with a symmetric load. We could use symmetry to determine the loads in the remaining members. We will continue, and use symmetry as a check. Joint E : FH DF C FG cos EF cos D 0 Fy : 50 EF sin FG sin D 0 Solving: FH D 225 kN C FG D 90.1 kN C Thus, we have DF D 300 kN C EF D 0 FG D 90.1 kN C y EF DE Fx : Note-symmetry holds! θ θ CE E EG x D 33.69° CE D 300 kN T DE D 0 Fx : EG CE C EF cos DE cos D 0 Fy : DE sin C EF sin D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 407 Problem 6.24 The Pratt bridge truss supports five forces (F D 300 kN). The dimension L D 8 m. Determine the axial forces in members BC, BI, and BJ. L L L L L L B C D E G I J K L M L A H F Solution: Find support reactions at A and H. From the free body and Fx D AX D 0, L 8 MA D 68HY 3008 C 16 C 24 C 32 C 40 D 0. L 8 L K L 8 L 8 F L 8 L H HY Joint I y TAB TBI θ TAI F F F F = 300 kN y A M L 8 Joint A From the geometry, the angle D 45° J F F L=8m Fy D AY C HY 5300 D 0, From these equations, AY D HY D 750 kN. F G θ I A AY Joint A: From the free body diagram, F B diagram, F I x TIJ TAI x F Fx D AX C TAB cos C TAI D 0, AY Joint B y Fy D TAB sin C AY D 0. TBC From these equations, TAB D 1061 kN and TAI D 750 kN. θ θ x TBJ TAB TBI Joint I: From the free body diagram, Fx D TIJ TAI D 0, Fy D TBI 300 D 0. From these equations, TBI D 300 kN and TIJ D 750 kN. Joint B: From the free body diagram, Fx D TBC C TBJ cos TAB cos D 0, Fy D TBI TBJ sin TAB sin D 0. From these equations, TBC D 1200 kN and TBJ D 636 kN. 408 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.25 For the roof truss shown, determine the axial forces in members AD, BD, DE, and DG. Model the supports at A and I as roller supports. 10 kN 8 kN E 6 kN C 8 kN F B 6 kN H D 3m 3m Solution: Use the whole structure to find the reaction at A. Next go to joint C MI : 6 kN3 m C 8 kN6 m C 10 kN9 m C 8 kN12 m C 6 kN15 m C A18 m D 0 ) A D 19 kN G 3m 3m 3m 3m Fy : 8 kN FCD C FCE FBC sin 21.8° D 0 Fx : FCE FBC cos 21.8° D 0 Solving: FCD D 8 kN, FCE D 43.1 kN 10 kN 6 kN 3.6 m I A 8 kN 8 kN 8 kN FCD 6 kN C FCD FBC A I Finally examine joint D Now work with joint A Fx : FAD C FDG FBD cos 21.8° C FDE cos 50.19° D 0 Fy : FAB sin 21.8° C A D 0 ) FAB D 51.2 kN Fx : FAD C FAB cos 21.8° D 0 ) FAD D 47.5 kN Solving: Fy : FBD sin 21.8° C FCD C FDE sin 50.19° D 0 FDE D 14.3 kN, FDG D 30.8 kN FCD FAB FDE FBD 21.8° A FAD 50.19° FAD A FDG In Summary Next use joint B D Fx : FAB C FBC C FBD cos 21.8° D 0 FAD D 47.5 kNT, FBD D 8.08 kNC, FDE D 14.32 kNT, FDG D 30.8 kNT Fy : Solving: FAB C FBC FBD sin 21.8° 6 kN D 0 FBC D 43.1 kN, FBD D 8.08 kN 6 kN FBC B FAB FBD c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 409 Problem 6.26 The Howe truss helps support a roof. Model the supports at A and G as roller supports. Determine the axial forces in members AB, BC, and CD. 800 lb 600 lb 600 lb D 400 lb 400 lb C E 8 ft B F A G H I 4 ft 4 ft J 4 ft Solution: The strategy is to proceed from end A, choosing joints 8 12 BH D 4 1400 lb BI α Pitch from which the angle ˛HIB D tan1 CI D 8 2.6667 4 D 33.7° . from which the angle ˛IJC D tan1 5.333 4 CI AH HI Joint H 600 lb CD 400 lb α Pitch BC α Pitch AB BH BI Joint B α Pitch α IJC BC CI CJ Joint C Joint H : BH AH HI IJ Joint I 8 D 5.3333 ft, 12 600 lb 400 lb G α Pitch Joint A D 2.6667 ft, 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft 4 ft D 33.7° . AB 8 12 4 ft A The length of the vertical members: L 800 lb The interior angles HIB and HJC differ. The pitch angle is ˛Pitch D tan1 4 ft 600 lb 400 lb with only one unknown axial force in the x- and/or y-direction, if possible, and if not, establish simultaneous conditions in the unknowns. K Fy D BH D 0, or, BH D 0. Fx D AH C HI D 0, D 53.1° . from which HI D 2100 lb T The moment about G: MG D 4 C 20400 C 8 C 16600 C 12800 24A D 0, 33600 D 1400 lb. Check: The total load is 2800 lb. 24 From left-right symmetry each support A, G supports half the total load. check. from which A D Joint B : Fx D AB cos ˛Pitch C BC cos ˛Pitch C BI cos ˛Pitch D 0, from which BC C BI D AB The method of joints: Denote the axial force in a member joining two points I, K by IK. Joint A: Fy D AB sin ˛P C 1400 D 0, from which AB D 1400 D 2523.9 lb C sin ˛p Fx D AB cos ˛Pitch C AH D 0, from which AH D 2523.90.8321 D 2100 lb T 410 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6.26 (Continued ) Fy D 400 AB sin ˛Pitch C BC sin ˛Pitch BI sin ˛Pitch D 0, from which BC BI D AB C 400 . sin ˛Pitch Solve the two simultaneous equations in unknowns BC, BI: BI D and 400 D 360.56 lb C, 2 sin ˛Pitch BC D AB BI D 2163.3 lb C Joint I : Fx D BI cos ˛Pitch HI C IJ D 0, from which IJ D 1800 lb T Fy D CBI sin ˛Pitch C CI D 0, from which CI D 200 lb (T) Joint C: Fx D BC cos ˛Pitch C CD cos ˛Pitch C CJ cos ˛IJC D 0, from which CD0.8321 C CJ0.6 D 1800 Fy D 600 CI BC sin ˛Pitch C CD sin ˛Pitch CJ sin ˛IJC D 0, from which CD0.5547 CJ0.8 D 400 Solve the two simultaneous equations to obtain CJ D 666.67 lb C, and CD D 1682.57 lb C c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 411 Problem 6.27 The plane truss forms part of the supports of a crane on an offshore oil platform. The crane exerts vertical 75-kN forces on the truss at B, C, and D. You can model the support at A as a pin support and model the support at E as a roller support that can exert a force normal to the dashed line but cannot exert a force parallel to it. The angle ˛ D 45° . Determine the axial forces in the members of the truss. C B D 1.8 m 2.2 m F G H A α E 3.4 m 3.4 m 3.4 m 3.4 m Solution: The included angles D tan1 ˇ D tan1 D tan1 4 3.4 2.2 3.4 1.8 3.4 75 kN 75 kN 75 kN D 49.64° , D 32.91° , AY 3.4 m D 27.9° . AB AX Ex cos 45° C Ey cos 45° with this relation and the fact that obtain Ex D 112.5 kN and Ey D 112.5 kN. From D 0, we FAy D Ay 375 C Ey D 0, The method of joints: Denote the axial force in a member joining two points I, K by IK. Joint A: 3.4 m EY 3.4 m BF DE EX AF γ β CD θ DG EY Joint E 75 kN γ DH Joint D DE β AF DH FG Joint F 75 kN BC CD GH β EH Joint H CG Joint C FAx D Ax C Ex D 0, AX D EX D 112.5 kN. from which Ay D 112.5 kN. Thus the reactions at A and E are symmetrical about the truss center, which suggests that symmetrical truss members have equal axial forces. β γ EH AY Joint A 75 kN BC θ γ BF BG AB Joint B MA D 753.41 C 2 C 3 C 43.4Ey D 0. 3.4 m The complete structure as a free body: The sum of the moments about A is EX AX Fy D AB sin C Ay C AF sin ˇ D 0, from which two simultaneous equations are obtained. AF D 44.67 kN C , and and EH D 44.67 kNC , DE D 115.8 kNC Joint F : Fx D AF cos ˇ C FG D 0, from which FG D 37.5 kN C Fx D AB cos C Ax C AF cos ˇ D 0, Solve: Solve: Fy D AF sin ˇ C BF D 0, from which BF D 24.26 kN C Joint H: Fx D EH cos ˇ GH D 0, AB D 115.8 kN C Joint E: Fy D DE cos C Ex EH cos ˇ D 0. Fy D DE sin C Ey C EH sin ˇ D 0, from which two simultaneous equations are obtained. 412 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 6.27 (Continued ) from which GH D 37.5 kN C from which DG D 80.1 kN T Fy D EH sin ˇ C DH D 0, Fx D DE cos CD DG cos D 0, from which DH D 24.26 kN C from which CD D 145.8 kN C Joint B: Joint C : Fy D AB sin BF C BG sin 75 D 0, Fx D CD BC D 0, from which BG D 80.1 kN T from which CD D BC Check. Fx D AB cos C BC C BG cos D 0, from which BC D 145.8 kN C Fy D CG 75 D 0, from which CG D 75 kN C Joint D: Fy D DE sin DH DG sin 75 D 0, Problem 6.28 (a) Design a truss attached to the supports A and B that supports the loads applied at points C and D. (b) Determine the axial forces in the members of the truss you designed in (a) 1000 lb 2000 lb C D 4 ft A 5 ft 5 ft 5 ft Obstacle Problem 6.29 (a) Design a truss attached to the supports A and B that goes over the obstacle and supports the load applied at C. (b) Determine the axial forces in the members of the truss you designed in (a). 2 ft B C 4m 2m B A 6m 3.5 m 10 kN 4.5 m 1m Solution: This is a design problem with many possible solutions. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 413 Problem 6.30 Suppose that you want to design a truss supported at A and B (Fig. a) to support a 3-kN downward load at C. The simplest design (Fig. b) subjects member AC to 5-kN tensile force. Redesign the truss so that the largest force is less than 3 kN. A A 1.2 m C C B B 3 kN 3 kN 1.6 m (a) Solution: There are many possible designs. To better understand D 36.87° the problem, let us calculate the support forces in A and B and the forces in the members in Fig. (b). Ay Fx : (b) BC AC cos D 0 Fy : AC sin 3 kN D 0 Solving: BC D 4 kN C AC D 5 kN C Ax A Thus, AC is beyond the limit, but BC (in compression) is not, Joint B : 1.2 m θ AB C B Bx x 1.6 m BX BC 3 kN 1.2 1.6 tan D D 36.87° sin D 0.6 Fx : Ax C Bx D 0 Fy : Ay 3 kN D 0 MA : 1.2Bx 1.63 D 0 C Fy : AB D 0 Solving, BC and Bx are both already known. We get AB D 0 cos D 0.8 Fx : Bx C BC D 0 Thus, we need to reduce the load in AC. Consider designs like that shown below where D is inside triangle ABC. Move D around to adjust the load. A Solving, we get Ax D 4 kN Bx D 4 kN D Ay D 3 kN Note: These will be the external reactions for every design that we produce (the supports and load do not change). B C Reference Solution (Fig. (b)) Joint C : However, the simplest solution is to place a second member parallel to AC, reducing the load by half. AC θ BC 3 kN 414 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.31 The bridge structure shown in Example 6.2 can be given a higher arch by increasing the 15° angles to 20° . If this is done, what are the axial forces in members AB, BC, CD, and DE? F F b F b F b F b 2b (1) F F G F F I J H b a b 15⬚ b 15⬚ K b 2b C D B F a A E (2) Solution: Follow the solution method in Example 6.3. F is known For joint C, Joint B : y F Fx : TBC cos 20° C TCD cos 20° D 0 Fy : F TBC sin 20° TCD sin 20° D 0 TBC TBC D TCD D 1.46F C For joint B. 20° x α TAB Fx : TBC cos 20 TAB cos ˛ D 0 Fy : TBC sin 20° F TAB sin ˛ D 0 Solving, we get ˛ D 47.5° and TAB D 2.03F C Joint C : For the new truss (using symmetry) F C 20° TBC 20° TCD Members Forces AG, BH, CI, DJ, EK F AB, DE 2.03F (C) BC, CD 1.46F (C) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 415 Problem 6.32 In Active Example 6.3, use the method of sections to determine the axial forces in members BC, BI and HI. A B C D E F G H I J K L 1m M 100 kN Solution: The horizontal members of the truss are each 1 m in length. We cut through the relevant members and draw a free-body diagram of the section to the right of the cut. We will use equilibrium equations for this section that are designed to allow us to easily solve for the unknowns. The equilibrium equations MI : TBC 1 m 100 kN4 m D 0 ) TBC D 400 kN MB : THI 1 m 100 kN5 m D 0 ) THI D 500 kN Fy : TBI sin 45° 100 kN D 0 ) TBI D 141 kN In summary we have BC : 400 kN (T), BI : 141 kN (T), HI : 500 kN (C) Problem 6.33 In Example 6.4, obtain a section of the truss by passing planes through members BE, CE, CG, and DG. Using the fact that the axial forces in members DG and BE have already been determined, use your section to determine the axial forces in members CE and CG. G D J L C I L A B E F L H 2F L K F L L Solution: From Example 6.4 we know that TDG D F, TBE D F Ax D 0, Ay D 2F We make the indicated cuts and isolate the section to the left of the cuts. The equilibrium equations are Fx : TDG C TBE C TCG cos 45° C TCE cos 45° D 0 Fy : Ay F C TCG sin 45° TCE sin 45° D 0 F F Solving yields TCE D p , TCG D p 2 2 We have 416 F CE : p T, 2 F CG : p C 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.34 The truss supports a 100-kN load at J. The horizontal members are each 1 m in length. (a) (b) Use the method of joints to determine the axial force in member DG. Use the method of sections to determine the axial force in member DG. A B C D E F G H 1m J 100 kN Solution: (a) We draw free-body diagrams of joints J, H, and D. From joint J we have Fy : TDJ sin 45° 100 kN D 0 ) TDJ D 141 kN From joint H we have Fy : TDH D 0 From joint D we have Fy : TDG sin 45° TDH TDJ sin 45° D 0 Solving yields TDG D 141 kN (b) We cut through CD, DG and GH. The free-body diagram of the section to the right of the cut is shown. From this diagram we have Fy : TDG sin 45° 100 kN D 0 ) TDG D 141 kN In summary (a), (b) DG : 141 kN (C) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 417 Problem 6.35 For the truss in Problem 6.34, use the method of sections to determine the axial forces in members BC, CF, and FG. BC Solution: Fx : C D 45° BC CF cos 45 FG D 0 1m CF CF sin 45° 100 D 0 Fy : J F MC : FG G 1FG 2100 D 0 H 1m 1m 100 kN Solving BC D 300 kN T CF D 141.4 kN C FG D 200 kN C Problem 6.36 Use the method of sections to determine the axial forces in members AB, BC, and CE. 1m 1m A B 1m D 1m G C E F 2F Solution: First, determine the forces at the supports Θ = 45° D B Method of Sections: y AY = 1. 33 F AX AY C 1m F E 1m 1m θ 1m AX = 0 AX = 0 Fx : Ax D 0 Fy : Ay C Gy 3F D 0 MA : 1F 22F C 3Gy D 0 C 1m AY 1m C CE x F Fx : CE C AB D 0 Fy : BC C Ay F D 0 MB : 1Ay C 1CE D 0 Solving B BC GY 2F AB Ax D 0 Gy D 1.67F Ay D 1.33F C Solving, we get AB D 1.33F C CE D 1.33F T BC D 0.33F C 418 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 18 kN Problem 6.37 Use the method of sections to determine the axial forces in members DF, EF, and EG. 24 kN C E G H A 300 mm B F D 400 mm 400 mm 400 mm 400 mm Solution: We will first use the free-body diagram of the entire structure to find the reaction at F. MB : 18 kN 400 mm 24 kN 1200 mm C F 800 mm D 0 ) F D 27 kN Next we cut through DF, EF, EG and look at the section to the right of the cut. The angle ˛ is given by ˛ D tan1 3/4 D 36.9° The equilibrium equations are MF : TEG 300 mm 24 kN 400 mm D 0 ME : TDF 300 mm 24 kN 800 mm C F400 mm D 0 Fy : F 24 kN C TEF sin ˛ D 0 Solving yields TDF D 28 kN, TEF D 5 kN, TEG D 32 kN Thus DF : 28 kN (C), EF : 5 kN (C), EG : 32 kN (T) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 419 Problem 6.38 The Pratt bridge truss is loaded as shown. Use the method of sections to determine the axial forces in members BD, BE, and CE. B D F A H C 17 ft E 8 ft G 10 kip 30 kip 20 kip 17 ft 17 ft 17 ft Solution: Use the whole structure to find the reaction at A. MH : 20 kip17 ft C 30 kip34 ft C 10 kip51 ft A68 ft D 0 10 kip A 20 kip 30 kip H ) A D 27.5 kip B Now cut through BD, BE, CE and use the left section FBD 8 MB : A17 ft C FCE 8 ft D 0 ) FCE D 58.4 kip 17 FBE ME : 10 kip17 ft A34 ft FBD 8 ft D 0 C A FCE ) FBD D 95.6 kip Fy : A 10 kip p 8 353 FBE D 0 ) FBE D 41.1 kip 10 kip In Summary A FCE D 58.4 kipT, FBD D 95.6 kipC, FBE D 41.1 kipT Problem 6.39 The Howe bridge truss is loaded as shown. Use the method of sections to determine the axial forces in members BD, CD, and CE. B D F A H C 17 ft E 10 kip 17 ft G 30 kip 17 ft 20 kip 17 ft B Solution: Use the whole structure to find the reaction at A (same FBD as 6.38) A D 27.5 kip Now cut through BD, CD, and CE and use the left section. MC : A17 ft FBD 8 ft D 0 ) FBD D 58.4 kip 17 FCD 8 C A FCE MD : A34 ft C 10 kip17 ft C FCE 8 ft D 0 ) FCE D 95.6 kip 8 ft Fy : A 10 kip C p 10 kip 8 353 A FCD D 0 ) FCD D 41.1 kip In Summary FBD D 58.4 kipC, FCE D 95.6 kipT, FCD D 41.1 kipC 420 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.40 For the Howe bridge truss in Problem 6.39, use the method of sections to determine the axial forces in members DF, DG, and EG. D Solution: Same truss as 6.39. 17 FDG MD : A34 ft C 10 kip17 ft C FEG 8 ft D 0 E ) FEG D 95.6 kip FDF 8 Cut through DF, DG, and EG and use left section FEG MG : A51 ft C 10 kip34 ft C 30 kip17 ft FDF 8 ft A 30 kip 10 kip D 0 ) FDF D 69.1 kip 8 FDG D 0 ) FDG D 29.4 kip Fy : A 10 kip 30 kip p 353 In summary FEG D 95.6 kipT, FDF D 69.1 kipC, FDG D 29.4 kipC Problem 6.41 The Pratt bridge truss supports five forces F D 340 kN. The dimension L D 8 m. Use the method of sections to determine the axial force in member JK. L L L L L L B C D E G I J K L M L A H Solution: First determine the external support forces. F AX L L L F AY F L F L F HY F = 340 kN, L = 8 M C Solving: F F F D 45° L F F L D 8M F D 340 kN Ay D 850 kN Fx : Ax D 0 Fy : Ay 5F C Hy D 0 MA : 6LHy LF 2LF 3LF 4LF 5LF D 0 Ax D 0, C Ay D 850 kN Fx : CD C JK C CK cos D 0 Fy : Ay 2F CK sin D 0 MC : LJK C LF 2LAy D 0 Solving, JK D 1360 kN T Also, CK D 240.4 kN T Hy D 850 kN Note the symmetry: Method of sections to find axial force in member JK. C B θ I A L CD D 1530 kN C CD D CK J K L AY F F JK c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 421 Problem 6.42 For the Pratt bridge truss in Problem 6.41, use the method of sections to determine the axial force in member EK. Solution: From the solution to Problem 6.41, the support forces L are Ax D 0, Ay D Hy D 850 kN. L L L L L B C D E G I J K L M Method of Sections to find axial force in EK. L E DE A G H θ F EK EK D 240.4 kN T Also, KL D 1360 kN T F F F F HY DE D 1530 kN C Solution: F L KL F Fx : DE EK cos KL D 0 Fy : Hy 2F EK sin D 0 ME : LKL LF C 2LHy D 0 Problem 6.43 The walkway exerts vertical 50-kN loads on the Warren truss at B, D, F, and H. Use the method of sections to determine the axial force in member CE. B D F H 2m A C 6m Solution: First, find the external support forces. By symmetry, Solving: Ay D Iy D 100 kN (we solved this problem earlier by the method of joints). Also, B y 50 kN BD 6m C 6m G 6m I 6m CE D 300 kN T BD D 225 kN C CD D 90.1 kN C D 2m A E CD θ CE x AY tan D 2 3 D 33.69° 422 Fx : BD C CD cos C CE D 0 Fy : Ay 50 C CD sin D 0 MC : 6Ay C 350 2BD D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.44 Use the method of sections to determine the axial forces in members AC, BC, and BD. 600 lb E 4 ft C D 4 ft A B 3 ft 3 ft Solution: Obtain a section by passing a plane through members AC, BC, and BD, isolating the part of the truss above the planes. The angle between member AC and the horizontal is ˛ D tan1 4/3 D 53.3° The equilibrium equations are MC : 600 lb 4 ft TBD cos ˛ 3 ft D 0 MB : 600 lb 8 ft C TAC sin ˛ 4 ft D 0 Fy : TBC TAC cos ˛ TBD cos ˛ D 0 Solving yields TBD D 1000 lb, TAC D 2000 lb, TBC D 800 lb BD : 100 lb (T), AC : 2000 lb (C), BC : 800 lb (T) Thus c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 423 Problem 6.45 Use the method of sections to determine the axial forces in member FH, GH, and GI. I 300 mm C E G H 300 mm A B D F 6 kN 400 mm 400 mm 400 mm 4 kN 400 mm Solution: The free-body diagram of the entire truss is used to find the force I. MA : I600 mm 4 kN 1200 mm 6 kN 800 mm D 0 ) I D 16 kN Obtain a section by passing a plane through members FH, GH, and GI, isolating the part of the truss to the right of the planes. The angle ˛ is ˛ D tan1 3/4 D 36.9° The equilibrium equations for the section are MH : TGI cos ˛ 300 mm C I300 mm D 0 MG : I300 mm TFH cos ˛ 400 mm D 0 Fx : TGH TGI sin ˛ TFH sin ˛ D 0 Solving yields TGI D 20 kN, TFH D 20 kN, TGH D 16 kN Thus 424 GI : 20 kN (C), FH : 20 kN (T), GH : 16 kN (C) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.46 Use the method of sections to determine the axial forces in member DF, DG, and EG. I 300 mm C E G H 300 mm A B D F 6 kN 400 mm 400 mm 400 mm 4 kN 400 mm Solution: The free-body diagram of the entire truss is used to find the force I. MA : I600 mm 4 kN 1200 mm 6 kN 800 mm D 0 ) I D 16 kN Obtain a section by passing a plane through members DF, DG, and EG, isolating the part of the truss to the right of the planes. The angle ˛ is ˛ D tan1 3/4 D 36.9° The equilibrium equations for the section are MG : I 300 mm TDF 300 mm D 0 MD : TEG 300 mm C I600 mm 4 kN400 mm D 0 Fy : TDG sin ˛ 4 kN D 0 Solving yields TDF D 16 kN, TEG D 26.7 kN, TDG D 6.67 kN Thus DF : 16 kN (T), EG : 26.7 kN (C), DG : 6.67 kN (C) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 425 Problem 6.47 The Howe truss helps support a roof. Model the supports at A and G as roller supports. 2 kN 2 kN 2 kN (a) (b) Use the method of joints to determine the axial force in member BI. Use the method of sections to determine the axial force in member BI. D 2 kN 2 kN C E 4m B F A G Solution: The pitch of the roof is ˛ D tan1 H 4 D 33.69° . 6 2m I 2m J K 2m L 2m 2m 2m F F This is also the value of interior angles HAB and HIB. The complete structure as a free body: The sum of the moments about A is F F = 2 kN F MA D 221 C 2 C 3 C 4 C 5 C 62G D 0, G A 30 from which G D D 5 kN. The sum of the forces: 6 FY D A 52 C G D 0, 2m 2m 2m 2m 2m 2m BH AB from which A D 10 5 D 5 kN. The method of joints: Denote the axial force in a member joining I, K by IK. (a) (a) A α AH Joint A AH HI 2 kN BH Joint B Joint H Joint A: F B Fy D A C AB sin ˛ D 0, α (b) 5 A D D 9.01 kN (C). from which AB D sin ˛ 0.5547 Fx D AB cos ˛ C AH D 0, BC α BI α AB A BC α α BI HI 2m from which AH D AB cos ˛ D 7.5 kN (T). Joint H : Fy D BH D 0. Joint B : Fx D AB cos ˛ C BI cos ˛ C BC cos ˛ D 0, Fy D 2 AB sin ˛ BI sin ˛ C BC sin ˛ D 0. Solve: BI D 1.803 kN C , BC D 7.195 kN C (b) Make the cut through BC, BI and HI. The section as a free body: The sum of the moments about B: MB D A2 C HI2 tan ˛ D 0, from which HI D 3 A D 7.5 kNT. The sum of the forces: 2 Fx D BC cos ˛ C BI cos ˛ C HI D 0, Fy D A F C BC sin ˛ BI sin ˛ D 0. Solve: BI D 1.803 kN C . 426 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.48 Consider the truss in Problem 6.47. Use the method of sections to determine the axial force in member EJ. Solution: From the solution to Problem 6.47, the pitch angle is ˛ D 36.69° , and the reaction G D 5 kN. The length of member EK is LEK D 4 tan ˛ D 16 D 2.6667 m. 6 The interior angle KJE is ˇ D tan1 LEK 2 DE β F E F EJ α JK 2m G 2m D 53.13° . Make the cut through ED, EJ, and JK. Denote the axial force in a member joining I, K by IK. The section as a free body: The sum of the moments about E is ME D C4G 2F JK2.6667 D 0, from which JK D 20 4 D 6 kN T. 2.6667 The sum of the forces: Fx D DE cos ˛ EJ cos ˇ JK D 0. Fy D DE sin ˛ EJ sin ˇ 2F C G D 0, from which the two simultaneous equations: 0.8321DE C 0.6EJ D 6, 0.5547DE 0.8EJ D 1. Solve: EJ D 2.5 kN C . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 427 C Problem 6.49 Use the method of sections to determine the axial forces in member CE, DE, and DF. E 4 ft G B D F 12 kip 4 ft H A 4 ft 4 ft 4 ft Solution: The free-body diagrams for the entire structure and the section to the right of the cut are shown. From the entire structure: MA : 12 kip 4 ft H 12 ft D 0 ) H D 4 kip Using the section to the right of the cut we have ME : H4 ft TDF 4 ft D 0 MD : H8 ft C TCE 4 ft D 0 Fy : H TDE sin 45° D 0 Solving yields TDF D 4 kip, TCE D 8 kip, TDE D 5.66 kip Thus we have DF : 4 kip (T) CE : 8 kip (C) DE : 5.66 kip (T) 428 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.50 For the bridge truss shown, use the method of sections to determine the axial forces in members CE, CF, and DF. 200 kN 200 kN 200 kN D B 200 kN 200 kN F H J E G C 3m 7m 4m I A 5m 5m 5m 5m Solution: From the entire structure we find the reactions at A Now we cut through DF, CF, and CE and use the left section. Fx : Ax D 0 ) FDF D 375 kN MI : 200 kN5 m C 200 kN10 m C 200 kN15 m C 200 kN20 m Ay 20 m D 0 ) Ay D 500 kN 200 kN 200 kN 200 kN 200 kN MC : 200 kN5 m Ay 5 m C Ax 3 m FDF 4 m D 0 200 kN MF : 200 kN10 m C 200 kN5 m Ay 10 m C Ax 7 m 5 1 C p FCE 4 m p FCE 5 m D 0 ) FCE D 680 kN 26 26 5 5 Fx : Ax C FDF C p FCE C p FCF D 0 26 41 ) FCF D 374 kN 200 kN 200 kN Ax D I FDF Ay FCF 5 4 FCE 1 C 5 Ax Ay Summary: FDF D 375 kNC, FCE D 680 kNT, FCF D 374 kNC c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 429 Problem 6.51 The load F D 20 kN and the dimension L D 2 m. Use the method of sections to determine the axial force in member HK. L L A B C F Strategy: Obtain a section by cutting members HK, HI, IJ, and JM. You can determine the axial forces in members HK and JM even though the resulting freebody diagram is statically indeterminate. L E D F G L I H J K M L Solution: The complete structure as a free body: The sum of the 2L F moments about K is MK D FL2 C 3 C ML2 D 0, from which 5F D 50 kN. The sum of forces: MD 2 FY D KY C M D 0, L F 2L from which KY D M D 50 kN. FX D KX C 2F D 0, KX M KY from which KX D 2F D 40 kN. The section as a free body: Denote the axial force in a member joining I, K by IK. The sum of the forces: from which HI IJ D Kx . Sum moments about K to get MK D ML2 C JML2 IJL C HIL D 0. HK KX Fx D Kx HI C IJ D 0, Substitute HI IJ D Kx , to obtain JM D M HI IJ JM L M KY 2L Kx D 30 kN C. 2 Fy D Ky C M C JM C HK D 0, from which HK D JM D 30 kNT 430 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.52 The weight of the bucket is W D 1000 lb. The cable passes over pulleys at A and D. D A C (a) Determine the axial forces in member FG and HI. (b) By drawing free-body diagrams of sections, explain why the axial forces in members FG and HI are equal. F B H 3 ft 6 in J 3 ft E 3 ft 3 in L G I 35° 3 ft W K Solution: The truss is at angle ˛ D 35° relative to the horizontal. The angles of the members FG and HI relative to the horizontal are ˇ D 45° C 35° D 80° . (a) Make the cut through FH, FG, and EG, and consider the upper section. Denote the axial force in a member joining, ˛, ˇ by ˛ˇ. W FH β 3.25 ft The section p as a free body: The perpendicular distance from point F is LFW D 3 2 sin ˇ C 3.5 D 7.678 ft. 3 ft The sum of the moments about F is MF D WLFW C W3.25 jEGj3 D 0, from which EG D 1476.1 lb C. W α FG EG W 3.5 ft W The sum of the forces: FY D FG sin ˇ FH sin ˛ EG sin ˛ W sin ˛ W D 0, JH HI GI FX D FG cos ˇ FH cos ˛ EG cos ˛ W cos ˛ D 0, from which the two simultaneous equations: 0.9848FG 0.5736FH D 726.9, and 0.1736FG 0.8192FH D 389.97. Solve: FG D 1158.5 lb C , and FH D 721.64 lb T. Make the cut through JH, HI, and GI, and consider the upper section. (b) Choose a coordinate system with the y axis parallel to JH. Isolate a section by making cuts through FH, FG, and EG, and through HJ, HI, and GI. The free section of the truss is shown. The sum of the forces in the x- and y-direction are each zero; since the only external x-components of axial force are those contributed by FG and HI, the two axial forces must be equal: The section as a free body: The perpendicular distance from point p H to the line of action of the weight is LHW D 3 cos ˛ C 3 2 sin ˇ C 3.5 D 10.135 ft. The sum of the moments about H is MH D WL jGIj3 C W3.25 D 0, from which jGIj D 2295 lb C. Fx D HI cos 45° FG cos 45° D 0, from which HI D FG FY D HI sin ˇ JH sin ˛ GI sin ˛ W sin ˛ W D 0, FX D HI cos ˇ JH cos ˛ GI cos ˛ W cos ˛ D 0, from which the two simultaneous equations: 0.9848HI 0.5736JH D 257.22, and 0.1736HI 0.8192JH D 1060.8. Solve: and HI D 1158.5 lbC , JH D 1540.6 lbT . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 431 Problem 6.53 Consider the truss in Problem 6.52. The weight of the bucket is W D 1000 lb. The cable passes over pulleys at A and D. Determine the axial forces in members IK and JL. Solution: Make a cut through JL, JK, and IK, and consider the upper section. Denote the axial force in a member joining, ˛, ˇ by ˛ˇ. The section as a free body: The perpendicular distance p from point J to the line of action of the weight is L D 6 cos ˛ C 3 2 sin ˇ C 3.5 D 12.593 ft. The sum of the moments about J is MJ D WL C W3.25 IK3 D 0, from which IK D 3114.4 lbC. The sum of the forces: W W β JL α 3.5 ft 3.25 ft 3 ft Fx D JL cos ˛ IK cos ˛ JK IK W cos ˛ JK cos ˇ D 0, and Fy D JL sin ˛ IK sin ˛ W sin ˛ W JK sin ˇ D 0, from which two simultaneous equations: 0.8192JL C 0.1736JK D 1732 and 0.5736JL C 0.9848JK D 212.75. Solve: and JL D 2360 lbT , JK D 1158.5 lbC . Problem 6.54 The truss supports loads at N, P, and R. Determine the axial forces in members IL and KM. 2m 2m 2m 2m 2m K M O Q I L N P 1 kN 2 kN 1m J R 2m Solution: The strategy is to make a cut through KM, IM, and IL, and consider only the outer section. Denote the axial force in a member joining, ˛, ˇ by ˛ˇ. H G 1 kN 2m F E 2m The section as a free body: The moment about M is D MM D IL 21 42 61 D 0, from which C 2m A B IL D 16 kN C . 6m The angle of member IM is ˛ D tan1 0.5 D 26.57° . The sums of the forces: KM α Fy D IM sin ˛ 4 D 0, 1m IM IL 4 D 8.944 kN (C). from which IM D sin ˛ 1 kN 2m 2 kN 2m 1 kN 2m Fx D KM IM cos ˛ IL D 0, from which KM D 24 kNT 432 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.55 Consider the truss in Problem 6.54. Determine the axial forces in members HJ and GI. Solution: The strategy is to make a cut through the four members AJ, HJ, HI, and GI, and consider the upper section. The axial force in AJ can be found by taking the moment of the structure about B. 1m AJ The complete structureas a free body: The angle formed by AJ with the 4 D 26.57° . The moment about B is MB D vertical is ˛ D tan1 8 6AJ cos ˛ 24 D 0, from which AJ D 4.47 kN (T). I HJ αβ γ HI GI 2m 2m 1 kN 2m 2 kN 2m 1 kN 2m The section as a free body: The of members HJ and HI angles relative 2 1.5 to the vertical are ˇ D tan1 D 14.0° , and D tan1 D 8 2 ° 36.87 respectively. Make a cut through the four members AJ, HJ, HI, and GI, and consider the upper section. The moment about the point I is MI D 24 C 2AJ cos ˛ C 2HJ cos ˇ D 0. From which HJ D 8.25 kN T . The sums of the forces: Fx D AJ sin ˛ C HJ sin ˇ HI sin D 0, from which HI D 22 AJ sin ˛ HJ sin ˇ D D 0. sin sin FY D AJ cos ˛ HJ cos ˇ HI cos GI 4 D 0, from which GI D 16 kN C Problem 6.56 Consider the truss in Problem 6.54. By drawing free-body diagrams of sections, explain why the axial forces in members DE, FG, and HI are zero. Solution: Define ˛, ˇ to be the interior angles BAJ and ABJ respectively. The sum of the forces in the x-direction at the base yields AX C BX D 0, from which Ax D Bx . Make a cut through AJ, BD and BC, from which the sum of forces in the x-direction, Ax BD sin ˇ D 0. Since Ax D AJ sin ˛, then AJ sin ˛ BD sin ˇ D 0. A repeat of the solution to Problem 6.55 shows that this result holds for each section, where BD is to be replaced by the member parallel to BD. For example: make a cut through AJ, FD, DE, and CE. Eliminate the axial force in member AJ as an unknown by taking the moment about A. Repeat the solution process in Problem 6.55, obtaining the result that DE D AJ sin ˛ DF sin ˇ D0 cos DE where DE is the angle of the member DE with the vertical. Similarly, a cut through AJ, FH, FG, and EG leads to FG D AJ sin ˛ FH sin ˇ D 0, cos FG and so on. Thus the explanation is that each member BD, DF, FH and HJ has equal tension, and that this tension balances the x-component in member AJ c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 433 y Problem 6.57 In Active Example 6.5, draw the freebody diagram of joint B of the space truss and use it to determine the axial forces in members AB, BC, and BD. 1200 lb A (5, 3, 2) ft B D (10, 0, 0) ft x z C (6, 0, 6) ft Solution: From Active Example 6.5 we know that the vertical reaction force at B is 440 lb. The free-body diagram of joint B is shown. We have the following position vectors. rBA D 5i C 3j C 2k ft rBC D 6i C 6k ft rBD D 10i ft The axial forces in the rods can then be written as TAB rBA D TAB 0.811i C 0.487j C 0.324k jrBA j TBC rBC D TBC 0.707i C 0.707k jrBC j TBD rBD D TBD i jrBD j The components of the equilibrium equations are Fx : 0.811TAB C 0.707TBC C TBD D 0 Fy : 0.487TAB C 440 lb D 0 Fz : 0.324TAB C 0.707TBC D 0 Solving yields TAB D 904 lb, TBC D 415 lb, TBD D 440 lb Thus 434 AB : 904 lb (C), BC : 415 lb (T), BD : 440 lb (T) c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.58 The space truss supports a vertical 10kN load at D. The reactions at the supports at joints A, B, and C are shown. What are the axial forces in the members AD, BD, and CD? y 10 kN D (4, 3, 1) m Ay Ax A Cy C (6, 0, 0) m Az By z Solution: Consider the joint D only. The position vectors parallel to the members from D are Cz B (5, 0, 3) m x 10 kN rDA D 4i 3j k, rDB D i 3j C 2k, TDA TDC TDB rDC D 2i 3j k. The unit vectors parallel to the members from D are: eDA D rDA D 0.7845i 0.5883j 0.1961k jrDA j eDB D rDB D 0.2673i 0.8018j C 0.5345k jrDB j eDC D rDC D 0.5345i 0.8018j 0.2673k jrDC j The equilibrium conditions for the joint D are F D TDA eDA C TDB eDB C TDC eDC FD D 0, from which Fx D 0.7845TDA C 0.2673TDB C 0.5345TDC D 0 Fy D 0.5883TDA 0.8018TDB 0.8108TDC 10 D 0 Fz D 0.1961TDA C 0.5345TDB 0.2673TDC D 0. Solve: TDA D 4.721 kN C , TDB D 4.157 kN C TDC D 4.850 kN C c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 435 Problem 6.59 Consider the space truss in Problem 6.58. The reactions at the supports at joints A, B, and C are shown. What are the axial forces in members AB, AC, and AD? Solution: The reactions at A are required for a determination of the equilibrium conditions at A. The complete structure as a free body: The position vectors are rAB D 5i C 3k, rAC D 6i, rAD D 4i C 3j C k. The sum of the forces: and Ay Ax TAC Az Fx D Ax D 0, TAD TAB Fy D Ay C Cy C By 10 D 0, Fz D Az C Cz D 0. The moments due to the reactions: M D rAB ð FB C rAC ð FC C rAD ð FD D 0 i M D 5 0 j 0 By k i 3 C 6 0 0 j 0 Cy k i 0 C 4 Cz 0 j 3 10 k 1 D 0 0 D 3By C 10i 6Cz j C 5By C 6Cy 40k D 0. These equations for the forces and moments are to be solved for the unknown reactions. The solution: Ax D Cz D 0, Ay D 2.778 kN, By D 3.333 kN, and Cy D 3.889 kN The method of joints: Joint A: The position vectors are given above. The unit vectors are: eAB D 0.8575i C 0.5145k, eAC D i, eAD D 0.7845i C 0.5883j C 0.1961k. The equilibrium conditions are: F D TAB eAB C TAC C eAC C TAD eAD C A D 0, from which Fx D 0.8575TAB C TAC C 0.7845TAD D 0 Fy D 0TAB C 0TAC C 0.5883TAD C 2.778 D 0 Fz D 0.5145jTAB j C 0jTAC j C 0.1961jTAD j D 0. Solve: TAB D 1.8 kN T , TAC D 2.16 kN T TAD D 4.72 kN C 436 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.60 The space truss supports a vertical load F at A. Each member is of length L, and the truss rests on the horizontal surface on roller supports at B, C, and D. Determine the axial forces in members AB, AC, and AD. F A B D C Solution: By symmetry, the axial forces in members AB, AC, and AD are equal. We just need to determine the angle between each of these members and the vertical: we see that b D tan 30° L 2 F and A TAB TAD = TAB from which we obtain TAC = TAB 1 Ltan 60° tan 30° . 2 c Then D arcsin L cD θ θ θ D 35.26° F C 3TAB cos D 0, so TAB D TAC D TAD D F . 3 cos and From the top view, L bCc D tan 60° , L 2 TAB D TAC D TAD D F 3 cos 35.26° D 0.408F. C 60° b 30° L /2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 437 Problem 6.61 For the truss in Problem 6.60, determine the axial forces in members AB, BC, and BD. Solution: See the solution of Problem 6.60. The axial force in From the equilibrium equation member AB is TAB D 0.408F, and the angle between AB and the vertical is D 35.26° . The free-body diagram of joint B is TAB sin C 2TBC cos 30° D 0, we obtain TAB TBC D TBD D 0.136F. θ TBD = TBC 30° 30° TBC Problem 6.62 The space truss has roller supports at B, C, and D and supports a vertical 800-lb load at A. What are the axial forces in members AB, AC, and AD? y 800 lb A (4, 3, 4) ft B D (6, 0, 0) ft x z Solution: The position vectors of the points A, B, C, and D are rA D 4i C 3j C 4k, rC D 5i C 6k, rD D 6i. C (5, 0, 6) ft The equilibrium conditions at point A: The position vectors from joint A to the vertices are: Fx D 0.6247TAB C 0.2673TAC C 0.3714TAD D 0 Fy D 0.4685TAB 0.8018TAB 0.5570TAD 800 D 0 Fz D 0.6247TAB C 0.5345TAC 0.7428TAD D 0. rAB D rB rA D 4i 3j 4k, 800 lb rAC D rC rA D 1i 3j C 2k, rAD D rD rA D 2i 3j 4k Joint A: The unit vectors parallel to members AB, AC, and AD are eAB D rAB D 0.6247i 0.4685j 0.6247k, jrAB j eAC D rAC D 0.2673i 0.8018j C 0.5345k, jrAC j and eAD D rAD D 0.3714i 0.5570j 0.7428k. jrAD j TAB TAC Solve: and 438 TAD TAB D 379.4 lb C , TAC D 665.2 lb C , TAD D 159.6 lb C c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.63 The space truss shown models an airplane’s landing gear. It has ball and socket supports at C, D, and E. If the force exerted at A by the wheel is F D 40j (kN), what are the axial forces in members AB, AC, and AD? y E (0, 0.8, 0) m D 0.4 m B x (1, 0, 0) m 0.6 m A (1.1, –0.4, 0) m C z F Solution: The important points in this problem are A (1.1, 0.4, 0), B (1, 0, 0), C (0, 0, 6), and D (0, 0, 0.4). We do not need point E as all of the needed unknowns converge at A and none involve the location of point E. The unit vectors along AB, AC, and AD are y E (0, 0.8, 0) m uAB D 0.243i C 0.970j C 0k, uAC D 0.836i C 0.304j C 0.456k, and uAD D 0.889i C 0.323j 0.323k. D 0.4 m 0.6 m TAD C The forces can be written as z B x (1, 0, 0) m TAB TAC F A (1.1, −0.4, 0) m TRS D TRS uRS D TRSX i C TRSY j C TRSZ k, where RS takes on the values AB, AC, and AD. We now have three forces written in terms of unknown magnitudes and known directions. The equations of equilibrium for point A are Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0, and Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0, Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0, where F D FX i C FY j C FZ k D 40j kN. Solving these equations for the three unknowns, we obtain TAB D 45.4 kN (compression), TAC D 5.26 kN (tension), and TAD D 7.42 kN (tension). c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 439 Problem 6.64 If the force exerted at point A of the truss in Problem 6.63 is F D 10i C 60j C 20k (kN), what are the axial forces in members BC, BD and BE? Solution: The important points in this problem are A (1.1, 0.4, 0), B (1, 0, 0), C (0, 0, 0.6), D (0, 0, 0.4), and E (0, 0.8, 0). The unit vectors along AB, AC, AD, BC, BD, and BE are y E (0, 0.8, 0) m uAB D 0.243i C 0.970j C 0k, uAC D 0.836i C 0.304j C 0.456k, 0.4 m 0.6 m uAD D 0.889i C 0.323j 0.323k, TBC C z uBC D 0.857i C 0j C 0.514k, DT TDE AD F B x (1, 0, 0) m TAB A (1.1, −0.4, 0) m uBD D 0.928i C 0j 0.371k, and uBE D 0.781i C 0.625j C 0k. The forces can be written as TRS D TRS uRS D TRSX i C TRSY j C TRSZ k, where RS takes on the values AB, AC, and AD when dealing with joint A and AB, BC, BD, and BD when dealing with joint B. We now have three forces written in terms of unknown magnitudes and known directions. Joint A: The equations of equilibrium for point A are, and Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0, Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0, Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0, where F D FX i C FY j C FZ k D 10i C 60j C 20k kN. Solving these equations for the three unknowns at A, we obtain TAB D 72.2 kN (compression), TAC D 13.2 kN (compression), and TAD D 43.3 kN (tension). Joint B: The equations of equilibrium at B are and Fx D TAB uABX C TBC uBCX C TBD uBDX C TBE uBEX D 0, Fy D TAB uABY C TBC uBCY C TBD uBDY C TBE uBEY D 0, Fz D TAB uABZ C TBC uBCZ C TBD uBDZ C TBE uBEZ D 0. Since we know the axial force in AB, we have three equations in the three axial forces in BC, BD, and BE. Solving these, we get TBC D 32.7 kN (tension), TBD D 45.2 kN (tension), and TBE D 112.1 kN (compression). 440 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.65 The space truss is supported by roller supports on the horizontal surface at C and D and a ball and socket support at E. The y axis points upward. The mass of the suspended object is 120 kg. The coordinates of the joints of the truss are A: (1.6, 0.4, 0) m, B: (1.0, 1.0, 0.2) m, C: (0.9, 0, 0.9) m, D: (0.9, 0, 0.6) m, and E: (0, 0.8, 0) m. Determine the axial forces in members AB, AC, and AD. y B E D A C x z Solution: The important points in this problem are A: (1.6, 0.4, 0) m, B: (1, 1, 0.2) m, C: (0.9, 0, 0.9) m, and D: (0.9, 0, 0.6) m. We do not need point E as all of the needed unknowns converge at A and none involve the location of point E. The unit vectors along AB, AC, and AD are y E TAB D uAB D 0.688i C 0.688j 0.229k, TAD uAC D 0.579i 0.331j C 0.745k, C and uAD D 0.697i 0.398j 0.597k. The forces can be written as TRS D TRS uRS D TRSX i C TRSY j C TRSZ k, where RS takes on the values AB, AC, and AD. We now have three forces written in terms of unknown magnitudes and known directions. The equations of equilibrium for point A are and B z A mg x TAC L Fx D TAB uABX C TAC uACX C TAD uADX C FX D 0, Fy D TAB uABY C TAC uACY C TAD uADY C FY D 0, Fz D TAB uABZ C TAC uACZ C TAD uADZ C FZ D 0, where F D FX i C FY j C FZ k D mgj D 1177j N. Solving these equations for the three unknowns, we obtain TAB D 1088 N (tension), TAC D 316 N (compression), and TAD D 813 N (compression). c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 441 Problem 6.66 The free-body diagram of the part of the construction crane to the left of the plane is shown. The coordinates (in meters) of the joints A, B, and C are (1.5, 1.5, 0), (0, 0, 1), and (0, 0, 1), respectively. The axial forces P1 , P2 , and P3 are parallel to the x axis. The axial forces P4 , P5 , and P6 point in the directions of the unit vectors e4 D 0.640i 0.640j 0.426k, e5 D 0.640i 0.640j 0.426k, e6 D 0.832i 0.555k. The total force exerted on the free-body diagram by the weight of the crane and the load it supports is Fj D 44j (kN) acting at the point (20, 0, 0) m. What is the axial force P3 ? Strategy: Use the fact that the moment about the line that passes through joints A and B equals zero. y A F z B P6 P2 C P5 P1 P4 P3 x Solution: The axial force P3 and F are the only forces that exert moments about the line through A and B. The moment they exert about pt B is i j MB D 20 0 0 44 k i 1 C 0 0 P3 j 0 0 k 2 0 D 44i 2P3 j C 880k (kN-m). The position vector from B to A is rBA D 1.5i C 1.5j k (m), and the unit vector that points from B toward A is eBA D rBA D 0.640i C 0.640j 0.426k. jrBA j From the condition that eBA Ð MB D 0.64044 C 0.6402P3 0.426880 D 0, we obtain P3 D 315 kN. 442 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.67 In Problem 6.66, what are the axial forces P1 , P4 , and P5 ? Strategy: Write the equilibrium equations for the entire free-body diagram. Solution: The equilibrium equations are Fx D P1 C P2 C P3 C 0.64P4 C 0.64P5 C 0.832P6 D 0, Fy D 0.64P4 0.64P5 44 D 0, Fz D 0.426P4 C 0.426P5 0.555P6 D 0, i MB D 20 0 j 0 44 i C 1.5 P1 j 1.5 0 i C 1.5 0.64P4 i C 1.5 0.64P5 k i 1 C 0 0 P3 j 0 0 k 2 0 k 1 0 j 1.5 0.64P4 k 1 0.426P4 j 1.5 0.64P5 k 1 D 0. 0.426P5 The components of the moment equation are MBx D 44 1.279P4 0.001P5 D 0, MBy D 2P3 P1 0.001P4 1.279P5 D 0, MBz D 880 1.5P1 1.92P4 1.92P5 D 0. Solving these equations, we obtain P1 D 674.7 kN, P2 D P3 D 315.3 kN, P4 D P5 D 34.4 kN, and P6 D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 443 Problem 6.68 The mirror housing of the telescope is supported by a 6-bar space truss. The mass of the housing is 3 Mg (megagrams), and its weight acts at G. The distance from the axis of the telescope to points A, B, and C is 1 m, and the distance from the axis to points D, E, and F is 2.5 m. If the telescope axis is vertical (˛ D 90° ), what are the axial forces in the members of the truss? A G F C B 60° 4m 60° G B 60° Solution: A cut through the 6-bar space truss leads to six equations 60° C E x 60° y 60° 60° D A F G 60° E F A 60° D α 1m END VIEW y D in the unknowns (see Problem 6.59). However for this problem an alternate strategy based on reasonable assumptions about the equality of the tensions is used to get the reactions. Assume that each support carries one-third of the weight, which is equally divided between the two bars at the support. y Mirror housing z 60° 60° Mirror housing y C B E x z 60° A F G B The coordinate system has its origin in the upper platform, with the x axis passing though the point C. The coordinates of the points are: C D α 1m A cos 60° , sin 60° , 0 D 0.5, 0.866, 0, E 4m B cos 60° , sin 60° , 0 D 0.5, 0.866, 0, r=1m A C1, 0, 0, y C B x D2.5, 0, 4, 4m E2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4, R = 2.5 m F2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4. Consider joint B in the upper housing. The position vectors of the points E and D relative to B are D E rBD D 2i C 0.866j 4k, rBE D 1.75i 1.299j 4k. The unit vectors are eBD D 0.4391i C 0.1901j 0.8781k, and eBE D 0.3842i 0.2852j 0.8781k. The weight is balanced by the z components: Fz D W 0.8781TBD 0.8781TBE D 0. 3 Assume that the magnitude of the axial force is the same in both members BD and BE, TBE D TBD . The weight is W D 39.81 D 29.43 kN. Thus the result: TBE D TBD D 5.5858 kN C . From symmetry (and the assumptions made above) the axial force is the same in all members. 444 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.69 Consider the telescope described in Problem 6.68. Determine the axial forces in the members of the truss if the angle ˛ between the horizontal and the telescope axis is 20° . Solution: The coordinates of the points are, y F A cos 60° , sin 60° , 0 D 0.5, 0.866, 0 m, A B cos 60° , sin 60° , 0 D 0.5, 0.866, 0 m, x C D B C1, 0, 0 m, E D2.5, 0, 4 m, E2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4 m, F2.5 cos 60° , 2.5 sin 60° , 4 D 1.25, 2.165, 4 m. 20000 The coordinates of the center of gravity are G (0, 0, 1) (m). Make a cut through the members just below the upper platform supports, such that the cut members have the same radial distance from the axis as the supports. Consider the upper section. A x i a l The section as a free body: The strategy is to sum the forces and moments to obtain six equations in the six unknown axial forces. The axial forces and moments are expressed in terms of unit vectors. The position vectors of the points E, D, and F relative to the points A, B, and C are required to obtain the unit vectors parallel to the members. The unit vectors are obtained from these vectors. The vectors and their associated unit vectors are given in Table I. Note: While numerical values are shown below to four significant figures, the calculations were done with the full precision permitted (15 digits for TK Solver Plus.) Vector rAD rAF rBD rBE rCE rCF x y z 2 1.75 2 1.75 0.25 0.25 0.866 1.299 0.866 1.299 2.165 2.165 4 4 4 4 4 4 Table I Unit Vector eAD eAF eBD eBE eCE eCF Axial Forces in Bars 25000 |AF| & |CF| 15000 10000 |AD| & |BD| 5000 0 F −5000 , −10000 N −15000 −20000 |CE| & |BD| −25000 −100 −50 0 50 100 alpha, deg x y z 0.4391 0.3842 0.4391 0.3842 0.0549 0.0549 0.1901 0.2852 0.1901 0.2852 0.4753 0.4753 0.8781 0.8781 0.8781 0.8781 0.8781 0.8781 The equilibrium condition for the forces is jTAB jeAD C jTAF jeAF C jTBD jeBD C jTBE jeBE C jTCE jeCE C jTCF jeCF C W D 0. This is three equations in six unknowns. The unit vectors are given in Table I. The weight vector is W D jWjj cos ˛ k sin ˛, where ˛ is the angle from the horizontal of the telescope housing. The remaining three equations in six unknowns are obtained from the moments: rA ð TAD C TAF C rB ð TBD C TBE C rC ð TCE C TCF C rG ð W D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 445 6.69 (Continued ) Carry out the indicated operations on the moments to obtain the vectors defining the moments: rA ð TAD i D jTAD j 0.5 0.4391 k 0 0.8781 j 0.866 0.1901 The six equations in six unknowns are: jTAD jeADx C jTAF jeAFx C jTBD jeBDx C jTBE jeBEx C jTCE jeCEx C jTCF jeCFx C Wx D 0 jTAD jeADy C jTAF jeAFy C jTBD jeBDy C jTBE jeBEy C jTCE jeCEy D jTAD j0.7605i 0.4391j C 0.4753 C jTCF jeCFy C Wy D 0 D jTAD jiuADx C juADy C juADz rA ð TAF i D jTAF j 0.5 0.3842 j 0.866 0.2852 k 0 0.8781 jTAD jeADz C jTAF jeAFz C jTBD jeBDz C jTBE jeBEz C jTCE jeCEz C jTCF jeCFz C Wz D 0 jTAD juADx C jTAF juAFx C jTBD juBDx C jTBE juBEx C jTCE juCEx D jTAF j0.7605i 0.4391j 0.4753k D jTAF jiuAFx C juAFy C kuAFz rB ð TBD i D jTBD j 0.5 0.4391 j 0.866 0.1901 k 0 0.8781 D jTBD j0.7605i 0.4391j 0.4753k D jTBD jiuBDx C juBDy C kuBDz rB ð TBE rC ð TCE i D jTBE j 0.5 0.3842 j 0.866 0.2852 k 0 0.8781 jTAD juADy C jTAF juAFy C jTBD juBDy C jTBE juBEy C jTCE juCEy C jTCF juCFy D 0, jTAD juADz C jTAF juAFz C jTBD juBDz C jTBE juBEz C jTCE juCEz C jTCF juCFz D 0 This set of equations was solved by iteration using TK Solver 2. For ˛ D 20° the results are: jTAD j D jTBD j D 1910.5 N C , D jTBE j0.7605i 0.4391j 0.4753k jTAF j D jTCF j D 16272.5 N T , D jTBE jiuBEx C juBEy C kuBEz jTBE j D jTCE j D 19707 N C . i D jTCE j 1 0.0549 j 0 0.4753 k 0 0.8781 D jTCE j0i C 0.8781j 0.4753k D jTCE jiuCEx C juCEy C kuCEz rC ð TCF C jTCF juCFx C MWx D 0 i D jTCF j 1 0.0549 j 0 0.4753 k 0 0.8781 D jTCF j0i C 0.8781j C 0.4753k D jTCF jiuCFx C juCFy C kuCFz i rG ð W D jWj 0 0 j 0 cos ˛ k 1 sin ˛ Check: For ˛ D 90° , the solution is jTAD j D jTAF j D jTBD j D jTBE j D jTCE j D jTCF j D 5585.8 N C, which agrees with the solution to Problem 6.68, obtained by another method. check. Check: The solution of a six-by-six system by iteration has risks, since the matrix of coefficients may be ill-conditioned. As a reasonableness test for the solution process, TK Solver Plus was used to graph the axial forces in the supporting bars over the range 90° < ˛ < 90° . The graph is shown. The negative values are compression, and the positive values are tension. When ˛ D 90° , the telescope platform is pointing straight down, and the bars are in equal tension, as expected. When ˛ D 90° the telescope mount is upright and the supporting bars are in equal compression, as expected. The values of compression and tension at the two extremes are equal and opposite in value, and the values agree with those obtained by another method (see Problem 6.58), as expected. Since the axial forces go from tension to compression over this range of angles, all axial forces must pass through zero in the interval. check. D jWji cos ˛ j0 C k0 D iMWx 446 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.70 In Active Example 6.6, suppose that in addition to being loaded by the 200 N-m couple, the frame is subjected to a 400-N force at C that is horizontal and points toward the left. Draw a sketch of the frame showing the new loading. Determine the forces and couples acting on members AB of the frame. A B 200 N-m 400 mm C Solution: The sketch of the frame with the new loading is shown. We break the frame into separate bars and draw the free-body diagram of each bar. 600 mm 400 mm Starting with bar BC, we have the equilibrium equations MB : C400 mm 400 N400 mm 200 N-m D 0 Fy : C By D 0 Fx : Bx 400 N D 0 Now using bar AB we have the equilibrium equations Fx : Ax C Bx D 0 Fy : Ay C By D 0 MA : MA C By 600 mm D 0 Solving these six equations yields C D 900 N and Ax D 400 N, Ay D 900 N Bx D 400 N, By D 900 N MA D 540 N-m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 447 Problem 6.71 The object suspended at E weighs 200 lb. Determine the reactions on member ACD at A and C. D 3 ft E C B 5 ft A 4 ft 6 ft Solution: We start with the free-body diagram of the entire frame. We have the equilibrium equations: Fx : Ax D 0 Fy : Ay 200 lb D 0 MA : MA 200 lb 6 ft D 0 Next we use the free-body diagram of the post ACD. Notice that BD is a two-force body and the angle ˛ is ˛ D tan1 3/4 D 36.9° The equilibrium equations are MC : MA C Ax 5 ft C TBD cos ˛ 3 ft D 0 Fx : Ax C Cx TBD cos ˛ D 0 Fy : Ay C Cy TBD sin ˛ D 0 Solving these six equations we find TBD D 500 lb and Ax D 0, Ay D 200 lb Cx D 400 lb, Cy D 500 lb MA D 1200 ft-lb 448 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.72 The mass of the object suspended at G is 100 kg. Determine the reactions on member CDE at C and E. E B F G 800 mm D 200 mm A C 400 mm 400 mm 800 mm 400 mm Solution: The free-body diagram of the entire frame and of member CDE are shown. The angle ˛ is ˛ D tan1 4/8 D 26.6° The equilibrium equations are MC : TAB cos ˛ 400 mm C TAB sin ˛ 800 mm 981 N1200 mm D 0 Fx : Cx TAB sin ˛ D 0 Fy : Cy TAB cos ˛ 981 N D 0 The free-body diagram for bar CDE is shown. Note that DF is a two-force member. The angle ˇ is ˇ D tan1 3/4 D 36.9° The equilibrium equations are ME : TDF cos ˇ600 mm C Cx 800 mm D 0 Fx : TDF cos ˇ C Ex C Cx D 0 Fy : TDF sin ˇ C Ey C Cy D 0 Solving these six equations, we find TAB D 1650 N, Cx D 736 N, Ex D 245 N, TDF D 1230 N and Cy D 2450 N Ey D 1720 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 449 Problem 6.73 The force F D 10 kN. Determine the forces on member ABC, presenting your answers as shown in Fig. 6.25. F E D A B C 1m 1m Solution: The complete structure as a free body: The sum of the MG D C3F 5A D 0, 2m 1m F moments about G: G GY GX A 2m 3F D 6 kN which is the reaction of the floor. The 5 sum of the forces: 3m from which A D F Fy D Gy F C A D 0, D 1m A Fx D Gx D 0. 1m Element DEG: The sum of the moments about D M D F C 3E C 4Gy D 0, from which E D 10 16 F 4Gy D D 2 kN. 3 3 2m E 1m F from which Gy D F A D 10 6 D 4 kN. GY A 1m C = −E B = −D 8 kN 2 kN B C 3m 6 kN The sum of the forces: Fy D Gy F C E C D D 0, from which D D F E Gy D 10 C 2 4 D 8 kN. Element ABC : Noting that the reactions are equal and opposite: B D D D 8 kN , and C D E D 2 kN . The sum of the forces: Fy D A C B C C D 0, from which A D 8 2 D 6 kN. Check 450 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.74 In Example 6.7, suppose that the frame is redesigned so that the distance from point C to the attachment point E of the two-force member BE is increased from 8 in to 10 in. Determine the forces acting at C on member ABCD. D 6 in 3 in G E C 6 in B W 6 in A 8 in 8 in Solution: The analysis of the free-body diagram of the entire structure as presented in Example 6.7 is unchanged. From the example we know that Ax D 42.2 lb, Ay D 40 lb, D D 42.4 lb The free-body diagram for ABCD is shown. Note that BE is a two-force body. The angle ˛ is now ˛ D tan1 6/10 D 31.0° The equilibrium equations are MC : TBE cos ˛ 6 in C D 6 in C Ax 6 in D 0 Fx : TBE cos ˛ C Cx C Ax D D 0 Fy : TBE sin ˛ C Cy C Ay D 0 Solving yields TBE D 124 lb and Cx D 66.7 lb, Cy D 24 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 451 Problem 6.75 The tension in cable BD is 500 lb. Determine the reactions at A for cases (1) and (2). E G 6 in D 6 in A B C 300 lb 8 in 8 in (1) G E 6 in D 6 in A B C 300 lb 8 in 8 in Solution: Case (a) The complete structure as a free body: The sum (2) of the moments about G: Gy MG D 16300 C 12Ax D 0, (a) 12 in from which Ax D 400 lb . The sum of the forces: Ey Gy Ay Ax Gx 16 in Ex 300 lb Fx D Ax C Gx D 0, Ay from which Gx D 400 lb. Gx (b) B α Ax Fy D Ay 300 C Gy D 0, 8 in 8 in Cy Cx 300 lb from which Ay D 300 Gy . Element GE : The sum of the moments about E: ME D 16Gy D 0, from which Gy D 0, and from above Ay D 300 lb. Case (b) The complete structure as a free body: The free body diagram, except for the position of the internal pin, is the same as for case (a). The sum of the moments about G is MG D 16300 C 12Ax D 0, from which Ax D 400 lb . Element ABC : The tension at the lower end of the cable is up and to the right, so that the moment exerted by the cable tension about point C is negative. The sum of the moments about C: MC D 8B sin ˛ 16Ay D 0, noting that B D 500 lb and ˛ D tan1 then 452 6 D 36.87° , 8 Ay D 150 lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.76 Determine the reactions on member ABCD at A, C, and D. B A 0.4 m E C 600 N 0.4 m D 0.6 m 0.4 m 0.4 m Solution: Consider the entire structure first MA : Dy 0.6 m 600 N1.0 m D 0 ) Dy D 1000 N Fx : Ax D 0 Fy : Ay C Dy 600 N D 0 ) Ay D 400 N Ax Ay E C 600 N Dy Now examine bar CE. Note that the reactions on ABD are opposite to those on CE. ME : 600 N0.4 m C Cy 0.8 m D 0 ) Cy D 300 N MB : Cx 0.4 m 600 N0.4 m D 0 ) Cx D 600 N T Cy Cx E 600 N In Summary we have Ax D 0, Ay D 400 N Cx D 600 N, Cy D 300 N Dx D 0, Dy D 1000 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 453 Problem 6.77 Determine the forces exerted on member ABC at A and C. D 400 lb 2 ft 1 ft A B C 100 lb 1 ft E 2 ft 2 ft 2 ft Solution: We start with the free-body diagram of the entire frame. Two of the equilibrium equations for the whole frame are Fx : Ax C 100 lb D 0 ME : Ax 2 ft Ay 4 ft 100 lb 1 ft 400 lb 2 ft D 0 Next we examine the free-body diagram of bar ABC. Note that BD is a two-force body and that the angle ˛ D 45° . The equilibrium equations are MC : Ay 4 ft TBD sin ˛ 2 ft 400 lb 2 ft D 0 Fx : Ax C TBD cos ˛ C Cx D 0 Fy : Ay C TBD sin ˛ C Cy 400 lb D 0 Solving, we find that TBD D 70.7 lb and Ax D 100 lb, Ay D 175 lb Cx D 150 lb, Cy D 625 lb 454 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.78 An athlete works out with a squat thrust machine. To rotate the bar ABD, she must exert a vertical force at A that causes the magnitude of the axial force in the two-force member BC to be 1800 N. When the bar ABD is on the verge of rotating, what are the reactions on the vertical bar CDE at D and E? 0.6 m 0.6 m C A 0.42 m B D 1.65 m E Solution: Member BC is a two force member. The force in BC is along the line from B to C. C y FBC Ay 0.6 m 0.6 m tan Θ = Dy 0.42 m θ D 0.42 0.6 Dx x (FBC = 1800 N) Θ = 34.990 FBC D 1800 N tan D 0.42 D 34.99° . 0.6 C Fx : Dx FBC cos D 0 Fy : Ay FBC sin C Dy D 0 MD : Solving, we get 1.2Ay C 0.6FBC sin D 0 Dx D 1475 N Dy D 516 N Ay D 516 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 455 Problem 6.79 The frame supports a 6-kN load at C. Determine the reactions on the frame at A and D. 6 kN 0.4 m A B 1.0 m C 0.5 m D E F Solution: Note that members BE and CF are two force members. Consider the 6 kN load as being applied to member ABC. Ay Ax 0.4 m 0.4 m 6 kN 1.0 m B 0.8 m C FCF FBE φ θ tan D 0.5 0.4 D 51.34° tan D 0.5 0.2 D 68.20° Member DEF FBE θ Dx 0.8 m FCF E F φ 0.4 m Dy Equations of equilibrium: Member ABC: Fx : Ax C FBE cos FCF cos D 0 Fy : Ay FBE sin FCF sin 6 D 0 MA : 0.4FBE sin 1.4FCF sin 1.46 D 0 C Member DEF: C Fx : Dx FBE cos C FCF cos D 0 Fy : Dy C FBE sin C FCF sin D 0 MD : 0.8FBE sin C 1.2FCF sin D 0 Unknowns Ax , Ay , Dx , Dy , FBE , FCF we have 6 eqns in 6 unknowns. Solving, we get Also, Ax Ay Dx Dy D 16.8 kN D 11.25 kN D 16.3 kN D 5.25 kN FBE D 20.2 kN T FCF D 11.3 kN C 456 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.80 The mass m D 120 kg. Determine the forces on member ABC, presenting your answers as shown in Fig. 6.25. A B C 300 m m D E m 200 mm Solution: The equations of equilibrium for the entire frame are AX AY CX CX DY FY D AY 2mg D 0, CY DY EX MA D 0.3EX 0.2mg 0.4mg D 0. Solving yields AX D 2354 N, AY D 2354 N, and EX D 2354 N. Member ABC: The equilibrium equations are and CY BY and summing moments at A, 200 mm BY FX D AX C EX D 0, m 2354 N B 2354 N A 4708 N 4708 N 2354 N C 2354 N B FX D AX C CX D 0, 2354 N 2354 N 4708 N FY D AY BY C CY D 0, MA D 0.2BY C 0.4CY D 0. We have three equations in the three unknowns BY , CX , and CY . Solving, we get BY D 4708 N, CX D 2354 N, and CY D 2354 N. This gives all of the forces on member ABC. A similar analysis can be made for each of the other members in the frame. The results of solving for all of the forces in the frame is shown in the figure. 4708 N E 2354 N C D 1177 N 1177 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 457 30 lb Problem 6.81 Determine the reactions on member BCD. F D G 8 in. E 40 lb 8 in. C 8 in. B A 18 in. 12 in. 8 in. Solution: We will use frame ADG, bar DFG and bar BCD. The free-body diagrams ares shown. The angle ˛ D tan1 18/24 D 36.9° From ADG we have MD : Bx 24 in C 40 lb8 in 30 lb 20 in D 0 MA : By 18 in 40 lb16 in 30 lb38 in D 0 Fy : By 30 lb TAD cos ˛ D 0 From DFG we have MF : Dy C TAD cos ˛ 12 in 30 lb8 in D 0 And finally from BCD we have Fy : By C Cy C Dy D 0 MD : Bx 24 in C Cx 16 in D 0 Fx : Dx C Cx C Bx D 0 Solving these seven equations, we find TAD D 86.1 lb and Bx D 11.7 lb, By D 98.9 lb Cx D 17.5 lb, Cy D 50 lb Dx D 5.83 lb, 458 Dy D 48.9 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.82 The weight of the suspended object is W D 50 lb. Determine the tension in the spring and the reactions at F. (The slotted member DE is vertical.) A B 4 in E 6 in W C 10 in F D 8 in 8 in 10 in Solution: Start with member AB Finally examine DCE 1 MA : 50 lb8 in C p FB 16 in D 0 ) FB D 35.4 lb 2 10 in MD : T16 in C FC 10 in D 0 ) T D 62.5 lb FB T 1 1 Ax FC Ay 50 lb Now examine BCF p MF : FB 20 2 in FC 10 in D 0 ) FC D 100 lb Dx 1 Fx : p FB C FC C Fx D 0 ) Fx D 75 lb 2 1 Fy : p FB C Fy D 0 ) Fy D 25 lb 2 Dy Summary Tension in Spring D 62.5 lb Fx D 25 lb, Fy D 75 lb 1 1 FB FC Fx Fy c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 459 Problem 6.83 The mass m D 50 kg. Bar DE is horizontal. Determine the forces on member ABCD, presenting your answers as shown in Fig. 6.25. 1m 1m D E 1m C m 1m B 1m A F Solution: The weight of the mass hanging is W D mg D 509.81 D 490.5 N The complete structure as a free body: The sum of the moments about A: MA D 2W C Fy D 0, from which Cy D 981 N. Fy D Ey C Cy D 0, Fx D Ex C Cx D 0, from which Fy D 981 N. The sum of the forces: from which Cx D 981 N, and Fy D Ay C Fy W D 0, from which Ay D 490.5 N, Element ABCD: All reactions on ABCD have been determined above. The components at B and C have the magnitudes p B D C D 9812 C 9812 D 1387 N , at angles of 45° . Fx D Ax C Fx D 0, from which Ax D Fx . Element BF: The sum of the moments about F: MF D Bx By D 0, from which By D Bx . The sum of the forces: Dy Dy Cy Cx Cy Bx By Fy D By C Fy D 0, Ey Ey W Ex By Cx Bx Fy Ay Ax from which By D 981 N, and Bx D 981 N. Ex Dx Dx Fx Fx D Bx C Fx D 0, 490.5 N from which Fx D 981 N, and from above, Ax D 981 N , Element DE: The sum of the moments about D: MD D Ey 2W D 0, D C 981 N 1387 N 45° B from which Ey D 981 N. The sum of the forces: Fy D Dy Ey W D 0, A 45° 1387 N 981 N 490.5 N from which Dy D 490.5 N . Fx D Dx Ex D 0, from which Dx D Ex . Element CE : The sum of the moments about C: MC D Ey Ex D 0, from which Ex D 981 N, and from above Dx D 981 N . 460 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.84 Determine the forces on member BCD. 400 lb 6 ft B A 4 ft C 4 ft D E 8 ft Solution: The following is based on free body diagrams of the elements: The complete structure as a free body: The sum of the moments about D: MD D 6400 C 8Ey D 0, from which Ey D 300 lb. The sum of the forces: Fx D Dx D 0. MA D 8By 6400 D 0, from which By D 300 lb. The sum of forces: The reactions are now known: By D 300 lb , Bx D 400 lb , Cy D 200 lb , Dx D 0 , Dy D 100 lb , where negative sign means that the force is reversed from the direction shown on the free body diagram. Fy D Ey C Dy 400 D 0, from which Dy D 100 lb. Element AB: The sum of the moments about A: Element BCD: Fy D By Ay 400 D 0, 400 lb Ay Ax Ax By Bx Cy Cx E Ay Bx Cy Dy Cx By Dx from which Ay D 100 lb. Fx D Ax Bx D 0, from which (1) Ax C Bx D 0 Element ACE: The sum of the moments about E: ME D 8Ax C 4Cx 8Ay C 4Cy D 0, from which (2) 2Ax C Cx 2Ay C Cy D 0. The sum of the forces: Fy D Ay C Ey Cy D 0, from which Cy D 200 lb . Fx D Ax Cx D 0, from which (3) Ax D Cx . The three numbered equations are solved: Ax D 400 lb, Cx D 400 lb , and Bx D 400 lb . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 461 Problem 6.85 Determine the forces on member ABC. E 6 kN 1m D C 1m A B 2m Solution: The frame as a whole: The equations of equilibrium are EX FX D AX C EX D 0, FY D AY C EY 6000 N D 0, and, with moments about E, EY E DY DX D DX D DY BX B BY BX B AX A BY AY 2m CY 1m 6 kN C CY C ME D 2AX 56000 D 0. Solving for the support reactions, we get AX D 15,000 N and EX D 15,000 N. We cannot yet solve for the forces in the y direction at A and E. Member ABC: The equations of equilibrium are FX D AX BX D 0, FY D AY BY CY D 0, and summing moments about A, MA D 2BY 4CY D 0. Member BDE: The equations of equilibrium are FX D EX C DX C BX D 0, FY D EY C DY C BY D 0, and, summing moments about E, ME D 1DY C 1DX C 2BY C 2BX D 0. Member CD: The equations of equilibrium are FX D DX D 0, FY D DY C CY 6000 D 0, and summing moments about D, MD D 46000 C 3CY D 0. Solving these equations simultaneously gives values for all of the forces in the frame. The values are AX D 15,000 N, AY D 8,000 N, BX D 15,000 N, BY D 16,000 N, CY D 8,000 N, DX D 0, and DY D 2,000 N. 462 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.86 Determine the forces on member ABD. 8 in 8 in 8 in A 60 lb 8 in 60 lb B E 8 in C Solution: The equations of equilibrium for the frame as a AY whole are AX FX D AX C CX D 0, and BX DX FY D AY 60 60 D 0, D BY BX B 60 lb EX EY DY DY CX MA D 16CX 1660 2460 D 0. 60 lb BY EY EX DX Solving these three equations yields AX D 150 lb, AY D 120 lb, and CX D 150 lb. Member ABD: The equilibrium equations for this member are: FX D AX BX DX D 0, and FY D AY BY DY D 0, MA D 8BY 8DY 8BX 16DX D 0. Member BE: The equilibrium equations for this member are: FX D BX C EX D 0, and FY D BY C EY 60 60 D 0, MB D 860 1660 C 16EY D 0. Member CDE: The equilibrium equations for this member are: and FX D CX C DX EX D 0, FY D DY EY D 0, MD D 8EX 16EY D 0. Solving these equations, we get BX D 180 lb, BY D 30 lb, DX D 30 lb, DY D 90 lb, EX D 180 lb, and EY D 90 lb. Note that we have 12 equations in 9 unknowns. The extra equations provide a check. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 463 Problem 6.87 The mass m D 12 kg. Determine the forces on member CDE. A 200 mm 100 mm E B 200 mm C D 200 mm Solution: Start with a free-body diagram of the entire frame. m 400 mm Ax Eq. for entire frame: C ! C " C Ay Fx D 0: Ax C Cx D 0 ) Ax D Cx 1 B Fy D 0: Mc D 0: Ay 117.7 D 0 ) Ay D 117.7 N Ax 0.4 117.70.7 D 0 Ax D 206 N Cx W = (9.81) (12) W = 117.7 D ∴ Ax D Cx D 206 N. Now look at free-body diagram ABD. Ax = 206 Eq. for ABD: C MB D 0: Ay = 117.7 T = 117.7 By Dx 0.2 117.70.2 C 2060.2 117.70.1 D 0 Bx Dx D 29.45 N C ! Fx D 0: Dx 206 C 117.7 C Bx 29.45 D 0 C " Bx D 117.75 N Dy Fy D 0: 117.7 By C Dy D 0 Draw free-body diagram of CDE Ey Ex Eq. for CDE: C ! Fx D 0: 206 C 29.45 C Ex D 0 C Dx = 29.45 Ex D 235.45 or Ex D 235.45 Cx = 206 MD D 0: Ex 0.2 C Ey 0.4 D 0 Ey D Dy 235.450.2 Ex 0.2 D 0.4 0.4 Ey D 117.7 or Ey D 117.7 N C " Fy D 0: Ey Dy D 0 or Dy D Ey D 117.7 464 Dy D 117.7 N# c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.88 The weight W D 80 lb. Determine the forces on member ABCD. 11 in 5 in 12 in 3 in B A D C 8 in W E Solution: The complete structure as a free body: The sum of the moments about A: Ay Fx D Ex C Ax D 0, Cx By Cx Ey F Dy Dx Cy Bx Ax MA D 31W C 8Ex D 0, from which Ex D 310 lb. The sum of the forces: F Cy W Ex from which Ax D 310 lb . Fy D Ey C Ay W D 0, from which (1) Ey C Ay D W. Element CFE: The sum of the forces parallel to x: Fx D Ex Cx D 0, from which Cx D 310 lb . The sum of the moments about E: ME D 8F 16Cy C 8Cx D 0. For frictionless pulleys, F D W, and thus Cy D 195 lb . The sum of forces parallel to y: Fy D Ey Cy C F D 0, from which Ey D 115 lb . Equation (1) above is now solvable: Ay D 35 lb . Element ABCD: The forces exerted by the pulleys on element ABCD are, by inspection: Bx D W D 80 lb , By D 80 lb , Dx D 80 lb , and Dy D 80 lb , where the negative sign means that the force is reversed from the direction of the arrows shown on the free body diagram. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 465 Problem 6.89 The woman using the exercise machine is holding the 80-lb weight stationary in the position shown. What are the reactions at the built-in support E and the pin support F? (A and C are pinned connections.) 2 ft 2 in B A 9 in 1 ft 6 in 2 ft D C 60⬚ 6 ft 80 lb E Solution: The complete structure as a free body: The sum of the 26 in moments about E: F 42 in 60° M D 26W 68W sin 60° C 50Fy 81W cos 60° C ME D 0 W from which (1) 50Fy C ME D 10031. The sum of the forces: 81 in Fx D Fx C W cos 60° C Ex D 0, ME from which (2) Fx C Ex D 40. W Ey Ex Fx 50 in Fy D W W sin 60° C Ey C Fy D 0, from which (3) Ey C Fy D 149.28 Ay Cy Element CF: The sum of the moments about F: Fy Ax Cx M D 72Cx D 0, from which Cx D 0. The sum of the forces: ME Fx D Cx C Fx D 0, Fx from which Fx D 0 . From (2) above, Ex D 40 lb Element AE: The sum of the moments about E: Fy Ex Ey M D ME 72Ax D 0, . from which (4) ME D 72Ax . The sum of the forces: Fy D Ey C Ay D 0, from which (5) Ey C Ay D 0. Fx D Ax C Ex D 0; from which Ax D 40 lb, and from (4) ME D 2880 in lb D 240 ft lb . From (1) Fy D 143.0 lb , and from (2) Ey D 6.258 lb . This completes the determination of the 5 reactions on E and F. 466 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.90 Determine the reactions on member ABC at A and B. 80 lb E 9 in B C 8 in D A 13 in Solution: We first examine the entire structure. Next examine body ABC MD : Ay 13 in C 80 lb21 in D 0 Solving: Ay D 129.2 lb 80 lb 4 in MB : Ax 8 in Ay 13 in C 80 lb 4 in D 0 Fx : Ax C Bx D 0 Fy : Ay C By C 80 lb D 0 80 lb Bx By Dx Ax Ax Ay Dy Ay Solving and summarizing we have Ax D 170 lb, Ay D 129.2 lb Bx D 170 lb, By D 209 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 467 Problem 6.91 The mass of the suspended object is m D 50 kg. Determine the reactions on member ABC. 0.2 m A B 0.6 m E D 0.8 m C 0.2 m 0.6 m m Solution: Begin with an examination of the pulley at B. Finally look at member ABC 1 Fx : Bx C p 490.5 N D 0 ) Bx D 347 N 2 1 Fy : By 490.5 N p 490.5 N D 0 ) By D 837 N 2 1 By MC : Ax 0.6 m Ay 1.4 m Bx 0.6 m By 0.6 m D 0 ) Ay D 771 N 1 Fx : Ax C Bx C Cx D 0 ) Cx D 961 N Fy : Ay C By C Cy D 0 ) Cy D 66.6 N By Ay Bx 490.5 N Ax Bx 490.5 N Cy Now examine the entire structure MD : 490.5 N1.6 m Ax 0.6 m D 0 ) Ax D 1308 N Cx In Summary Ay Ax D 1308 N, Ay D 771 N Ax Bx D 347 N, By D 837 N Cx D 961 N, Cy D 66.6 N Dy Dx 490.5 N 468 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.92 The unstretched length of the string is LO . Show that when the system is in equilibrium the angle ˛ satisfies the relation sin ˛ D 2LO 2F/kL. F 1– L 4 1– L 4 k 1– L 2 α α Solution: Since the action lines of the force F and the reaction E are co-parallel and coincident, the moment on the system is zero, and the system is always in equilibrium, for a non-zero force F. The object is to find an expression for the angle ˛ for any non-zero force F. The complete structure as a free body: The solution for angle ˛: The spring force is Cy D T D k L sin ˛ LO , 2 L sin ˛ LO D 2F. 2 2F 2 LO k Solve: sin ˛ D L from which k The sum of the moments about A MA D FL sin ˛ C EL sin ˛ D 0, from which E D F. The sum of forces: F Fx D Ax D 0, L from which Ax D 0. α Fy D Ay C E F D 0, from which Ay D 0, which completes a demonstration that F does not exert a moment on the system. The spring C: The elongation of the L spring is s D 2 sin ˛ LO , from which the force in the spring is 4 TDk L sin ˛ LO 2 Ax Ay E By Cy Bx L 4 α E L 4 Element BE: The strategy is to determine Cy , which is the spring force on BE. The moment about E is L L L ME D Cy cos ˛ By cos ˛ Bx cos ˛ D 0, 4 2 2 from which Cy C By D Bx . The sum of forces: 2 Fx D Bx D 0, from which Bx D 0. Fy D Cy C By C E D 0, from which Cy C By D E D F. The two simultaneous equations are solved: Cy D 2F, and By D F. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 469 Problem 6.93 The pin support B will safely support a force of 24-kN magnitude. Based on this criterion, what is the largest mass m that the frame will safely support? C 500 mm 100 mm E D B 300 mm m A 300 mm Solution: The weight is given by W D mg D 9.81 g Sum the forces in the x-direction: Fx D Ax D 0, Cx Cx W By W Bx By Bx W Ay from which Ax D 0 Element ABC: The sum of the moments about A: 400 mm 400 mm Cy Cy The complete structure as a free body: F Ey Ex Ey Ex F Ax MA D C0.3Bx C 0.9Cx 0.4W D 0, from which (1) 0.3Bx C 0.9Cx D 0.4W. The sum of the forces: Fx D Bx Cx C W C Ax D 0, from which (2) Bx C Cx D W. Solve the simultaneous equations (1) 5 and (2) to obtain Bx D W 6 Element BE : The sum of the moments about E: ME D 0.4W 0.7By D 0, from which By D jBj D W 4 W. The magnitude of the reaction at B is 7 2 2 4 5 C D 1.0104W. 6 7 24 D 23.752 kN is the 1.0104 maximum load that can be carried. Thus, the largest mass that can be supported is m D W/g D 23752 N/9.81 m/s2 D 2421 kg. For a safe value of jBj D 24 kN, W D 470 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.94 Determine the reactions at A and C. C A 3 ft 72 ft-lb 36 lb 3 ft B 18 lb 4 ft Solution: The complete structure as a free body: The sum of the moments about A: 8 ft Cy Ay Ax Cx 3 ft 72 ft-lb 36 lb MA D 418 C 336 C 12Cy 72 D 0, 18 lb from which Cy D 3 lb. The sum of the forces: 8 ft 4 ft Ay Fy D Ay C Cy 18 D 0, Ax from which Ay D 15 lb. Fx D Ax C Cx C 36 D 0, 72 ft-lb By 6 ft Bx 18 lb from which (1) Cx D Ax 36 Element AB: The sum of the forces: Fy D Ay By 18 D 0, from which By D 3 lb. The sum of the moments: MA D 6Bx 418 4By 72 D 0, from which Bx D 22 lb. The sum of the forces: Fx D Ax C Bx D 0, from which Ax D 22 lb From equation (1) Cx D 14 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 471 Problem 6.95 Determine the forces on member AD. 200 N 130 mm D 400 mm C A B 400 mm Solution: Denote the reactions of the support by Rx and Ry . The 400 mm Dy 200 N complete structure as a free body: from which Rx D 400 N. The sum of moments: MA D 800C 400930 C 400530 400200 D 0, Ay Ax Ry Dy 400 N Ay Fx D Rx 400 D 0, 400 N By 400 N Ax Rx Dx Dx Bx By Bx C from which C D 300 N. Fy D C C Ry 400 200 D 0, from which Ry D 300 N. Element ABC : The sum of the moments: MA D 4By C 8C D 0, from which By D 600 N. Element BD: The sum of the forces: Fy D By Dy 400 D 0, from which Dy D 200 N. Element AD: The sum of the forces: Fy D Ay C Dy 200 D 0, from which Ay D 0: Element AD: The sum of the forces: and Fx D Ax C Dx D 0 MA D 400200 C 800Dy 400Dx D 0 Ax D 200 N, and Dx D 200 N. Element BD: The sum of forces: Fx D Bx Dx 400 D 0 from which Bx D 600 N. This completes the solution of the nine equations in nine unknowns, of which Ax , Ay , Dx , and Dy are the values required by the Problem. 472 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.96 The frame shown is used to support high-tension wires. If b D 3 ft, ˛ D 30° , and W D 200 lb, what is the axial force in member HJ? A B α C D α E G F W H α I J α W W b Solution: Joints B and E are sliding joints, so that the reactions are normal to AC and BF, respectively. Member HJ is supported by pins at each end, so that the reaction is an axial force. The distance h D b tan ˛ D 1.732 ft Member ABC. The sum of the forces: B b b Ay h Ax B Dy Dx E b G y W Fx D Ax C B sin ˛ D 0, b Gx H W W Fy D Ay W B cos ˛ D 0. The sum of the moments about B: MB D bAy hAx C bW D 0. These three equations have the solution: Ax D 173.21 lb, Ay D 100 lb, and B D 346.4 lb. Member BDEF: The sum of the forces: Fx D Dx B sin ˛ E sin ˛ D 0, Fy D Dy W C B cos ˛ E cos ˛ D 0. The sum of the moments about D: MD D 2bW bE cos ˛ hE sin ˛ bB cos ˛ C hB sin ˛ D 0. These three equations have the solution: Dx D 259.8 lb, Dy D 350 lb, E D 173.2 lb. Member EGHI: The sum of the forces: Fx D Gx C E sin ˛ H cos ˛ D 0, Fy D Gy W C E cos ˛ C H sin ˛ D 0. The sum of the moments about H: MH D bGy hGx C bW C 2bE cos ˛ 2hE sin ˛ D 0. These three equations have the solution: Gx D 346.4 lb, Gy D 200 lb, and H D 300 lb. This is the axial force in HJ. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 473 Problem 6.97 Determine the force exerted on the ball by the bolt cutters and the magnitude of the axial force in the two-force member AB. 20 lb A 20 in B 3 in 6 in 4 in 20 lb Solution: Free-body diagrams of the top head and the top handle are shown. From the head we learn that Fx : Cx D 0 From the handle we have MD : 20 lb20 in C Cy 4 in D 0 ) Cy D 100 lb Now we return to the head MA : Cy 6 in F3 in D 0 Fy : F TAB C Cy D 0 Solving yields Force on the ball D F D 200 lb, 474 Axial force D TAB D 300 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.98 The woman exerts 20-N forces to the pliers as shown. (a) What is the magnitude of the forces the pliers exert on the bolt at B? (b) Determine the magnitude of the force the members of the pliers exert on each other at the pinned connection C. 25 mm 80 mm B C 50 mm 45⬚ 20 N 20 N Solution: Look at the piece that has the lower jaw of the pliers (a) B D 73.5 N (b) Cy MC : B25 mm 20 N cos 45° 80 mm 20 N sin 45° 50 mm D 0 B Cx 45° Fx : Cx 20 N sin 45° D 0 ) Cx D 14.14 N 20 N Fy : Cy B 20 N cos 45° D 0 ) Cy D 87.7 N Thus the magnitude is CD Cx 2 C Cy 2 D 88.8 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 475 Problem 6.99 Figure a is a diagram of the bones and biceps muscle of a person’s arm supporting a mass. Tension in the biceps muscle holds the forearm in the horizontal position, as illustrated in the simple mechanical model in Fig. b. The weight of the forearm is 9 N, and the mass m D 2 kg. (a) (b) Determine the tension in the biceps muscle AB. Determine the magnitude of the force exerted on the upper arm by the forearm at the elbow joint C. B 290 mm (a) A 50 mm C 9N m 200 mm 150 mm (b) Solution: Make a cut through AB and BC just above the elbow joint C. The angle formed by the biceps muscle with respect to the 290 D 80.2° . The weight of the mass is W D forearm is ˛ D tan1 50 29.81 D 19.62 N. T W 9N 200 mm α Cy 50 150 mm mm Cx The section as a free body: The sum of the moments about C is MC D 50T sin ˛ C 1509 C 350W D 0, from which T D 166.76 N is the tension exerted by the biceps muscle AB. The sum of the forces on the section is FX D Cx C T cos ˛ D 0, from which Cx D 28.33 N. FY D Cy C T sin ˛ 9 W D 0, from which Cy D 135.72. The magnitude of the force exerted by the forearm on the upper arm at joint C is FD 476 C2x C C2y D 138.65 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.100 The bones and the tendons in a horse’s rear leg are shown in Fig. a. A biomechanical model of the leg is shown in Fig. b. If the horse is stationary and the normal force everted on its leg by the ground is N D 1200 N, determine the tension s in the superficial digital flexor BC and the patellar ligament DF. 6 cm 6 cm 6 cm C D F E 40 cm A B 3 cm 72 cm N (a) 8 10 8 cm cm cm (b) Solution: The free-body diagrams for AB and AE are shown. The angle ˛ D tan1 18/58 D 17.2° The equilibrium equations for AB are MA : 1200 N10 cm C TBC cos ˛8 cm C TBC sin ˛3 cm D 0 Fx : Ax TBC sin ˛ D 0 Fy : Ay C TBC cos ˛ C 1200 N D 0 One of the equilibrium equations for AE is ME : TDF 8 cm Ax 49 cm Ay 10 cm D 0 Solving these four equations yields Ax D 417 N, Ay D 2540 N TBC D 1410 N, TDF D 625 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 477 Problem 6.101 The pressure force exerted on the piston is 2 kN toward the left. Determine the couple M necessary to keep the system in equilibrium. B 300 mm 350 mm 45° A C M 400 mm Solution: From the diagram, the coordinates of point B are d, d b can be determined from the where d D 0.3 cos45° . The distance Pythagorean Theorem as b D 0.352 d2 . From the diagram, the angle D 37.3° . From these calculations, the coordinates of points B and C are B (0.212, 0.212), and C (0.491, 0) with all distances being measured in meters. All forces will be measured in Newtons. B 0.3 m 45° A 0.35 m d d θ b C The unit vector from C toward B is uCB D 0.795i C 0.606j. y The equations of force equilibrium at C are and FX D FBC cos 2000 D 0, FBC FBCY c 2000 N x FBCX N FY D N FBC sin D 0. Solving these equations, we get N D 1524 Newtons(N), and FBC D 2514 N. The force acting at B due to member BC is FBC uBC D 2000i C 1524j N. B y FBC uCB M A rAB x The position vector from A to B is rAB D 0.212i C 0.212j m, and the moment of the force acting at B about A, calculated from the cross product, is given by MFBC D 747.6k N-m (counter - clockwise). The moment M about A which is necessary to hold the system in equilibrium, is equal and opposite to the moment just calculated. Thus, M D 747.6k N-m (clockwise). 478 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.102 In Problem 6.101, determine the forces on member AB at A and B. Solution: In the solution of Problem 6.101, we found that the force FBC uCB acting at point B of member AB was FBC uBC D 2000i C 1524j N, and that the moment acting on member BC about point A was given by M D 747.6k N-m (clockwise). Member AB must be in equilibrium, and we ensured moment equilibrium in solving Problem 6.101. y B M From the free body diagram, the equations for force equilibrium are and AX A x FX D AX C FBC uBCX D AX 2000 N D 0, AY FY D AY C FBC uBCY D AY C 1524 N D 0. Thus, AX D 2000 N, and AY D 1524 N. Problem 6.103 In Example 6.8, suppose that the object being held by the plier’s is moved to the left so that the horizontal distance from D to the object at E decreases from 30 mm to 20 mm. Draw a sketch of the pliers showing the new position of the object. What forces are exerted on the object at E as a result of the 150-N forces on the pliers? 150 N A C E B 150 N 30 mm 70 mm 30 mm D 30 mm 30 mm Solution: The analysis of the bottom grip of the pliers (member 3) is unchanged. The reactions Dx D 1517 N, Dy D 500 N. From the free-body diagram of the lower jaw (member 2) we obtain MC : E20 mm Dx 30 mm D 0 Therefore E D 2280 N 2280 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 479 Problem 6.104 The shovel of the excavator is supported by a pin support at E and the two-force member BC. The 300-lb weight W of the shovel acts at the point shown. Determine the reactions on the shovel at E and the magnitude of the axial force in the two-force member BC. Hydraulic cylinder Shovel 15 in A B 3 in E C 12 in D W 12 in 7 in 20 in Solution: The angle ˛ D tan1 3/15 D 11.3° The equilibrium equations for the shovel are Fx : Ex TBC cos ˛ D 0 Fy : Ey C TBC sin ˛ 300 lb D 0 MC : 300 lb20 in C Ex 12 in Ey 7 in D 0 Solving yields Ex D 604 lb, Thus 480 Ex D 604 lb, Ey D 179 lb, Ey D 179 lb, TBC D 616 lb axial force D 616 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.105 The shovel of the excavator has a pin support at E. The position of the shovel is controlled by the horizontal hydraulic piston AB, which is attached to the shovel through a linkage of the two-force members BC and BD. The 300-lb weight W of the shovel acts at the point shown. What is the magnitude of the force the hydraulic piston must exert to hold the shovel in equilibrium? Hydraulic cylinder Shovel 15 in A B 3 in E C 12 in D W 12 in 7 in 20 in Solution: From the solution to Problem 6.104 we know that the force in member BC is TBC D 616 lb We draw a free-body diagram of joint B and note that AB is horizontal. The angles are ˛ D tan1 3/15 D 11.3° ˇ D tan1 4/15 D 14.9° The equilibrium equations for joint B are Fx : TBC cos ˛ TAB TBD sin ˇ D 0 Fy : TBD cos ˇ TBC sin ˛ D 0 Solving yields TAB D 637 lb, TBD D 125 lb 637 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 481 Problem 6.106 The woman exerts 20-N forces on the handles of the shears. Determine the magnitude of the forces exerted on the branch at A. 20 N D C B A E 36 mm 25 mm 25 mm 65 mm Solution: Assume that the shears are symmetrical. Consider the 2 pieces CD and CE 20 N Now examine CD by itself MC D 20 N90 mm C Dy 25 mm D 0 ) Dy D 72 N 20 N Fx D 0 ) Dx D Ex Dy Fy D 0 ) Dy D Ey Cy MC D 0 ) Dx D Ex D 0 Dx = 0 20 N Dy Cx Finally examine DBA Dx MB : A36 mm Dy 50 mm D 0 A C Dx = 0 By Ex Ey Bx 20 N Dy Solving we find 482 A D 100 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.107 The person exerts 40-N forces on the handles of the locking wrench. Determine the magnitude of the forces the wrench exerts on the bolt at A. A B 40 N 8 mm 40 mm E C D 50 mm 30 mm 75 mm Solution: Recognize that DE is a 2-force member. Examine FDE 75 part CD 40 N 8 75 D Cx FDE C Cx D 0 Fx : p 5689 Cy 8 FDE C Cy C 40 N D 0 Fy : p 5689 MC : p 8 5689 40 N FDE 30 mm C 40 N105 mm D 0 By Solving we find Cx D 1312.5 N, Cy D 100 N, FDE D 1320 N A Bx Now examine ABC MB : A50 mm Cx 40 mm D 0 Fx : Bx Cx D 0 Cx Fy : By Cy A D 0 Solving: Answer: Cy A D 1050 N, Bx D 1312.5 N By D 1150 N A D 1050 N Problem 6.108 In Problem 6.107, determine the magnitude of the force the members of the wrench exert on each other at B and the axial force in the two-force member DE. Solution: From the previous problem we have BD Bx 2 C By 2 D 1745 N FDE D 1320 NC c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 483 Problem 6.109 The device is designed to exert a large force on the horizontal bar at A for a stamping operation. If the hydraulic cylinder DE exerts an axial force of 800 N and ˛ D 80° , what horizontal force is exerted on the horizontal bar at A? 90° D m m α 25 0m B 0m 25 25 0m m A E C 400 mm Solution: Define the x-y coordinate system with origin at C. The projection of the point D on the coordinate system is Fy B Ry D 250 sin ˛ D 246.2 mm, and Rx D 250 cos ˛ D 43.4 mm. Py D Px Fx Cx Cy The angle formed by member DE with the positive x axis is D Ry 180 tan1 D 145.38° . The components of the force 400 Rx produced by DE are Fx D F cos D 658.3 N, and Fy D F sin D 454.5 N. The angle of the element AB with the positive x axis is ˇ D 180 90 ˛ D 10° , and the components of the force for this member are Px D P cos ˇ and Py D P sin ˇ, where P is to be determined. The angle of the arm BC with the positive x axis is D 90 C ˛ D 170° . The projection of point B is Lx D 250 cos D 246.2 mm, and Ly D 250 sin D 43.4 mm. Sum the moments about C: MC D Rx Fy Ry Fx C Lx Py Ly Px D 0. Substitute and solve: P D 2126.36 N, and Px D P cos ˇ D 2094 N is the horizontal force exerted at A. 484 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.110 This device raises a load W by extending the hydraulic actuator DE. The bars AD and BC are 4 ft long, and the distances b D 2.5 ft and h D 1.5 ft. If W D 300 lb, what force must the actuator exert to hold the load in equilibrium? b W A B h D C Solution: The angle ADC is ˛ D sin1 h D 22.02° . The 4 (1) dCx hCy dB D 0. The sum of the forces: distance CD is d D 4 cos ˛. E Fx D Cx Ex D 0, from which The complete structure as a free body: The sum of the forces: (2) Fy D W C Cy C Dy D 0. Ex Cx D 0, Fx D Cx C Dx D 0. Fy D Cy Ey C B D 0, from which The sum of the moments about C: W MC D bW C dDy D 0. A These have the solution: B Ex B A Cy Cy D 97.7 lb, Ey Cx E Ex Dy D 202.3 lb, Dx and Cx D Dx . Divide the system into three elements: the platform carrying the weight, the member AB, and the member BC. (3) Dy Cy Ey C B D 0 Element AD: The sum of the moments about E: The Platform: (See Free body diagram) The moments about the point A: MA D bW dB D 0. The sum of the forces: Fy D A C B C W D 0. ME D d h d Dy C Dx A D 0, 2 2 2 from which (4) dDy C hDx dA D 0. These are four equations in the four unknowns: EX , EY , Dx , CX and DX These have the solution: Solving, we obtain Dx D 742 lb. B D 202.3 lb, and A D 97.7 lb. Element BC: The sum of the moments about E is MC D h d d Cy C Cx C B D 0, from which 2 2 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 485 Problem 6.111 The four-bar linkage operates the forks of a fork lift truck. The force supported by the forks is W D 8 kN. Determine the reactions on member CDE. 0.7 m 0.15 m 0.2 m W C 0.15 m B D E Forks 0.2 m 0.3 m A F 0.2 m Solution: Consider body BC. Note that AB is a 2-force body. W = 8 kN Fx : Cx D 0 MB : Cy 0.2 m 8 kN0.9 m D 0 Cx ) Cx D 0, Cy D 36 kN Now examine CDE. Note that DF is a 2-force body. Cy 3 ME : Cy 0.15 m Cx 0.15 m C p FDF 0.15 m D 0 13 2 Fx : Cx C Ex C p FDF D 0 13 FAB Cy 3 Fy : Cy C Ey p FDF D 0 13 Solving we find Note that Cx FDF D 43.3 kN, Ex D 24 kN, Ey D 0 2 3 Dx D p FDF , Dy D p FDF 13 13 Summary: Cx D 0, Cy D 36 kN Dx D 24 kN, Dy D 36 kN Ex D 24 kN, Ey D 0 Ey Ex D 3 2 FDF 486 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.112 If the horizontal force on the scoop is F D 2000 lb, what is the magnitude of the axial force in the hydraulic actuator AC? C 38 in B 28 in Scoop D 10 in A F 10 in 20 in 12 in Solution: We start with the free-body diagram of the scoop. Note that BC is a two-force body. The angle ˛ D tan1 38/32 D 49.9° We have the following equilibrium equation MA : 2000 lb 10 in C TBC cos ˛28 in C TBC sin ˛12 in D 0 ) TBC D 735 lb Now we work with the free-body diagram of joint C. The angles ˇ D tan1 20/66 D 16.9° D tan1 10/38 D 14.7° The equilibrium equations are Fx : TBC cos ˛ C TAC sin ˇ TCD sin D 0 Fy : TBC sin ˛ TAC cos ˇ TCD cos D 0 Solving yields TAC D 1150 lb, TCD D 553 lb Thus 1150 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 487 Problem 6.113 A 10-kip horizontal force acts on the bucket of the excavator. Determine the reactions on member ACF at A and F. 9 ft 2 ft E D 1 ft 4 in F C 2 ft 4 ft 4 in A 1 ft 8 in B 5 ft 6 in Bucket 2 ft 3 ft 10 kip Solution: We start with the free-body diagram of the entire structure. The angle ˛ D tan1 36/72 D 26.6° The equilibrium equations are Fx : Ax TBC sin ˛ C 10 kip D 0 Fy : Ay TBC cos ˛ D 0 MA : 10 kip 66 in C TBC sin ˛ 52 in TBC cos ˛ 60 in D 0 Next we examine the free-body diagram of the member on the right. The angle ˇ D tan1 84/4 D 87.3° The equilibrium equations are Fx : Fx C 10 kip TDE sin ˇ D 0 Fy : Fy TDE cos ˇ D 0 MF : TDE sin ˇ 24 in C 10 kip 120 in D 0 Solving these six equations we find TBC D 21.7 kip, TDE D 49.2 kip Ax D 0.294 kip, Fx D 59.2 kip, 488 Ay D 19.4 kip Fy D 2.34 kip c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.114 The structure shown in the diagram (one of the two identical structures that support the scoop of the excavator) supports a downward force F D 1800 N at G. Members BC and DH can be treated as two-force members. Determine the reactions on member CDK at K. 320 mm C Shaft 100 mm Scoop 260 mm H D J 160 mm L Solution: Start with the scoop Now examine CDK 1800 N0.2 m D 0 ) FBC D 873 N FBC 1 380 mm 1120 mm 200 mm 56 4 FDH 0.26 m p FBC 0.52 m D 0 MK : p 3161 17 56 4 FDH C p FBC C Kx D 0 Fx : p 3161 17 5 1 FDH p FBC C Ky D 0 Fy : p 3161 17 Solving we find 4 G F K 1040 mm 4 1 MJ : p FBC 0.44 m p FBC 0.06 m 17 17 260 mm B 180 mm Kx D 847 N, Ky D 363 N 4 1800 N 1 FBC Jx 56 5 Jy FDH Kx Ky c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 489 Problem 6.115 The loads F1 D 440 N and F2 D 160 N. Determine the axial forces in the members. Indicate whether they are in tension (T) or compression (C). F1 A F2 400 mm C 200 mm B 700 mm Solution: The sum of the moments about C is MC D 0.7BY C 0.7F1 C 0.4F2 D 0, from which By D 0.7F1 C 0.4F2 D 531.43 N . 0.7 The axial loads at joint B are AB D By D 531.4 N C , and BC D 0 . Similarly, the sum of the forces at the joint A is FAx D F2 C AC cos ˇ D 0, from which AC D 490 F2 D 184.3 N T cos ˇ c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. B Problem 6.116 The truss supports a load F D 10 kN. Determine the axial forces in the members AB, AC, and BC. 3m C A D 4m 3m F Solution: Find the support reactions at A and D. Fx : Ax D 0 3m C Fy : Ay C Dy 10 D 0 AX 4m MA : 410 C 7Dy D 0 3m DY AY Solving, 10 kN Ax D 0, Ay D 4.29 kN FAB Dy D 5.71 kN y θ Joint A: tan D 3 4 FAC D 36.87° x AY Ay D 4.29 kN Fx : FAB cos C FAC D 0 Fy : Ay C FAB sin D 0 Solving, FAB D 7.14 kN C FBC FAC FCD 10 kN FAC D 5.71 kN T Joint C: Fx : FCD FAC D 0 Fy : FBC 10 kN D 0 Solving FBC D 10 kN T FCD D C5.71 kN T c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 491 Problem 6.117 Each member of the truss shown in Problem 6.116 will safely support a tensile force of 40 kN and a compressive force of 32 kN. Based on this criterion, what is the largest downward load F that can safely be applied at C? Solution: Assume a unit load F and find the magnitudes of the tensile and compressive loads in the truss. Then scale the load F up (along with the other loads) until either the tensile limit or the compressive limit is reached. B Fx : Ax D 0 (1) Fy : Ay C Dy F D 0 (2) 2 5 1 External Support Loads: C A MA : 4F C 7Dy D 0 D 3 4 4m 3m 3m F (3) Joint A: 3 2 tan D D 36.87° Fx : AX FAC C FAB cos D 0 (4) F Fy : FAB sin C Ay D 0 (5) y Joint C DY AY FAB Fx : FCD FAC D 0 (6) Fy : FBC F D 0 (7) θ x FAC Joint D tan D 3 3 AY D 45° y Fx : FCD FBD cos D 0 (8) Fy : FBD sin C Dy D 0 (9) FBD φ Setting F D 1 and solving, we get the largest tensile load of 0.571 in AC and CD. The largest compressive load is 0.808 in member BD. x FCD Largest Tensile is in member BC. BC D F D 1 DY The compressive load will be the limit 32 Fmax D 1 0.808 y FBC Fmax D 40 kN FCD FAC x F 492 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.118 The Pratt bridge truss supports loads at F, G, and H. Determine the axial forces in members BC, BG, and FG. B C D 4m E A F G 60 kN 80 kN 20 kN 4m 4m 4m 4m Solution: The angles of the cross-members are ˛ D 45° . H 4m 4m 4m 4m The complete structure as a free body: Ay MA D 604 808 2012 C 16E D 0, from which E D 70 kN. The sum of the forces: Fx D Ax D 0. 4m Ax The sum of the moments about A: Ay AB α AF Joint A 60 kN 80 kN 20 kN BF AF FG 60 kN Joint F E α α BC AB BF BG Joint B Fy D Ay 60 80 20 C E D 0, from which Ay D 90 kN The method of joints: Joint A: FY D Ay C AB sin ˛ D 0, from which AB D 127.3 kN C, Fx D AB cos ˛ C AF D 0, from which AF D 90 kN T. Joint F: Fx D AF C FG D 0, from which FG D 90 kN T . Fy D BF 60 D 0, from which BF D 60 kN C. Joint B: and Fx D AB cos ˛ C BC C BG cos ˛ D 0, Fy D AB sin ˛ BF BG sin ˛ D 0, from which: AB sin ˛ BF BG sin ˛ D 0. Solve: BG D 42.43 kN T , and AB cos ˛ C BC C BG cos ˛ D 0, from which BC D 120 kN C c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 493 Problem 6.119 Consider the truss in Problem 6.118. Determine the axial forces in members CD, GD, and GH. Solution: Use the results of the solution of Problem 6.130: BC D 120 kN C, BG D 42.43 kN T, and FG D 90 kN T. BC CD CG Joint C BG α CG GD α GH 80 kN Joint G The angle of the cross-members with the horizontal is ˛ D 45° . Joint C: Fx D BC C CD D 0, from which CD D 120 kN C FY D CG D 0, from which CG D 0. Joint G: Fy D BG sin ˛ C GD sin ˛ C CG 80 D 0, from which GD D 70.71 kN T . Fy D BG cos ˛ C GD cos ˛ FG C GH D 0, from which GH D 70 kN T 494 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.120 The truss supports loads at F and H. Determine the axial forces in members AB, AC, BC, BD, CD, and CE. 200 lb F 100 lb 4 in D 4 in H B E C 4 in J G A I 6 in 6 in 6 in Solution: The complete structure as a free body: The sum of the moments about I: 6 in 200 lb 100 lb MA D 1006 C 20012 24AY D 0, from which AY D 125 lb. The sum of forces: Ax Fx D Ax D 0. The method of joints: The angles of the inclined members with the horizontal are 12 in 6 in 6 in CD AB BD α BC AC α ˛ D tan1 0.6667 D 33.69° Joint A: I Ay Ay Joint A AB Joint B CE BC α AC Joint C Fx D AC cos ˛ D 0, from which AC D 0. Fy D Ay C AB C AC sin ˛ D 0, from which AB D 125 lb C Joint B : Fyt D AB C BD sin ˛ D 0, from which BD D 225.3 lb C . Fx D BD cos ˛ C BC D 0, from which BC D 187.5 lb T Joint C : Fx D BC AC cos ˛ C CE cos ˛ D 0, from which CE D 225.3 lb T Fy D AC sin ˛ C CD C CE sin ˛ D 0, from which CD D 125 lb C c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 495 Problem 6.121 Consider the truss in Problem 6.120. Determine the axial forces in members EH and FH. Solution: Use the results from the solution to Problem 6.132: DF α DE CE D 225.3 lb T, BD CD D 125 lb C, 200 lb α α FH DF CD Joint D EF Joint F EF DE α CE EH α EG Joint E BD D 225.3 lb C. The method of joints: The angle of inclined members with the horizontal is ˛ D 33.69° . Joint D: Fy D BD sin ˛ CD C DF sin ˛ D 0, from which DF D 450.7 lb C. Fx D DF cos ˛ C DE BD cos ˛ D 0, from which DE D 187.5 lb T Joint F : Fx D DF cos ˛ C FH cos ˛ D 0, from which FH D 450.7 lb C Fy D 200 DF sin ˛ FH sin ˛ EF D 0, from which EF D 300 lb T Joint E : Fy D CE sin ˛ C EF EG sin ˛ D 0, from which EG D 315 lb T Fx D DE C EH CE cos ˛ C EG cos ˛ D 0, from which EH D 112.5 lb T 496 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.122 Determine the axial forces in members BD, CD, and CE. 10 kN A 2m 14 kN C B 2m D E 2m G F 2m I H 6m Solution: Use the method of sections y 10 kN A 2m Θ B 1.5 m Θ 14 kN FBD C x Θ FCD FCE D tan D 2 1.5 D 53.13° Fx : FCE cos FCD cos C 24 D 0 Fy : FBD FCD sin FCE sin D 0 MB : 210 1.5FCD sin 1.5FCE sin D 0 3 eqns-3 unknowns. Solving FBD D 13.3 kN, FCD D 11.7 kN, FCE D 28.3 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 497 Problem 6.123 For the truss in Problem 6.122, determine the axial forces in members DF, EF, and EG. Solution: Use method of sections A 10 kN 2m A 14 kN 10 kN C B 2m 2m 14 kN D E 2m 2m D FDF G F 2m 3 E I H φ Θ 2 FEG FEF 3 tan D 6m 1.5 2 1.5 D 53.13° tan D 2 3 D 33.69° Fx : 24 C FEG cos FEF cos D 0 Fy : FDF FEF sin FEG sin D 0 ME : 3FDF 214 410 D 0 Solving, FEG D 32.2 kN C FDF D 22.67 kN T FEF D 5.61 kN T 498 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.124 The truss supports a 400-N load at G. Determine the axial forces in members AC, CD, and CF. 400 N A C E G 300 mm 600 mm H F D B 300 mm Solution: The complete structure as a free body: The sum of the moments about A: 400 N Ax MA D 900400 C 600B D 0, 600 mm B Fx D Ax C B D 0, AB B from which Ax D 600 N. 300 mm 900 mm Ay from which B D 600 N. The sum of forces: 300 mm from which Ay D 400 N. The method of joints: The angle from the horizontal of element BD is 300 900 AC αAD AD AB Joint A AC Joint D CE αCF CF CD Joint C D 18.43° . The angle from the horizontal of element AD is ˛AD D 90 CD AD αAD DF θ BD AY AX Joint B Fy D Ay 400 D 0, D tan1 θ BD tan1 300 600 300 tan Joint D: D 59.04° . Fx D AD cos ˛AD BD cos C DF cos D 0, from which DF D 505.96 N C The angle from the horizontal of element CF is ˛CF D 90 tan1 300 6001 tan Fy D AD sin ˛AD C CD BD sin C DF sin D 0, D 53.13° . Joint B: Fx D B C BD cos D 0, from which CD D 240 N C Joint C : Fy D CD CF sin ˛CF D 0, from which BD D 632.5 N C from which CF D 300 N T Fy D AB C BD sin D 0, from which AB D 200 N T Joint A: Fy D Ay AD sin ˛AD AB D 0, from which AD D 233.2 N T Fx D Ax C AC C AD cos ˛AD D 0, from which AC D 480 N T c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 499 Problem 6.125 Consider the truss in Problem 6.124. Determine the axial forces in members CE, EF, and EH. Solution: Use the results of the solution of Problem 6.124: AC CE αCF AC D 480 N T, CD CF D 300 N T, CF Joint C CF αCF θ DF EF EG αEH CE FH EF Joint F EH Joint E DF D 505.96 N C, Joint F : D 18.4° , ˛CF D 53.1° . The method of joints: The angle from the horizontal of element EH is ˛EH D 90 tan1 300 600 900 tan Fy D CF cos ˛CF DF cos C FH cos D 0, from which FH D 316.2 N C D 45° Fy D EF C CF sin ˛CF DF sin C FH sin D 0, from which EF D 300 N C Joint C: Joint E : Fx D AC C CE C CF cos ˛CF D 0, Fy D EH sin ˛EH EF D 0, from which CE D 300 N T from which EH D 424.3 N T Problem 6.126 Consider the truss in Problem 6.124. Which members have the largest tensile and compressive forces, and what are their values? Solution: The axial forces for all members have been obtained in Problems 6.124 and 6.125 except for members EG and GH. These are: CE Joint E: EF Fx D CE C EG C EH cos ˛EH D 0, EG αEH Joint E EH 400 N EG GH Joint G from which EG D 0 Joint G: Fy D GH 400 D 0, from which GH D 400 N C. This completes the determination for all members. A comparison of tensile forces shows that AC D 480 N T is the largest value, and a comparison of compressive forces shows that BD D 632.5 N C is the largest value. 500 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.127 The Howe truss helps support a roof. Model the supports at A and G as roller supports. Use the method of joints to determine the axial forces in members BC, CD, CI, and CJ. 6 kN 4 kN 4 kN D 2 kN 2 kN C E 4m B F A G H I 2m Solution: The free body diagrams for the entire truss and the required joints are shown. The whole truss: The equations of equilibrium for the entire truss are: FX D 0, 2m J K 2m L 2m 2m 2m 6 kN y 4 kN D E 4 kN C 2 kN F 2 kN 4m B A G x H I J K L AY 12 m GY FY D AY C GY 18 kN D 0. y Instead of using the moment equation here (it would work), we see that the loading is symmetric. Thus, AY D GY D 9 kN. A TAH We need unit vectors along AB, BC, CD, (note that these are the same), and along BI, and CJ. We get E uAB D uBC D uCD D 0.832i C 0.555j, TDF F uBI D 0.832i 0.555j, y TAB TFH TEF TFG TBX THI x TAH H THI y 4 kN y TCX I x TBH x TIJ TBC C TCD x TCJ TCI and uCJ D 0.6i 0.8j. Joint C: Joint A: The equations of equilibrium are and FX D TAB uABX C TAH D 0 FY D TAB uABY C AY D 0. Joint H: The equations of equilibrium are and FX D TAH C THI D 0, FY D TBH D 0. Joint B: FX D TBC uBCX C TCJ uCJX C TCD uCDX D 0, FY D TBC uBCY C TCJ uCJY C TCD uCDY TCI 4 D 0. Solving these equations in sequence (we can solve at each joint before going to the next), we get TAB D 16.2 kN, TAH D 13.5 kN, TBH D 0 kN, THI D 13.5 kN, TBC D 14.4 kN, TBI D 1.80 kN, TIJ D 12.0 kN, TCI D 1.00 kN, TCJ D 4.17 kN, and TCD D 11.4 kN. FX D TAB uABX C TBC uBCX C TBI uBIX D 0, FY D TAB uABY C TBC uBCY C TBI uBIY TBH 2 D 0, Joint I: and FX D THI C TIJ TBI uBIX D 0, FY D TCI TBI uBIY D 0, c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 501 Problem 6.128 For the roof truss in Problem 6.127, use the method of sections to determine the axial forces in members CD, CJ, and IJ. Solution: The free body diagram of the section is shown at the right. The support force at A is already known from the solution to Problem 6.139. The equations of equilibrium for the section are and FX D TCD uCDX C TCJ uCJX C TIJ D 0, FY D TCD uCDY C TCJ uCJY C AY D 0, MC D yC TIJ 4AY D 0. Solving, we get TIJ D 12.0 kN, TCJ D 4.17 kN, and TCD D 11.4 kN. Note that these values check with the values obtained in Problem 6.139. 4 kN TCD 2 kN C I D TCJ TIJ J AY 502 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.129 A speaker system is suspended from the truss by cables attached at D and E. The mass of the speaker system is 130 kg, and its weight acts at G. Determine the axial forces in members BC and CD. 0.5 m 0.5 m 0.5 m 0.5 m 1m C E A 1m B D G Solution: The speaker as a free body: The weight of the speaker Cy 1m is W D 1309.81 D 1275.3 N. Make a cut through the suspension cables D, E, the sum of the moments about cable D is A Cx B 0.5 m The structure as a free body: The sum of the moments about C is D D W Fy D D C E W D 0, from which D D 425.1 N. E E MD D 1W C 1.5E D 0, from which E D 850.2 N. The sum of the forces: 2m 1 m 0.5 m CY CE α CD E DE BD MC D C1A 0.5D 2E D 0, Joint E β α DE D Joint D AC CE β β BC CD Joint C from which A D 1912.95 N. The sum of the forces: Joint D: Fy D A C Cy W D 0, Fy D CD sin ˇ C DE sin ˛ D D 0, from which Cy D 637.65 N and from which CD D 1425.8 N T Fx D Cx D 0. Joint C : The method of joints: angle of member DE relative to the hori The 1 D 33.69° . The angles of members AB, BC, zontal is ˛ D tan1 1.5 1 and CD are ˇ D 90 tan 0.5 D 63.43° . Fy D CD sin ˇ BC sin ˇ C Cy D 0, from which BC D 2138.7 N T Joint E : Fy D E DE sin ˛ D 0, from which DE D 1532.72 N C. Fx D CE DE cos ˛ D 0, from which CE D 1275.3 N T c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 503 y Problem 6.130 The mass of the suspended object is 900 kg. Determine the axial forces in the bars AB and AC. D (0, 4, 0) m Strategy: Draw the free-body diagram of joint A. A (3, 4, 4) m B (0, 0, 3) m C (4, 0, 0) m x z Solution: The free-body diagram of joint A is. TAD TAB TAC (900) (9.81) N The position vectors from pt A to pts B, C, and D are rAB D 3i 4j k (m), rAC D i 4j 4k (m), rAD D 3i 4k (m). Dividing these vectors by their magnitudes, we obtain the unit vectors eAB D 0.588i 0.784j 0.196k, eAC D 0.174i 0.696j 0.696k, eAD D 0.6i 0.8k. From the equilibrium equation TAB eAB C TAC eAC C TAD eAD 9009.81j D O, We obtain the equations 0.588TAB C 0.174TAC 0.6TAD D 0, 0.784TAB 0.696TAC 9009.81 D 0, 0.196TAB 0.696TAC 0.8TAD D 0. Solving, we obtain TAB D 7200 N, TAC D 4560 N, TAD D 5740 N. 504 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.131 Determine the forces on member ABC, presenting your answers as shown in Fig. 6.25. Obtain the answer in two ways: (a) When you draw the free-body diagrams individual members, place the 400-lb load free-body diagram of member ABC. (b) When you draw the free-body diagrams individual members, place the 400-lb load free-body diagram of member CD. of the on the 1 ft 1 ft 200 lb C D 400 lb 1 ft of the on the B 1 ft E Solution: The angle of element BE relative to the horizontal is ˛ D tan1 1 2 1 ft D 26.57° . A The complete structure as a free body: The sum of the moments about A: Cy MA D 3400 1200 C 2Fy D 0 from which Fy D 700 lb. The sum of forces: F 400 lb C Cx y B 200 lb Dx Dy Dy Dx B Fy D Ay C Fy 200 D 0, Ax from which Ay D 500 lb . Cx B Ay B Fx Fy Fx D Ax C Fx C 400 D 0. 100 lb (a) 400 lb Element CD: The sum of the moments about D: 26.6° MD D 200 C 2Cy D 0, 1341 lb from which Cy D 100 lb . 400 lb 400 lb 500 lb Fy D Dy Cy 200 D 0, from which Dy D 100. from which Check: Fx D Cx Dx D 0, BD from which Dx D Cx . 500 C 100 D 1341.6 lb. sin ˛ Element DEF: The sum of the moments about F: check. From above: Cx D Dx D 400 lb . MF D 3Dx C B cos ˛ D 0, cos ˛ . from which Dx D B 3 Fy D Fy C B sin ˛ C Dy D 0, from which BD 700 C 100 D 1341.6 lb , and Dx D sin ˛ Fx D 400 C Cx C B cos ˛ C Ax D 0, from which Ax D 400 lb . (b) When the 400 lb load is applied to element CD instead, the following changes to the equilibrium equations occur: Element CD: 400 lb Fx D Cx Dx C 400 D 0, Element ABC: from which Cx C Dx D 400. Element ABC: MA D 2B cos ˛ 3400 3Cx D 0. The sum of the forces Fy D Cy B sin ˛ C Ay D 0, Fx D Cx C Ax B cos ˛ D 0. Element DEF : No changes. The changes in the solution for Element ABC Cx D 800 lb when the external load is removed, instead of Cx D 400 lb when the external load is applied, so that the total load applied to point C is the same in both cases. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 505 Problem 6.132 The mass m D 120 kg. Determine the forces on member ABC. A B C 300 mm D m E 200 mm Solution: The weight of the hanging mass is given by m W D mg D 120 kg 9.81 2 D 1177 N. s FX D AX C EX D 0, Cx B W B Cx B Cy B Ex and Cy Ay Ax The complete structure as a free body: The equilibrium equations are: 200 mm FY D AY W D 0, MA D 0.3EX 0.4W D 0. Solving, we get AX D 1570 N, AY D 1177 N, and EX D 1570 N. Element ABC: The equilibrium equations are and: FX D Ax C CX D 0, FY D AY C CY BY W D 0, MA D 0.2BY C 0.4cY 0.4W D 0. Solution gives BY D 2354 N (member BD is in tension), CX D 1570 N, and CY D 2354 N. 506 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.133 Determine the reactions on member ABC at B and C. 4 kN A 0.2 m D B 2 kN-m 0.2 m C E 0.2 m 0.2 m Solution: We draw free-body diagrams for the entire structure, and for members BD and ABC. From the entire structure: Fx : Cx C 4 kN D 0 ME : Cy 0.4 m 4 kN0.4 m 2 kN-m D 0 From body ABC MA : Bx 0.2 m C Cx 0.4 m D 0 And from body BD MD : By 0.2 m 2 kN-m D 0 Solving these four equations yields Bx D 8 kN, By D 10 kN Cx D 4 kN, Cy D 9 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 507 Problem 6.134 The truck and trailer are parked on a 10° slope. The 14,000-lb weight of the truck and the 8000-lb weight of the trailer act at the points shown. The truck’s brakes prevent its rear wheels at B from turning. The truck’s front wheels at C and the trailer’s wheels at A can turn freely, which means they do not exert friction forces on the road. The trailer hitch at D behaves like a pin support. Determine the forces exerted on the truck at B, C, and D. 2 ft y 9 ft 3 ft 14 ft D 4 ft 3 ft 8 kip 6 ft 10⬚ B 14 kip 5 ft 6 in x C A Solution: We separate the two vehicles and draw a free-body diagram of each. Starting with the trailer we have MA : 8 kip cos 10° 4 ft C 8 kip sin 10° 6 ft C Dx 5.5 ft Dy 16 ft D 0 Fx : 8 kip sin 10° Dx D 0 Now we use the free-body diagram for the truck MB : C11 ft C Dy 2 ft Dx 5.5 ft 14 kip cos 10° 8 ft C 14 kip sin 10° 3 ft D 0 Fx : Bx C Dx 14 kip sin 10° D 0 Fy : By C Dy C C 14 kip cos 10° D 0 Solving yields Bx D 3820 lb, By D 6690 lb Dx D 1390 lb, Dy D 1930 lb, 508 C D 9020 lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 6.135 The 600-lb weight of the scoop acts at a point 1 ft 6 in to the right of the vertical line CE. The line ADE is horizontal. The hydraulic actuator AB can be treated as a two-force member. Determine the axial force in the hydraulic actuator AB and the forces exerted on the scoop at C and E. B C 2 ft A Solution: The free body diagrams are shown at the right. Place the coordinate origin at A with the x axis horizontal. The coordinates (in ft) of the points necessary to write the needed unit vectors are A (0, 0), B (6, 2), C (8.5, 1.5), and D (5, 0). The unit vectors needed for this problem are uBA D 0.949i 0.316j, E D 5 ft TCB 1 ft 6 in 1 ft 2 ft 6 in Scoop C 1.5 ft 1.5 ft G EX E EY uBC D 0.981i 0.196j, 600 lb and uBD D 0.447i 0.894j. y The scoop: The equilibrium equations for the scoop are and FX D TCB uBCX C EX D 0, TBA x TCB FY D TCB uBCY C EY 600 D 0, TBD MC D 1.5EX 1.5600 lb D 0. Solving, we get EX D 600 lb, EY D 480 lb, and TCB D 611.9 lb. Joint B: The equilibrium equations for the scoop are and FX D TBA uBAX C TBD uBDX C TCB uBCX D 0, FY D TBA uBAY C TBD uBDY C TCB uBCY D 0. Solving, we get TBA D 835 lb, and TBD D 429 lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 509 Problem 6.136 Determine the force exerted on the bolt by the bolt cutters. 100 N A 75 mm 40 mm C 55 mm B D 90 mm 60 mm 65 mm 300 mm 100 N Solution: The equations of equilibrium for each of the members will be developed. AY AX F Member AB: The equations of equilibrium are: and 100 N λ 40 mm B 55 mm 75 mm A FX D AX C BX D 0, BX BY FY D AY C BY D 0, 90 mm MB D 90F 75AX 425100 D 0 60 mm 65 mm 300 mm Member BD: The equations are AY AX CY FX D BX C DX D 0, 40 mm 75 mm and FY D BY C DY C 100 D 0, MB D 15DX C 60DY C 425100 D 0. 90 mm Member AC: The equations are and FX D AX C CX D 0, FY D AY C CY C F D 0, 60 mm 65 mm 300 mm BY DX B BX D DY MA D 90F C 125CY C 40CX D 0. Member CD: The equations are: D F C 55 mm X C 60 mm 65 mm 300 mm 100 N FX D CX DX D 0, CY FY D CY DY D 0. Solving the equations simultaneously (we have extra (but compatible) equations, we get F D 1051 N, AX D 695 N, AY D 1586 N, BX D 695 N, BY D 435 N, CX D 695 N, CY D 535 N, DX D 695 N, and Dy D 535 N C CX DX D DY Problem 6.137 For the bolt cutters in Problem 6.136, determine the magnitude of the force the members exert on each other at the pin connection B and the axial force in the two-force member CD. Solution: From the solution to 6.136, we know BX D 695 N, and BY D 435 N. We also know that CX D 695 N, and CY D 535 N, from which the axial load in member CD can be calculated. The load in CD is given by TCD D C2X C C2Y D 877 N 510 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.1 In Active Example 7.1, Suppose that the triangular area is oriented as shown. Use integration to determine the x and y coordinates of its centroid. (Notice that you already know the answers based on the results of Active Example 7.1.) Solution: Theheight of the vertical strip is h h/b x so the area is dA D h h x dx. Use this expression to evaluate Eq. (7.6). b The x coordinate of the centroid is 2 b x3 x h h x h x dx b 2 3b 0 b D 0 b x D A D b D 3 2 h x dA h x dx h x A b 0 2b 0 b xdA y 1 h The y coordinate of the midpoint of the vertical strip is h h x D 2 b 1 h h C x . We let this value be the value of y in Eq. (7.7): 2 b h x b b ydA y D A D 0 yD 2h 3 dA A b , 3 xD Problem 7.2 In Example 7.2, suppose that the area is redefined as shown. Determine the x coordinate of the centroid. 1 2 b x3 h2 h h x 2 hC x h x dx 2h 2 3b 0 b b D D b 2 b 3 h x h x dx h x b 0 2b 0 Solution: The height of the vertical strip is 1 x2 , so the area is dA D 1 x2 dx. Use this expression to evaluate Eq. (7.6). The x coordinate of the centroid is 1 x4 x2 3 2 4 0 x D A D D 01 D 3 1 8 x 2 dA 1 x dA x A 0 3 0 y 1 xdA y⫽1 (1, 1) x1 x2 x 3 8 xD y ⫽ x2 x Problem 7.3 In Example 7.2, suppose that the area is redefined as shown. Determine the y coordinate of the centroid. y y⫽1 Solution: The height of the vertical strip is 1 x2 , so the area is dA D 1 x2 dx. The y coordinate of the midpoint of the vertical strip is 1 1 1 1 x2 D 1 C x2 . 2 2 We let this be the value of y in Eq. (7.7): (1, 1) ydA y D A D dA A y ⫽ x2 yD 0 1 1 x5 x 1 1 C x2 1 x2 x 3 2 10 0 2 D D 1 3 1 5 x 2 1 x dA x 0 3 0 3 5 x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 511 Problem 7.4 Determine the centroid of the area. Solution: The height of a vertical strip of width dx is x2 x C 1, y so the area is dA D x2 x C 1 dx. Use this expression to evaluate Eq. (7.6). The x coordinate of the centroid is 2 x3 x2 x4 C 4 3 2 0 x D A D 01 D 2 D 1.25 3 2 x x 2 dA x x C 1 dA C x A 0 3 2 0 1 xdA y ⫽ x2 ⫺ x ⫹ 1 xx 2 x C 1 x The y coordinate of the midpoint of the vertical strip is We let this be the value of y in Eq. (7.7): 1 ydA x 2 y D A D 0 dA A 1 1 2 x x C 1. 2 1 2 x x C 12 x 2 x 2 x C 1 dA 0 2 2x4 1 x5 C x3 x2 C x 2 5 4 0 D 0.825 D 2 3 x2 x Cx 3 2 0 x D 1.25, Problem 7.5 Determine the coordinates of the centroid of the area. y D 0.825 y Solution: Use a vertical strip - The equation of the line is y D 8 2x 3 6 9 xy dx 3 xD 9 y dx 3 yD 9 3 y dx 3 9 2 x 8 x dx 11 3 D D 3 9 2 2 8 x dx 3 3 1 yy dx 2 9 2 3 9 x 1 2 2 8 x dx 13 2 3 D 3 9 D 6 2 8 x dx 3 3 9 x D 5.5 y D 2.17 512 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.6 Determine the x coordinate of the centroid of the area and compare your answers to the values given in Appendix B. Solution: 0 0 y = cx n xD a x cxn dy dx D y 0 b AD 1 A b cbnC1 provided that n > 1 nC1 cxn bn C 1 nC2 x dy dx D 0 0 Matches the appendix Problem 7.7 Determine the y coordinate of the centroid of the area and compare your answer to the value given in Appendix B. Solution: See solution to 7.6 yD 1 A b cxn y dy dx D 0 0 bn cn C 1 4n C 2 Matches the appendix Problem 7.8 Suppose that an art student wants to paint a panel of wood as shown, with the horizontal and vertical lines passing through the centroid of the painted area, and asks you to determine the coordinates of the centroid. What are they? Solution: The area: y The x-coordinate: 1 AD x C x 3 dx D 0 1 xx C x 3 dx D 0 y = x + x3 x2 x4 C 2 4 x3 x5 C 3 5 Divide by the area: x D 1 D 0 1 D 0 3 . 4 8 . 15 32 D 0.711 45 The y-coordinate: The element of area is dA D 1 x dy. Note that dy D 1 C 3x2 dx, hence dA D 1 x1 C 3x2 dx. Thus yA D y dA D A 0 1 ft x 1 x C x 3 1 x1 C 3x2 dx, 0 from which 1 x x 2 C 4x3 4x4 C 3x5 3x6 dx 0 D 1 4 4 3 3 1 C C D 0.4381. 2 3 4 5 6 7 Divide by A y D 0.5841 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 513 Problem 7.9 Determine the value of the constant c so that the y coordinate of the centroid of the area is y D 2. What is the x coordinate of the centroid? y Solution: The height of a vertical strip of width dx is cx2 so the area is dA D cx 2 dx. The y coordinate of the midpoint of the vertical strip is 12 cx 2 . We let this be the value of y in Eq. (7.7): 4 4 c x5 1 2 cx cx2 dx 2 5 2 D 4 2 D 5.31c 4 x3 cx 2 dx 2 3 2 ydA y D A y ⫽ cx 2 D dA 2 A D 2 ) c D 0.376 0 2 4 x The x coordinate of the centroid is 4 xdA x D A D xcx 2 dx 2 4 dA D 2 cx dx A 2 x4 4 x3 3 4 2 4 D 3.21 2 Notice that the value of x does not depend on the value of c. c D 0.376, Problem 7.10 Determine the coordinates of the centroid of the metal plate’s cross-sectional area. y Solution: Let dA be a vertical strip: The area dA D y dx D where 4 1 y = 4 – – x 2 ft 4 Therefore 4 1 2 x 4 dx. The curve intersects the x axis 1 2 x D 0, or x D š4. 4 4 4 1 3 2 x 2x x 4x dx x dA 16 4 4 D 44 D 0. x D A D 3 4 1 x 2 dA 4 x dx 4x A 4 4 12 4 x x D 3.21 4 To determine y, let y in equation (7.7) be the height of the midpoint of the vertical strip: 4 y dA y D A D 4 1 2 dA A 1 1 4 x2 4 x 2 dx 4 4 4 1 2 4 x dx 4 4 4 x5 x3 1 4 C 8x x 8 x2 C dx 3 532 4 32 D 4 4 D 3 4 x2 x 4 dx 4x 4 4 12 4 D 4 34.1 D 1.6 ft. 21.3 y dA y x x 514 dx c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 4 3 y, m Problem 7.11 An architect wants to build a wall with the profile shown. To estimate the effects of wind loads, he must determine the wall’s area and the coordinates of its centroid. What are they? y = 2 + 0.02x2 2 1 0 0 2 4 6 8 10 x, m Solution: 10 Area D 10 y dx D 0 2 C 0.02x2 dx 0 10 x3 Area D 2x C 0.02 D 26.67 m2 3 0 dA D y dx D 2 C 0.02x2 dx Y X 10 10 x dA x D 0 10 2x C 0.02x3 dx 0 D 26.67 dA 0 2 10 x x 4 2 C 0.02 2 4 0 m xD 26.67 104 100 C 0.02 150 4 xD D 26.67 26.67 x D 5.62 m yD 0 10 y 2 y dx D 10 dA 1 2 10 2 C 0.02x2 2 dx 0 26.67 0 yD 1 226.67 10 4 C 0.08x2 C 0.0004x4 dx 0 3 5 10 x x 4x C 0.08 C 0.0004 3 5 0 yD 226.67 yD 74.67 53.34 y D 1.40 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 515 Problem 7.12 Determine the coordinates of the centroid of the area. y y⫽⫺ 1 2 x ⫹ 4x ⫺ 7 4 x Solution: Use a vertical strip. We first need to find the x intercepts. 1 y D x 2 C 4x 7 D 0 ) x D 2, 14 4 xy dx 2 xD 14 y dx 2 14 2 yD 1 x x 2 C 4x 7 dx 4 D 2 14 D8 1 2 x C 4x 7 dx 4 2 14 1 yy dx 2 14 y dx 2 14 2 1 1 x 2 C 4x 7 dx 18 2 4 D 2 14 D 5 1 2 x C 4x 7 dx 4 2 14 xD8 y D 3.6 Problem 7.13 Determine the coordinates of the centroid of the area. y y⫽⫺ 1 2 x ⫹ 4x ⫺ 7 4 y⫽5 x Solution: Use a vertical strip. We first need to find the x intercepts. 1 y D x 2 C 4x 7 D 5 ) x D 4, 12 4 xy dx 4 xD 12 y dx 4 1 x x 2 C 4x 7 5 dx 4 D 4 12 D8 1 2 x C 4x 7 5 dx 4 4 12 12 12 yc y dx 4 yD 12 y dx 4 D 12 4 1 2 1 1 x 2 C 4x 7 5 dx x 2 C 4x 7 C 5 33 4 4 D 12 5 1 2 x C 4x 7 5 dx 4 4 x D 8, y D 6.6 516 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.14 Determine the x coordinate of the centroid of the area. y y = x3 y=x x Solution: Work this problem like Example 7.2 1 1 x dA 0 xD y = x3 0 D 1 1 dA 0 xx x 3 dx y y=x x x 3 dx 0 1 1 1 2 3 5 15 0 D D 0.533 xD 1 D 1 1 1 x4 x2 4 2 4 2 4 0 x5 x3 3 5 dA x x D 0.533 Problem 7.15 Determine the y coordinate of the centroid of the area shown in Problem 7.14. Solution: Solve this problem like example 7.2. 1 y dA y D A D dA A 0 1 x C x 3 x x 3 dx 2 1 x x 3 dx 0 1 x3 x7 1 3 7 0 D y D 0 1 1 2 2 x x4 3 x x dx 2 0 2 4 0 1 x 2 x 6 dx 1 1 4 8 3 7 21 D D D 0.381 yD 1 1 2 21 2 2 4 y D 0.381 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 517 Problem 7.16 Determine the x component of the centroid of the area. y Solution: The value of the function y D x2 x C 1 at x D 0 is y D 1, and its value at x D 2 is y D 3. We need a function describing a straight line that passes through those points. Let y D ax C b. Determining the constants a and b from the conditions that y D 1 when x D 0 and y D 3 when x D 2, we obtain a D 1 and b D 1. The straight line is described by the function y D x C 1. The height of the vertical strip of width dx is x C 1 x2 x C 1 D 2x x2 , so the area is dA D 2x x3 dx. Using this expression to evaluate Eq. (7.6). 2 xdA x D A D 0 x2x x3 dx 2 dA D 3 2x x dx A 0 x5 2x3 3 5 x4 2x2 2 4 y ⫽ x2 ⫺ x ⫹ 1 2 2 0 2 D 1 x x D 1. 0 y Problem 7.17 Determine the x coordinate of the centroid of the area. y = x 2 – 20 y=x x Solution: The intercept of the straight line with the parabola occurs at the roots of the simultaneous equations: y D x, and y D x2 20. This is equivalent to the solution of the quadratic x2 x 20 D 0, x1 D 4, and x2 D 5. These establish the limits on the integration. The area: Choose a vertical strip dx wide. The length of the strip is x x 2 C 20, which is the distance between the straight line y D x and the parabola y D x2 20. Thus the element of area is dA D x x 2 C 20 dx and C5 x x 2 C 20 dx D AD 4 x2 x3 C 20x 2 3 C5 D 121.5. 4 The x-coordinate: xA D A D xD 518 C5 x 2 x 3 C 20x dx x dA D 4 x3 x4 C 10x2 3 4 C5 D 60.75. 4 60.75 D 0.5 121.5 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.18 Determine the y coordinate of the centroid of the area in Problem 7.17. Solution: Use the results of the solution to Problem 7.17 in the following. The y-coordinate: The centroid of the area element occurs at the midpoint of the strip enclosed by the parabola and the straight line, and the y-coordinate is: yDx 1 1 x x 2 C 20 D x C x 2 20. 2 2 yA D y dA D A yD 5 1 x C x 2 20x x2 C 20 dx 2 4 D C5 1 x 4 C 41x2 400 dx 2 4 D 5 5 1 41x3 x 400x D 923.4. C 2 5 3 4 923.4 D 7.6 121.5 Problem 7.19 What is the x coordinate of the centroid of the area? y y⫽⫺ 1 2 x ⫹ 2x 6 2 Solution: Use vertical strips, do an integral for the parabola then x subtract the square 6 2 First find the intercepts 1 y D x 2 C 2x D 0 ) x D 0, 12 6 1 x x 2 C 2x dx 722 65 6 x D 0 12 D 11 1 2 x C 2x dx 22 6 0 12 x D 5.91 Problem 7.20 What is the y coordinate of the centroid of the area in Problem 7.19? Solution: Use vertical strips, do an integral for the parabola then subtract the square First find the intercepts 1 y D x 2 C 2x D 0 ) x D 0, 12 6 2 1 x 2 C 2x dx 122 139 6 y D 0 12 D 55 1 2 x C 2x dx 22 6 0 12 1 2 y D 2.53 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 519 Problem 7.21 An agronomist wants to measure the rainfall at the centroid of a plowed field between two roads. What are the coordinates of the point where the rain gauge should be placed? y 0.5 mi Solution: The area: The element of area is the vertical strip yt yb long and dx wide, where yt D mt x C bt and yb D mb x C bb are the two straight lines bounding the area, where 0.8 0.3 D 0.3846, 1.3 0 mt D 0.3 mi 0.3 mi x 0.5 mi 0.6 mi 0.2 mi and bt D 0.8 1.3 mt D 0.3. Similarly: 0.3 0 D 0.2308, 1.3 0 mb D and bb D 0. The element of area is dA D yt yb dx D mt mb x C bt bb dx D 0.1538x C 0.3 dx, from which 1.1 0.1538x C 0.3 dx AD 0.5 1.1 x2 D 0.1538 C 0.3x D 0.2538 sq mile. 2 0.5 The x-coordinate: 1.1 0.1538x C 0.3x dx x dA D A 0.5 1.1 x2 x3 D 0.2058. D 0.1538 C 0.3 3 2 0.5 x D 0.8109 mi The y-coordinate: The y-coordinate of the centroid of the elemental area is y D yb C 12 yt yb D 12 yt C yb D 0.3077x C 0.15. Thus, yA D y dA A 1.1 0.3077x C 0.150.1538x C 0.3 dx D 0.5 1.1 D 0.0473x2 C 0.1154x C 0.045 dx 0.5 1.1 x3 x2 D 0.0471 C 0.1153 C 0.045x D 0.1014. 3 2 0.5 Divide by the area: y D 520 0.1014 D 0.3995 mi 0.2538 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.22 The cross section of an earth-fill dam is shown. Determine the coefficients a and b so that the y coordinate of the centroid of the cross section is 10 m. y y = ax – bx3 x 100 m Solution: The area: The elemental area is a vertical strip of length y and width dx, where y D ax bx3 . Note that y D 0 at x D 100, thus b D a ð 104 . Thus 100 dA D a AD x 104 x 3 dx 0 A D 0.5a[x2 0.5 ð 104 x 4 ]100 0 D 0.5a ð 104 0.25b ð 108 , and the area is A D 0.25a ð 104 . The y-coordinate: The y-coordinate of the centroid of the elemental area is y D 0.5ax bx3 D 0.5ax 104 x 3 , from which yA D y dA A 100 D 0.5a2 x 104 x 3 2 dx 0 100 D 0.5a2 x 2 2104 x 4 C 108 x 6 dx 0 D 0.5a2 x3 2x5 x7 104 C 108 3 5 7 100 0 D 3.81a2 ð 104 . Divide by the area: yD 3.810a2 ð 104 D 15.2381a. 0.25a ð 104 For y D 10, a D 0.6562 , and b D 6.562 ð 105 m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 521 Problem 7.23 The Supermarine Spitfire used by Great Britain in World War II had a wing with an elliptical profile. Determine the coordinates of its centroid. y y2 = 1 x2 + — — a2 b2 Solution: y a2 x 2b y2 = 1 x2 + — — b2 b a x b a By symmetry, y D 0. From the equation of the ellipse, yD bp 2 a x2 a By symmetry, the x centroid of the wing is the same as the x centroid of the upper half of the wing. Thus, we can avoid dealing with š values for y. y = ab a2 – x2 y b dA = y dx 0 a x dx b a x a2 x 2 dx a 0 a D xD b dA a2 x 2 dx a 0 x dA Using integral tables x a2 x 2 dx D a2 x 2 dx D a2 x 2 3/2 3 p x x a2 x 2 a2 C sin1 2 2 a Substituting, we get a2 x 2 3/2 a /3 0 xD p x a x a2 x 2 a2 C sin1 2 2 a 0 0 C a3 /3 a3 /3 xD D 2 2 a /4 a 00 0C 2 2 xD 522 4a 3 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.24 Determine the coordinates of the centroid of the area. y Strategy: Write the equation for the circular boundary in the form y D R2 x 2 1/2 and use a vertical “strip” of width dx as the element of area dA. R Solution: The area: The equation of the circle is x2 Cpy 2 D R2 . 2 2 x Take the elemental area to be a vertical strip of height p y D R x and width dx, hence the element of area is dA D R2 x 2 dx. The Acircle R2 area is A D D . The x-coordinate: 4 4 xA D x 0 A xD R R2 x 2 3/2 R3 R2 x 2 dx D D : 3 3 0 R x dA D 4R 3 The y-coordinate: The y-coordinate of the centroid of the element of area is at the midpoint: p y D 12 R2 x 2 , hence yA D y dA D A D yD R 1 R2 x 2 dx 2 0 R x3 R3 1 R2 x D 2 3 0 3 4R 3 Problem 7.25* If R D 6 and b D 3, what is the y coordinate of the centroid of the area? Solution: We will use polar coordinates. First find the angle ˛ ˛ D cos1 y ˛ b 3 D cos1 D 60° D R 6 3 R /3 6 rdrd D 6 rdrd D AD 0 yD 1 A 0 /3 0 0 6 0 0 r 2 sin drd D 6 D 1.910 R x b R α b Problem 7.26* What is the x coordinate of the centroid of the area in Problem 7.25? Solution: See the solution to 7.25 1 xD A /3 0 0 6 p 6 3 D 3.31 r cos drd D 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 523 Problem 7.27 In Active Example 7.3, suppose that the area is placed as shown. Let the dimensions R D 6 in, c D 14 in, and b D 18 in. Use Eq. (7.9) to determine the x coordinate of the centroid. y R x Solution: Let the semicircular area be area 1, let the rectangular b c area be area 2, and let the triangular area be area 3. The areas and the x coordinates of their centroids are A1 D 1 2 R , 2 x1 D A2 D c 2R, A3 D x2 D 1 b 2R, 2 4R , 3 1 c, 2 x3 D c C 1 b 3 The x coordinate of the centroid of the composite area is xD x 1 A1 C x 2 A2 C x 3 A3 A1 C A2 C A3 D 4R 3 1 2 R 2 C 1 c 2 1 2cR C c C b bR 3 1 2 R C 2cR C bR 2 Substituting the values for R, b, and c yields x D 9.60 in Problem 7.28 In Example 7.4, suppose that the area is given a second semicircular cutout as shown. Determine the x coordinate of the centroid. y Solution: Let the rectangular area without the cutouts be area 1, let the left cutout be area 2, and let the right cutout be area 3. The areas and the x coordinates of their centroids are A1 D 200 280 mm2 , x 1 D 100 mm, 1 A2 D 1002 mm2 , 2 x2 D 1 A3 D 502 mm2 , 2 x 3 D 200 100 mm x 50 mm 4100 mm, 3 450 mm. 3 140 mm 140 mm 200 mm The x coordinated of the centroid of the composite area is xD x 1 A1 C x 2 A2 C x 3 A3 A1 C A2 C A3 450 1 1 1002 C 200 502 2 3 2 1 1 200 280 1002 502 2 2 100 [200 280] C D 4100 3 D 116 mm. x D 116 mm 524 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.29 Determine the coordinates of the centroids. Solution: Break into a rectangle, a triangle and a circular hole xD y yD 5[108] C 12 108 C 1 1 2 86 4[108] C 13 8 108 C 2 86 1 2 86 4[22 ] 22 3[22 ] 1 2 2 86 2 D 6.97 in D 3.79 in 2 in 8 in x D 6.97 in y D 3.79 in 3 in x 4 in 6 in 10 in Problem 7.30 Determine the coordinates of the centroids. Solution: The strategy is to find the centroid for the half circle area, and use the result in the composite algorithm. The area: The element of area is a vertical strip y high and dx wide. From the equation p of the circle, y D š R2 x 2 . The p height of the strip will be twice the positive value, so that dA D 2 R2 x 2 dx, from which R dA D 2 AD R2 x 2 1/2 dx 10 in x 20 in 0 A y p R R2 R2 x R2 x 2 1 x C sin D D2 2 2 R 2 0 The x-coordinate: R x dA D 2 A x R2 x 2 dx 0 R R2 x 2 3/2 2R3 . D2 D 3 3 0 Divide by A: x D 4R 3 The y-coordinate: From symmetry, the y-coordinate is zero. 420 D 8.488 in. For 3 the inner half circle x2 D 4.244 in. The areas are The composite: For a complete half circle x1 D A1 D 628.32 in2 and A2 D 157.08 in2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 525 Problem 7.31 Determine the coordinates of the centroids. y Solution: Use a big triangle and a triangular hole xD 2 1 2 1 3 1.0 2 1.00.8 0.6 C 3 0.4 2 0.40.8 1 1 2 1.00.8 2 0.40.8 yD 1 1 3 0.8 2 1.00.8 1 2 1.00.8 0.8 m D 0.533 1 1 3 0.8 2 0.40.8 1 2 0.40.8 D 0.267 x 0.6 m x D 0.533 m y D 0.267 m 1.0 m y Problem 7.32 Determine the coordinates of the centroid. Solution: Let the area be divided into parts as shown. The areas 2 and the coordinates are A1 D 40 50 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 20 30 in2 , x 2 D 10 in, y 2 D 40 C 15 in, A3 D 1 302 in2 , 4 x 3 D 20 C 430 in. 3 y 3 D 40 C 3 30 in 1 40 in 430 in. 3 x 20 in The x coordinate of the centroid of the composite area is xD x 1 A1 C x 2 A2 C x 3 A3 A1 C A2 C A3 430 1 25 [40 50] C [10] [20 30] C 20 C 302 3 4 D 1 40 50 C 20 30 C 302 4 D 23.9 in. The y coordinate of the centroid of the composite area is yD y 1 A1 C y 2 A2 C y 3 A3 A1 C A2 C A3 430 1 302 20 [40 50] C [55] [20 30] C 40 C 3 4 D D 33.3 in. 1 2 40 50 C 20 30 C 30 4 x D 23.9 in, y D 33.3 in. 526 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.33 Determine the coordinates of the centroids. y Solution: Break into 4 pieces (2 rectangles, a quarter circle, and 400 mm a triangle) 4[0.4] [0.4]2 0.2[0.40.3] C 0.4 3 4 1 C 0.550.30.7 C 0.8 0.30.7 2 xD [0.4]2 1 0.40.3 C C 0.30.7 C 0.30.7 4 2 300 mm x 300 mm 300 mm [0.4]2 4[0.4] 0.15[0.40.3] C 0.3 C 3 4 1 1 0.7 0.30.7 C 0.350.30.7 C 3 2 yD 2 1 [0.4] C 0.30.7 C 0.30.7 0.40.3 C 4 2 x D 0.450 m, y D 0.312 m y Problem 7.34 Determine the coordinates of the centroid. 3 2 Solution: Let the area be divided into parts as shown. The areas and the coordinates are A1 D 4 3 ft2 , x 1 D 2 ft, A2 D 1 4 4 ft2 , 2 x2 D A3 D 1 22 ft2 , 2 x3 D 4 C y 1 D 1.5 ft, 2 4 ft, 3 y2 D 3 C 3 ft 1 4 ft, 3 2 ft 1 x 4 ft 42 ft. y 3 D 3 C 2 ft. 3 The x coordinate of the centroid of the composite area is xD x 1 A1 C x 2 A2 C x 3 A3 A1 C A2 C A3 1 42 1 4 4 C 4 C 22 2 3 2 D 2.88 ft. 1 1 2 4 3 C 4 4 C 2 2 2 2 [4 3] C D 2 4 3 The y coordinate of the centroid of the composite area is yD y 1 A1 C y 2 A2 C y 3 A3 A1 C A2 C A3 1 1 1 1.5 [4 3] C 3 C 4 4 4 C [3 C 2] 22 3 2 2 D D 3.20 ft. 1 1 2 4 3 C 4 4 C 2 2 2 x D 2.88 ft, y D 3.20 ft. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 527 y Problem 7.35 Determine the coordinates of the centroids. 20 mm 30 mm 20 mm 10 mm 30 mm x 90 mm Solution: Determine this result by breaking the compound object into parts For the composite: y m 20 m m 50 m m 10 m 30 mm A1 = A2 + + A3 – A4 xD x1 A1 C x2 A2 C x3 A3 x4 A4 A1 C A2 C A3 A4 xD 155782 D 35.3 mm 4414.2 yD y1 A1 C y2 A2 C y3 A3 y4 A4 A1 C A2 C A3 A4 yD 146675 D 33.2 mm 4414.2 40 mm 20 mm 30 mm 90 mm A1 : x A1 D 3090 D 2700 mm2 x1 D 45 mm The value for y is not the same as in the new problem statement. This value seems correct. (The x value checks). y1 D 15 mm A2 : (sits on top of A1 ) A2 D 4050 D 2000 mm2 x2 D 20 mm y2 D 30 C 25 D 55 mm A3 : A3 D 1 2 r D 202 D 628.3 mm2 2 0 2 x3 D 20 mm y3 D 80 mm C A4 : 4r0 D 88.49 mm 3 A4 D 3020 C ri2 A4 D 600 C 102 D 914.2 mm2 x4 D 20 mm y4 D 50 C 15 D 65 mm Area (composite) D A1 C A2 C A3 A4 D 4414.2 mm2 528 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.36 Determine the coordinates of the centroids. y 5 mm 15 mm 50 mm 5 mm 5 mm 15 mm x 15 mm 10 15 15 10 mm mm mm mm y Solution: Comparison of the solution to Problem 7.29 and our areas 1, 2, and 3, we see that in order to use the solution of Problem 7.29, we must set a D 25 mm, R D 15 mm, and r D 5 mm. If we do this, we find that for this shape, measuring from the y axis, x D 18.04 mm. The corresponding areas for regions 1, 2, and 3 is 1025 mm2 . The centroids of the rectangular areas are at their geometric centers. By inspection, we how have the following information for the five areas Area 1: Area1 D 1025 mm2 , x1 D 18.04 mm, and y1 D 50 mm. Area 2: Area2 D 1025 mm2 , x2 D 18.04 mm, and y2 D 0 mm. Area 3: Area3 D 1025 mm2 , x3 D 18.04 mm, and y3 D 0 mm. 1 5 mm 5 15 15 mm 50 mm y 4 5 mm 3 2 15 15 mm 5 mm x 15 mm 10 15 15 10 mm mm mm mm Area 4: Area4 D 600 mm2 , x4 D 0 mm, and y4 D 25 mm. Area 5: Area5 D 450 mm2 , x5 D 7.5 mm, and y5 D 50 mm. Combining the properties of the five areas, we can calculate the centroid of the composite area made up of the five regions shown. AreaTOTAL D Area1 C Area2 C Area3 C Area4 C Area5 D 4125 mm2 . Then, x D x1 Area1 C x2 Area2 C x3 Area3 C x4 Area4 C x5 Area5 /AreaTOTAL D 3.67 mm, and y D y1 Area1 C y2 Area2 C y3 Area3 C y4 Area4 C y5 Area5 /AreaTOTAL D 21.52 mm. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 529 y Problem 7.37 The dimensions b D 42 mm and h D 22 mm. Determine the y coordinate of the centroid of the beam’s cross section. 200 mm h 120 mm x b Solution: Work as a composite shape y 100 mm 100 mm A2 A 1 h 120 mm x b b D 42 mm h D 22 mm A1 D 120 b mm2 D 5040 mm2 x1 D 0 y1 D 60 mm by symmetry A2 D 200 h D 4400 mm2 x2 D 0 y2 D 120 C xD h D 131 mm 2 A1 x1 C A2 x2 0C0 D A1 C A2 9440 xD0 yD 530 A1 y1 C A2 y2 D 93.1 mm A1 C A2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.38 If the cross-sectional area of the beam shown in Problem 7.37 is 8400 mm2 and the y coordinate of the centroid of the area is y D 90 mm, what are the dimensions b and h? Solution: From the solution to Problem 7.37 A1 D 120 b, A2 D 200 h and y D y1 A1 C y2 A2 A1 C A2 h 60120 b C 120 C 200 h 2 yD 120 b C 200 h where y1 D 60 mm y D 90 mm A1 C A2 D 8400 mm2 Also, y2 D 120 C h/2 Solving these equations simultaneously we get h D 18.2 mm b D 39.7 mm 200 mm h A2 A1 120 mm b c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 531 y Problem 7.39 Determine the y coordinate of the centroid of the beam’s cross section. 5 in 2 in 8 in x 3 in 5 in 3 in 5 in Solution: Take advantage of the symmetry. Work with only one half of the structure. Break into 2 rectangles, a quarter circle, and a quarter circular hole. [5]2 4[5] 438 C 11.533 C 8 C 3 4 4[2] [2]2 8C 3 4 yD [2]2 [5]2 38 C 33 C 4 4 y D 7.48 in y Problem 7.40 Determine the coordinates of the centroid of the airplane’s vertical stabilizer. 11 m 48° x 70° 12.5 m Solution: We work with a rectangle and two triangular holes y e We have d D 12.5 m C 11 m cot 70° D 16.50 m 11 m e D 11 m tan 48° D 12.22 m 48° In the x direction 1 2 d[d11] xD yD x 13 e 12 e11 12.5 C 23 d 12.5 1 2 d 12.511 70° 12.5 m d d11 12 e11 12 d 12.511 1 2 11[d11] 23 11 1 2 e11 d11 12 e11 1 1 3 11 2 d 12.5 m11 m 12 d 12.511 Solving we find x D 9.64 m, y D 4.60 m 532 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.41 The area has elliptical boundaries. If a D 30 mm, b D 15 mm, and ε D 6 mm, what is the x coordinate of the centroid of the area? y Solution: The equation of the outer ellipse is x2 y2 C D1 a C ε2 b C ε2 b and for the inner ellipse y2 x2 C 2 D1 2 a b x a We will handle the problem by considering two solid ellipses For any ellipse ˇ ˛ x ˛2 x 2 dx x dA ˛ 0 D xD ˇ dA ˛2 x 2 dx ˛ y ε From integral tables b x ˛2 x 2 3/2 dx D 3 p x ˛2 x ˛2 x 2 C sin1 ˛2 x 2 dx D 2 2 ˛ ˛2 x2 ˛2 x 2 ε a 3/2 ˛ A1 = 0 Substituting x D p x ˛ ˛2 x ˛2 x 2 C sin 3 2 ˛ – x A2 0 0 C ˛3 /3 ˛3 /3 xD D 2 2 ˛ /4 ˛ 00 0C 2 2 xD dA D ˇ ˛ ˛2 C x 2 dx 0 ˛2 2 2 x For the composite ˛ 0 ˇ ˛ β α p x ˛ ˛2 ˇ x ˛2 x 2 C sin1 D ˛ 2 2 ˛ Area D β y = α √α 2 – x2 dA = y dx 4˛ 3 Also Area D y D ˛ˇ/4 (The area of a full ellipse is ˛ˇ so this checks. Now for the composite area. xD x1 A1 x2 A2 A1 A2 Substituting, we get x1 D 15.28 mm x2 D 12.73 mm A1 D 2375 mm2 A2 D 1414 mm2 and x D 19.0 mm For the outer ellipse, ˛ D a C ε ˇ D b C ε and for the inner ellipse ˛Da ˇDb Outer ellipse x1 D 4a C ε 3 A1 D a C εb C ε 4 Inner Ellipse x2 D 4a 3 A2 D ab 4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 533 Problem 7.42 By determining the x coordinate of the centroid of the area shown in Problem 7.41 in terms of a, b, and ε, and evaluating its limit as ε ! 0, show that the x coordinate of the centroid of a quarter-elliptical line is xD 4aa C 2b . 3a C b Solution: From the solution to 7.41, we have x1 D 4a C ε 3 A1 D 4a x2 D 3 so x1 A1 D x2 A2 D a C εb C ε 4 C a2 ε C 2aε2 C ε3 a2 b x1 A1 x2 A2 D 13 2ab C a2 ε ab A2 D 4 a x1 A1 x2 A2 D 13 a2 b C 2abε C bε2 C ε2 b C C 2a C bε2 C ε3 ε Finally x D 3 a2 b 3 A1 A2 D ab C aε C bε C ε2 ab 4 A1 A2 D aε C bε C ε2 4 x1 A1 x2 A2 A1 A2 1 2ab C a2 C 2a C bε C ε2 ε 3 xD [a C b C ε] ε 4 xD 42a C bε 4 2 4aa C 2b C C ε 3a C b 3 3 Taking the limit as ε ! 0 xD Problem 7.43 Three sails of a New York pilot schooner are shown. The coordinates of the points are in feet. Determine the centroid of sail 1. 4aa C 2b 3a C b Solution: Divide the object into three areas: (1) The triangle with altitude 21 ft and base 20 ft. (2) The triangle with altitude 21 ft and base 20 16 D 4ft, and (3) the composite sail. The areas and coordinates are: (1) A1 D 210 ft2 , x1 D y1 D 1 2 3 (2) 2 20 D 13.33 ft, 3 1 21 D 7 ft. 3 A2 D 42 ft2 , (a) x2 D 16 C y y y y2 D 7 ft. (14, 29) (12.5, 23) (20, 21) (3, 20) 1 (16, 0) (3) (3.5, 21) 2 The composite area: A D A1 A2 D 168 ft2 . The composite centroid: 3 x x x (10, 0) (23, 0) 2 4 D 18.67 ft, 3 xD A1 x1 A2 x2 D 12 ft , A yD A1 y1 A2 y2 D 7 ft A (b) 534 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.44 Determine the centroid of sail 2 in Problem 7.43. Solution: Divide the object into five areas: (1) a triangle on the (3) A3 D left with altitude 20 ft and base 3 ft, (2) a rectangle in the middle 23 ft by 9.5 ft, (3) a triangle at the top with base of 9.5 ft and altitude of 3 ft. (4) a triangle on the right with altitude of 23 ft and base of 2.5 ft. (5) the composite sail. The areas and centroids are: (1) A1 D x1 D y1 D (2) x3 D 3 C 320 D 30 ft2 , 2 2 3 D 2 ft, 3 (4) A4 D y2 D 9.5 2 y4 D D 7.75 ft, 2 3 D 22 ft 3 1 2.523 D 28.75 ft2 , 2 x4 D 10 C A2 D 239.5 D 218.5 ft2 , 1 9.5 D 6.167 ft, 3 y3 D 20 C 1 20 D 6.67 ft. 3 x2 D 3 C 1 39.5 D 14.25 ft2 , 2 2 2.5 D 11.67 ft, 3 1 23 D 7.66 ft 3 The composite area: A D A1 C A2 A3 A4 D 205.5 ft2 . The composite centroid: (5) 23 D 11.5 ft 2 xD A1 x1 C A2 x2 A3 x3 A4 x4 D 6.472 ft , A yD A1 y1 C A2 y2 A3 y3 A4 y4 D 10.603 ft A Problem 7.45 Determine the centroid of sail 3 in Problem 7.43. Solution: Divide the object into six areas: (1) The triangle Oef, with base 3.5 ft and altitude 21 ft. (2) The rectangle Oabc, 14 ft by 29 ft. (3) The triangle beg, with base 10.5 ft and altitude 8 ft. (4) The triangle bcd, with base 9 ft and altitude 29 ft. (5) The rectangle agef 3.5 ft by 8 ft. (6) The composite, Oebd. The areas and centroids are: (1) a f g b e A1 D 36.75 ft2 , x1 D 1.167 ft, o c d y1 D 14 ft. (2) A2 D 406 ft2 , (5) A5 D 28 ft2 , x5 D 1.75 ft, x2 D 7 ft, y5 D 25 ft. y2 D 14.5 ft. (6) (3) The composite area: A3 D 42 ft2 , A D A1 C A2 A3 C A4 A5 D 429.75 ft2 . x3 D 7 ft, The composite centroid: y3 D 26.33 ft (4) A4 D 130.5 ft2 , xD A1 x1 C A2 x2 A3 x3 C A4 x4 A5 x5 D 10.877 ft A yD A1 y1 C A2 y2 A3 y3 C A4 y4 A5 y5 D 11.23 ft A x4 D 17 ft, y4 D 9.67 ft. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 535 Problem 7.46 In Active Example 7.5, suppose that the distributed load is modified as shown. Determine the reactions on the beam at A and B. 60 N/m A B 8m Solution: We can treat the distributed load as two triangular distributed loads. Using the area analogy, the magnitude of the left one is 12 8 m60 N/m D 240 N, and the magnitude of the right one is 1 2 4 m60 N/m D 120 N. They must be placed at the centroids of the triangular distributions. The equilibrium equations are Fx : Ax D 0, Fy : Ay C B 240 N 120 N D 0, 4m MA : 12 m B [ 23 8 m]240 N [12 m 2 3 4 m]120 N D 0. Solving yields Ax D 0, Problem 7.47 Determine the reactions at A and B. Ay D 160 N, B D 200 N. 6 ft 200 lb/ft B A 6 ft 4 ft 200 lb/ft Solution: From the free-body diagram of the bar (with the distributed loads represented by equivalent force), the equilibrium equations are Fx : Ax 600 lb D 0 Fy : Ay C B 800 lb D 0 MA : 800 lb 2 ft 600 lb 4 ft C B10 ft D 0 Solving yields Ax D 600 lb, 536 Ay D 400 lb, B D 400 lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.48 In Example 7.6, suppose that the distributed loads are modified as shown. Determine the reactions on the beam at A and B. 400 N/m 6m 600 N/m A 6m 400 N/m B 6m Solution: The distributed loads are represented by three equivalent forces. The equilibrium equations are Fx : 1200 N C Ax D 0 Fy : Ay C B 2400 N 600 N D 0 MA : 1200 N 4 m D 600 N 2 m 2400 N 3 m C B 6 m D 0 Solving yields Ax D 1200 N, Ay D 800 N, B D 2200 N Problem 7.49 In Example 7.7, suppose that the distributed load acting on the beam from x D 0 to x D 10 ft is given by w D 350 C 0.3x3 lb/ft. (a) Determine the downward force and the clockwise moment about A exerted by the distributed load. (b) Determine the reactions at the fixed support. y 2000 lb w 10,000 ft-lb x A 10 ft 10 ft Solution: (a) The force and moment are 10 350 C 0.3x3 dx D 4250 lb RD 0 R D 4250 lb, M D 23,500 ft-lb 10 3 x350 C 0.3x dx D 23500 ft-lb MD 0 (b) The equilibrium equations are Fx : Ax D 0, Fy : Ay 4250 lb C 2000 lb D 0, MA : MA 23500 ft-lb C 2000 lb20 ft C 10,000 ft-lb D 0. Solving yields Ax D 0, Ay D 2250 lb, MA D 26,500 ft-lb c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 537 Problem 7.50 Determine the reactions at the fixed support A. y = 3(1 – x 2/25) kN/m x A 5m Solution: The free-body diagram of the beam is: The downward = 3(1 – x 2/25) kN/m force exerted by the distributed load is 5 w dx D 0 L x2 3 1 dx 25 5 x3 D 10 kN. D3 x 75 0 Ma 5m Ax x Ay The clockwise moment about the left end of the beam due to the distributed load is 5 xw dx D 0 L D3 x3 3 x dx 25 x4 x2 2 100 5 D 18.75 kN-m. 0 From the equilibrium equations Fx D Ax D 0, Fy D Ay 10 D 0, mleftend D Ma C 5Ay 18.75 D 0, we obtain Ax D 0, Ay D 10 kN, and Ma D 31.25 kN-m. 538 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.51 exerted by the foundation and distributed load (a) (b) An engineer measures the forces soil on a 10-m section of a building finds that they are described by the w D 10x x2 C 0.2x 3 kN/m. Solution: (a) The total force is 12 FD Determine the magnitude of the total force exerted on the foundation by the distributed load. Determine the magnitude of the moment about A due to the distributed load. 10x C x2 0.2x3 dx 0 10 0.2 4 x3 C x D 5x2 3 4 0 y jFj D 333.3 kN 2m 10 m (b) The moment about the origin is A x 10 MD 10x C x2 0.2x3 x dx 0 1 0.2 5 10 10 x , D x3 x4 C 3 4 5 0 jMj D 1833.33 kN. The distance from the origin to the equivalent force is dD jMj D 5.5 m, F from which jMA j D d C 2F D 2500 kN m. Problem 7.52 Determine the reactions on the beam at A and B. Solution: Replace the distributed load with three equivalent single forces. The equilibrium equations 3 kN/m 2 kN/m A B 4m 2m Fx : Ax D 0 Fy : Ay C B 8 kN 2 kN 3 kN D 0 MA : B4 m 8 kN2 m 2 kN 2 34 m 3 kN 4 m C 13 2 m D 0 Ax D 0, Ay D 4.17 kN B D 8.83 kN 2 kN 8 kN 3 kN Ax Ay B c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 539 Problem 7.53 The aerodynamic lift of the wing is described by the distributed load y p w D 300 1 0.04x 2 N/m. The mass of the wing is 27 kg, and its center of mass is located 2 m from the wing root R. x R 2m (a) (b) Determine the magnitudes of the force and the moment about R exerted by the lift of the wing. Determine the reactions on the wing at R. 5m Solution: w (a) The force due to the lift is 5 F D w D MR 3001 0.04x2 1/2 dx, mg 0 FD 300 5 F D 60 5 FR 2m 25 x2 1/2 dx 3m 0 5 p 25 1 x x 25 x2 C sin D 375 N, 2 2 5 0 jFj D 1178.1 N. The moment about the root due to the lift is 5 M D 300 1 0.04x2 1/2 x dx, 0 M D 60 25 x2 3/2 3 5 D 0 60253/2 D 2500 3 jMj D 2500 Nm. (b) The sum of the moments about the root: M D MR C 2500 27g2 D 0, from which MR D 1970 N-m. The sum of forces Fy D FR C 1178.1 27g D 0, from which FR D 1178.1 C 27g D 913.2 N 540 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.54 Determine the reactions on the bar at A and B. 400 lb/ft B y 2 ft 600 lb/ft 400 lb/ft 2 ft x 4 ft A 4 ft Solution: First replace the distributed loads with three equivalent forces. The equilibrium equations Fx : Bx C 800 lb D 0 MB : 800 lb1 ft A4 ft C 1600 lb6 ft C 400 lb6.67 ft D 0 Fy : A C By 1600 lb 400 lb D 0 Solving: A D 3267 lb, Bx D 800 lb, By D 1267 lb Bx 800 lb By 400 lb 1600 lb A c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 541 Problem 7.55 Determine the reactions on member AB at A and B. 300 lb/ft A B 6 ft 6 ft 6 ft C 300 lb/ft Solution: From the free-body diagram of the entire structure (with the distributed loads represented by equivalent forces), one of the equilibrium equations is MC : Ax 6 ft 3600 lb 6 ft 900 lb 4 ft 1800 lb 9 ft D 0 From the free-body diagram of member AB we have the equilibrium equations Fx : Ax C Bx D 0 Fy : Ay C By 3600 lb D 0 MA : By 12 ft 3600 lb 6 ft D 0 Solving yields Ax D 6900 lb, Bx D 6900 lb, 542 Ay D 1800 lb, By D 1800 lb. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.56 Determine the axial forces in members BD, CD, and CE of the truss and indicate whether they are in tension (T) or compression (C). 2m 2m B A 2m 2m H F D 2m C G E 4 kN/m 8 kN/m Solution: We start by analyzing the horizontal bar with the distributed load MG : 16 kN0.667 m C 32 kN2 m FC FG FC 4 m D 0 ) FC D 18.67 kN Fy D FC C FG 32 kN 16 kN D 0 32 kN ) FG D 29.33 kN Now work with the whole structure in order to find the reactions at A Fx : Ax D 0 ) Ax D 0 16 kN Ax MH : FG 2 m C FC 6 m Ay Ay 8 m D 0 ) Ay D 21.3 kN H Finally, cut through the truss and look at the left section MC : Ax 2 m Ay 2 m BD2 m D 0 FC MD : Ay 4 m C FC 2 m C CE2 m D 0 Ax FG B BD 1 Fy : Ay FC C p CD D 0 2 CD Solving we find BD D 21.3 kN, CD D 3.77 kN, CE D 24 kN Ay C CE In summary: FC BD D 21.3 kNC, CD D 3.77 kNC, CE D 24 kNT c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 543 Problem 7.57 Determine the reactions on member ABC at A and B. 400 N/m 200 N/m C 160 mm B D 240 mm E A 400 N/m 160 mm Solution: Work on the entire structure first to find the reactions at A. Replace the distributed forces with equivalent concentrated forces 160 mm 160 mm 48 N 96 N Fx : Ax C 160 N D 0 ME : 96 N0.08 m C 48 N0.16 m 160 N0.2 m Ay 0.32 m D 0 Solving: Ax D 160 N, Ay D 52 N Now look at body ABC. Take advantage of the two-force body CD. 160 N MB : Ax 0.24 m C 160 N0.04 m 48 N0.16 m 96 N0.24 m 4 7 p CD0.32 m C p CD0.16 m D 0 65 65 Ax E Ay 4 Fx : Ax C Bx C 160 N p CD D 0 65 48 N 7 Fy : Ay C By 48 N 96 N p CD D 0 65 96 N Solving: Ax D 160 N, Ay D 52 N Bx D 157 N, By D 78.4 N C 7 By 4 Bx CD 160 N Ax Ay 544 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.58 Determine the forces on member ABC of the frame. A 1m 3 kN/m B 1m C 2m Solution: The free body diagram of the member on which the distributed load acts is 1m (4 m)(3 kN/m) = 12 kN BX From the equilibrium equations 2m 2m BY Fx D Bx D 0, 1m E AX Fy D By C E 12 D 0, 4 kN AY mleftend D 3E 212 D 0, CX we find that Bx D 0, By D 4 kN, and E D 8 kN. From the lower fbd, writing the equilibrium equation mleftend D 2Cy 48 D 0, CX DX we obtain Cy D 16 kN. Then from the middle free body diagram, we write the equilibrium equations 8 kN CY DY CY 2m 2m Fx D Ax C Cx D 0, Fy D Ay 4 16 D 0, mrightend D 2Ax 2Ay C 14 D 0 obtaining Ax D 18 kN, Ay D 20 kN, Cx D 18 kN. Problem 7.59 Use the method described in Active Example 7.8 to determine the centroid of the truncated cone. y Solution: Just as in Active Example 7.8, the volume of the disk element is dV D R x h 2 z dx x the x coordinate of the centroid is 2 R x x h 45 h/2 h D h x D V 2 D 56 R dV x x V h h/2 xdV h 2 h x R 45 h xD 56 h 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 545 Problem 7.60 A grain storage tank has the form of a surface of revolution with the profile shown. The height of the tank is 7 m and its diameter at ground level is 10 m. Determine the volume of the tank and the height above ground level of the centroid of its volume. y y = ax1/2 7m 10 m x Solution: O y y = ax1/2 y dx dV = π y2dx x dV D y 2 dx 7 x D 0 7 y 2 dx 7 D 0 y 2 dx 0 a2 x dx 7 a2 x dx 0 7 x 3 /3 0 xD 7 D 4.67 m x2 /2 0 The height of the centroid above the ground is 7 m x h D 2.33 m The volume is 7 VD a2 x dx D a2 0 49 2 m3 To determine a, y D 5, m when x D 7 m. p y D ax1/2 , 5 D a 7 p a D 5/ 7a2 D 25/7 VD 25 7 49 2 D 275 m3 V D 275 m3 546 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 7.61 The object shown, designed to serve as a pedestal for a speaker, has a profile obtained by revolving the curve y D 0.167x 2 about the x axis. What is the x coordinate of the centroid of the object? z x 0.75 m 0.75 m Solution: y = 0.167 x2 dV = π y2dx x dv 1.50 xdV x D V D 0.75 1.50 dV V x0.167x2 2 dx 0.167x2 2 dx 0.75 1.5 0.1672 Ð 0.75 xD 1.5 0.1672 x 5 dx x 4 dx 1.5 x 6 /6 0.75 D 1.5 x5 /5 0.75 0.75 x D 1.27 m y Problem 7.62 The volume of a nose cone is generated by rotating the function y D x 0.2x2 about the x axis. (a) (b) What is the volume of the nose cone? What is the x coordinate of the centroid of the volume? z x 2m Solution: (a) 2 m VD 0 (b) xD 2 m 2 m y 2 dx D x 0.2x2 2 dx D 4.16 m3 0 xy 2 dx 0 V D 2 xx 0.2x2 2 dx 0 4.155 m3 D 1.411 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 547 Problem 7.63 Determine the centroid of the hemispherical volume. y R z x Solution: The equation of the surface of a sphere is x2 C y 2 C z 2 D R2 . The volume: The element of volume is a disk of radius and thickness dx. The radius of the disk at any point within the hemisphere is 2 D y 2 C z2 . From the equation of the surface of the sphere, 2 D R2 x 2 . The area is 2 , and the element of volume is dV D R2 x 2 dx, from which Vsphere 2 3 D R . 2 3 VD The x-coordinate is: R x dV D R2 x 2 x dx 0 V D D x4 R2 x 2 2 4 R 0 4 R . 4 Divide by the volume: xD R4 4 3 2R3 D 3 R. 8 By symmetry, the y- and z-coordinates of the centroid are zero. y x R 548 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.64 The volume consists of a segment of a sphere of radius R. Determine its centroid. y x R R 2 z Solution: The volume: The element of volume is a disk of radius and thickness dx. The area of the disk is 2 , and the element of volume is 2 dx. From the equation of the surface of a sphere (see solution to Problem 7.63) 2 D R2 x 2 , from which the element of volume is dV D R2 x 2 dx. Thus VD R dV D V R2 x 2 dx R/2 R x3 5 R3 . D D R2 x 3 R/2 24 The x-coordinate: R x dV D V R2 x 2 x dx R/2 D R2 x 2 x4 2 4 R D R/2 9 4 R . 64 Divide by the volume: xD 9R4 64 24 5R3 D 27 R D 0.675R. 40 By symmetry the y- and z-coordinates are zero. y R – 2 x R c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 549 Problem 7.65 A volume of revolution is obtained x2 y2 by revolving the curve 2 C 2 D 1 about the x axis. a b Determine its centroid. y y2 x 2 + –– –– =1 a2 b2 z x Solution: The volume: The element of volume is a disk of radius y and thickness dx. The area of the disk is y2 . From the equation for the surface of the ellipse, y x2 + y2 = 1 a2 b2 x2 y 2 D b2 1 2 a and dV D y 2 dx D b2 x2 1 2 a x dx, from which a dV D b2 VD 0 V x2 1 2 dx a a x3 2b2 a . D b2 x 2 D 3a 0 3 The x-coordinate: a x dV D b2 0 V D b2 x2 1 2 x dx a x4 x2 2 2 4a a D 0 b2 a2 . 4 Divide by volume: xD b2 a2 4 3 2b2 a D 3 a. 8 By symmetry, the y- and z-coordinates of the centroid are zero. 550 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.66 In Example 7.9, determine the y coordinate of the centroid of the line. y (1, 1) y ⫽ x2 L x Solution: The expression derived in Example 7.9 for the element dL of the line in terms of x is dL D y (1, 1) 1 C 4x2 dx dL The y coordinate of the centroid is 1 ydL y D L D x2 D 0.410 1 dL dy 1 C 4x2 dx 0 1 C 4x2 dx L x x dx 0 y D 0.410 y Problem 7.67 Determine the coordinates of the centroid of the line. y ⫽x2 Solution: 2 2 dy x 1C dx xds x 1 C 2x2 dx dx 1 1 D x D 12 D 2 2 2 dy ds 1 C 2x2 dx 1 C dx 1 1 dx 1 2 2 2 2 dy y 1C dx x 2 1 C 2x2 dx dx 1 1 D y D 12 D 2 2 2 dy ds 1 C 2x2 dx dx 1 C 1 1 dx 1 2 2 yds ⫺1 2 x x D 0.801 y D 1.482 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 551 Problem 7.68 Determine the x coordinate of the centroid of the line. Solution: The length: Noting that of length is y 1C dL D dy dx 2 dx D dy D x 11/2 , the element dx p x dx from which 2 y = – (x – 1)3/2 3 LD 5 dL D x1/2 dx D 1 L 2 3/2 x 3 5 D 6.7869. 1 The x-coordinate: 5 x 3/2 dx D x dL D 0 5 0 L x Divide by the length: x D Problem 7.69 Determine the x coordinate of the centroid of the line. 2 5/2 x 5 5 D 21.961. 1 21.961 D 3.2357 6.7869 Solution: The length: Noting that length is y 1C dL D dy dx 2 dx D dy D x1/2 the element of dx p 1 C x dx from which 2 y = – x 3/2 3 2 dL D LD 1 C x1/2 dx D 0 L 2 1 C x3/2 3 2 D 2.7974 0 The x-coordinate: 0 2 2 x dL D x L x1 C x1/2 dx D 2 0 1 C x5/2 1 C x3/2 5 3 2 0 33/2 1 1 35/2 C D 3.0379. D2 5 3 5 3 Divide by the length: x D 1.086 Problem 7.70 Use the method described in Example 7.10 to determine the centroid of the circular arc. y Solution: The length of the differential line element of the circular arc is dL D Rd. The coordinates of the centroid are ˛ R cos Rd xdL R sin ˛ D 0 ˛ D x D L ˛ dL Rd L 0 ydL y D L D dL 0 552 x R ˛ R sin Rd R 1 cos ˛ ˛ D ˛ Rd L Thus a 0 xD R sin ˛ , ˛ yD R1 cos ˛ ˛ c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 7.71 In Active Example 7.11, suppose that the cylinder is hollow with inner radius R/2 as shown. If the dimensions R D 6 in, h D 12 in, and b D 10 in, what is the x coordinate of the centroid of the volume? z R h Solution: Let the cone be volume 1, let the solid cylinder be volume 2, and let the cylindrical hole be volume 3. The volumes and the x coordinates of the their centroids are V1 D 1 3 R2 h, V2 D R2 b, x1 D 3 4 x2 D h C b 1 2 x R 2 h, b, V3 D 12 R2 b, x 3 D h C 1 2 b The x coordinate of the centroid of the composite volume is xD x 1 V1 C x 2 V2 C x 3 V3 V1 C V2 C V3 3 h 4 D 2 1 1 1 1 R2 h C h C b R2 b C h C b R b 3 2 2 2 2 1 1 R2 h C R2 b R b 3 2 Substituting the values for R, h, and b, we have x D 14.2 in. Problem 7.72 Use the procedure described in Example 7.12 to determine the x component of the centroid of the volume. y y 25 mm Solution: Let the rectangular part without the cutout be volume 1, let the semicylindrical part be volume 2, and let the cylindrical hole be volume 3. The volumes and the x coordinates of their centroids are V1 D 60 50 20 mm3 , V2 D 1 252 20 mm3 , 2 V3 D 102 20 mm3 , x z 10 mm x 1 D 30 mm, x 2 D 60 C 425 mm, 3 60 mm 20 mm x 3 D 60 mm. The x coordinate of the centroid of the composite volume is xD x 1 V1 C x 2 V2 C x 3 V3 V1 C V2 C V3 425 1 252 20 C 60 102 20 30 [60 50 20] C 60 C 3 2 D 1 60 50 20 C 252 20 102 20 2 D 38.3 mm. x D 38.3 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 553 Problem 7.73 Determine the centroids of the volumes. y Solution: The object will be divided into a cone and a hemisphere. From symmetry y D z D 0 Using tables we have in the x direction xD 3 4R 4 3R 1 2 2R3 R [4R] C 4R C 83R 3 8 3 D 24 1 2 2R3 R [4R] C 3 3 z R x 4R In summary xD 83R , y D 0, z D 0 24 Problem 7.74 Determine the centroids of the volumes. y Solution: We have a hemisphere and a hemispherical hole. From symmetry y D z D 0 200 mm In the x direction we have 3[300 mm] 2[300 mm]3 8 3 2[200 mm]3 3[200 mm] 8 3 xD 2[200 mm]3 2[300 mm]3 3 3 300 mm z x We have x D 128 mm, y D 0, z D 0 y Problem 7.75 Determine the centroids of the volumes. Solution: This is a composite shape. Let us consider a solid cylinder and then subtract the cone. Use information from the appendix Volume Cylinder Cone R2 L 1 2 3 r h Volume (mm3 ) ð 107 1.1706 1.3572 ð 106 x x (mm) L/2 L-h/4 230 370 z 60 mm 90 mm R D 90 mm L D 460 mm 360 mm x 460 mm r D 60 mm h D 360 mm xD XCyL VCyL XCONE VCONE VCyL VCONE x D 211.6 mm y D z D 0 mm 554 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.76 Determine the centroids of the volumes. y 20 mm 25 mm 75 mm x 120 mm 25 mm 100 mm z Solution: Break the composite object into simple shapes, find the volumes and centroids of each, and combine to find the required centroid. Object Volume (V) x y z 1 2 LWH hWD 0 0 L/2 D/2 3 R2 D/2 0 4 r 2 D 0 H/2 H C h/2 4R HChC 3 H C h y x (H) 25 mm + D/2 D/2 12 0m (L) m 100 mm (W) y where R D W/2. For the composite, z x1 V1 C x2 V2 C x3 V3 x4 V4 V1 C V2 C V3 V4 + with similar eqns for y and z – 50 xD mm 1 3 x The dimensions, from the figure, are L D 120 mm W D 100 mm H D 25 mm r D 20 mm h D 75 mm D D 25 mm R D 50 mm Object V mm3 x (mm) y (mm) z (mm) C1 C2 C3 4 300000 187500 98175 31416 0 0 0 0 12.5 62.5 121.2 100 60 12.5 12.5 12.5 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 555 7.76 (Continued ) y Substituting into the formulas for the composite, we get xD0 100 mm 25 mm (D) y D 43.7 mm mm 75 z D 38.2 mm x (h) 2 H y z r = 20 mm x 4 z Problem 7.77 Determine the centroids of the volumes. y 1.75 in 1 in 5 in z 4 in 1 in x Solution: Divide the object into six volumes: (1) A cylinder 5 in long of radius 1.75 in, (2) a cylinder 5 in long of radius 1 in, (3) a block 4 in long, 1 in thick, and 21.75 D 3.5 in wide. (4) Semicylinder 1 in long with a radius of 1.75 in, (5) a semi-cylinder 1 in long with a radius of 1.75 in. (6) The composite object. The volumes and centroids are: 1 in x 1.75 in z z x 5 in Volume V1 V2 V3 V4 V5 Vol, cu in 48.1 15.7 14 4.81 4.81 x, in 0 0 2 0.743 0 y, in 2.5 2.5 0.5 0.5 4.743 z, in 0 0 0 0 0 4 in y 1 in x The composite volume is V D V1 V2 C V3 V4 C V5 D 46.4 in3 . The composite centroid: xD V1 x1 V2 x2 C V3 x3 V4 x4 C V5 x5 D 1.02 in, V yD V1 y1 V2 y2 C V3 y3 V4 y4 C V5 y5 D 1.9 in, V zD0 556 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 7.78 Determine the centroids of the volumes. 30 mm 60 mm z x 180 mm Solution: Consider the composite volume as being made up of three volumes, a cylinder, a large cone, and a smaller cone which is removed Cone 1 Cone 2 Cylinder Cone 1 Cone 2 V r 2 L/2 1 2 R L 3 1 2 L r 3 2 y x 2R Object Cylinder 180 mm 2r L/4 3L/4 L/2 3(L/2)/4 (mm3 ) (mm) 5.089 ð 105 1.357 ð 106 1.696 ð 105 90 270 135 L/2 y 60 mm Cylinder L D 360 mm 1 x r D 30 mm R D 60 mm y For the composite shape xCyl VCyL C x1 V1 x2 V2 VCyL C V1 V2 120 mm cone + xD 360 mm 2 x x D 229.5 mm y cone – 60 mm 3 x 180 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 557 Problem 7.79 The dimensions of the Gemini spacecraft (in meters) are a D 0.70, b D 0.88, c D 0.74, d D 0.98, e D 1.82, f D 2.20, g D 2.24, and h D 2.98. Determine the centroid of its volume. y g e b a Solution: The spacecraft volume consists of three truncated cones and a cylinder. Consider the truncated cone of length L with radii at the ends R1 and R2 , where R2 > R1 . Choose the origin of the x –y coordinate system at smaller end. The radius of the cone is a linear function of the length; from geometry, the length of the cone before truncations was R2 L with volume R2 R1 (1) HD (2) R22 H . The length of the truncated portion is 3 (3) (4) D R1 L with volume R2 R1 R12 . The volume of the truncated cone is the difference of the 3 two volumes, L (5) V D 3 cone is R23 R13 R2 R1 . The centroid of the removed part of the 3 , and the centroid of the complete cone is 4 (6) x D (7) 3 H, measured from the pointed end. From the 4 composite theorem, the centroid of the truncated cone is (8) 558 c f d h x Beginning from the left, the volumes are (1) a truncated cone, (2) a cylinder, (3) a truncated cone, and (4) a truncated cone. The algorithm and the data for these volumes were entered into TK Solver Plus and the volumes and centroids determined. The volumes and x-coordinates of the centroids are: Volume V1 V2 V3 V4 Composite Vol, cu m 0.4922 0.5582 3.7910 11.8907 16.732 x, m 0.4884 1.25 2.752 4.8716 3.999 The last row is the composite volume and x-coordinate of the centroid of the composite volume. The total length of the spacecraft is 5.68 m, so the centroid of the volume lies at about 69% of the length as measured from the left end of the spacecraft. Discussion: The algorithm for determining the centroid of a system of truncated cones may be readily understood if it is implemented for a cone of known dimensions divided into sections, and the results compared with the known answer. Alternate algorithms (e.g. a Pappus-Guldinus algorithm) are useful for checking but arguably do not simplify the computations End discussion. xh D Vh xh V x C x, where x is the x-coordinate of the left V hand edge of the truncated cone in the specific coordinate system. These eight equations are the algorithm for the determination of the volumes and centroids of the truncated cones forming the spacecraft. xD c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 7.80 Two views of a machine element are shown. Determine the centroid of its volume. y 24 mm Solution: We divide the volume into six parts as shown. Parts 3 and 6 are the “holes”, which each have a radius of 8 mm. The volumes are 8 mm 18 mm 60 mm V1 D 604850 D 144,000 mm3 , V2 D 12 242 50 D 45, 239 mm3 , x V3 D 82 50 D 10, 053 mm3 , V4 D 163620 D 11, 520 8 mm z 20 mm mm3 , 16 mm 50 mm V5 D 12 182 20 D 10, 179 mm3 , y V6 D 82 20 D 4021 mm3 . The coordinates of the centroids are 2 x1 D 25 mm, 3 y1 D 30 mm, 5 x2 D 25 mm, y2 D 60 C 1 6 z1 D 0, 4 z 424 D 70.2 mm, 3 z5 D 24 C 16 C z2 D 0, x3 D 25 mm, x6 D 10 mm, y3 D 60 mm, y6 D 18 mm, z3 D 0, 418 D 47.6 mm, 3 z6 D 24 C 16 D 40 mm. x4 D 10 mm, The x coordinate of the centroid is y4 D 18 mm, xD z4 D 24 C 8 D 32 mm, x5 D 10 mm, x1 V1 C x2 V2 x3 V3 C x4 V4 C x5 V5 x6 V6 D 23.65 mm. V1 C V2 V3 C V4 C V5 V6 Calculating the y and z coordinates in the same way, we obtain y D 36.63 mm and z D 3.52 mm y5 D 18 mm, y Problem 7.81 In Example 7.13, suppose that the circular arc is replaced by a straight line as shown. Determine the centroid of the three-segment line. Solution: Let the new straight-line segment be line 1 and let the segment in the x-z plane be line 2. Let the other line segment be line 3. The centroid locations of the parts and their lengths are x 1 D 0, x 2 D 2 m, y 1 D 1 m, y 2 D 1 m, z1 D 1 m, z2 D 2 m, L1 D 2.83 m, L2 D 4 m, x 3 D 2 m, y 3 D 1 m, z3 D 1 m, L2 D 4.90 m. (0, 2, 0) m (0, 0, 2) m z x (4, 0, 2) m Applying Eqs. (7.18) yields x D 1.52 m, y D 0.659 m, z D 1.34 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 559 Problem 7.82 Determine the centroids of the lines. y Solution: The object is divided into two lines and a composite. 3m (1) (2) (3) L1 D 6 m, x1 D 3 m, y1 D 0. 6 L2 D 3 m, x2 D 6 C m (Note: See Example 7.13) y2 D 3. The composite length: L D 6 C 3 m. The composite centroid: xD L1 x1 C L2 x2 D 6 m, L x 6m 3 yD D 1.83 m 2C y Problem 7.83 Determine the centroids of the lines. Solution: Break the composite line into three parts (the quarter circle and two straight line segments) (see Appendix B). Part 1 Part 2 Part 3 xD xi yi Li 2R/ 3 m 0 2R/ 0 3 m R/2 2 m 2 m x 1 L 1 C x2 L 2 C x3 L 3 D 1.4 m L1 C L2 C L3 2m (R D 2 m) 2m x 2m 2m y1 L1 C y2 L2 C y3 L3 yD D 1.4 m L1 C L2 C L3 560 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.84 The semicircular part of the line lies in the x –z plane. Determine the centroid of the line. y 100 mm 160 mm x 120 mm z Solution: The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The composite length is: 3 LD y 100 mm 2 3 160 mm Li . iD1 z The composite coordinates are: 3 xD and y D 120 mm x Li xi iD1 L 3 1 , Li yi iD1 L Segment Length, mm x, mm 240 0 y, mm L1 120 L2 100 L3 188.7 80 50 Composite 665.7 65.9 21.7 z, mm 0 120 50 0 0 68.0 Problem 7.85 Determine the centroid of the line. y Solution: Break into a straight line and an arc. 2/3 mm tan 60° 2 cos 30° C 0 1 200 mm2 C cos d cos 60° xD D 332 mm 2/3 200 mm tan 60° C 200 mm d 1 2 200 200 mm 60⬚ x 0 1 ° 2 200mm tan 60 200 2/3 2 mm sin 60° 200 mm sin d D 118 mm 2/3 200 mm tan 60° C 200 mm d C yD 0 0 x D 332 mm, y D 118 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 561 y Problem 7.86 Use the method described in Active Example 7.14 to determine the area of the curved part of the surface of the truncated cone. z R x h 2 h 2 Solution: Work with the solid line shown. The surface area is given by A D 2yL D 2 AD 3R 4 2 2 h R C 2 2 3R p 2 h C R2 4 Problem 7.87 Use the second Pappus–Guldinus theorem to determine the volume of the truncated cone. Solution: Work with the trapezoidal area A D R/2h/2 C 1/2R/2h/2 D 3Rh 8 R R/2 yD 7R R/2h/2R/4 C 1/2R/2h/2[1/3R/2 C R/2] D A 18 V D 2yA D 2 VD 7R 18 3Rh 8 D h/2 h/4 h/4 7R2 h 24 7R2 h 24 Problem 7.88 The area of the shaded semicircle is 1 R2 . The volume of a sphere is 43 R3 . Extend the 2 approach described in Example 7.15 to the second Pappus–Guldinus theorem and determine the centroid y S of the semicircular area. Solution: The semicrcular area is A D 1 2 R2 , and y s is the y coordinate of its centroid. Rotating the area about the x axis generates the volume of a sphere. The second Pappus–Guldinus theorem states that the volume of the sphere is y _ yS x R V D 2y s A 4 3 R3 D 2y s 1 2 R2 Solving for ys yields 562 ys D 4R 3 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.89 Use the second Pappus–Guldinus theorem to determine the volume generated by revolving the curve about the y axis. Solution: The x coordinate of the centroid:. The element of area is the vertical strip of height 1 y and width dx. Thus y 1 1 1 y dx D AD 0 1 x2 dx. 0 Integrating, (1, 1) 1 2 x3 D . AD x 3 0 3 y ⫽ x2 1 x dA D A x x x 3 dx D 0 divide by the area: x D x2 x4 2 4 1 D 0 1 , 4 3 . The volume is V D 2xA D 8 2 Problem 7.90 The length of the curve is L D 1.479, and the area generated by rotating it about the x axis is A D 3.810. Use the first Pappus–Guldinus theorem to determine the y coordinate of the centroid of the curve. Solution: The surface area is A D 2yL, from which yD A D 0.41 2L Problem 7.91 Use the first Pappus–Guldinus theorem to determine the area of the surface generated by revolving the curve about the y axis. Solution: The length of the line is given in Problem 7.90. L D 1.479. The elementary length of the curve is dy dx 2 dL D 1C Noting dy D 2x, the element of line is dL D 1 C 4x2 1/2 . dx dx. The x-coordinate: 1 x dL D x1 C 4x2 1/2 dx 0 L D 1 1 53/2 1 1 C 4x2 3/2 0 D D 0.8484. 12 12 Divide by the length to obtain x D 0.5736. The surface area is A D 2xL D 5.33 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 563 y Problem 7.92 A nozzle for a large rocket engine is designed by revolving the function y D 23 x 13/2 about the y axis. Use the first Pappus–Guldinus theorem to determine the surface area of the nozzle. y = _2 (x – 1)3 / 2 3 x 5 ft dy D x 11/2 , the element dx Solution: The length: Noting that of length is 1C dL D dy dx 2 dx D p x dx from which 5 dL D LD x1/2 dx D 1 L 2 3/2 x 3 5 D 6.7869 ft 1 The x-coordinate: 5 x dL D x 3/2 dx D 1 L 2 5/2 x 5 5 D 21.961. 1 Divide by the length: x D 3.2357. The area A D 2xL D 138 ft2 Problem 7.93 The coordinates of the centroid of the line are x D 332 mm and y D 118 mm. Use the first Pappus-Guldinus theorem to determine the area of the surface of revolution obtained by revolving the line about the x axis. y 200 mm 60⬚ x Solution: L D 200 mm tan 60° C 200 mm 120° 180° D 765 mm A D 2yL D 20.118 m0.765 m D 0.567 m2 564 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.94 The coordinates of the centroid of the area between the x axis and the line in Problem 7.93 are x D 355 mm and y D 78.4 mm. Use the second PappusGuldinus theorem to determine the volume obtained by revolving the area about the x axis. Solution: The area is AD 1 0.2 m0.2 m tan 60° C 2 120° 0.2 m2 D .0765 m2 360° V D 2yA D 20.0784 m0.0765 m2 D 0.0377 m3 Problem 7.95 The volume of revolution contains a hole of radius R. (a) (b) R+a Use integration to determine its volume. Use the second Pappus–Guldinus theorem to determine its volume. R h Solution: (a) The element of volume is a disk of radius y and thickness dx. The area of the disk is y 2 R2 . The radius is yD a h from which dV D Denote m D a h x C R, 2 xCR dx R2 dx. a , dV D m2 x 2 C 2mRx dx, h from which h dV D m VD mx 2 C 2Rx dx 0 V 3 h x mh CR D m m C Rx 2 D mh2 3 3 0 D ah (b) a 3 CR . The area of the triangle is A D 12 ah. The y-coordinate of the centroid is y D R C 13 a. The volume is V D 2yA D ahR C 13 a c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 565 Problem 7.96 Determine the volume of the volume of revolution. Solution: The area of the semicircle is A D y DRC 4r . The volume is 3 V D 2 r 2 2 RC 4r 3 r 2 . The centroid is 2 4r D 2 r 2 R C . 3 For r D 40 mm and R D 140 mm, V D 2.48 ð 103 m3 140 mm 80 mm Problem 7.97 Determine the surface area of the volume of revolution in Problem 7.96. Solution: The length and centroid of the semicircle is Lo D r, yDRC y D R. 2r . The length and centroid of the inner line is Li D 2r, and 2r A D 2r R C C 22rR D 2rR C 2r C 2R. For r D 40 mm and R D 140 mm, A D 0.201 m2 Problem 7.98 The volume of revolution has an elliptical cross section. Determine its volume. 230 mm 130 mm 180 mm Solution: Use the second theorem of Pappus-Guldinus. The centroid of the ellipse is 180 mm from the axis of rotation. The area of the ellipse is ab where a D 115 mm, b D 65 mm. 2b The centroid moves through a distance jdj D 2R D 2 (180 mm) as the ellipse is rotated about the axis. 2a V D Ad D abd D 2.66 ð 107 mm3 v D 0.0266 m3 566 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.99 Suppose that the bar in Active Example 7.16 is replace with this 100-kg homogeneous bar. (a) What is the x coordinate of the bar’s center of mass? (b) Determine the reactions at A and B. y 0.5 m B 1m A x 1m Solution: (a) Let the new horizontal segment of the bar part 3. The x coordinate of the centroid of the bar’s axis, which coincides with its center of mass is xD x 1 L1 C x 2 L2 C x 3 L3 0.51 C 11 C 0.750.5 D 0.75 m D L1 C L2 C L3 0.5 C 1 C 0.5 x D 0.75 m (b) The equilibrium equations are MA : B1 m 981 N 0.75 m D 0, Fx : Ax B D 0, Fy : Ay 981 N D 0. Solving yields Ax D 736 N, Ay D 981 N, B D 736 N. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 567 Problem 7.100 The mass of the homogeneous flat plate is 50 kg. Determine the reactions at the supports A and B. 100 mm 400 mm 200 mm A B 600 mm 800 mm 600 mm Solution: Divide the object into three areas and the composite. Since the distance to the action line of the weight is the only item of importance, and since there is no horizontal component of the weight, it is unnecessary to determine any centroid coordinate other than the xcoordinate. The areas and the x-coordinate of the centroid are tabulated. The last row is the composite area and x-coordinate of the centroid. 500 N AX X B AY Area A, sq mm x Rectangle 3.2 ð 105 400 Circle 3.14 ð 104 600 Triangle 1.2 ð 105 1000 Composite 4.09 ð 105 561 1400 mm The composite area is A D Arect Acirc C Atriang . The composite xcoordinate of the centroid is xD Arect xrect Acirc xcirc C Atriang xtriang . A The sum of the moments about A: MA D 500561 C 1400B D 0, from which B D 200 N. The sum of the forces: Fy D Ay C B 500 D 0, from which Ay D 300 N. 568 Fx D Ax D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.101 The suspended sign is a homogeneous flat plate that has a mass of 130 kg. Determine the axial forces in members AD and CE. (Notice that the y axis is positive downward.) A 2m 4m C 1m E B x D y y = 1 + 0.0625x2 Solution: The strategy is to determine the distance to the action line of the weight (x-coordinate of the centroid) from which to apply the equilibrium conditions to the method of sections. The area: The element of area is the vertical strip of length y and width dx. The element of area dA D y dx D 1 C ax2 dx, where a D 0.0625. Thus 4 dA D AD 0 A 4 ax 3 1 C ax2 dx D x C D 5.3333 sq ft. 3 0 The x-coordinate: 4 x dA D A x1 C ax2 dx D 0 Divide A: x D ax 4 x2 C 2 4 4 D 12. 0 12 D 2.25 ft. 5.3333 The equilibrium conditions: The angle of the member CE is ˛ D tan1 14 D 14.04° . The weight of the sign is W D 1309.81 D 1275.3 N. The sum of the moments about D is MD D 2.25W C 4CE sin ˛ D 0, from which CE D 2957.7 N T . Method of sections: Make a cut through members AC, AD and BD and consider the section to the right. The angle of member AD is ˇ D tan1 12 D 26.57° . The section as a free body: The sum of the vertical forces: FY D AD sin ˇ W D 0 from which AD D 2851.7 N T c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 569 Problem 7.102 The bar has a mass of 80 kg. What are the reactions at A and B? A 2m 2m B y Solution: Break the bar into two parts and find the masses and centers of masses of the two parts. The length of the bar is X1 m1g m2g AX L D L1 C L2 D 2 m C 2R/4R D 2 m X2 L D2C m AY Lengthi (m) Part 1 2 2 m1 D 31.12 kg x1 D 1 m m2 D 48.88 kg x2 D 3.27 m Fx : Ax D 0 Fy : Ay C By m1 g m2 g D 0 MA : x1 m1 g x2 m2 g C 4By D 0 Massi (kg) 2 80 2C 80 2C xi (m) x 4m BY 1 2C 2R Solving Ax D 0, Ay D 316 N, B D 469 N 570 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.103 The mass of the bar per unit length is 2 kg/m. Choose the dimension b so that part BC of the suspended bar is horizontal. What is the dimension b, and what are the resulting reactions on the bar at A? A 1m 30⬚ B b Solution: We must have C Ay MA : g1.0 m0.5 m cos 30° gb b 1.0 m cos 30° 2 Ax D0 ) b D 2.14 m Then Fx : Ax D 0 ρg(1.0 m) ρ gb Fy : Ay g1.0 m gb D 0 ) Ax D 0, Ay D 61.6 N, b D 2.14 m Problem 7.104 The semicircular part of the homogeneous slender bar lies in the x –z plane. Determine the center of mass of the bar. Solution: The bar is divided into three segments plus the composite. The lengths and the centroids are given in the table: The composite length is: y 3 LD Li . iD1 The composite coordinates are: 10 in 3 16 in 12 in xD Li xi iD1 L , z x 3 and y D Li yi iD1 L Segment Length, in 0 12 10 x, in 24 0 L1 12 L2 y, in z, in 5 0 L3 18.868 8 5 0 Composite 66.567 6.594 2.168 6.796 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 571 y Problem 7.105 The density of the cone is given by the equation D 0 1 C x/h, where 0 is a constant. Use the procedure described in Example 7.17 to show that the mass of the cone is given by m D 7/40 V, where V is the volume of the cone, and that the x coordinate of the center of mass of the cone is x D 27/35h. z R x h Solution: Consider an element of volume dV of the cone in the form of a “disk” of width dx. The radius of such a disk at position x is (R/h)x, so dV D [R/hx]2 dx. The mass of the cone is h 7 7 dV D 0 1 C x/h [R/hx]2 dx D 0 R2 h D 0 V. mD 12 3 V 0 The x coordinate of the center of mass is h xdV x0 1 C x/h [R/hx]2 dx 27 h. D 0h x D V D 35 2 dV 0 1 C x/h [R/h x] dx V 572 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 7.106 A horizontal cone with 800-mm length and 200-mm radius has a built-in support at A. Its density is D 60001 C 0.4x 2 kg/m3 , where x is in meters. What are the reactions at A? 200 mm A x 800 mm Solution: The strategy is to determine the distance to the line of action of the weight, from which to apply the equilibrium conditions. The mass: The element of volume is a disk of radius y and thickness dx. y varies linearly with x: y D 0.25x. Denote a D 0.4. The mass of the disk is dm D y 2 dx D 60001 C ax2 0.25x2 dx y 200 mm x A 800 mm D 3751 C ax2 x 2 dx, from which 0.8 m D 375 1 C ax2 x 2 dx D 375 0 x5 x3 Ca 3 5 0.8 0 D 231.95 kg The x-coordinate of the mass center: x dm D 375 0.8 1 C ax2 x 3 dx D 375 0 m x6 x4 Ca 4 6 0.8 0 D 141.23. Divide by the mass: x D 0.6089 m The equilibrium conditions: The sum of the moments about A: M D MA mgx D 0, from which MA D mgx D 231.949.810.6089 D 1385.4 N-m . The sum of the vertical forces: FY D AY mg D 0 from which AY D 2275.4 N . The horizontal component of the reaction is zero, FX D 0. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 573 y Problem 7.107 In Active Example 7.18, suppose that bar 1 is replaced by a bar with the same dimensions that consists of aluminum alloy with a density of 2600 kg/m3 . Determine the x coordinate of the center of mass of the machine part. 240 mm 1 40 mm 2 80 mm Solution: The mass of bar 1 is 4 m D 7.68 ð 10 3 80 mm z 3 240 mm m 2600 kg/m D 2.00 kg x The x coordinate of the center of mass is xD x 1 m1 C x 2 m2 40 mm 2.00 kg C 200 mm 5.99 kg D 160 mm D m 1 C m2 2.00 kg C 5.99 kg x D 160 mm y Problem 7.108 The cylindrical tube is made of aluminum with mass density 2700 kg/m3 . The cylindrical plug is made of steel with mass density 7800 kg/m3 . Determine the coordinates of the center of mass of the composite object. z x y y Tube A Plug 20 mm x z 35 mm 100 mm 100 mm A Section A-A Solution: The volume of the aluminum tube is VAl D 0.0352 0.022 0.2 D 5.18 ð 104 m3 . The mass of the aluminum tube is mAl D 2700VAl D 1.4 kg. The centroid of the aluminum tube is xAL D 0.1 m, yAl D zAl D 0. The volume of the steel plug is VFe D 0.022 0.1 D 1.26 ð 104 m3 . The mass of the steel plug is mFe D 7800VFe D 0.9802 kg. The centroid of the steel plug is xFe D 0.15 m, yFe D zFe D 0. The composite mass is m D 2.38 kg. The composite centroid is xD mAl 0.1 C mFe 0.15 D 0.121 m m yDzD0 574 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.109 In Example 7.19, suppose that the object is redesigned so that the radius of the hole in the hollow cylinder is increased from 2 in to 3 in. What is the x coordinate of the center of mass of the object? Front View Side View y y 2 in x z 4 in 10 in 5 in 5 in 12 in Solution: The volume of the cylinder is Vcylinder D 12 in [4 in2 3 in2 ] D 264 in3 D 0.153 ft3 Its weight is Wcylinder D 0.153 ft3 530 lb/ft3 D 80.9 lb. The x coordinate of the center of mass (same as center of weight) is xD D x bar Wbar C x cylinder Wcylinder Wbar C Wcylinder 1.86 in 15.6 lb C 10 in 80.9 lb D 8.68 in. 15.6 lb C 80.9 lb x D 8.68 in. Problem 7.110 A machine consists of three parts. The masses and the locations of the centers of mass of two of the parts are: Part 1 2 Mass (kg) 2.0 4.5 x (mm) 100 150 y (mm) 50 70 z (mm) 20 0 The mass of part 3 is 2.5 kg. The design engineer wants to position part 3 so that the center of mass location of the machine is x D 120 mm, y D 80 mm, and z D 0. Determine the necessary position of the center of mass of part 3. Solution: The composite mass is m D 2.0 C 4.5 C 2.5 D 9 kg. The location of the third part is x3 D 1209 2100 4.5150 D 82 mm 2.5 y3 D 809 250 4.570 D 122 mm 2.5 z3 D 220 D 16 mm 2.5 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 575 Problem 7.111 Two views of a machine element are shown. Part 1 is aluminum alloy with density 2800 kg/m3 , and part 2 is steel with density 7800 kg/m3 . Determine the coordinates of its center of mass. y y 1 24 mm 2 8 mm 18 mm x 20 mm 60 mm 8 mm z 16 mm 50 mm Solution: The volumes of the parts are V1 D 6048 C 12 242 82 50 D 179, 186 mm3 D 17.92 ð 105 m3 , V2 D 1636 C 12 182 82 20 D 17, 678 mm3 D 1.77 ð 105 m3 , so their masses are m1 D S1 V1 D 280017.92 ð 105 D 0.502 kg, m2 D S2 V2 D 78001.77 ð 105 D 0.138 kg. The x coordinates of the centers of mass of the parts are x1 D 25 mm, x2 D 10 mm, so xD 576 x1 m1 C x2 m2 D 21.8 mm m1 C m 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.112 The loads F1 D F2 D 25 kN. The mass of the truss is 900 kg. The members of the truss are homogeneous bars with the same uniform cross section. (a) What is the x coordinate of the center of mass of the truss? (b) Determine the reactions at A and G. y D F1 3m F2 B E 3m G A x C 4m 4m Solution: (a) The center of mass of the truss is located at the centroid of the composite line of the axes of the members. The lengths of the diagonal members are 4 m2 C 3 m2 D 5 m. The lengths and x coordinates of the centroids of the axes of the members are Member AB AC BC BD BE BG CG DE EG Length 5m 4m 3m 5m 4m 5m 4m 3m 3m x coordinate 2m 2m 4m 6m 6m 6m 6m 8m 8m The x coordinate of the centroid of the composite line, which is coincident with the center of mass of the truss, is xD xi Li 2 5 C 4 C 4 3 C 6 5 C 4 C 5 C 4 C 8 3 C 3 D D 5.17 m Li 5C4C3C5C4C5C4C3C3 x D 5.17 m (b) The equilibrium equations for the truss are Fx : Ax C 25 kN C 25 kN, Fy : Ay C G 9009.81N D 0, MA : 25 kN 3 m 25 kN 6 m 900 9.81 N5.17 m C G8 m D 0. Solving yields Axc D 50 kN, Ay D 25.0 kN, G D 33.8 kN. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 577 Problem 7.113 With its engine removed, the mass of the car is 1100 kg and its center of mass is at C. The mass of the engine is 220 kg. E Suppose that you want to place the center of mass E of the engine so that the center of mass of the car is midway between the front wheels A and the rear wheels B. What is the distance b? (b) If the car is parked on a 15° slope facing up the slope, what total normal force is exerted by the road on the rear wheels B? C (a) 0.6 m 0.45 m A B 1.14 m b 2.60 m Solution: (a) The composite mass is m D mC C mE D 1320 kg. The xcoordinate of the composite center of mass is given: xD 2.6 D 1.3 m, 2 from which the x-coordinate of the center of mass of the engine is xE D b D 1.3 m 1.14 mC D 2.1 m. mE The y-coordinate of the composite center of mass is yD (b) 0.45 mC C 0.6 mE D 0.475 m. m Assume that the engine has been placed in the new position, as given in Part (a). The sum of the moments about B is MA D 2.6A C ymg sin15° 2.6 xmg cos15° D 0, from which A D 5641.7 N. This is the normal force exerted by the road on A. The normal force exerted on B is obtained from; FN D A mg cos15° C B D 0, from which B D 6866 N 578 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.114 The airplane is parked with its landing gear resting on scales. The weights measured at A, B, and C are 30 kN, 140 kN, and 146 kN, respectively. After a crate is loaded onto the plane, the weights measured at A, B, and C are 31 kN, 142 kN, and 147 kN, respectively. Determine the mass and the x and y coordinates of the center of mass of the crate. B 6m A x Solution: The weight of the airplane is WA D 30 C 140 C 146 D 316 kN. The center of mass of the airplane: 10 m Myaxis D 3010 xA WA D 0, from which xA D 0.949 m. C 6m y Mxaxis D 140 1466 C yA WA D 0, from which yA D 0.114 m. The weight of the loaded plane: W D 31 C 142 C 147 D 320 kN. The center of mass of the loaded plane: Myaxis D 3110 xW D 0, from which x D 0.969 m. Mxaxis D 142 1476 C yW D 0, from which y D 0.0938 m. The weight of the crate is Wc D W WA D 4 kN. The center of mass of the crate: xc D Wx WA xA D 2.5 m, Wc yc D Wy WA yA D 1.5 m. Wc The mass of the crate: mc D Wc ð 103 D 407.75 kg 9.81 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 579 Problem 7.115 A suitcase with a mass of 90 kg is placed in the trunk of the car described in Example 7.20. The position of the center of mass of the suitcase is x s D 0.533 m, y s D 0.762 m, and z s D 0.305 m. If the suitcase is regarded as part of the car, what is the new position of the car’s center of mass? Solution: In Example 7.20, the following results were obtained The new center of mass is at for the car without the suitcase xN D Wc D 17303 N xc Wc C xs Ws Wc C Ws with similar eqns for yN and zN xc D 1.651 m Solving, we get yc D 0.584 m xN D 1.545 m, yN D 0.593 m, zN D 0.717 m zc D 0.769 m For the suitcase Ws D 90 g, y D 0.762 m, xs D 0.533 m, z D 0.305 m. Problem 7.116 A group of engineering students constructs a miniature device of the kind described in Example 7.20 and uses it to determine the center of mass of a miniature vehicle. The data they obtain are shown in the following table: Wheelbase = 36 in Track = 30 in Left front wheel, NLF Right front wheel, NRF Left rear wheel, NLR Right rear wheel, NRR Measured Loads (lb) ˛D0 ˛ D 10° 35 32 36 33 27 34 29 30 Determine the center of mass of the vehicle. Use the same coordinate system as in Example 7.20. Solution: The weight of the go-cart: W D 35 C 36 C 27 C 29 D 127 lb. The sum of the moments about the z axis With the go-cart in the tilted position, the sum of the moments about the z axis Mzaxis D WheelbaseNLF C NRF xW D 0, Mzaxis D WheelbaseNLF C NRF C yW sin10° xW cos10° D 0, from which xD 3635 C 36 D 20.125 in. W The sum of the moments about the x axis: Mxaxis D zW TrackNRF C NRR D 0, from which yD xW cos10° 3632 C 33 W sin10° D 8.034 in. from which zD 580 3036 C 29 D 15.354 in. W c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.117 Determine the centroid of the area by letting dA be a vertical strip of width dx. y (1, 1) y = x2 x Solution: The area: The length of the vertical strip is 1 y, so that the elemental area is dA D 1 y dx D 1 x2 dx. The area: 1 dA D 0 A 1 2 x3 1 1 x2 dx D x D1 D . 3 0 3 3 The x-coordinate: xA D 1 x dA D x1 x2 dx D 0 A x4 x2 2 4 1 D 0 3 1 : xD 4 8 The y-coordinate: The y-coordinate of the centroid of each element of area is located at the midpoint of the vertical dimension of the area element. y D y C 12 1 x2 . Thus 1 y dA D x2 C 0 A D yD 1 1 x2 1 x2 dx 2 1 1 2 x5 D . x 2 5 0 5 3 5 Problem 7.118 Determine the centroid of the area in Problem 7.117 by letting dA be a horizontal strip of height dy. Solution: The area: The length of the horizontal strip is x, hence Divide by the area: x D the element of area is 3 8 The y-coordinate: dA D x dy D y 1/2 dy. yA D Thus A 1 AD 0 y 1/2 dy D 1 2y 3/2 3 0 D 2 3 1 y dA D D 0 1 y y 1/2 dy 0 y 3/2 dy D 2y 5/2 5 Check: The x-coordinate: The x-coordinate of the centroid of each element of area is x D 12 x D 12 y 1/2 . Thus Divide by the area: y D 1 D 0 2 . 5 3 5 1 2 1 1 1 1 1 y y dy D D . y 1/2 dA D 2 2 2 0 4 A 2 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 581 y Problem 7.119 Determine the centroid of the area. 60 cm x 80 cm Solution: The strategy is to develop useful general results for the triangle and the rectangle. Divide by the area: y D The rectangle: The area of the rectangle of height h and width w is xD w AD 60 cm a 20 cm. The composite: 3 xR AR C xT AT 404800 C 1001800 D AR C AT 4800 C 1800 h dx D hw D 4800 cm2 . 0 D 56.36 cm The x-coordinate: w x2 hx dx D h 2 0 w 0 1 D hw2 . 2 yD 304800 C 201800 4800 C 1800 D 27.27 cm Divide by the area: x D w D 40 cm 2 The y-coordinate: w 1 1 h2 dx D h2 w. 2 2 0 Divide by the area: y D 1 h D 30 cm 2 The triangle: The area of the triangle of altitude a and base b is (assuming that the two sides a and b meet at the origin) b b yx dx D AD 0 0 b a ax 2 ax x C a dx D b 2b 0 ab ab C ab D D 1800 cm2 D 2 2 Check: This is the familiar result. check. The x-coordinate: b 0 b a ax 3 ax2 ab2 C . D x C a x dx D b 3b 2 0 6 Divide by the area: x D b D 20 cm 3 The y-coordinate: y dA D A b 2 a 1 x C a dx 2 b 0 D 582 3 b ba2 b a . D xCa 6a b 6 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 7.120 Determine the centroid of the area. 40 mm 20 mm Solution: Divide the object into five areas: 40 mm (1) (2) (3) (4) (5) The rectangle 80 mm by 80 mm, The rectangle 120 mm by 80 mm, the semicircle of radius 40 mm, The circle of 20 mm radius, and the composite object. The areas and centroids: 80 mm (1) A1 D 6400 mm2 , x1 D 40 mm, y1 D 40 mm, (2) A2 D 9600 mm2 , x2 D 120 mm, y2 D 60 mm, (3) A3 D 2513.3 mm2 , x3 D 120 mm, y3 D 136.98 mm, (4) A4 D 1256.6 mm2 , x4 D 120 mm, y4 D 120 mm. (5) The composite area: A D A1 C A2 C A3 A4 D 17256.6 mm2 . The composite centroid: x 120 mm 160 mm xD A1 x1 C A2 x2 C A3 x3 A4 x4 D 90.3 mm . A yD A1 y1 C A2 y2 C A3 y3 A4 y4 D 59.4 mm A y Problem 7.121 The cantilever beam is subjected to a triangular distributed load. What are the reactions at A? Solution: The load distribution is a straight line with intercept w D 200 N/m at x D 0, and slope 200 10 D 20 N/m2 . The sum 200 N/m of the moments is x A 10 20x C 200x dx D 0, M D MA 10 m 0 from which 10 20 D 3333.3 N-m. MA D x 3 C 100x2 3 0 The sum of the forces: 200 N/m AX MA AY 10 m 10 20x C 200 dx D 0, Fy D Ay 0 from which 10 Ay D 10x2 C 200x 0 D 1000 N, and Fx D Ax D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 583 Problem 7.122 What is the axial load in member BD of the frame? C 100 N/m 5m B D 5m A E 10 m Solution: The distributed load is two straight lines: Over the Over the interval 5 y 10, the load is a constant w D 100 N/m. The moment about the origin E due to the load is 5 ME D 10 100y dy, 20yy dy C 0 Cy Cx interval 0 y 5 the intercept is w D 0 at y D 0 and the slope is 100 D 20. C 5 Cx By Bx Ay Cy Bx By Dy Dx Dx Dy Ey Ex 5 from which ME D 20 3 y 3 5 C 0 100 2 y 2 10 D 4583.33 N-m. 5 Check: The area of the triangle is F1 D 12 5100 D 250 N. The area of the rectangle: F2 D 500 N. The centroid distance for the triangle is d1 D 23 5 D 3.333 m. The centroid distance of the rectangle is d2 D 7.5 m. The moment about E is ME D d1 F1 C d2 F2 D 4583.33 Nm check. The Complete Structure: The sum of the moments about E is M D 10AR C ME D 0, where AR is the reaction at A, from which AR D 458.33 N. The element ABC : Element BD is a two force member, hence By D 0. The sum of the moments about C: MC D 5Bx 10Ay D 0, where Ay is equal and opposite to the reaction of the support, from which Bx D 2Ay D 2AR D 916.67 N. Since the reaction in element BD is equal and opposite, Bx D 916.67 N, which is a tension in BD. 584 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.123 An engineer estimates that the maximum wind load on the 40-m tower in Fig. a is described by the distributed load in Fig. b. The tower is supported by three cables A, B, and C from the top of the tower to equally spaced points 15 m from the bottom of the tower (Fig. c). If the wind blows from the west and cables B and C are slack, what is the tension in cable A? (Model the base of the tower as a ball and socket support.) 200 N/m B N A 40 m 15 m C 400 N/m (a) Solution: The load distribution is a straight line with the intercept w D 400 N/m, and slope 5. The moment about the base of the tower due to the wind load is 40 5y C 400y dy, MW D (b) (c) 200 N/m θ 40 m TA 0 40 5 D 213.33 kN-m, MW D y 3 C 200y 2 3 0 clockwise about the base, looking North. The angle formed by the cable with the horizontal at the top of the tower is D 90° tan1 15 40 Fx Fy 400 N/m D 69.44° . The sum of the moments about the base of the tower is M D MW C 40TA cos D 0, from which TA D 1 40 cos MW D 15.19 kN c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 585 Problem 7.124 Determine the reactions on member ABCD at A and D. 2 kN/m 2 kN/m D E 1m 1m C 1m B 1m A F 1m Solution: First, replace the distributed forces with equivalent 2 kN / m concentrated forces, then solve for the loads. Note that BF and CE are two force members. y Distributed Load on ABCD, F1 By area analogy, concentrated load is applied at y D šm. The load is 1 2 (2)(3) kN 3m F1 D 3 kN By the area analogy, F2 D 4 kN applied at x D 1 m Assume FCE and FBF are tensions For ABCD: Fx : Ax C FBF cos 45° C FCE cos 45° C Dx C 3 kN D 0 Fy : Ay C Dy FBF sin 45° C FCE sin 45° D 0 x y 2 kN / m 2m MA : 1FBF cos 45° 2FCE cos 45° 23 3Dx D 0 For DE: 586 Fx : Dx FCE cos 45° D 0 Fy : Dy FCE sin 45° 4 D 0 ME : 1Dy D 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 7.124 (Continued ) Solving, we get Ax D 7 kN Ay D 6 kN Dx D 4 kN Dy D 0 also FBF D 14.14 kN(c) FCE D 5.66 kN(c) DY DX FCE 45° 3 kN C B 2m FBF 45° AX AY 1 4 kN 1m DX E DY 45° FCE 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 587 Problem 7.125 Estimate the centroid of the volume of the Apollo lunar return configuration (not including its rocket nozzle) by treating it as a cone and a cylinder. y x 12.8 ft Nozzle 10 ft 14 ft Solution: The volume of the cone is V1 D R2 h D 428.93 ft3 . 3 The x-coordinate of the centroid from the nose of the cone is x1 D 3h D 7.5 ft. The volume of the cylinder is V2 D R2 L D 1801.5 ft3 . 4 The x-coordinate of the centroid from the nose of the cone is x2 D L h C D 17 ft. The composite volume is V D V1 C V2 D 2230.4 ft3 . 2 The x-coordinate of the composite centroid is xD V1 x1 C V2 x2 D 15.2 ft . V The y- and z-coordinates are zero, from symmetry. 588 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.126 The shape of the rocket nozzle of the Apollo lunar return configuration is approximated by revolving the curve shown around the x axis. In terms of the coordinate system shown, determine the centroid of the volume of the nozzle. y y = 0.350 + 0.435x – 0.035x 2 x Solution: x dV xD . dV 2.83 m Let dV be a disk of radius y and thickness dx Thus, dV D y 2 dx, where y 2 D 0.350 C 0.435x 0.035x2 2 y 2 D a C bx C cx 2 2 y 2 D a2 C 2abx C 2ac C b2 x 2 C 2bcx3 C c2 x 4 a D 0.350 b D 0.435 c D 0.035 2.83 a2 x C 2abx2 C 2ac C b2 x 3 C 2bcx4 C c2 x 5 dx 0 xD 2.83 a2 C 2abx C 2ac C b2 x 2 C 2bcx3 C c2 x 4 dx 0 4 2.83 3 x x C 2ab C 2ac C b2 3 4 6 5 x x 2 C2bc Cc 5 6 xD 3 2.830 x2 x 2 2 C 2ac C b a x C 2ab 2 3 5 4 x x 2 Cc C2bc 4 5 0 a2 x2 2 Evaluating, xD 4.43 D 1.87 m 3.37 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 589 y Problem 7.127 Determine the coordinates of the centroid of the volume. 120 mm 40 mm 100 mm z 30 mm 20 mm x Solution: From symmetry y D z D 0 xD 0.042 0.032 0.120.06 C 0.032 0.022 0.220.11 0.042 0.032 0.12 C 0.032 0.022 0.22 x D 0.0884 m D 88.4 mm, y D z D 0 Problem 7.128 Determine the surface area of the volume of revolution in Problem 7.127. 5 in 9 in 6 in Solution: The outer surface: The length of the line is L1 D p 42 C 62 D 7.2111 in. The y-coordinate of the centroid is y D 5 C 2 D 7 in. The surface area is A1 D 2L1 y1 D 317.16 in2 . The side surface: The length of the line is L2 D 4 in. The ycoordinate of the centroid is y2 D 5 C 2 D 7 in. The surface area is A2 D 2L2 y2 D 87.96 in2 . The inner surface: The length of the line is L3 D 6 in. The y-coordinate is y3 D 5 in. The surface area is A3 D 2L3 y3 D 188.5 in2 . The total surface: A D A1 C A2 C A3 D 681.6 in2 590 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.129 Determine the y coordinate of the center of mass of the homogeneous steel plate. y 20 mm 10 mm 20 mm 20 mm x 80 mm Solution: Divide the object into five areas: (1) The lower rectangle 20 by 80 mm, (2) an upper rectangle, 20 by 40 mm, (3) the semicircle of radius 20 mm, (4) the circle of radius 10 mm, and (5) the composite part. The areas and the centroids are tabulated. The last row is the composite and the centroid of the composite. The composite area is AD 3 Ai A4 . 1 The centroid: 3 xD , A 3 and y D Ai xi A4 x4 1 Ai yi A4 y4 1 A . The following relationships were used for the centroids: For a rectangle: the centroid is at half the side and half the base. For a semicircle, 4R from the base. For a circle, the centroid is on the centerline and at 3 the centroid is at the center. Area A, sq mm x, mm y, mm A1 1600 40 10 A2 800 60 30 A3 628.3 60 48.5 A4 314.2 60 40 48.2 21.3 Composite 2714 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 591 Problem 7.130 Determine the x coordinate of the center of mass of the homogeneous steel plate. y 220 mm x2 Solution: The quarter circle: The equation of the circle is C y 2 DpR2 . Take the elemental area to be a vertical strip of height ypD R2 x 2 and width dx, hence the element of area is dA D R2 x 2 dx, and the area is R R2 x 2 dx D AD 0 150 mm p x R R2 R2 x R2 x 2 C sin1 D 2 2 R 4 x 0 50 mm The x-coordinate: xC A D A xC D R x dA D x 0 R R2 x 2 3/2 R3 R2 x 2 dx D D 3 3 0 4R 3 The rectangle: The area is A D 50150 D 7500 mm2 . The xcoordinate of the centroid is xR D 25 mm. The composite: The area of the quarter circle is AC D 2202 D 3.8013 ð 104 mm. 4 The area of the rectangle is AR D 50150 D 0.75 ð 104 mm2 . The composite area is A D AC AR D 3.0513 ð 104 mm2 . The centroid: AC xC AR xR xD D 110 mm A Problem 7.131 The area of the homogeneous plate is 10 ft2 . The vertical reactions on the plate at A and B are 80 lb and 84 lb, respectively. Suppose that you want to equalize the reactions at A and B by drilling a 1-ft-diameter hole in the plate. What horizontal distance from A should the center of the hole be? What are the resulting reactions at A and B? A B 5 ft Solution: The weight of the plate is W D 80 C 84 D 164 lb. From the sum of moments about A, the centroid is xD 845 D 2.56 ft. W A The weight density is wD 5 ft W D 16.4 lb/ft2 . 10 The weight of the cutout is WC D 0.52 w D 12.88 lb. The new weight of the plate is W2 D W WC D 151.12 lb. The new centroid must be at x2 D 5 D 2.5 ft for the reactions to be equal. 2 B X A X2 B XC Therefore the x-coordinate of the center of the circle will be xC D Wx W2 x2 D 3.26 ft. WC The reactions at A and B will be ADBD 592 W2 D 75.56 lb 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.132 The plate is of uniform thickness and is made of homogeneous material whose mass per unit area of the plate is 2 kg/m2 . The vertical reactions at A and B are 6 N and 10 N, respectively. What is the x coordinate of the centroid of the hole? 1m A B 2m Solution: Choose an origin at A. The basic relation is WC xC D Wx WH xH , where WC is the weight of the composite plate (the one with the hole), W is the weight of the plate without the hole, WH is the weight of the material removed from the hole, and xC , x, and xH are the x-coordinates of the centroids of the composite plate, the plate without the hole, and the hole, respectively. The composite weight: FY D A C B WC D 0, WC A WH B XC XH 2m from which WC D 16 N. The x-coordinate of the centroid: MA D WC xC C 2B D 0, from which xC D 1.25 m. The weight of the plate without the hole and the x-coordinate of the centroid: W D Ag D 12 2129.81 D 19.62 N, and x D 23 2 D 1.3333 m. The weight of the material removed from the hole: WH D W WC D 3.62 N. Solve: xH D Wx WC xC D 1.702 m WH c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 593 y Problem 7.133 Determine the center of mass of the homogeneous sheet of metal. Solution: Divide the object into four parts: (1) The lower plate, x (2) the left hand plate, (3) the semicircular plate, and (4) the composite plate. The areas and centroids are found by inspection: (1) Area: A1 D 912 D 108 in2 , x1 D 0.5 in, y1 D 8 in, z1 D 6 in. (2) A2 D 812 D 96 in2 , x2 D 4 in, y2 D 4 in, z2 D 6 in. (3) 4 in 8 in z 12 in 9 in A3 D 412 D 150.8 in2 , 24 x3 D 0, y D D 2.546 in, z D 6 in. The composite area is AD 3 Ai D 354.796 in2 . 1 The centroid for the composite: 3 xD A 3 yD 594 D 0.930 in Ai yi 1 A 3 zD Ai xi 1 D 2.435 in Ai zi 1 A D 6 in c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.134 Determine the center of mass of the homogeneous object. 60 mm z 10 mm y (1) A triangular solid 30 mm altitude, 60 mm base, and 10 mm thick. (2) A rectangle 60 by 70 mm by 10 mm. (3) A semicircle with radius 20 mm and 10 mm thick. The volumes and their centroids are determined by inspection: (1) V1 D y1 D 10 C z1 D (2) 30 mm z x x y 1 306010 D 9000 mm3 , 2 x1 D 5 mm, 60 mm 20 mm Solution: Divide the object into three parts and the composite: 30 mm z 10 mm 30 D 20 mm, 3 60 D 20 mm 3 V2 D 607010 D 42000 mm3 , x2 D 35 mm, y2 D 5 mm, z2 D 30 mm (3) V3 D 202 10 D 6283.2 mm3 , 2 x3 D 70 420 D 61.51 mm, 3 y3 D 5 mm, z3 D 30 mm. The composite volume is V D V1 C V2 V3 D 44716.8 mm3 . The centroid is xD V1 x1 C V2 x2 V3 x3 D 25.237 mm V yD V1 y1 C V2 y2 V3 y3 D 8.019 mm V zD V1 z1 C V2 z2 V3 z3 D 27.99 mm V c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 595 Problem 7.135 Determine the center of mass of the homogeneous object. 5 in 1.5 in y x z y Solution: Divide the object into five parts plus the composite. (1) A solid cylinder with 1.5 in radius, 3 in long. (2) A rectangle 3 by 5 by 1 in (3) A solid cylinder with radius 1.5 in, 2 in long. (4) A semicircle with radius 1.5 in, 1 inch thick, (5) a semicircle with radius 1.5 in, 1 inch thick. The volumes and centroids are determined by inspection. These are tabulated: Part No Vol, cu in x, in V1 21.205 0 V2 15 2.5 V3 14.137 5 V4 3.534 0.6366 3.534 V5 Composite 43.27 y, in z, in 1 0 0 0 0.5 0 0 0 4.363 0 0 2.09 0.3267 0 Top View z 1 in x 3 in 2 in x Side View The composite is VD 3 Vi 5 1 Vi . 4 The centroid: 3 xD Vi xi 1 5 4 V Vi xi , with a corresponding expression for y. The z-coordinate is zero because of symmetry. 596 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 7.136 The arrangement shown can be used to determine the location of the center of mass of a person. A horizontal board has a pin support at A and rests on a scale that measures weight at B. The distance from A to B is 2.3 m. When the person is not on the board, the scale at B measures 90 N. (a) (b) y When a 63-kg person is in position (1), the scale at B measures 496 N. What is the x coordinate of the person’s center of mass? When the same person is in position (2), the scale measures 523 N. What is the x coordinate of his center of mass? B x A (1) y B x A (2) Solution: (a) WB 1.15 m W D mg D 63 g D 618 N Unloaded Beam (assume uniform beam) Fy : Ay C By WB D 0 MA : 1.15WB C 2.3By D 0 X Fy : Ay C By WB W D 0 By = 90 N W WB Solving, Ay D 90 N, WB D 180 N 1.15 m (b) 2.3 m Ay Ay 2.3 m By MA : 2.3By 1.15WB xW D 0 W D 618 N, WB D 180 N For (a), By D 496 N. Solving the equations for this case yields x D 1.511 m For (b), By D 523 N. Solving the equations for this case yields x D 1.611 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 597 Problem 7.137 If a string is tied to the slender bar at A and the bar is allowed to hang freely, what will be the angle between AB and the vertical? B 4 in A 8 in Solution: When the bar hangs freely, the action line of the weight will pass through the mass center. With a homogenous, slender bar, the mass center corresponds to the centroid of the lines making up the bar. Choose the origin at A, with the x axis parallel to the lower bar. Divide the bar into three segments plus the composite: (1) The segment from A to the semi circle, (2) the segment AB, and (3) the semicircle. (1) L1 D 8 in, x1 D 4 in, y1 D 0. (2) L2 D (3) L3 D 4 D 12.566 in, x3 D 8 C p 82 C 82 D 11.314 in, x2 D 4, y2 D 4 24 D 10.546 in, y3 D 4. The composite length LD 3 Li D 31.88 in. 1 The composite centroid: xD L1 x1 C L2 x2 C L3 x3 D 6.58 in, L yD L1 y1 C L2 y2 C L3 y3 D 2.996 in L The angle from the point A to the centroid relative to the lower bar is ˛ D tan1 y x D 24.48° . The angle between AB and the lower bar is 45° , hence the angle between the line from A to the centroid and AB is ˇ D 45 ˛ D 20.52° Since the line from A to the centroid will be vertical, this is the angle between AB and the vertical. 598 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 7.138 When the truck is unloaded, the total reactions at the front and rear wheels are A D 54 kN and B D 36 kN. The density of the load of gravel is D 1600 kg/m3 . The dimension of the load in the z direction is 3 m, and its surface profile, given by the function shown, does not depend on z. What are the total reactions at the front and rear wheels of the loaded truck? y = 1.5 – 0.45x + 0.062x2 x A B 2.8 m 3.6 m 5.2 m Solution: First, find the location of the center of mass of the unloaded truck (and its mass). Then find the center of mass and mass of the load. Combine to find the wheel loads on the loaded truck. Unloaded Truck Fx : no forces Fy : 54000 C 36000 mT g D 0N MA : xT mT g C 5.236 D 0 Now we can find the wheel loads on the loaded truck Fx : no forces Fy : Ay C By mT g mL g D 0 MA : 5.2By xT mT g dL mL g D 0 Solving Ay D 80.7 kN, By D 171.6 kN Solving xT D 2.08 m, mT D 9174 kg Next, find xL and mL (for the load) x dm Num m D xL D L mL dm XT 5.2 m mL dm, Num D mL 0 mL g y 3.6 3y dx D 3 mL D x dm mL 3.6 B 36 kN 54 kN where mL D mTg 2 1.5 0.45x C 0.062x dx y = 1.5 − 0.45x + 0.062x2 dm = p(3)y dx 0 3 3.6 2 x x C 0.062 mL D 3 1.5x 0.45 2 3 0 XL mL D 16551 kg 3.6 Num D 3 1.5x 0.45x2 C 0.062x3 dx 0 3.6 m x 0 2 3 4 3.6 x x x Num D 3 1.5 0.45 C 0.062 2 3 4 0 dL XT Num D 25560 kg Ð m xL D mL g mT g Num D 1.544 m mL measured from the front of the load AY 5.2 m BY The horizontal distance from A to the center of mass of the load is dL D xL C 2.8 m D 4.344 m c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 599 Problem 7.139 The mass of the moon is 0.0123 times the mass of the earth. If the moon’s center of mass is 383,000 km from the center of mass of the earth, what is the distance from the center of mass of the earth to the center of mass of the earth–moon system? Solution: mE mM xmE C mM D 383,000 mM so xD mM 383,000 mE C m M X mM /mE 383,000 D 1 C mM /mE D 383,000 km 0.0123 383,000 1 C 0.0123 D 4650 km. (The earth’s radius is 6370 km, so the center of mass of the earth-moon system is within the earth.) 600 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.1 Use the method described in Active Example 8.1 to determine IY and ky for the rectangular area. Solution: The height of the vertical strip of width dx is 0.6 m, so the area is dA D (0.6 m) dx. We can use this expression to determine Iy . y x 2 dx D (0.6 m) Iy D 0.4 m x 2 dx 0.2m A D (0.6 m) x3 3 0.4 m D 0.0416 m4 0.2 m The radius of gyration about the y axis is 0.6 m ky D Iy D A 0.0416 m4 D 0.416 m (0.4 m) (0.6 m) x 0.2 m Iy D 0.0416 m4 , ky D 0.416 m 0.4 m Problem 8.2 Use the method described in Active Example 8.1 to determine Ix and kx for the rectangular area. y Solution: It was shown in Active Example 8.1 that the moment of inertia about the x axis of a vertical strip of width dx and height f(x) is 1 Ix strip D [fx]3 dx. 3 For the rectangular strip, fx D 0.6 m. Integrating to determine Ix for the rectangular area., 0.4m Ix D 0.2m 1 1 4 0.6 m3 dx D 0.6 m3 [x]0.4m 0.2m D 0.0288 m 3 3 The radius of gyration about the x axis is 0.6 m kx D Ix D A 0.0288 m4 D 0.346 m (0.4 m)(0.6 m) x 0.2 m Ix D 0.0288 m4 , ky D 0.346 m 0.4 m Problem 8.3 In Active Example 8.1, suppose that the triangular area is reoriented as shown. Use integration to determine Iy and ky . Solution: The height of a vertical strip of width dx is h h/bx, so the area dA D h h x b y h dx. We can use this expression to determine Iy : b x 2 dA D Iy D x2 h 0 A h x b dx D h x3 3 x4 4b x b D 0 1 3 hb 12 b The radius of gyration about the y axis is ky D Iy D Iy D A 1 3 hb b 12 D p 1 6 hb 2 hb3 b , ky D p . 12 6 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 601 Problem 8.4 (a) Determine the moment of inertia Iy of the beam’s rectangular cross section about the y axis. Solution: (b) Determine the moment of inertia Iy 0 of the beam’s cross section about the y 0 axis. Using your numerical values, show that Iy D Iy 0 C d2x A, where A is the area of the cross section. (a) 40 mm 0 x 2 dydx D 1.28 ð 106 mm4 0 (b) 60 mm Iy D 20 mm 30 mm Iy 0 D 20 mm x 2 dydx D 3.2 ð 105 mm4 30 mm y Iy D Iy 0 C dx 2 A y dx 1.28 ð 106 mm4 D 3.2 ð 105 mm4 C 20 mm2 [40 mm60 mm] 60 mm x O dy x O 40 mm Problem 8.5 (a) Determine the polar moment of inertia JO of the beam’s rectangular cross section about the origin O. (b) Determine the polar moment of inertia JO0 of the beam’s cross section about the origin O0 . Using your numerical values, show that JO D JO0 C d2x C d2y A, where A is the area of the cross section. Solution: (a) 40 mm 0 (b) 60 mm JO D x 2 C y 2 dydx D 4.16 ð 106 mm4 0 20 mm 30 mm JO 0 D 20 mm x 2 C y 2 dydx D 1.04 ð 106 mm4 30 mm JO D JO0 C dx 2 C dy 2 A (c) 4.16 ð 106 mm4 D 1.04 ð 106 mm4 C [20 mm2 C 30 mm2 ][40 mm60 mm] Problem 8.6 Determine Iy and ky . y Solution: A D 0.3 m1 m C 1 m 0.3 mC0.3x Iy D 0 0.6 m 0.3 m ky D x 1 0.3 m1 m D 0.45 m2 2 0 Iy D A x 2 dydx D 0.175 m4 0.175 m4 D 0.624 m 0.45 m2 1m 602 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.7 Determine JO and kO . Solution: A D 0.3 m1 m C 1m 0.3 mC0.3x JO D 0 1 0.3 m1 m D 0.45 m2 2 x 2 C y 2 dydx D 0.209 m4 0 0.209 m4 D 0.681 m 0.45 m2 kO D Problem 8.8 Determine Ixy . Solution: 1m 0.3 mC0.3x Ixy D 0 xydydx D 0.0638 m4 0 Problem 8.9 Determine Iy . y Solution: The height of a vertical strip of width dx is 2 x2 , so y 2 x2 the area is dA D 2 x2 dx. We can use this expression to determine Iy : x 2 dA D Iy D 1 x 2 2 x2 dx D 0 A 2x3 x5 3 5 1 0 D 0.467. Iy D 0.467. 1 Problem 8.10 Determine Ix . Solution: It was shown in Active Example 8.1 that the moment of inertia about the x axis of a vertical strip of width dx and height fx is 1 Ix strip D [fx]3 dx. 3 x y y 2 x2 In this problem fx D 2 x3 . Integrating to determine Ix for the area, 1 Ix D 0 D D 1 3 1 3 1 2 x2 3 dx 3 1 8 12x2 C 6x4 x 6 dx 0 1 12x3 6x5 x7 8x C D 1.69. 3 5 7 0 1 x Ix D 1.69. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 603 Problem 8.11 Determine JO . y y 2 x2 Solution: See the solutions to Problems 8.9 and 8.10. The polar moment of inertia is JO D Ix C Iy D 1.69 C 0.467 D 2.15. JO D 2.15. 1 Problem 8.12 Determine Ixy . x y y 2 x2 Solution: It was shown in Active Example 8.1 that the product of inertia of a vertical strip of width dx and height f(x) is Ixy strip D 1 [fx]2 xdx. 2 In this problem, fx D 2 x2 . Integrating to determine Ixy for the area, 1 1 2 x2 2 xdx 2 Ixy D 0 D D 1 2 1 4x 4x3 C x 5 x 0 1 4x4 x6 1 4x2 C D 0.583. 2 2 4 6 0 1 x Ixy D 0.583. Problem 8.13 Determine Iy and ky . y y 1 2 x 4x 7 4 Solution: First we need to locate the points where the curve intersects the x axis. 4 š 1 x 2 C 4x 7 D 0 ) x D 4 Now 14 x x2 /4C4x7 dydx D 72 AD 2 0 14 x2 /4C4x7 Iy D 2 ky D 604 p 16 41/47 D 2,14 21/4 0 Iy D A x 2 dydx D 5126 5126 D 8.44 72 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.14 Determine Ix and kx . Solution: See Solution to Problem 8.13 14 x2 /4C4x7 Ix D 2 0 Ix D A kx D y 2 dydx D 1333 1333 D 4.30 72 Problem 8.15 Determine JO and kO . Solution: See Solution to 8.13 and 8.14 JO D Ix C Iy D 1333 C 5126 D 6459 JO 6459 D D 9.47 kO D A 72 Problem 8.16 Determine Ixy . Solution: 14 x2 /4C4x7 xydydx D 2074 Ixy D 2 0 Problem 8.17 Determine Iy and ky . y y 1 2 x 4x 7 4 y5 Solution: First we need to locate the points where the curve intersects the line. 4 š 1 2 x C 4x 7 D 5 ) x D 4 12 p 16 41/412 D 4,12 21/4 x x2 /4C4x7 dydx D 21.33 AD 4 5 12 x2 /4C4x7 Iy D 4 ky D 5 Iy D A x 2 dydx D 1434 1434 D 8.20 21.33 Problem 8.18 Determine Ix and kx . Solution: See Solution to Problem 8.17 12 x2 /4C4x7 Ix D 4 kx D 5 Ix D A y 2 dydx D 953 953 D 6.68 21.33 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 605 Problem 8.19 (a) Determine Iy and ky by letting dA be a vertical strip of width dx. (b) The polar moment of inertia of a circular area with its center at the origin is JO D 12 R4 . Explain how you can use this information to confirm your answer to (a). y x R Solution: The equation of the circle is x2 C y 2 D R2 , from which p y D š pR2 x 2 . The strip dx wide and y long has the elemental area dA D 2 R2 x 2 dx. The area of the semicircle is AD R2 Iy D 2 R x 2 dA D 2 x2 R2 x 2 dx 0 A R R2 xR2 x 2 1/2 R4 x xR2 x 2 3/2 C C sin1 D2 4 8 8 R 0 (b) If the integration were done for a circular area with the center at the origin, the limits of integration for the variable x would be from R to R, doubling the result. Hence, doubling the answer above, Iy D R4 . 4 By symmetry, Ix D Iy , and the polar moment would be JO D 2Iy D D R4 8 ky D R4 , 2 which is indeed the case. Also, since kx D ky by symmetry for the full circular area, R Iy D A 2 kO D Iy Ix C D A A 2 Iy D A JO A as required by the definition. Thus the result checks. Problem 8.20 (a) Determine Ix and kx for the area in Problem 8.19 by letting dA be a horizontal strip of height dy. (b) The polar moment of inertia of a circular area with its center at the origin is JO D 12 R4 . Explain how you can use this information to confirm your answer to (a). Solution: Use the results of the solution to Problem 8.19, A D R2 . The equation for the circle is x2 C y 2 D R2 , from which x D 2 š R2 y 2 . The horizontal strip is from 0 to R, hence the element of area is CR y 2 dA D Ix D A y2 D kx D 606 and JO D 2Ix D R2 y 2 dy R R4 R4 C D 8 2 8 2 R4 . 4 R y R2 yR2 y 2 1/2 R4 yR2 y 2 3/2 C C sin1 D 4 8 8 R R Ix D By symmetry Iy D Ix , R2 y 2 dy. dA D (b) If the area were circular, the strip would be twice as long, and the moment of inertia would be doubled: R4 R4 , 2 which is indeed the result. Since kx D ky by symmetry for the full circular area, the kO D Iy Ix C D A A 2 Ix D A JO A 8 as required by the definition. This checks the answer. R Ix D . A 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.21 Use the procedure described in Example 8.2 to determine the moment of inertia Ix and Iy for the annular ring. y Ro x Ri Solution: We first determine the polar moment of inertia JO by integrating in terms of polar coordinates. Because of symmetry and the relation JO D Ix C Iy , we know that Ix and Iy each equal 12 JO . Integrating as in Example 8.2, the polar moment of inertia for the annular ring is r 2 dA D JO D A Therefore Ix D Iy D Ro r 2 2rdr D Ri 1 Ro4 Ri4 2 1 Ro4 Ri4 4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 607 y Problem 8.22 What are the values of Iy and ky for the elliptical area of the airplane’s wing? y2 x2 21 b a2 Solution: a x 2 dA D Iy D A 0 a y y x 2 dy dx y 0 a Iy D 2 0 5m x2 b1 2 1/2 a [x2 y]0 y dx 0 1/2 x2 x2 b 1 2 dx a a x 2m x 2 dy dx Iy D 2 0 a Iy D 2 x2 + y2 = 1 a2 b2 2b x x2 Iy D 2b 1 0 x2 a2 dx a Rewriting Iy D 2b a a x2 y = b 1– x2 a2 a2 x 2 dx 0 p xa2 x 2 3/2 a2 x a2 x 2 2b Iy D C a 4 8 x a4 C sin1 8 a a 0 D a D 0 p 0 2b aa2 a2 3/2 a3 a2 a2 Iy D C a 4 8 a a4 sin1 8 a x2 b 1 dx a 2b a a a2 x 2 1/2 dx 0 p a x x a2 x 2 a2 C sin1 2 2 a 0 (from the integral tables) C AD2 2b a 0 p a2 a 2b a 0 C sin1 D a 2 2 a 0 0a2 3/2 p a2 0 0 a C sin1 2 2 a 4 0p a2 Ð 0 a2 a4 01 0 C C sin 8 8 a AD ab 2b a2 D a 2 2 2 Evaluating, we get A D 7.85 m2 Iy D a4 2b a 8 2 Finally Iy D A 2a3 b Iy D 8 ky D Evaluating, we get ky D 2.5 m 49.09 7.85 Iy D 49.09 m4 The area of the ellipse (half ellipse) is a x2 b 1 a AD2 0 608 1/2 dy dx 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.23 What are the values of Ix and kx for the elliptical area of the airplane’s wing in Problem 8.22? Solution: y y 2 dA D 2 Ix D a Ix D 2 0 b 3 a y 3 p b yD a y2 x2 — + —2 = 1 a2 b p a2 x2 y 2 dy dx 0 0 A 2b a y = b a2 – x2 a dx 0 b3 2 a x 2 3/2 dx 3a3 Ix D 2 0 2b3 3a3 p 3a2 x a2 x 2 xa2 x 2 3/2 C 4 8 3 4 1 x a sin 8 a C 2b3 Ix D 3 3a 3 a4 8 2 Ix D 2b3 Ð 3a3 Ix D 3ab3 ab3 D 3.8 8 a Evaluating (a D 5, b D 1) 0 p 3a3 0 3 a0 C C a4 4 8 8 2 Ix D p 3a2 Ð 0 a2 0a2 C0 4 8 5 D 1.96 m4 8 From Problem 8.22, the area of the wing is A D 7.85 m2 x a2 x2 a Ix D a kx D Ix D A 1.96 7.85 Problem 8.24 Determine Iy and ky . kx D 0.500 m y Solution: The straight line and curve intersect where x D x2 20. Solving this equation for x, we obtain xD 1š y = x 2 – 20 p 1 C 80 D 4, 5. 2 y=x If we use a vertical strip: the area x dA D [x x2 20] dx. Therefore A D 5 x 2 dA D Iy D x 2 x x 2 C 20 dx 4 x4 x5 20x3 C 4 5 3 5 D 522. The area is y = x2 – 20 A ky D y=x x x 2 C 20 dx x 4 x2 x3 C 20x 2 3 So 5 dA D AD D y 4 Iy D A dA 5 D 122. 4 dx 522 D 2.07. 122 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 609 Problem 8.25 Determine Ix and kx for the area in Problem 8.24. Solution: Let us determine the moment of inertia about the x axis of a vertical strip holding x and dx fixed: y 2 dAs D Ix strip D As x y y 2 dx dy D dx x2 20 y3 3 x y=x y= x2 20 x2 – 20 dAs x dx x 6 C 60x4 C x 3 1200x2 C 8000. 3 D Integrating this value from x D 4 to x D 5 (see the solution to Problem 8.24), we obtain Ix for the entire area: 5 x 1 x 6 C 60x4 C x 3 1200x2 C 8000 dx 3 Ix D 4 dx 7 5 x x4 400x3 8000x D D 10,900. C 4x5 C C 21 12 3 3 4 From the solution to Problem 8.24, A D 122 so kx D Ix D A 10,900 D 9.45. 122 Problem 8.26 A vertical plate of area A is beneath the surface of a stationary body of water. The pressure of the water subjects each element dA of the surface of the plate to a force p0 C y dA, where p0 is the pressure at the surface of the water and is the weight density of the water. Show that the magnitude of the moment about the x axis due to the pressure on the front face of the plate is Mx x A y D p0 yA C Ix , axis where y is the y coordinate of the centroid of A and Ix is the moment of inertia of A about the x axis. Solution: The moment about the x axis is dM D yp0 C y dA integrating over the surface of the plate: p0 C yy dA. MD A Noting that p0 and are constants over the area, y dA C M D p0 y 2 dA. A By definition, y dA A yD A y 2 dA, and Ix D A then M D p0 yA C IX , which demonstrates the result. 610 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.27 Using the procedure described in Active Example 8.3, determine Ix and kx for the composite area by dividing it into rectangles 1 and 2 as shown. Solution: Using results from Appendix B and applying the parallel-axis theorem, the moment on inertia about the x axis for area 1 is Ix 1 D Ix C d2y A D y 1m 1 4m 2 1m 1 (1 m)(3 m)3 C (2.5 m)[(1 m)(3 m)] 12 x 3m D 21.0 m4 The moment of inertia about the x axis for area 2 is Ix 2 D 1 (3 m)(1 m)3 D 1 m4 . 3 The moment of inertia about the x axis for the composite area is Ix D Ix 1 C Ix 2 D 22.0 m4 . The radius of gyration about the x axis is kx D Ix D A 22.0 m4 D 1.91 m 6 m2 Ix D 22.0 m4 , kx D 1.91 m. Problem 8.28 Determine Iy and ky for the composite area by dividing it into rectangles 1 and 2 as shown. Solution: Using results from Appendix B, the moment of inertia about the y axis for area 1 is Iy 1 D 1m 1 4m 1 (3 m)(1 m)3 D 1 m4 . 3 The moment of inertia about the y axis for area 2 is Iy 2 D y 2 1m 3m x 1 (1 m)(3 m)3 D 9 m4 . 3 The moment of inertia about the y axis for the composite area is Iy D Iy 1 C Iy 2 D 10 m4 . The radius of gyration about the y axis is ky D Iy D A 10 m4 D 1.29 m 6 m2 Iy D 10 m4 , kx D 1.29 m. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 611 Problem 8.29 Determine Ix and kx . y Solution: Break into 3 rectangles Ix D 1 1 0.60.23 C 0.20.63 C 0.20.60.52 3 12 0.8 m 1 0.80.23 C 0.80.20.92 D 0.1653 m4 12 C 0.2 m A D 0.20.6 C 0.60.2 C 0.80.2 D 0.4 m2 0.6 m kx D Ix D A 0.1653 m4 D 0.643 m 0.4 m2 0.2 m x Ix D 0.1653 m4 ) 0.2 m kx D 0.643 m 0.6 m Problem 8.30 In Example 8.4, determine Ix and kx for the composite area. y Solution: The area is divided into a rectangular area without the 20 mm cutout (part 1), a semicircular areas without the cutout (part 2), and the circular cutout (part 3). x Using the results from Appendix B, the moment of inertia of part 1 about the x axis is Ix 1 D 1 120 mm80 mm3 D 5.12 ð 106 mm4 , 12 40 mm 120 mm the moment of inertia of part 2 is Ix 2 D 1 40 mm4 D 1.01 ð 106 mm4 , 8 and the moment of inertia of part 3 is Ix 3 D 1 20 mm4 D 1.26 ð 105 mm4 . 4 The moment of inertia of the composite area is Ix D Ix 1 C Ix 2 C Ix 3 D 6.00 ð 106 mm4 . From Example 8.4, the composite area is A D 1.086 ð 104 mm4 , so the radius of gyration about the x axis is kx D Ix D A 6.00 ð 106 mm4 D 23.5 mm. 1.086 ð 104 mm2 Ix D 6.00 ð 106 mm4 , kx D 23.5 mm. 612 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.31 Determine Ix and kx . y 0.8 m 0.2 m x 0.6 m 0.2 m 0.2 m 0.6 m Solution: Break into 3 rectangles — See 8.29 First locate the centroid dD 0.60.20.1 C 0.20.60.5 C 0.80.20.9 D 0.54 m 0.60.2 C 0.20.6 C 0.80.2 1 0.20.63 C 0.20.6d 0.52 12 1 0.60.23 C 0.60.2d 0.12 C 12 1 0.80.23 C 0.80.20.9 d2 D 0.0487 m4 C 12 Ix 0.0487 m4 D kx D D 0.349 m A 0.4 m2 Ix D d Problem 8.32 Determine Iy and ky . Solution: Break into 3 rectangles — See 8.29 Iy D 1 1 1 0.60.23 C 0.20.63 C 0.20.83 12 12 12 D 0.01253 m4 Iy 0.01253 m4 ky D D 0.1770 m D A 0.4 m2 Problem 8.33 Determine JO and kO . Solution: See 8.29, 8.31 and 8.32 JO D Ix C Iy D 0.0612 m4 JO 0.0612 m4 D KO D D 0.391 m A 0.4 m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 613 y Problem 8.34 If you design the beam cross section so that Ix D 6.4 ð 105 mm4 , what are the resulting values of Iy and JO ? h Solution: The area moment of inertia for a triangle about the x base is Ix D 1 12 bh3 , h from which Ix D 2 1 12 60h3 D 10h3 mm4 , 30 mm 30 mm Ix D 10h3 D 6.4 ð 105 mm4 , from which h D 40 mm. Iy D 2 1 12 2h303 D from which Iy D 1 h303 3 1 40303 D 3.6 ð 105 mm4 3 and JO D Ix C Iy D 3.6 ð 105 C 6.4 ð 105 D 1 ð 106 mm4 614 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.35 Determine Iy and ky . 160 mm Solution: Divide the area into three parts: Part (1): The top rectangle. A1 D 16040 D 6.4 ð 103 mm2 , dx1 160 D 80 mm, D 2 Iyy1 D 1 12 40 mm 200 mm 40 mm 40 mm x 120 mm 401603 D 1.3653 ð 107 mm4 . From which Iy1 D d2x1 A1 C Iyy1 D 5.4613 ð 107 mm4 . Part (2): The middle rectangle: A2 D 200 8040 D 4.8 ð 103 mm2 , dx2 D 20 mm, Iyy2 D 1 12 120403 D 6.4 ð 105 mm4 . From which, Iy2 D d2x2 A2 C Iyy2 D 2.56 ð 106 mm4 . Part (3) The bottom rectangle: A3 D 12040 D 4.8 ð 103 mm2 , dx3 D 120 D 60 mm, 2 Iyy3 D 1 12 401203 D 5.76 ð 106 mm4 From which Iy3 D d2X3 A3 C Iyy3 D 2.304 ð 107 mm4 The composite: Iy D Iy1 C Iy2 C Iy3 D 8.0213 ð 107 mm4 ky D Iy D 70.8 mm. A1 C A2 C A3 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 615 Problem 8.36 Determine Ix and kx . Solution: Use the solution to Problem 8.35. Divide the area into from which three parts: Part (1): The top rectangle. A1 D 6.4 ð 103 mm2 , Ix2 D d2y2 A2 C Ixx2 D 5.376 ð 107 mm4 Part (3) The bottom rectangle: A3 D 4.8 ð 103 mm2 , dy1 D 200 20 D 180 mm, Ixx1 D 1 12 dy3 D 20 mm, 160403 D 8.533 ð 105 mm4 . Ixx3 D From which Ix1 D d2y1 A1 C Ixx1 D 2.082 ð 108 mm4 Part (2): The middle rectangle: A2 D 4.8 ð 103 mm2 , dy2 120 C 40 D 100 mm, D 2 Ixx2 D 1 12 1 12 120403 D 6.4 ð 105 mm4 and Ix3 D d2y3 A3 C Ixx3 D 2.56 ð 106 mm4 . The composite: Ix D Ix1 C Ix2 C Ix3 D 2.645 ð 108 mm4 kx D Ix D 128.6 mm A1 C A2 C A3 401203 D 5.76 ð 106 mm4 Problem 8.37 Determine Ixy . Solution: (See figure in Problem 8.35). Use the solutions in from which Problems 8.35 and 8.36. Divide the area into three parts: Part (1): A1 D 16040 D 6.4 ð 103 mm2 , dx1 160 D 80 mm, D 2 dy1 D 200 20 D 180 mm, Ixy2 D dx2 dy2 A2 D 9.6 ð 106 mm4 . Part (3): A3 D 12040 D 4.8 ð 103 mm2 , dx3 D 120 D 60 mm, 2 dy3 D 20 mm, Ixxyy1 D 0, from which from which Ixy1 D dx1 dy1 A1 C Ixxyy1 D 9.216 ð 107 mm4 . Ixy3 D dx3 dy3 A3 D 5.76 ð 106 . The composite: Part (2) A2 D 200 8040 D 4.8 ð 103 mm2 , Ixy D Ixy1 C Ixy2 C Ixy3 D 1.0752 ð 108 mm4 dx2 D 20 mm, dy2 D 616 120 C 40 D 100 mm, 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.38 Determine Ix and kx . y Solution: The strategy is to use the relationship Ix D d2 A C Ixc , 160 mm where Ixc is the area moment of inertia about the centroid. From this Ixc D d2 A C Ix . Use the solutions to Problems 8.35, 8.36, and 8.37. Divide the area into three parts and locate the centroid relative to the coordinate system in the Problems 8.35, 8.36, and 8.37. 40 mm x 200 mm 40 mm Part (1) A1 D 6.4 ð 103 mm2 , 40 mm 120 mm dy1 D 200 20 D 180 mm. Part (2) A2 D 200 8040 D 4.8 ð 103 mm2 , dx1 D 160 D 80 mm, 2 dy2 D 120 C 40 D 100 mm, 2 dx2 D 20 mm, Part (3) A3 D 12040 D 4.8 ð 103 mm2 , dx3 D 120 D 60 mm, 2 dy3 D 20 mm. The centroid coordinates are xD A1 dx1 C A2 dx2 C A3 dx3 D 56 mm, A yD A1 dy1 C A2 dy2 C A3 dy3 D 108 mm A from which Ixc D y2 A C Ix D 1.866 ð 108 C 2.645 ð 108 D 7.788 ð 107 mm4 The total area is A D A1 C A2 C A3 D 1.6 ð 104 mm2 . kxc D Ixc D 69.77 mm A Problem 8.39 Determine Iy and ky . Solution: The strategy is to use the relationship Iy D d2 A C Iyc , where Iyc is the area moment of inertia about the centroid. From this Iyc D d2 A C Iy . Use the solution to Problem 8.38. The centroid coordinates are x D 56 mm, y D 108 mm, from which Iyc D x2 A C Iy D 5.0176 ð 107 C 8.0213 ð 107 D 3.0 ð 107 mm4 , kyc D Iyc D 43.33 mm A Problem 8.40 Determine Ixy . Solution: Use the solution to Problem 8.37. The centroid coordinates are x D 56 mm, y D 108 mm, from which Ixyc D xyA C Ixy D 9.6768 ð 107 C 1.0752 ð 108 D 1.0752 ð 107 mm4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 617 Problem 8.41 Determine Ix and kx . Solution: Divide the area into two parts: y 3 ft 4 ft Part (1): a triangle and Part (2): a rectangle. The area moment of inertia for a triangle about the base is Ix D 1 12 3 ft x bh3 . The area moment of inertia about the base for a rectangle is Ix D 1 bh3 . 3 Part (1) Ix1 D Part (2) Ix2 D 1 12 433 D 9 ft2 . 1 333 D 27. 3 The composite: Ix D Ix1 C Ix2 D 36 ft4 . The area: AD 1 43 C 33 D 15 ft2 . 2 kx D Ix D 1.549 ft. A Problem 8.42 Determine JO and kO . Solution: (See Figure in Problem 8.41.) Use the solution to from which Problem 8.41. Part (1): The area moment of inertia about the centroidal axis parallel to the base for a triangle is Iyc D 1 36 bh3 D 1 36 343 D 5.3333 ft4 , from which Iy1 2 8 D A1 C Iyc D 48 ft4 . 3 Iy2 D 5.52 A2 C Iyc D 279 ft4 , where A2 D 9 ft2 . The composite: Iy D Iy1 C Iy2 D 327 ft4 , from which, using a result from Problem 8.41, JO D Ix C Iy D 327 C 36 D 363 ft4 and kO D JO D 4.92 ft A where A1 D 6 ft2 . Part (2): The area moment of inertia about a centroid parallel to the base for a rectangle is Iyc D 618 1 12 bh3 D 1 12 333 D 6.75 ft4 , c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.43 Determine Ixy . Solution: (See Figure in Problem 8.41.) Use the results of the solutions to Problems 8.41 and 8.42. The area cross product of the moment of inertia about centroidal axes parallel to the bases for a 1 2 2 b h , and for a rectangle it is zero. Therefore: triangle is Ix0 y 0 D 72 Ixy1 D 1 72 42 32 C 8 3 A1 D 18 ft4 3 3 and Ixy2 D 1.55.5A2 D 74.25 ft4 , Ixy D Ix0 y 0 1 C Ixy2 D 92.25 ft4 y Problem 8.44 Determine Ix and kx . Solution: Use the results of Problems 8.41, 8.42, and 8.43. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. The centroidal coordinate yD A1 1 C A2 1.5 D 1.3 ft. A 4 ft 3 ft 3 ft x From which Ixc D y2 A C Ix D 10.65 ft4 and kxc D Ixc D 0.843 ft A Problem 8.45 Determine JO and kO . Solution: Use the results of Problems 8.41, 8.42, and 8.43. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. The centroidal coordinate: A1 xD 8 C A2 5.5 3 D 4.3667 ft, A from which IYC D x2 A C IY D 40.98 ft4 . Using a result from Problem 8.44, JO D IXC C IYC D 10.65 C 40.98 D 51.63 ft4 and kO D JO D 1.855 ft A Problem 8.46 Determine Ixy . Solution: Use the results of Problems 8.41–8.45. The strategy is to use the parallel axis theorem and solve for the area moment of inertia about the centroidal axis. Using the centroidal coordinates determined in Problems 8.44 and 8.45, Ixy D xyA C Ixy D 85.15 C 92.25 D 7.1 ft4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 619 Problem 8.47 Determine Ix and kx . y 120 mm 20 mm 80 mm x 40 mm 80 mm Solution: Let Part 1 be the entire rectangular solid without the y hole and let part 2 be the hole. m 20 m Area D hb R2 D 80120 40 mm R2 y′ Ix1 D 13 bh3 where b D 80 mm h D 120 mm Ix1 D 13 801203 D 4.608 ð 107 mm4 40 mm 120 mm Part 2 x′ For Part 2, Ix0 2 D 14 R4 D 14 204 mm4 Ix0 2 D 1.257 ð 105 dy = 80 mm mm4 Ix2 D Ix0 2 C d2y A Part 1 x where A D R2 D 1257 mm2 80 mm d D 80 mm Ix2 D 1.257 ð 105 C 202 802 Ix2 D 0.126 ð 106 C 8.042 ð 106 mm4 D 8.168 ð 106 mm4 D 0.817 ð 107 mm4 Ix D Ix1 Ix2 D 3.79 ð 107 mm4 Area D 8343 mm2 kx D 620 Ix D 67.4 mm Area c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.48 Determine JO and kO . Solution: For the rectangle, y 40 mm JO1 D Ix1 C Iy1 D 13 bh3 C 13 hb3 JO1 D 4.608 ð 107 C 2.048 ð 107 mm4 y′ R = 20 mm JO1 D 6.656 ð 107 mm4 x′ A1 D bh D 9600 mm2 120 mm A2 For the circular cutout about x0 y 0 J0O2 D Ix0 2 C Iy 0 D 14 R4 C 14 R4 2 (h) 80 mm A1 J0O2 D 1.257 ð 105 C 1.257 ð 105 mm4 J0O2 D 2.513 ð 105 mm2 x Using the parallel axis theorem to determine JO2 (about x, y) 80 mm (b) JO2 D J002 C d2x C d2y A2 A2 D R2 D 1257 mm2 JO2 D 1.030 ð 107 mm4 JO D 5.63 ð 107 mm4 kO D JO D JO1 JO2 JO D Area JO A1 A2 kO D 82.1 mm JO D 6.656 ð 107 1.030 ð 107 mm4 Problem 8.49 Determine Ixy . Solution: A1 D 80120 D 9600 y 80 mm mm2 A2 D R2 D 202 D 1257 mm2 R = 20 mm y′ For the rectangle A1 x′ Ixy1 D 14 b2 h2 D 14 802 1202 Ixy1 D 2.304 ð 107 mm2 For the cutout A2 A1 120 mm A2 dy = 80 mm A1 Ix 0 y 0 2 D 0 and by the parallel axis theorem x dx = 40 mm Ixy2 D Ix0 y 0 2 C A2 dx dy Ixy2 D 0 C 12574080 Ixy2 D 4.021 ð 106 mm4 Ixy D Ixy1 Ixy2 Ixy D 2.304 ð 107 0.402 ð 107 mm4 Ixy D 1.90 ð 107 mm4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 621 y Problem 8.50 Determine Ix and kx . 20 mm 120 mm x 80 mm 40 mm 80 mm Solution: We must first find the location of the centroid of the total y area. Let us use the coordinates XY to do this. Let A1 be the rectangle and A2 be the circular cutout. Note that by symmetry Xc D 40 mm Y 80 mm Rectangle1 Circle2 Area Xc Yc 9600 mm2 40 mm 60 mm mm2 40 mm 80 mm 1257 R = 20 mm 120 mm A1 D 9600 mm2 A2 D 1257 mm2 x 80 mm For the composite, Xc D A1 Xc1 A2 Xc2 D 40 mm A1 A2 Yc D A1 Yc1 A2 Yc2 D 57.0 mm A1 A2 Now let us determine Ix and kx about the centroid of the composite body. Rectangle about its centroid (40, 60) mm Ix1 D 1 3 1 bh D 801203 12 12 Ix1 D 1.152 ð 107 mm3 , X 40 mm 40 mm Now to C ! dy2 D 80 57 D 23 mm Ixc2 D Ix2 C dy2 2 A2 Ixc2 D 7.91 ð 105 mm4 For the composite about the centroid Ix D Ixc1 Ixc2 Now to C Ix D 1.08 ð 107 mm4 Ixc1 D Ix1 C 60 Yc 2 A1 The composite Area D 9600 1257 mm2 Ixc1 D 1.161 ð 107 mm4 D 8343 mm2 Circular cut out about its centroid A2 D R2 D 202 D 1257 mm2 kx D Ix D 36.0 mm A Ix2 D 14 R4 D 204 /4 Ix2 D 1.26 ð 105 mm4 622 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.51 Determine Iy and ky . Solution: From the solution to Problem 8.50, the centroid of the y composite area is located at (40, 57.0) mm. The area of the rectangle, A1 , is 9600 mm2 . The area of the cutout, A2 , is 1257 mm2 . The area of the composite is 8343 mm2 . (1) Rectangle about its centroid (40, 60) mm. Iy1 D 1 3 1 hb D 120803 12 12 + x 80 mm Iy1 D 5.12 ð 106 mm4 dx1 D 0 (2) 40 mm Circular cutout about its centroid (40, 80) 80 mm Iy2 D R4 /4 D 1.26 ð 105 mm4 dx2 D 0 Since dx1 and dx2 are zero. (no translation of axes in the xdirection), we get Iy D Iy1 Iy2 Iy D 4.99 ð 106 mm4 Finally, ky D Iy D A1 A2 4.99 ð 106 8343 ky D 24.5 mm Problem 8.52 Determine JO and kO . y Solution: From the solutions to Problems 8.51 and 8.52, Ix D 1.07 ð 107 mm4 Iy D 4.99 ð 106 mm4 and A D 8343 mm2 20 mm JO D Ix C Iy D 1.57 ð 107 mm4 kO D JO D 43.4 mm A x 120 mm 80 mm c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 623 y Problem 8.53 Determine Iy and ky . 12 in x 20 in Solution: Treat the area as a circular area with a half-circular y y y cutout: From Appendix B, 1 Iy 1 D 14 204 in4 and Iy 2 D 18 124 in4 , 12 in. 20 in. 2 x x 2 in. x 20 in. so Iy D 14 204 18 124 D 1.18 ð 105 in4 . The area is A D 202 12 122 D 1030 in2 so, ky D Iy D A 1.18 ð 105 1.03 ð 103 D 10.7 in Problem 8.54 Determine JO and kO . Solution: Treating the area as a circular area with a half-circular cutout as shown in the solution of Problem 8.53, from Appendix B, JO 1 D Ix 1 C Iy 1 D 12 204 in4 and JO 2 D Ix 2 C Iy 2 D 14 124 in4 . Therefore JO D 12 204 14 124 D 2.35 ð 105 in4 . From the solution of Problem 8.53, A D 1030 in2 Ro D D 624 JO A 2.35 ð 105 D 15.1 in. 1.03 ð 103 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.55 Determine Iy and ky if h D 3 m. y Solution: Break the composite into two parts, a rectangle and a semi-circle. 1.2 m For the semi-circle Ix 0 c D 9 8 8 R4 h Iy 0 c D 1 4 4R R d D 8 3 y′ x y d x′ 1.2 m AC d = 4R 3π AR To get moments about the x and y axes, the dxc , dyc for the semicircle are dxc D 0, dyc D 3 m C 3m = h 4R 3 x and Ac D R2 /2 D 2.26 m2 Iy 0 c D To get moments of area about the x, y axes, dxR D 0, dyR D 1.5 m 1 4 R 8 and Iyc D Iy 0 c C d2xc A 0 IyR D Iy 0 R C dxR 2 bh dx D 0 1 32, 43 m4 12 Iyc D Iy 0 c D 1.24 /8 IyR D Iy 0 R D Iyc D 0.814 m4 IyR D 3.456 m2 For the Rectangle Ix 0 R D 1 3 bh 12 Iy 0 R D 1 3 hb 12 AR D bh D 7.2 m2 Iy D Iyc C IyR Iy D 4.27 m2 To find ky , we need the total area, A D AR C Ac AR D bh A D 7.20 C 2.26 m2 y′ y A D 9.46 m2 2.4 m ky D h 3m b Iy D 0.672 m A x′ x c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 625 Problem 8.56 Determine Ix and kx if h D 3 m. Solution: Break the composite into two parts, the semi-circle and the rectangle. From the solution to Problem 8.55, Ix 0 c D 9 8 8 dyc D 3C 4R 3 y Ac R4 m Ac D 2.26 m2 R = 1.2 m h=3m b = 2.4 m AR 3m h Ixc D Ix0 c C Ac d2yc Substituting in numbers, we get dyc D 3.509 m x 2.4 m Ix0 c D 0.0717 m4 yc′ and Ixc D Ix0 c C Ac d2y Ixc D 27.928 m2 xc′ R For the Rectangle h D 3 m, b D 2.4 m Area: AR D bh D 7.20 m2 Ix 0 R D 1 3 bh , dyR D 1.5 m 12 4R 3π IxR D Ix0 R C d2yR AR Substituting, we get Ix0 R D 5.40 m4 IxR D 21.6 m4 For the composite, Ix D IxR C Ixc Ix D 49.5 m4 Also kx D Ix D 2.29 m AR C Ac kx D 2.29 m 626 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.57 If Iy D 5 m4 , what is the dimension of h? Solution: From the solution to Problem 8.55, we have: y c′ y xc′ For the semicircle Iy 0 c D Iy D 1.24/8 D 0.814 m2 h For the rectangle Iy 0 R D IyR D 1.2 y ′R m h x ′R b 1 h2.43 m4 12 2.4 m x Also, we know IyR C Iyc D 5 m4 . Hence 0.814 C 1 h2.43 D 5 12 Solving, h D 3.63 m Problem 8.58 Determine Iy and ky . Solution: Let the area be divided into parts as shown. The areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, A3 D 1 302 D 707 in2 , 4 x 3 D 20 C 430 D 32.7 in, 3 y 3 D 40 C 430 D 52.7 in. 3 Using the results from Appendix B, the moments of inertia of the parts about the y axis are Iy 1 D 1 40 in50 in3 D 167 ð 104 in4 , 3 Iy 2 D 1 30 in20 in3 D 8.00 ð 104 in4 , 3 Iy 3 D 4 16 9 " 430 in 2 ! 30 in4 C 20 in C 30 in2 3 4 D 80.2 ð 104 in4 . The moment of inertia of the composite area about the y axis is Iy D Iy 1 C Iy 2 C Iy 3 D 2.55 ð 106 in4 . 2 The composite area is A D A1 C A2 C A3 D 3310 in . Iy D 27.8 in. The radius of gyration about the y axis is ky D A Iy D 2.55 ð 106 in4 , ky D 27.8 in. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 627 Problem 8.59 Determine Ix and kx . Solution: See the solution to Problem 8.58. Let the area be divided into parts as shown. The areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 A3 D 1 302 D 707 in2 , 4 y 3 D 40 C 430 D 52.7 in. 3 Using the results from Appendix B, the moments of inertia of the parts about the x axis are Ix 1 D 1 50 in40 in3 D 1.07 ð 106 in4 , 3 Ix 2 D 1 20 in30 in3 C 55 in2 600 in2 D 1.86 ð 106 in4 , 12 Ix 3 D 4 4 9 " 430 in 2 ! 30 in4 C 40 in C 30 in2 D 2.01 ð 106 in4 . 3 4 The moment of inertia of the composite area about the x axis is Ix D Ix 1 C Ix 2 C Ix 3 D 4.94 ð 106 in4 . 2 The composite area is A D A1 C A2 C A3 D 3310 in . Ix D 38.6 in. The radius of gyration about the y axis is kx D A Ix D 4.94 ð 106 in4 , kx D 38.6 in 628 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.60 Determine Ixy . Solution: See the solution to Problem 8.58. Let the area be divided into parts as shown. The areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 A3 D 1 302 D 707 in2 , 4 y 3 D 40 C 430 D 52.7 in. 3 Using the results from Appendix B, the products of inertia of the parts about are Ixy 1 D 1 50 in2 40 in2 D 10.0 ð 105 in4 , 4 Ixy 2 D 10 in55 in600 in2 D 3.30 ð 105 in4 , Ixy 3 D 4 1 8 9 430 in 430 in 30 in4 C 20 in C 40 in C [707 in2 ] 3 3 D 12.1 ð 105 in4 . The product of inertia of the composite area is Ixy D Ixy 1 C Ixy 2 C Ixy 3 D 2.54 ð 106 in4 . Ixy D 2.54 ð 106 in4 . y Problem 8.61 Determine Iy and ky . Solution: See the solution to Problem 8.58. In terms of the coordinate system used in Problem 8.58, the areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , x 1 D 25 in, y 1 D 20 in, A2 D 2030 D 600 in2 , x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 A3 D 1 302 D 707 in2 , 4 30 in 40 in x 430 D 52.7 in. y 3 D 40 C 3 20 in The composite area is A D A1 C A2 C A3 D 3310 in2 . x 1 A1 C x 2 A2 C x 3 A3 The x coordinate of its centroid is x D D 23.9 in A The moment of inertia about the y axis in terms of the coordinate system used in Problem 8.58 is Iy D 2.55 ð 106 in4 . Applying the parallel axis theorem, the moment of inertia about the y axis through the centroid of the area is Iy D 2.55 ð 106 in4 23.9 in2 3310 in2 D 6.55 ð 105 in4 . The radius of gyration about the y axis is ky D Iy D 14.1 in. A Iy D 6.55 ð 105 in4 , ky D 14.1 in. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 629 y Problem 8.62 Determine Ix and kx . Solution: See the solution to Problem 8.59. In terms of the coordinate system used in Problem 8.59, the areas and the coordinates of their centroids are A1 D 4050 D 2000 in2 , A2 D 2030 D 600 in2 , A3 D 1 302 D 707 in2 , 4 x 1 D 25 in, y 1 D 20 in, x 2 D 10 in, y 2 D 55 in, x 3 D 20 C 430 D 32.7 in, 3 30 in 40 in x 430 D 52.7 in. 3 y 3 D 40 C 20 in 2 The composite area is A D A1 C A2 C A3 D 3310 in . y A1 C y 2 A2 C y 3 A3 The x coordinate of its centroid is y D 1 D 33.3 in A The moment of inertia about the x axis in terms of the coordinate system used in Problem 8.59 is Ix D 4.94 ð 106 in4 . Applying the parallel axis theorem, the moment of inertia about the x axis through the centroid of the area is Ix D 4.94 ð 106 in4 33.3 in2 3310 in2 D 1.26 ð 106 in4 . The radius of gyration about the x axis is kx D Ix D 19.5 in. A Ix D 1.26 ð 106 in4 , kx D 19.5 in. Problem 8.63 Determine Ixy . Solution: See the solution to Problem 8.60. Ixy D [0 C 1.0 m0.8 mdx 0.5 mdy 0.4 m] [0 C 0.2 m2 dx 0.4 mdy 0.3 m] C 1 1 0.8 m2 0.6 m2 0.8 m0.6 m0.2 m 24 2 C Solving: 0.8 m 3 0.8 m 1 0.8 m0.6 mdx 1.2 m dy 2 3 Ixy D 0.0230 m4 Check using the noncentroidal product of inertia from Problem 8.60 we have Ixy D Ixy 0 Adx dy D 0.2185 m4 0.914 m2 0.697 m0.379 m D 0.0230 m4 630 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.64 Determine Iy and ky . y 18 in x 6 in 6 in Solution: Divide the area into three parts: The composite area: Part (1) The rectangle 18 by 18 inches; Part (2) The triangle with base 6 in and altitude 18 in; Part (3) The semicircle of 9 in radius. AD 3 # 6 in Ai D 397.23 in2 . 1 Part (1): A1 D 1818 D 324 in2 , x1 D 9 in, Iy D x21 A1 C Iyy1 x22 A2 Iyy2 C x23 A3 C Iyy3 , y1 D 9 in, Ixx1 D Iyy1 D 1 12 1 12 The area moment of inertia: Iy D 4.347 ð 104 in4 , 18183 D 8748 in4 , ky D Iy D 10.461 in A 18183 D 8748 in4 . 1 186 D 54 in2 , 2 Part (2): A2 D x2 D 9 in, y2 D 1 18 D 6 in, 3 Ixx2 D 1 36 6183 D 972 in4 , Iyy2 D 1/181833 D 27 in4 . Part (3) A3 D 92 D 127.23 in2 , 2 x3 D 9 in, y3 D 18 C Ixx3 D Iyy3 D 49 3 D 21.82 in, 1 49 2 A3 D 720.1 in4 , 94 8 3 1 94 D 2576.5 in4 . 8 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 631 Problem 8.65 Determine Ix and kx . Solution: Use the results of the solution to Problem 8.64. IX D y21 A1 C IXX1 y22 A2 IXX2 C y23 A3 C IXX3 , Ix D 9.338 ð 104 in4 , kx D Ix D 15.33 in A Problem 8.66 Determine Ixy . Solution: Use the results of the solutions to Problems 8.63 and 8.64. Ixy D x1 y1 A1 x2 y2 A2 C x3 y3 A3 Ixy D 4.8313 ð 104 in4 Problem 8.67 Determine Iy and ky . y Solution: We divide the composite area into a triangle (1), rect- 6 in angle (2), half-circle (3), and circular cutout (4): 2 in Triangle: Iy 1 D 14 1283 D 1536 in4 x Rectangle: Iy 2 D 8 in 1 1283 C 122 812 D 14,336 in4 . 12 8 in y Half-Circle: Iy 3 D 8 8 9 2 46 2 1 62 D 19,593 in4 64 C 16 C 3 2 3 4 1 x Circular cutout: Iy 4 D 14 24 C 162 22 D 3230 in4 . y 8 in. y Therefore Iy D Iy 1 C Iy 2 C Iy 3 Iy 4 D 3.224 ð 104 in4 . 12 in. 2 1 12 in. The area is x A D A1 C A2 C A3 A4 D 1 1 128 C 812 C 62 22 D 188 in2 , 2 2 so ky D Iy D A x 8 in. 12 in. y y 3 6 in. 4 3.224 ð 104 D 13.1 in. 188 6 in. x 16 + 4(6) in. 3π 632 2 in. x 16 in. c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.68 Determine JO and kO . Solution: Iy is determined in the solution to Problem 8.67. We will determine Ix and use the relation JO D Ix C Iy . Using the figures in the solution to Problem 8.67, Triangle: Ix 1 D 1 8123 D 1152 in4 . 12 Rectangle: Ix 2 D 13 8123 D 4608 in4 . Half Circle: Ix 3 D 18 64 C 62 21 62 D 2545 in4 . Circular Cutout: Ix 4 D 14 24 C 62 22 D 465 in4 . Therefore Ix D Ix 1 C Ix 2 C Ix 3 Ix 4 D 7840 in4 . Using the solution of Problem 8.67, JO D Ix C Iy D 0.784 ð 104 C 3.224 ð 104 D 4.01 ð 104 in4 . From the solution of Problem 8.67, A D 188 in2 , so R0 D JO D A 4.01 ð 104 D 14.6 in. 188 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 633 Problem 8.69 Determine Iy and ky . y 4 in Solution: Divide the area into four parts: Part (1) The rectangle 8 in by 16 in. Part (2): The rectangle 4 in by 8 in. Part (3) The semicircle of radius 4 in, and Part (4) The circle of radius 2 in. Part (1): A1 D 168 D 128 in2 , x1 D 8 in, 2 in 4 in 8 in y1 D 4 in, Ixx1 D Iyy1 D 1 12 1 12 x 1683 D 682.67 in4 , 16 in 8163 D 2730.7 in4 . Part (2): A2 D 48 D 32 in2 , Iyy2 D 1 12 1 12 843 D 42.667 in4 , AD 42 D 25.133 in2 , 2 Ixx3 D 24 D 12.566 in4 , 3 # Ai A4 D 172.566 in2 . 1 The area moment of inertia: Iy D x21 A1 C Iyy1 C x22 A2 C Iyy2 C x23 A3 C Iyy3 x24 A4 Iyy4 44 3 D 13.698 in. The area moments of inertia about the centroid of the semicircle are Iyy3 D 4 Iyy4 D Ixx4 D 12.566 in4 . 483 D 170.667 in4 . x3 D 12 in. $1% The composite area: Part (3): A3 D y3 D 12 C x4 D 12 in, Ixx4 D y2 D 10 in, Part (4): A4 D 22 D 12.566 in2 , y4 D 12 in, x2 D 12 in, Ixx2 D 12 in 1 44 D 100.53 in4 , 8 Iy D 1.76 ð 104 in4 , ky D Iy D 10.1 in A 1 44 2 A3 D 28.1 in4 . 44 8 3 Check: Ixx3 D 0.1098R4 D 28.1 in4 . check. Problem 8.70 Determine Ix and kx . Solution: Use the results in the solution to Problem 8.69. Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 y24 A4 Ixx4 Ix D 8.89 ð 103 in4 kx D 634 Ix D 7.18 in A c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.71 Determine Ixy . Solution: Use the results in the solution to Problem 8.69. Ixy D x1 y1 A1 C x2 y2 A2 C x3 y3 A3 x4 y4 A4 , Ixy D 1.0257 ð 104 in4 Problem 8.72 Determine Iy and ky . y 4 in 2 in 4 in Solution: Use the results in the solutions to Problems 8.69 to 8.71. The centroid is xD x1 A1 C x2 A2 C x3 A3 x4 A4 A 1024 C 384 C 301.6 150.8 D 9.033 in, D 172.567 x 8 in 12 in 16 in from which Iyc D x2 A C Iy D 1.408 ð 104 C 1.7598 ð 104 D 3518.2 in4 kyc D Iyc D 4.52 in A Problem 8.73 Determine Ix and kx . Solution: Use the results in the solutions to Problems 8.69 to 8.71. The centroid is yD y1 A1 C y2 A2 C y3 A3 y4 A4 D 5.942 in, A from which Ixc D y2 A C Ix D 6092.9 C 8894 D 2801 in4 kxc D Ixc D 4.03 in A Problem 8.74 Determine Ixy . Solution: Use the results in the solutions to Problems 8.69–8.71. Ixyc D xyA C Ixy D 9.263 ð 103 C 1.0257 ð 104 D 994.5 in4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 635 Problem 8.75 Determine Iy and ky . y Solution: We divide the area into parts as shown: Iy 1 D 5 mm 1 50 C 15 C 15303 D 180,000 mm4 12 Iy 2 D Iy 3 D Iy 4 D 15 mm 50 mm 1 30103 C 202 1030 12 5 mm 5 mm D 122,500 mm4 Iy 5 D Iy 6 D Iy 7 D 8 8 9 x 15 mm 15 mm 154 10 15 15 10 mm mm mm mm 415 2 1 152 D 353,274 mm4 C 25 C 3 2 y 1 Iy 8 D Iy 9 D Iy 10 2 5 8 1 D 54 C 252 52 D 49,578 mm4 . 4 Therefore, 50 mm 7 4 10 Iy D Iy 1 C 3Iy 2 C 3Iy 5 3Iy 8 D 1.46 ð 106 3 6 9 mm4 . x 10 15 mm mm The area is A D A1 C 3A2 C 3A5 3A8 D 3080 C 31030 C 3 1 152 352 2 D 4125 mm2 so ky D Iy D A 1.46 ð 106 D 18.8 mm 4125 Problem 8.76 Determine JO and kO . Solution: Iy is determined in the solution to Problem 8.75. We will determine Ix and use the relation JO D Ix C Iy . Dividing the area as shown in the solution to Problem 8.75, we obtain Ix 1 D 1 30803 C 252 3080 D 2, 780, 000 mm4 12 Ix 2 D 1 10303 C 502 1030 D 772, 500 mm4 12 Ix 3 D Ix 4 D Ix 5 D 1 10303 D 22, 500 mm4 12 Therefore Ix D Ix 1 C Ix 2 C 2Ix 3 C Ix 5 C 2Ix 6 Ix 8 2Ix 9 D 4.34 ð 106 mm4 and JO D Ix C Iy D 5.80 ð 106 mm4 . From the solution to Problem 8.75, A D 4125 mm2 so kO D 1 1 154 C 502 152 D 903,453 mm4 8 2 D Ix 6 D Ix 7 D Ix 8 D 5.80 ð 106 4125 D 37.5 mm. 1 54 C 52 502 , 4 Ix 9 D Ix 10 D 636 1 154 D 19,880 mm4 , 8 JO A 1 54 D 491 mm4 . 4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.77 Determine Ix and Iy for the beam’s cross section. 5 in 2 in Solution: Use the symmetry of the object 1 1 Ix D 3 in.8 in3 C 3 in3 in3 C 3 in2 11.5 in2 2 3 12 C 5 in4 5 in2 16 4 C 4[5 in] 3 2 5 in2 4 8 in 4[5 in] 2 8 in C 3 x 3 in in4 2 16 in2 2 4 C 4[2 in] 3 5 in 3 in 5 in 2 2 in2 4 4[2 in] 2 8 in C 3 Solving we find Ix D 7016 in4 Iy 1 1 D 3 in3 in3 C 8 in3 in3 C 8 in3 in6.5 in2 2 3 12 C 5 in4 5 in2 16 4 4[5 in] 3 2 5 in2 C 4 2 in2 2 in4 16 4 C 4[2 in] 3 4[5 in] 3 in C 3 2 2 2 in2 4 3 in C 4[2 in] 3 2 Solving we find Iy D 3122 in4 y Problem 8.78 Determine Ix and Iy for the beam’s cross section. 5 in 2 in x Solution: Use Solution 8.77 and 7.39. From Problem 7.39 we know that y D 7.48 in, A D 98.987 in2 8 in Ix D 7016 in4 Ay 2 D 1471 in4 Iy D 3122 in4 A02 D 3122 in4 3 in 5 in 5 in 3 in c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 637 Problem 8.79 The area A D 2 ð 104 mm2 . Its moment of inertia about the y axis is Iy D 3.2 ð 108 mm4 . Determine its moment of inertia about the yO axis. ŷ y Solution: Use the parallel axis theorem. The moment of inertia A about the centroid of the figure is Iyc D x 2 A C Iy D 1202 2 ð 104 C 3.2 ð 108 D 3.20 ð 107 mm4 . x, x̂ The moment of inertia about the yO axis is 100 mm 120 mm IyO D x2 A C Iyc IyO D 2202 2 ð 104 C 3.2 ð 107 D 1 ð 109 mm4 Problem 8.80 The area A D 100 in2 and it is symmetric about the x 0 axis. The moments of inertia Ix0 D 420 in4 , Iy 0 D 580 in4 , JO D 11000 in4 , and Ixy D 4800 in4 . What are Ix and Iy ? y y' A x' O' O x Solution: The basic relationships: (1) (2) (3) (4) (5) (6) Ix D y 2 A C Ixc , Iy D x2 A C Iyc , JO D Ar 2 C Jc , JO D Ix C Iy , Jc D Ixc C Iyc , and Ixy D Axy C Ixyc , where the subscript c applies to the primed axes, and the others to the unprimed axes. The x, y values are the displacement of the primed axes from the unprimed axes. The steps in the demonstration are: From symmetry about the xc axis, the product of inertia Ixyc D 0. JO J c D 100 in2 , from which r 2 D x2 C From (3): r 2 D A y 2 D 100 in2 Ixy (iii) From (6) and Ixyc D 0, y D , from which x2 r 2 D x4 C Ax 2 Ixy . From which: x4 100x2 C 2304 D 0. A (iv) The roots: x12 D 64, and x22 D 36. The corresponding values of y p are found from y D r 2 x 2 from which x1 , y1 D 8, 6, and x2 , y2 D 6, 8. (v) Substitute these pairs to obtain the possible values of the area moments of inertia: (i) (ii) Ix1 D Ay12 C Ixc D 4020 in4 , Iy1 D Ax12 C Iyc D 6980 in4 Ix2 D Ay22 C Ixc D 6820 in4 , Iy2 D Ax22 C Iyc D 4180 in4 638 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.81 Determine the moment of inertia of the beam cross section about the x axis. Compare your result with the moment of inertia of a solid square cross section of equal area. (See Example 8.5.) y 20 mm x Solution: We first need to find the location of the centroid of the composite. Break the area into two parts. Use X, Y coords. 160 mm y 2 100 20 A1 = 2000 mm2 XC = 0 160 mm 1 A2 = 3200 mm2 XC2 = 0 20 mm YC2 = 80 mm 100 mm 1 YC1 = 170 mm x 100 mm y Y 20 mm Xc D Xc1 A1 C Xc2 A2 D0 A1 C A2 Yc D Yc1 A1 C Yc2 A2 A1 C A2 C1 1 For the composite 20 mm x 160 mm 2 C2 Substituting, we get Xc D 0 mm X Yc D 114.6 mm We now find Ix for each part about its center and use the parallel axis theorem to find Ix about C. 20 mm Part (1): b1 D 100 mm, h1 D 20 mm Finally, Ix D Ix1 C Ix2 1 1 b1 h13 D 100203 mm4 12 12 Ix D 1.686 ð 107 mm4 Ix 0 1 D for our composite shape. Ix0 1 D 6.667 ð 104 mm4 dy1 D Yc1 Yc D 55.38 mm Ix1 D Ix0 1 C dy1 2 A1 Now for the comparison. For the solid square with the same total area A1 C A2 D 5200 mm2 , we get a side of length l2 D 5200: l D 72.11 mm And for this solid section Ix1 D 6.20 ð 106 mm4 Part (2) b2 D 20 mm, h2 D 160 mm Ix 0 2 D 1 1 b2 h2 3 D 201603 mm4 12 12 IxSQ D 1 3 1 4 bh D l 12 12 IxSQ D 2.253 ð 106 mm4 1.686 ð 107 2.253 ð 106 Ix0 2 D 6.827 ð 106 mm4 Ratio D Ix /IxSQ D dy2 D Yc2 Yc D 34.61 mm Ratio D 7.48 Ix2 D Ix0 2 C dy2 A2 This matches the value in Example 8.5. Ix2 D 1.066 ð 107 mm4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 639 Problem 8.82 The area of the beam cross section is 5200 mm2 . Determine the moment of inertia of the beam cross section about the x axis. Compare your result with the moment of inertia of a solid square cross section of equal area. (See Example 8.5.) y x 20 mm Solution: Let the outside dimension be b mm, then the inside dimension is b 40 mm. The cross section is A D b2 b 402 D 5200 mm2 . Solve: b D 85 mm. Divide the beam cross section into two parts: the inner and outer squares. Part (1) A1 D 852 D 7225 mm2 , Ixx1 D 1 12 85853 D 4.35 ð 106 . Part (2) A2 D 452 D 2025 mm2 . Ixx2 D 1 12 45453 D 3.417 ð 105 . The composite moment of inertia about the centroid is Ix D Ixx1 Ixx2 D 4.008 ð 106 mm4 . For a square cross section of the same area, h D p 5200 D 72.111 mm. The area moment of inertia is Ixb D 1 12 72.11172.1113 D 2.253 ð 106 in4 . The ratio: RD 4.008 ð 106 D 1.7788 D 1.78 2.253 ð 106 which confirms the value given in Example 8.5. 640 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.83 If the beam in Fig. a is subjected to couples of magnitude M about the x axis (Fig. b), the beam’s longitudinal axis bends into a circular are whose radius R is given by y y z x (a) Unloaded. EIx RD , M where Ix is the moment of inertia of the beam’s cross section about the x axis. The value of the term E, which is called the modulus of elasticity, depends on the material of which the beam is constructed. Suppose that a beam with the cross section shown in Fig. c is subjected to couples of magnitude M D 180 N-m. As a result, the beam’s axis bends into a circular arc with radius R D 3 m. What is the modulus of elasticity of the beam’s material? (See Example 8.5.) Solution: The moment of inertia of the beam’s cross section about the x axis is & ' 1 1 mm4 393 C 2 933 C 62 93 Ix D 12 12 D 2170 mm4 D 2.17 ð 109 m4 . The modulus of elasticity is ED RM 3 m180 N-m D D 2.49 ð 1011 N/m2 Ix 2.17 ð 109 m4 M R M (b) Subjected to couples at the ends. y 3 mm x 9 mm 3 mm 3 mm 9 mm (c) Beam cross section. E D 2.49 ð 1011 N/m2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 641 Problem 8.84 Suppose that you want to design a beam made of material whose density is 8000 kg/m3 . The beam is to be 4 m in length and have a mass of 320 kg. Design a cross section for the beam so that Ix D 3 ð 105 m4 . (See Example 8.5.) Solution: The strategy is to determine the cross sectional area, and then use the ratios given in Figure 8.14 to design a beam. The volume of the beam is V D AL D 4A m3 . The mass of the beam is m D V8000 D 32000A D 320 kg, from which A D 0.01 m2 . The moment of inertia for a beam of square cross section with this area is Ixxb D 1 12 h 0.10.13 D 8.333 ð 106 m4 . The ratio is R D b ð 105 3 D 3.6. 8.333 ð 106 From Figure 8.6, this ratio suggests an I-beam of the form shown in the sketch. Choose an I-beam made up of three equal area rectangles, of dimensions b by hm in section. The moment of inertia about the centroid is Ix D y21 A1 C Ixx1 C y22 A2 C Ixx2 C y23 A3 C Ixx3 . Since all areas are equal, A1 D A2 D A3 D bh, and y1 D 0, and y3 D y1 , this reduces to Ix D bCh , y2 D 2 1 bCh 2 1 bh3 C 2 hb3 . hb C 6 2 12 A , where A is the known total cross section area. These 3 are two equations in two unknowns. Plot the function Note that bh D fb D 1 bCh 2 1 bh C bh3 C 2 hb3 Ix 6 2 12 Problem 8.85 The area in Fig. (a) is a C230ð30 American Standard Channel beam cross section. Its cross sectional area is A D 3790 mm2 and its moments of inertia about the x and y axes are Ix D 25.3 ð 106 mm4 and Iy D 1 ð 106 mm4 . Suppose that two beams with C230ð30 cross sections are riveted together to obtain a composite beam with the cross section shown in Fig. (b). What are the moments of inertia about the x and y axes of the composite beam? f ( b ) * E + 5 I - beam Flange dimension 1 .8 .6 .4 .2 0 –.2 –.4 –.6 –.8 –1 .03 .04 .05 .06 .07 .08 .09 .1 b A .The function was graphed using 3 TK Solver Plus. The graph crosses the zero axis at approximately b D 0.0395 m. and b D 0.09 m. The lower value is an allowable value for h and the greater value corresponds to an allowable value of b. Thus the I beam design has the flange dimensions, b D 90 mm and h D 39.5 mm. subject to the condition that hb D y y x x 14.8 mm (a) (b) Solution: Ix D 225.3 ð 106 mm4 D 50.6 ð 106 mm4 Iy D 2106 mm4 C [3790 mm2 ][14.8 mm]2 D 3.66 ð 106 mm4 642 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.86 The area in Fig. (a) is an L152ð102ð 12.7 Angle beam cross section. Its cross sectional area is A D 3060 mm2 and its moments of inertia about the x and y axes are Ix D 7.24 ð 106 mm4 and Iy D 2.61 ð 106 mm4 . Suppose that four beams with L152ð102ð12.7 cross sections are riveted together to obtain a composite beam with the cross section shown in Fig. (b). What are the moments of inertia about the x and y axes of the composite beam? y y 24.9 mm x Solution: x 50.2 mm (a) Ix D 47.24 ð 106 mm4 C [3060 mm2 ][50.2 mm]2 D 59.8 ð 106 mm4 Iy D 42.61 ð 106 mm4 C [3060 mm2 ][24.9 mm]2 D 18.0 ð 106 mm4 (b) Problem 8.87 In Active Example 8.6, suppose that the vertical 3-m dimension of the triangular area is increased to 4 m. Determine a set of principal axes and the corresponding principal moments of inertia. y Solution: From Appendix B, the moments and products of inertia of the area are Ix D 1 4 m4 m3 D 21.3 m4 , 12 Iy D 1 4 m3 4 m D 64 m4 , 4 Ixy D 1 4 m2 4 m2 D 32 m4 . 8 3m x 4m From Eq. (8.26), tan2p D 2Ixy 232 D 1.50 ) p D 28.2° . D Iy Ix 64 21.3 From Eqs. (8.23) and (8.24), the principal moments of inertia are Ix ‘ D Ix C Iy Ix Iy C cos 2p Ixy sin 2p 2 2 D 21.3 C 64 2 C 21.3 64 2 cos2[28.2° ] 32 sin2[28.2° ] D 4.21 m4 Iy ‘ D Ix Iy Ix C Iy cos 2p C Ixy sin 2p 2 2 D 21.3 C 64 2 21.3 64 2 cos2[28.2° ] C 32 sin2[28.2° ] D 81.8 m4 p D 28.2° , principal moments of inertia are 4.21 m4 , 81.1 m4 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 643 y Problem 8.88 In Example 8.7, suppose that the area is reoriented as shown. Determine the moments of inertia Ix0 , Iy 0 and Ix0 y 0 if D 30o . Solution: Based on Example 8.7, the moments and product of 1 ft 3 ft inertia of the reoriented area are 1 ft Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . x 4 ft Applying Eqs. (8.23)–(8.25), Ix 0 D D Iy 0 D D Ix 0 y 0 D D Ix Iy Ix C Iy C cos 2 Ixy sin 2 2 2 10 C 22 10 22 C cos 60° 6 sin 60° D 7.80 ft4 , 2 2 Ix Iy Ix C Iy C cos 2 C Ixy sin 2 2 2 10 C 22 10 22 cos 60° C 6 sin 60° D 24.2 ft4 , 2 2 Ix Iy sin 2 C Ixy cos 2 2 12 22 sin 60° C 6 cos 60° D 2.20 ft4 2 Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix0 y 0 D 2.20 ft4 . y Problem 8.89 In Example 8.7, suppose that the area is reoriented as shown. Determine a set of principal axes and the corresponding principal moments of inertia. Based on the result of Example 8.7, can you predict a value of p without using Eq. (8.26)? Solution: Based on Example 8.7, the moments and product of 1 ft 3 ft inertia of the reoriented area are 1 ft Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . x 2Ixy 26 D D 1 ) p D 22.5° Ix Iy 22 10 This value could have been anticipated from Example 8.7 by reorienting the axes. Substituting the angle into Eqs. (8.23) and (8.24), the principal moments of inertia are From Eq. (8.26), tan 2p D Ix 0 D D Iy 0 D D 4 ft Ix Iy Ix C Iy C cos 2p Ixy sin 2p 2 2 10 C 22 10 22 C cos 45° 6 sin 45° D 7.51 ft4 , 2 2 Ix Iy Ix C Iy cos 2p C Ixy sin 2p 2 2 10 22 10 C 22 cos 45° C 6 sin 45° D 24.5 ft4 , 2 2 p D 22.5° , principal moments of inertia are 7.51 ft4 , 24.5 ft4 . 644 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.90 The moment of inertia of the area are y y Ix D 1.26 ð 106 in4 , Iy D 6.55 ð 105 in4 , Ixy D 1.02 ð 105 in4 Determine the moments of inertia of the area Ix0 , Iy 0 and Ix0 y 0 if D 30° . x u x Solution: Applying Eqs. (8.23)–(8.25), Ix 0 D Ix Iy Ix C Iy C cos 2 Ixy sin 2Ð 2 2 D 1.26 C 0.655 1.26 0.655 C cos 60° 0.102 sin 60° ð 106 in4 2 2 D 1.20 ð 106 in4 Iy 0 D Ix Iy Ix C Iy cos 2 C Ixy sin 2 2 2 D 1.26 0.655 1.26 C 0.655 cos 60° C 0.102 sin 60° ð 106 in4 2 2 D 7.18 ð 105 in4 Ix 0 y 0 D Ix Iy sin 2 C Ixy cos 2 2 D 1.26 0.655 sin 60° C 0.102 cos 60° ð 106 in4 2 D 2.11 ð 105 in4 Ix0 D 1.20 ð 106 in4 , Iy 0 D 7.18 ð 105 in4 , Ix0 y 0 D 2.11 ð 105 in4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 645 Problem 8.91 The moment of inertia of the area are y y Ix D 1.26 ð 106 in4 , Iy D 6.55 ð 105 in4 , Ixy D 1.02 ð 105 in4 x u Determine a set of principal axes and the corresponding principal moments of inertia. x Solution: From Eq. (8.26), tan 2p D 2Ixy 2.102 D D 0.337 Iy Ix 0.655 1.26 ) p D 9.32° Substituting this angle into Eqs. (8.23) and (8.24), the principal moments of inertia are Ix 0 D Ix Iy Ix C Iy C cos 2p Ixy sin 2p 2 2 D 1.26 0.655 1.26 C 0.655 C cos 18.63° 0.102 sin 18.63° 2 2 ð 106 in4 D 1.28 ð 106 in4 Iy 0 D Ix C Iy Ix Iy cos 2p C Ixy sin 2p 2 2 D 1.26 0.655 1.26 C 0.655 cos 18.63° C 0.102 sin 18.63° 2 2 ð 106 in4 D 6.38 ð 105 in4 p D 9.32° , principal moments of inertia are 1.28 ð 106 in4 , 6.38 ð 105 in4 . 646 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.92* Determine a set of principal axes and the corresponding principal moments of inertia. 160 mm Solution: We divide the area into 3 rectangles as shown: In terms of the xO , yO coordinate system, the position of the centroid is x̂ D 40 mm x̂1 A1 C x̂2 A1 C x̂3 A3 A1 C A2 C A3 x D ŷ D 2040200 C 10012040 C 808040 D 56 mm, 40200 C 12040 C 8040 200 mm 40 mm ŷ1 A1 C ŷ2 A1 C ŷ3 A3 A1 C A2 C A3 40 mm 10040200 C 18012040 C 208040 D D 108 mm. 40200 C 12040 C 8040 120 mm The moments and products of inertia in terms of the xO , yO system are y y^ (Ix D (Ix 1 C (Ix 2 C (Ix 3 D 160 mm 2 1 1 402003 C 120403 C 1802 12040 3 12 C 1 80403 D 26.5 ð 107 mm4 , 3 200 mm x 40 mm (Iy D (Iy 1 C (Iy 2 C (Iy 3 40 mm 3 120 mm D 1 1 200403 C 401203 C 1002 12040 3 12 C 40 mm 1 x^ y y′ 1 40803 C 802 8040 D 8.02 ð 107 mm4 , 12 (Ixy D (Ixy 1 C (Ixy 2 C (Ixy 3 x 12.1° x′ D 2010040200 C 10018040120 C 20804080 D 10.75 ð 107 mm. The moments and product of inertia in terms of the xO , yO system are Ix D (Ix yO 2 A D 77.91 ð 106 mm4 , Iy D (Iy xO 2 A D 30.04 ð 106 mm4 , Ixy D (Ixy x̂ŷA D 10.75 ð 106 mm4 , from Equation (8.26), tan 2p D 2Ixy 210.75 ð 106 , D Iy Ix 30.04 ð 106 77.91 ð 106 we obtain p D 12.1° . We can orient the principal axes as shown: Substituting the values of Ix , Iy and Ixy into Equations (8.23) and (8.24) and setting D 12.1° , we obtain Ix1 D 80.2 ð 106 mm4 Iy 1 D 27.7 ð 106 mm4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 647 Problem 8.93 Solve Problem 8.87 by using Mohr’s Circle. y 3m x 4m Solution: The vertical 3-m dimension is increased to 4 m. From Problem 8.87, the moments and product of inertia for the unrotated system are Ix D 1 4 m4 m3 D 21.3 m4 , 12 Iy D 1 4 m3 4 m D 64 m4 , 4 Ixy D 1 4 m2 4 m2 D 32 m4 . 8 Mohr’s circle (shown) has a center and radius given by CD 21.3 C 64 D 42.7 m4 2 RD 21.3 64 2 2 C 322 D 38.5 m4 The angle and principal moments are now tan2p D 32 ) p D 28.2° , 64 42.7 I1 D C C R D 81.1 m4 , I2 D C R D 4.21 m4 . p D 28.2° , principal moments of inertia are 4.21 m4 , 81.1 m4 648 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.94 Solve Problem 8.88 by using Mohr’s Circle. 1 ft 3 ft 1 ft x 4 ft Solution: Based on Example 8.7, the moments and product of inertia of the reoriented area are Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . For Mohr’s circle we have the center, radius, and angle CD 10 C 22 D 16 ft4 , 2 RD p D 22 10 2 1 tan1 2 2 C 62 D 8.49 ft4 , 6 22 16 D 22.5° Now we can calculate the new inertias Ix D C R cos 15° D 7.80 ft4 Iy D C C R cos 15° D 24.2 ft4 Ixy D R sin 15° D 2.20 ft4 Ix0 D 7.80 ft4 , Iy 0 D 24.2 ft4 , Ix‘y‘ D 2.20 ft4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 649 y Problem 8.95 Solve Problem 8.89 by using Mohr’s Circle. 1 ft 3 ft 1 ft x 4 ft Solution: Based on Example 8.7, the moments and product of inertia of the reoriented area are Ix D 10 ft4 , Iy D 22 ft4 , Ixy D 6 ft4 . For Mohr’s circle we have the center, radius, and angle CD 10 C 22 D 16 ft4 , 2 RD p D 22 10 2 1 tan1 2 2 C 62 D 8.49 ft4 , 6 22 16 D 22.5° Now we can calculate the principal moments of inertias I1 D C C R D 24.5 ft4 I2 D C R D 7.51 ft4 p D 22.5° , principal moments of inertia are 7.51 ft4 , 24.5 ft4 . 650 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.96 Solve Problem 8.90 by using Mohr’s Circle. y y x u x Solution: For Mohr’s circle we have the center, radius, and angle CD 12.6 C 6.55 2 RD p D 12.6 6.55 2 1 tan1 2 ð 105 D 9.58 ð 105 in4 , 2 C 1.022 ð 105 D 3.19 in4 , 1.02 12.6 9.58 D 9.32° Now we can calculate the new inertias Ix0 D C C R cos 22.7° D 12.0 ð 105 in4 , Iy 0 D C R cos 22.7° D 7.18 ð 105 in4 , Ix0 y 0 D R sin 22.7° D 2.11 ð 105 in4 . Ix0 D 1.20 ð 106 in4 , Iy 0 D 7.18 ð 105 in4 , Ix0 y 0 D 2.11 ð 105 in4 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 651 Problem 8.97 Solve Problem 8.91 by using Mohr’s Circle. y y x u x Solution: For Mohr’s circle we have the center, radius, and angle CD 12.6 C 6.55 2 RD p D 12.6 6.55 2 1 tan1 2 ð 105 D 9.58 ð 105 in4 , 2 C 1.022 ð 105 D 3.19 in4 , 1.02 12.6 9.58 D 9.32° Now we can calculate the principal inertias I1 D C C R D 12.8 ð 105 in4 , Iy 0 D C R D 6.38 ð 105 in4 , p D 9.32° , principal moments of inertia are 1.28 ð 106 in4 , 6.38 ð 105 in4 . 652 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.98* Solve Problem 8.92 by using Mohr’s circle. Solution: The moments and product of inertia are derived in terms of the xy coordinate system in the solution of Problem 8.92: 50 ◊106 (+) Ix D 77.91 ð 106 mm4 Iy D 30.04 ð 106 mm4 1 2θp Ixy D 10.75 ð 106 mm4 . The Mohr’s circle is: Measuring the 2p, angle we estimate that p D 12° , and the principle moments of inertia are approximately 81 ð 106 mm4 and 28 ð 106 mm4 the orientation of the principal axes is shown in the solution of Problem 8.92. 2 (30.0, –10.8) (77.9,10.8) (+) 100 6 ◊10 - 50 ◊106 Problem 8.99 Derive Eq. (8.22) for the product of inertia by using the same procedure we used to derive Eqs. (8.20) and (8.21). Solution: Suppose that the area moments of inertia of the area A x′ are known in the coordinate system x, y, y 2 y dA, Ix D y′ A A 0 x 2 dA, Iy D x A and Ixy D xyA. A The objective is to find the product of inertia in the new coordinate system x 0 , y 0 in terms of the known moments of inertia. The new x 0 , y 0 system is formed from the old x, y system by rotation about the origin through a counterclockwise angle . By definition, Substitute into the definition: Ix0 y 0 D cos2 sin2 xy dA A A 0 0 x y dA. Ix 0 y 0 D x 2 dA , y 2 dA C cos sin A from which A From geometry, x 0 D x cos C y sin , Ix0 y 0 D cos2 sin2 Ixy C Ix Iy sin cos , which is the expression required. and y 0 D x sin C y cos . The product is x0 y 0 D xy cos2 xy sin2 C y 2 cos sin x2 cos sin . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 653 Problem 8.100 The axis LO is perpendicular to both segments of the L-shaped slender bar. The mass of the bar is 6 kg and the material is homogeneous. Use the method described in Example 8.10 to determine is moment of inertia about LO . Solution: Use Example 8.10 as a model for this solution. 1m LO 2m Introduce the coordinate system shown and divide the bar into two parts as shown y=1 y dy 2 r dx 1 0 2 r 2 dm D I0 1 D Ax 2 dx D A 0 I0 1 D x 2 x x3 3 2 0 8 A 3 However m1 D Al1 D 2A. Since part 1 is 2/3 of the length, its mass is 2/36 kg D 4 kg. Part 2 has mass 2 kg. For part 2, dm D A dy and 22 C y 2 rD 1 r 2 dm D I0 2 D I0 2 D A4y C I0TOTAL D A22 C y 2 dy 0 m2 y3 3 1 D A 0 13 3 13 8 21 A C A D A 3 3 3 I0 TOTAL D 7A The total mass D 3A D 6 kg I0TOTAL D 654 7 6 kg Ð m2 D 7 kg m2 6 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.101 Two homogenous slender bars, each of mass m and length l, are welded together to form the T-shaped object. Use integration to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars. l O l Solution: Divide the object into two pieces, each corresponding to a slender bar of mass m; the first parallel to the y axis, the second to the x axis. By definition l r 2 dm C ID 0 r 2 dm. m For the first bar, the differential mass is dm D A dr. Assume that the second bar is very slender, so that the mass is concentrated at a distance l from O. Thus dm D A dx, where x lies between the limits l l x . 2 2 The distance to a differential dx is r D becomes l I D A r 2 dr C A l 2 0 D A r3 3 D ml2 l 2 p l2 C x 2 . Thus the definition l2 C x 2 dx I l/2 x3 C A l2 x C 3 l/2 0 l 1 1 C1C 3 12 D 17 2 ml 12 Problem 8.102 The slender bar lies in the x –y plane. Its mass is 6 kg and the material is homogeneous. Use integration to determine its moment of inertia about the z axis. y 2m 50 x 1m Solution: 1 m 2 Iz D 6 kg D 2 kg/m The density is D 3m x 2 dx y 0 C m 2m [1 m C s cos 50° 2 C s sin 50° 2 ]ds 0 50° Iz D 15.14 kg m2 Problem 8.103 Use integration to determine the moment of inertia of the slender bar in Problem 8.102 about the y axis. x 1m Solution: See solution for 8.102 Iy D 0 1 m 2 x 2 dx C m 1 m C s cos 50° 2 ds D 12.01 kg m2 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 655 y Problem 8.104 The homogeneous thin plate has mass m D 12 kg and dimensions b D 2 m and h D 1 m. Use the procedure described in Active Example 8.9 to determine the moments of inertia of the plate about the x and y axes. h x Solution: From Appendix B, the moments of inertia about the x and y axes are 1 3 1 3 bh , Iy D hb . Ix D 36 36 b Therefore the moments of inertia of the plate about the x and y axes are 1 3 m m 1 1 b D mh2 D 12 kg1 m2 D 0.667 kg-m2 Ixaxis D Ix 1 A 36 18 8 bh 2 Iyaxis D m m Iy D 1 A bh 2 1 3 hb 36 D 1 1 mb2 D 12 kg2 m2 D 2.67 kg-m2 18 18 Ixaxis D 0.667 kg-m2 , Iyaxis D 2.67 kg-m2 . y Problem 8.105 The homogenous thin plate is of uniform thickness and mass m. (a) (b) Determine its moments of inertia about the x and z axes. Let Ri D 0, and compare your results with the values given in Appendix C for a thin circular plate. Ro Ri x Solution: (a) (b) The area moments of inertia for a circular area are Ix D Iy D R4 . For the plate with a circular cutout, Ix D Ro4 Ri4 . The 4 4 m area mass density is , thus for the plate with a circular cut, A m m , from which the moments of inertia D 2 A Ro2 Ri Ri Ro m mRo4 Ri4 D Ro2 C Ri2 4 4Ro2 Ri2 Ix axis D Iz axis D 2Ix axis D m 2 R C Ri2 . 2 o Let Ri D 0, to obtain Ix axis D m 2 R , 4 o Iz axis D m 2 R , 2 o which agrees with table entries. 656 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.106 The homogenous thin plate is of uniform thickness and weighs 20 lb. Determine its moment of inertia about the y axis. 1 y = 4 – – x 2 ft 4 Solution: y D4 1 2 x ft 4 x y The plate’s area is 4 4 AD 4 1 2 x 4 dx D 21.3 ft2 . y=4– 1 2 x ft 4 The plate’s density per unit area is υ D 20/32.2/21.3 D 0.0291 slug/ft2 . x –4 ft x The moment of inertia about the y axis is Iy axis D 4 ft dx 1 x 2 υ 4 x 2 dx 4 4 4 D 1.99 slug-ft2 . Problem 8.107 Determine the moment of inertia of the plate in Problem 8.106 about the x axis. Solution: See the solution of Problem 8.106. The mass of the strip so the moment of inertia of the plate about the x axis is element is mstrip 1 D υ 4 x2 4 dx. Ix axis 4 D 4 3 1 1 dx D 2.27 slug-ft2 . υ 4 x2 3 4 The moment of inertia of the strip about the x axis is Istrip D D 2 1 1 mstrip 4 x 2 3 4 3 1 1 υ 4 x2 dx, 3 4 Problem 8.108 The mass of the object is 10 kg. Its moment of inertia about L1 is 10 kg-m2 . What is its moment of inertia about L2 ? (The three axes lie in the same plane.) Solution: The strategy is to use the data to find the moment of inertia about L, from which the moment of inertia about L2 can be determined. IL D 0.62 10 C 10 D 6.4 m2 , 0.6 m L from which IL2 D 1.22 10 C 6.4 D 20.8 m2 0.6 m L1 L2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 657 Problem 8.109 An engineer gathering data for the design of a maneuvering unit determines that the astronaut’s center of mass is at x D 1.01 m, y D 0.16 m and that her moment of inertia about the z axis is 105.6 kg-m2 . Her mass is 81.6 kg. What is her moment of inertia about the z0 axis through her center of mass? y y x x Solution: The distance d from the z axis to the z0 axis is 1.012 C 0.162 dD D 1.0226 m. From the parallel-axis theorem, D Iz0 axis C d2 m : 105.6 D Iz0 axis C 1.02262 81.6. Iz axis Solving, we obtain Iz0 D 20.27 kg-m2 . axis Problem 8.110 Two homogenous slender bars, each of mass m and length l, are welded together to form the Tshaped object. Use the parallel axis theorem to determine the moment of inertia of the object about the axis through point O that is perpendicular to the bars. l O l Solution: Divide the object into two pieces, each corresponding to a bar of mass m. By definition l ID r 2 dm. 0 For the first bar, the differential mass is dm D A dr, from which the moment of inertia about one end is l I1 D A r 2 dr D A 0 r3 3 l D 0 ml2 . 3 For the second bar I2 D A l 2 l 2 r3 r dr D A 3 2 2l l 2 D ml2 12 is the moment of inertia about the center of the bar. From the parallel axis theorem, the moment of inertia about O is Io D 658 17 2 ml2 ml2 C l2 m C D ml . 3 12 12 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.111 Use the parallel-axis theorem to determine the moment of inertia of the T-shaped object in Problem 8.110 about the axis through the center of mass of the object that is perpendicular to the two bars. (See Active Example 8.11.) Solution: The location of the center of mass of the object is x D $ % l 2 C lm 3 D l. Use the results of Problem 8.110 for the moment 2m 4 of inertia of a bar about its center. For the first bar, m I1 D 2 l ml2 7 2 mC D ml . 4 12 48 For the second bar, I2 D 2 l ml2 7 2 mC D ml . 4 12 48 The composite: Ic D I1 C I2 D 7 2 ml 24 y Problem 8.112 The mass of the homogenous slender bar is 20 kg. Determine its moment of inertia about the z axis. y' x' Solution: Divide the object into three segments. Part (1) is the 1 m bar on the left, Part (2) is the 1.5 m horizontal segment, and Part (3) is the segment on the far right. The mass density per unit length is 1m x 20 m p D 5.11 kg/m. D D L 1 C 1.5 C 2 1.5 m 1m The moments of inertia about the centers of mass and the distances to the centers of mass from the z axis are: Part (1) I1 D l31 12 D m1 l21 D 0.426 kg-m2 , 12 m1 D 5.11 kg, d1 D 0.5 m, Part (2), I2 D l2 l32 D m2 2 D 1.437 kg- m2 , 12 12 m2 D 7.66 kg, d2 D Part (3) p 0.752 C 12 D 1.25 m I3 D p 22 l33 D m3 D 1.204 kg-m2 , 12 12 m3 D 7.23 kg, d3 D p 22 C 0.52 D 2.062 m. The composite: I D d21 m1 C I1 C d22 m2 C I2 C d23 m3 C I3 D 47.02 kg-m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 659 Problem 8.113 Determine the moment of inertia of the bar in Problem 8.112 about the z0 axis through its center of mass. Solution: The center of mass: xD D yD D x 1 m 1 C x2 m 2 C x3 m 3 20 0 C 0.757.66 C 27.23 D 1.01 m. 20 0.5m1 C 1m2 C 0.5m3 20 0.55.11 C 17.66 C 0.57.23 D 0.692 m. 20 The distance from the z axis to the center of mass is d D x2 C y2 D 1.224 m. The moment of inertia about the center o mass: Ic D d2 20 C Io D 17.1 kg-m2 Problem 8.114 The homogeneous slender bar weighs 5 lb. Determine its moment of inertia about the z axis. y y' Solution: The Bar’s mass is m D 5/32.2 slugs. Its length is L D L1 C L 2 C L 3 D 8 C 4 in p 82 C 82 C 4 D 31.9 in. The masses of the parts are therefore, L1 m1 D mD L m2 D L2 mD L m3 D L3 mD L 8 31.9 5 32.2 x' D 0.0390 slugs, p 264 5 D 0.0551 slugs, 31.9 32.2 4 31.9 5 32.2 x 8 in D 0.0612 slugs. y′ y The center of mass of part 3 is located to the right of its center C a distance 2R/ D 24/ D 2.55 in. The moment of inertia of part 3 about C is 2 C r 2 dm D m3 r 2 D 0.061242 D 0.979 slug-in2 . m3 1 The moment of inertia of part 3 about the center of mass of part 3 is therefore 4 in x′ 3 x 8 in I3 D 0.979 m3 2.552 D 0.582 slug-in2 . The moment of inertia of the bar about the z axis is Iz axis D 13 m1 L12 C 13 m2 L22 C I3 C m3 [8 C 2.552 C 42 ] D 11.6 slug-in2 D 0.0802 slug-ft2 . 660 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.115 Determine the moment of inertia of the bar in Problem 8.114 about the z0 axis through its center of mass. Solution: In the solution of Problem 8.114, it is shown that the moment of inertia of the bar about the z axis is Iz axis D 11.6 slug-in2 . The x and y coordinates of the center of mass coincide with the centroid of the axis: xD x1 L1 C x2 L2 C x3 L3 L1 C L2 C L3 p 24 4 48 C 4 82 C 82 C 8 C p D D 6.58 in, 8 C 82 C 82 C 4 yD D y1 L1 C y2 L2 C y3 L3 L1 C L2 C L3 p 0 C 4 82 C 82 C 44 p D 3.00 in. 8 C 82 C 82 C 4 The moment of inertia about the z axis is Iz0 axis D Iz axis x2 C y2 5 32.2 D 3.43 slug-in2 . y Problem 8.116 The rocket is used for atmospheric research. Its weight and its moment of inertia about the z axis through its center of mass (including its fuel) are 10 kip and 10,200 slug-ft2 , respectively. The rocket’s fuel weighs 6000 lb, its center of mass is located at x D 3 ft, y D 0, z D 0, and the moment of inertia of the fuel about the axis through the fuel’s center of mass parallel to z is 2200 slug-ft2 . When the fuel is exhausted, what is the rocket’s moment of inertia about the axis through its new center of mass parallel to z? Solution: Denote the moment of inertia of the empty rocket as IE about a center of mass xE , and the moment of inertia of the fuel as IF about a mass center xF . Using the parallel axis theorem, the moment of inertia of the filled rocket is x From which xE D 186.335 124.224 3 D 4.5 ft 2m , IR D IE C xE2 mE C IF C xF F is the new location of the center of mass. Substitute: about a mass center at the origin (xR D 0). Solve: 2m IE D IR xE2 mE IF xF F 2m . IE D IR xE2 mE IF xF F D 10200 2515.5 2200 1677.01 The objective is to determine values for the terms on the right from the data given. Since the filled rocket has a mass center at the origin, the mass center of the empty rocket is found from 0 D mE xE C mF xF , from which D 3807.5 slug-ft2 xE D mF mE xF . Using a value of g D 32.2 ft/s2 , mF D WF 6000 D D 186.34 slug, g 32.2 mE D 10000 6000 WR WF D D 124.23 slug. g 32.2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 661 Problem 8.117 The mass of the homogenous thin plate is 36 kg. Determine its moment of inertia about the x axis. y 0.4 m 0.4 m Solution: Divide the plate into two areas: the rectangle 0.4 m by 0.6 m on the left, and the rectangle 0.4 m by 0.3 m on the right. The m mass density is D . The area is A 0.3 m A D 0.40.6 C 0.40.3 D 0.36 m2 , 0.3 m x from which D 36 D 100 kg/m2 . 0.36 The moment of inertia about the x axis is Ix axis D $1% 3 0.40.63 C $1% 3 0.40.33 D 3.24 kg-m2 Problem 8.118 Determine the moment of inertia of the plate in Problem 8.117 about the z axis. Solution: The basic relation to use is Iz axis D Ix axis C Iy axis . The value of Ix axis is given in the solution of Problem 8.117. The moment of inertia about the y axis using the same divisions as in Problem 8.117 and the parallel axis theorem is Iy axis D 1 1 0.60.43 C 0.30.43 3 12 C 0.62 0.30.4 D 5.76 kg-m2 , from which Iz axis D Ix axis C Iy axis D 3.24 C 5.76 D 9 kg-m2. y Problem 8.119 The homogenous thin plate weighs 10 lb. Determine its moment of inertia about the x axis. 5 in 5 in Solution: Divide the area into two parts: the lower rectangle 5 in by 10 in and the upper triangle 5 in base and 5 in altitude. The mass W . The area is density is D gA A D 510 C $1% 2 Using g D 32 ft/s2 , the mass density is D 10 in 55 D 62.5 in2 . W D 0.005 slug/in2 . gA 5 in x Using the parallel axis theorem, the moment of inertia about the x axis is Ix axis D 1 1 1053 C 553 3 36 5 2 1 C 5C 55 D 4.948 slug-in2 3 2 Ix 662 axis D 0.03436 slug-ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.120 Determine the moment of inertia of the plate in Problem 8.119 about the y axis. Solution: Use the results of the solution in Problem 8.119 for the area and the mass density. Iy axis D 1 1 5103 C 553 3 36 10 2 1 C 5C 55 3 2 D 12.76 slug-in2 D 0.0886 slug-ft2 Problem 8.121 The thermal radiator (used to eliminate excess heat from a satellite) can be modeled as a homogenous, thin rectangular plate. Its mass is 5 slugs. Determine its moment of inertia about the x, y, and z axes. y 3 ft 6 ft 3 ft Solution: The area is A D 93 D 27 ft2 . The mass density is D 5 m D D 0.1852 slugs/ft2 . A 27 2 ft x The moment of inertia about the centroid of the rectangle is Ixc D Iyc D 1 12 1 12 933 D 3.75 slug-ft2 , 393 D 33.75 slug-ft2 . Use the parallel axis theorem: Ix axis D A2 C 1.52 C Ixc D 65 slug-ft2 , Iy axis D A4.5 32 C Iyc D 45 slug-ft2 . Iz axis D Ix axis C Iy axis D 110 slug-ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 663 Problem 8.122 The homogeneous cylinder has mass m, length l, and radius R. Use integration as described in Example 8.13 to determine its moment of inertia about the x axis. y x Solution: The volume of the disk element is R2 dz and its mass is dm D R2 dz, where is the density of the cylinder. From Appendix C, the moment of inertia of the disk element about the x0 axis is dIx0 axis D 1 1 dmR2 D R2 dzR2 . 4 4 R l Applying the parallel-axis theorem, the moment of inertia of the disk element about the x axis is dIxaxis z 1 D dIx0 axis C z2 dm D R2 dzR2 C z2 R2 dz 4 Integrating this expression from z D 0 to z D l gives the moment of inertia of the cylinder about the x axis. l Ixaxis D 0 1 1 1 R4 C R2 z2 dz D R4 l C R2 l3 . 4 4 3 In terms of the mass of the cylinder m D R2 l, Ixaxis D 1 2 1 2 mR C ml 4 3 Problem 8.123 The homogenous cone is of mass m. Determine its moment of inertia about the z axis, and compare your result with the value given in Appendix C. (See Example 8.13.) y x Solution: The differential mass dm D R 3m m r 2 dz D 2 r 2 dz. V R h The moment of inertia of this disk about the z axis is 12 mr 2 . The radius varies with z, rD h z R z, h from which Iz 664 axis D 3mR2 2h5 h 0 z4 dz D 3mR2 2h5 z5 5 h D 0 3mR2 10 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.124 Determine the moments of inertia of the homogenous cone in Problem 8.123 about the x and y axes, and compare your results with the values given in Appendix C. 3m m D . The differential V R2 h element of mass is dm D r 2 dz. The moment of inertia of this elemental disk about an axis through its center of mass, parallel to the x- and y-axes, is Solution: The mass density is D $1% dIx D 4 r 2 dm. Use the parallel axis theorem, $1% Ix D m 4 r 2 dm C z2 dm. m Noting that r D R z, then h r 2 dm D and z2 dm D R4 h4 R2 h2 z4 dz, z4 dz. Substitute: Ix D R4 4h4 h z4 dz C 0 R2 h2 h z4 dz. 0 Integrating and collecting terms: Ix D 3mR2 3m C 3 4h5 h z5 5 h D m 0 3 2 3 2 R C h . 20 5 By symmetry, Iy D Ix c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 665 Problem 8.125 The mass of the homogeneous wedge is m. Use integration as described in Example 8.13 to determine its moment of inertia about the z axis. (Your answer should be in terms of m, a, b, and h.) y x Solution: Consider a triangular element of the wedge of thickness dz. The mass of the element is the product of the density of the material and the volume of the element, dm D 12 bhdz. The moments of inertia of the triangular element about the x’ and y’ axes are given by Eqs. (8.30) and (8.31) in terms of the mass of the element, its triangular area, and the moments of inertia of the triangular area: dIx0 axis 1 bhdz 1 dm 0 1 2 I D rbh3 dz, bh3 D D 1 A x 12 12 bh 2 dIy 0 axis 1 bhdz 1 dm 0 1 I D 2 hb3 D rhb3 dz, D 1 A y 4 4 bh 2 h a z b The moment of inertia of this thin plate about the z axis is dIzaxis D dIx0 axis C dIy 0 axis D 1 1 bh3 dz C hb3 dz. 12 4 Integrating this expression from z D 0 to z D a gives the moment of inertia of the wedge about the z axis: a Izaxis D 0 1 1 1 1 bh3 C hb3 dz D bh3 a C hb3 a. 12 4 12 4 In terms of the mass m D 12 bha, Izaxis D 666 1 2 1 2 mh C mb . 6 2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.126 The mass of the homogeneous wedge is m. Use integration as described in Example 8.13 to determine its moment of inertia about the x axis. (Your answer should be in terms of m, a, b, and h.) y x Solution: Consider a triangular element of the wedge of thickness dz. The mass of the element is the product of the density of the material and the volume of the element, dm D 12 bhdz. The moments of inertia of the triangular element about the x’ axis is given by Eq. (8.30) in terms of the mass of the element, its triangular area, and the moments of inertia of the triangular area: dIx0 axis h a z b 1 bhdz 1 dm 1 Ix 0 D 2 rbh3 dz, bh3 D D 1 A 36 36 bh 2 Applying the parallel-axis theorem, the moment of inertia of the triangular element about the x axis is 1 dIxaxis D dIx0 axis C z2 C h2 dm 3 D 1 1 1 1 1 bh3 dz C [z2 C h2 ] bhdz D bh3 dz C bhz2 dz 36 3 2 12 2 Integrating this expression from z D 0 to z D a gives the moment of inertia of the wedge about the x axis: a Ixaxis D 0 1 1 1 1 bh3 C bhz2 dz D bh3 a C bha3 . 12 2 12 6 In terms of the mass m D 12 bha, Ixaxis D 1 2 1 2 mh C ma . 6 3 Problem 8.127 In Example 8.12, suppose that part of the 3-kg bar is sawed off so that the bar is 0.4 m long and its mass is 2 kg. Determine the moment of inertia of the composite object about the perpendicular axis L through the center of mass of the modified object. Solution: The mass of the disk is 2 kg. Measuring from the left 0.2 m 0.6 m L end of the rod, we locate the center of mass xD 2 kg0.2 m C 2 kg0.6 m D 0.4 m. 2 kg C 2 kg The center of mass is located at the point where the rod and disk are connected. The moment of inertia is ' & 1 1 I D 2 kg0.4 m2 C 2 kg0.2 m2 C 2 kg0.2 m2 3 2 I D 0.227 kg-m2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 667 y Problem 8.128 The L-shaped machine part is composed of two homogeneous bars. Bar 1 is tungsten alloy with mass density 14,000 kg/m3 , and bar 2 is steel with mass density 7800 kg/m3 . Determine its moment of inertia about the x axis. 240 mm 1 40 mm 2 80 mm Solution: The masses of the bars are m1 D 14,0000.240.080.04 D 10.75 kg 80 mm z 240 mm m2 D 78000.240.080.04 D 5.99 kg. Using Appendix C and the parallel axis theorem the moments of inertia of the parts about the x axis are Ix axis1 D 1 m1 [0.042 C 0.242 ] C m1 0.122 D 0.2079 kg-m2 , 12 Ix axis2 D 1 m2 [0.042 C 0.082 ] C m2 0.042 D 0.0136 kg-m2 . 12 x Therefore Ix 668 axis D Ix axis1 C Ix axis2 D 0.221 kg-m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.129 The homogeneous object is a cone with a conical hole. The dimensions R1 D 2 in, R2 D 1 in, h1 D 6 in, and h2 D 3 in. It consists of an aluminum alloy with a density of 5 slug/ft3 . Determine its moment of inertia about the x axis. y x R1 Solution: The density of the material is D 5 slug/ft3 1 ft 12 in 3 D 0.00289 slug/in3 . The volume of the conical object without the conical hole is V1 D R2 h1 z h2 1 2 1 R h1 D 2 in2 6 in D 25.1 in3 . 3 1 3 The mass of the conical object without the conical hole is m1 D V1 D 0.0727 slug. From Appendix C, the moment of inertia of the conical object without the conical hole about the x axis is Ix 1 D m1 3 2 3 2 h C R 5 1 20 1 D 0.0727 slug 3 3 6 in2 C 2 in2 D 1.61 slug-in2 5 20 The volume of the conical hole is V2 D 1 2 1 R h2 D 1 in2 3 in D 3.14 in3 . 3 2 3 The mass of the material that would occupy the conical hole is m2 D V2 D 0.00909 slug. The z coordinate of the center of mass of the material that would occupy the conical hole is z D h1 h2 C 3 3 h2 D 6 in 3 in C 3 in D 5.25 in. 4 4 Using Appendix C and applying the parallel-axis theorem, the moment of inertia about the x axis of the material that would occupy the conical hole is Ix 2 D m2 3 2 3 2 h C R C z2 m2 D 0.255 slug-in2 . 80 2 20 2 The moment of inertia of the conical object with the conical hole is Ix D Ix 1 Ix 2 D 1.36 slug-in2 . Ix D 1.36 slug-in2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 669 Problem 8.130 The circular cylinder is made of aluminum (Al) with density 2700 kg/m3 and iron (Fe) with density 7860 kg/m3 . Determine its moments of inertia about the x 0 and y 0 axes. y y Al z Fe 600 mm 200 mm z Solution: We have Al D 2700 kg/m3 , Fe D 7860 kg/m3 600 mm x, x We first locate the center of mass xD Al [0.1 m]2 [0.6 m]0.3 m C Fe [0.1 m]2 [0.6 m]0.9 m Al [0.1 m]2 [0.6 m] C Fe [0.1 m]2 [0.6 m] D 0.747 m We also have the masses mAl D Al 0.1 m2 0.6 m, mFe D Fe 0.1 m2 0.6 m Now find the moments of inertia 1 1 mAl 0.1 m2 C mFe 0.1 m2 D 0.995 kg m2 2 2 [0.1 m]2 [0.6 m]2 C D mAl C mAl x 0.3 m2 12 4 [0.1 m]2 [0.6 m]2 C mFe C C mFe 0.9 m x2 12 4 D 20.1 kg m2 Ix 0 D Iy 0 Problem 8.131 The homogenous half-cylinder is of mass m. Determine its moment of inertia about the axis L through its center of mass. Solution: The centroid of the half cylinder is located a distance R L T 4R from the edge diameter. The strategy is to use the parallel 3 axis theorem to treat the moment of inertia of a complete cylinder as the sum of the moments of inertia for the two half cylinders. From Problem 8.118, the moment of inertia about the geometric axis for a cylinder is IcL D mR2 , where m is one half the mass of the cylinder. of By the parallel axis theorem, IcL D 2 4R 3 2 m C IhL . Solve IhL D IcL 2 D mR2 D mR2 670 4R 3 2 2 16 mR mR2 m D 2 2 9 16 1 2 92 16 1 2 92 D 0.31987 mR2 D 0.32 mR2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.132 The homogeneous machine part is made of aluminum alloy with density D 2800 kg/m3 . Determine its moment of inertia about the z axis. y y 20 mm x z 40 mm 120 mm Solution: We divide the machine part into the 3 parts shown: (The dimension into the page is 0.04 m). The masses of the parts are y m1 D 28000.120.080.04 D 1.075 kg, m2 D 2800 12 0.042 0.04 D 0.281 kg, 40 mm 0.12 m 1 0.08 m x m3 D 28000.022 0.04 D 0.141 kg. y Iz axis1 0.12 m C 2 0.12 m 1 m1 [0.082 C 0.122 ] C m1 0.062 D 12 The moment of inertia of part 2 about the axis through the center C that is parallel to the z axis is x 0.02 m Iz axis2 D 0.000144 C m2 0.12 C 0.0172 D 0.00543 kg-m2 . The moment of inertia of the material that would occupy the hole 3 about the z axis is D 12 m2 0.042 . The distance along the x axis from C to the center of mass of part 2 is 40.04/3 D 0.0170 m. Iz axis3 Therefore Iz Therefore, the moment of inertia of part 2 about the z axis through its center of mass that is parallel to the axis is 1 2 2 m2 0.04 3 x – 0.04 m Using this result, the moment of inertia of part 2 about the z axis is D 0.00573 kg-m2 . 1 2 2 m2 R y + Using Appendix C and the parallel axis theorem the moment of inertia of part 1 about the z axis is axis D 12 m3 0.022 C m3 0.122 D 0.00205 kg-m2 . D Iz axis1 C Iz axis2 Iz axis3 D 0.00911 kg-m2 . m2 0.01702 D 0.000144 kg-m2 . Problem 8.133 Determine the moment of inertia of the machine part in Problem 8.132 about the x axis. Solution: We divide the machine part into the 3 parts shown in the solution to Problem 8.132. Using Appendix C and the parallel axis theorem, the moments of inertia of the parts about the x axis are: 1 m1 [0.082 C 0.042 ] D 0.0007168 kg-m2 12 Ix axis1 D Ix axis2 D m2 1 1 0.042 C 0.042 12 4 D 0.0001501 kg-m2 Ix axis3 D m3 1 1 0.042 C 0.022 12 4 D 0.0000328 kg-m2 . Therefore, Ix axis D Ix axis1 C Ix axis2 Ix axis3 D 0.000834 kg-m2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 671 Problem 8.134 The object consists of steel of density D 7800 kg/m3 . Determine its moment of inertia about the axis LO . 20 mm O 100 mm Solution: Divide the object into four parts: Part (1) The semicylinder of radius R D 0.02 m, height h1 D 0.01 m. Part (2): The rectangular solid L D 0.1 m by h2 D 0.01 m by w D 0.04 m. Part (3): The semi-cylinder of radius R D 0.02 m, h1 D 0.01 m. Part (4) The cylinder of radius R D 0.02 m, height h D 0.03 m. 10 mm 30 mm LO Part (1) m1 D I1 D R2 h1 D 0.049 kg, 2 m1 R 2 D 4.9 ð 106 kg-m2 , 4 Part (2): m2 D wLh2 D 0.312 kg, I2 D 1 12 m2 L 2 C w2 C m2 2 L D 0.00108 kg-m2 . 2 Part (3) m3 D m1 D 0.049 kg, I3 D 4R 3 2 4R 2 m2 C I 1 C m 3 L D 0.00041179 kg m2 . 3 Part (4) m4 D R2 h D 0.294 kg, I4 D $1% 2 m4 R2 C m4 L 2 D 0.003 kg m2 . The composite: ILo D I1 C I2 I3 C I4 D 0.003674 kg m2 Problem 8.135 Determine the moment of inertia of the object in Problem 8.134 about the axis through the center of mass of the object parallel to LO . Solution: The center of mass is located relative to LO 4R 4R m1 C m2 0.05 m3 0.1 C m4 0.1 3 3 xD m1 C m 2 m 3 C m 4 D 0.066 m, Ic D x2 m C ILo D 0.00265 C 0.00367 D 0.00102 kg m2 672 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.136 The thick plate consists of steel of density D 15 slug/ft3 . Determine its moment of inertia about the z axis. y y 4 in 2 in 2 in x Solution: Divide the object into three parts: Part (1) the rectangle 8 in by 16 in, Parts (2) & (3) the cylindrical cut outs. Part (1): m1 D 8164 D 4.444 slugs. I1 D 1 12 z 4 in 4 in 8 in 4 in 4 in m1 162 C 82 D 118.52 slug in2 . Part (2): m2 D 22 4 D 0.4363 slug, I2 D m2 22 C m2 42 D 7.854 slug in2 . 2 Part (3): m3 D m2 D 0.4363 slugs, I3 D I2 D 7.854 slug-in2 . The composite: Iz axis D I1 2I2 D 102.81 slug-in2 Iz axis D 0.714 slug-ft2 Problem 8.137 Determine the moment of inertia of the plate in Problem 8.136 about the x axis. Solution: Use Problem 8.136. the same divisions of the object as in Part (1): I1x axis D 1 12 m1 82 C 42 D 29.63 slug-in2 , Part (2): I2x axis D 1 12 m2 322 C 42 D 1.018 slug-in2 . The composite: Ix axis D I1x axis 2I2x axis D 27.59 slug in2 D 0.1916 slug ft2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 673 Problem 8.138 Determine Iy and ky . y (1, 1) Solution: 1 dA D dx dyA D dx Iy D A 1 x 2 dx D . 3 x2 x 2 dx 1 dy D 0 0 A ky D 0 1 x 2 dA D Iy D 1 dy D 0 0 y = x2 x2 x 4 dx D 0 x5 5 1 D 0 1 5 x 3 5 Problem 8.139 Determine Ix and kx . Solution: (See figure in Problem 8.138.) dA D dx dy, kx D [x7 ]10 D y 2 dy D 0 0 1 21 x2 dx A D 1 y 2 dA D Ix D 1 3 1 x 6 dx 0 1 21 Ix 1 D p A 7 Problem 8.140 Determine JO and kO . Solution: (See figure in Problem 8.138.) JO D Ix C Iy D kO D 1 26 1 C D , 5 21 105 kx2 C ky2 D 1 3 C D 5 7 26 35 Problem 8.141 Determine Ixy . Solution: (See figure in Problem 8.138.) dA D dx dy D 674 1 2 0 1 x2 x dx 0 A 1 xy dA D Ixy D x 5 dx D y dy 0 1 6 1 1 [x ]0 D 12 12 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.142 Determine Iy and ky . Solution: By definition, y y = x – 1– x 2 4 x 2 dA. Iy D A The element of area is dA D dx dy. The limits on the variable x are 0 x 4. The area is 4 dy D 0 0 4 Iy D x3 x2 2 12 xx2 /4 x 2 dx 4 0 x5 x4 4 20 4 D 2.6667 0 x dy D 0 D xx2 /4 dx AD x 0 x2 4 x 2 dx 4 D 12.8 0 from which ky D Iy D 2.19 A Problem 8.143 Determine Ix and kx . Solution: By definition, y 2 dA, Ix D A from which 4 0 Ix D xx2 /4 dx Ix D y 2 dy D 0 4 3 1 x2 dx x 3 4 0 4 4 1 3 6 x7 x 3 5 D 0.6095. x C x 3 4 20 96 448 0 From Problem 8.142, A D 2.667, kx D Ix D 0.4781 A Problem 8.144 Determine Ixy . Solution: Ixy D xy dA, A 4 x dx D 0 xx2 /4 y dy 0 D 4 2 1 x2 x dx x 2 4 0 D 4 4 x x5 x6 1 C D 2.1333 2 4 10 96 0 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 675 Problem 8.145 Determine Iy 0 and ky 0 . Solution: The limits on the variable x are 0 x 4. By definition, y dA D Ay D 4 y dy 0 4 2 1 x2 D dx x 2 4 0 D y' y = x – 1– x 2 4 xx2 /4 dx 0 A y x' x 3 4 x4 x5 1 x C D 1.06667. 2 3 8 80 0 From Problem 8.142 the area is A D 2.667, from which y D 0.3999 D 0.4. Similarly, 4 XX2 /4 x dx Ax D 0 dy 0 4 D 0 3 4 x2 x x4 x x D 5.3333, dx D 4 3 16 0 from which x D 1.9999 D 2. The area moment of inertia is Iyy D x2 A C Iy . Using the result of Problem 8.142, Iy D 12.8, from which the area moment of inertia about the centroid is Iy 0 D 10.6666 C 12.8 D 2.133 and ky 0 D Iy 0 D 0.8944 A Problem 8.146 Determine Ix0 and kx0 . Solution: Using the results of Problems 8.143 and 8.145, Ix D 0.6095 and y D 0.4. The area moment of inertia about the centroid is Ix0 D y2 A C Ix D 0.1828 and k D x0 Ix 0 D 0.2618 A Problem 8.147 Determine Ix0 y 0 . Solution: From Problems 8.143 and 8.144, Ixy D 2.133 and x D 2, y D 0.4. The product of the moment of inertia about the centroid is Ix0 y 0 D xyA C Ixy D 2.133 C 2.133 D 0 676 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. y Problem 8.148 Determine Iy and ky . Solution: Divide the section into two parts: Part (1) is the upper rectangle 40 mm by 200 mm, Part (2) is the lower rectangle, 160 mm by 40 mm. Part (1) A1 D 0.0400.200 D 0.008 m2 , y1 D 0.180 m 40 mm 160 mm x1 D 0, Iy1 D 1 12 0.040.23 D 2.6667 ð 105 m4 . x 80 mm 40 mm 80 mm Part (2): A2 D 0.040.16 D 0.0064 m2 , y2 D 0.08 m, x2 D 0, Iy2 D 1 12 0.160.043 D 8.5 ð 107 m4 . The composite: A D A1 C A2 D 0.0144 m2 , Iy D Iy1 C Iy2 , Iy D 2.752 ð 105 m4 D 2.752 ð 107 mm4 , and ky D Iy D 0.0437 m D 43.7 mm A Problem 8.149 Determine Ix and kx for the area in Problem 8.148. Solution: Use the results in the solution to Problem 8.148. Part (1) A1 D 0.0400.200 D 0.008 m2 , y1 D 0.180 m, Ix1 D 1 12 0.20.043 C 0.182 A1 D 2.603 ð 104 m4 . Part (2): A2 D 0.040.16 D 0.0064 m2 , y2 D 0.08 m, Ix2 D 1 12 0.040.163 C 0.082 A2 D 5.461 ð 105 m4 . The composite: A D A1 C A2 D 0.0144 m2 , The area moment of inertia about the x axis is Ix D Ix1 C Ix2 D 3.15 ð 104 m4 D 3.15 ð 108 mm4 , and kx D Ix D 0.1479 m D 147.9 mm A c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 677 y Problem 8.150 Determine Ix and kx . Solution: Use the results of the solutions to Problems 8.148– 40 mm 8.149. The centroid is located relative to the base at xc D x1 A1 C x2 A2 D 0, A yc D y1 A1 C y2 A2 D 0.1356 m. A x 160 mm The moment of inertia about the x axis is 80 mm Ixc D y2C A C IX D 5.028 ð 107 mm4 and kxc D 40 mm 80 mm Ixc D 59.1 mm A Problem 8.151 Determine JO and kO for the area in Problem 8.150. Solution: Use the results of the solutions to Problems 8.148– 8.149. The area moments of inertia about the centroid are Ixc D 5.028 ð 105 m4 and Iyc D Iy D 2.752 ð 105 m4 , from which JO D Ixc C Iyc D 7.78 ð 105 m4 D 7.78 ð 107 mm4 and kO D JO D 0.0735 m A D 73.5 mm Problem 8.152 Determine Iy and ky . y Solution: For a semicircle about a diameter: Iyy D Ixx D Iy D 1 1 44 24 D 44 24 D 94.25 ft4 , 8 8 8 ky D 678 1 R4 , 8 2 ft x 4 ft 2Iy D 2.236 ft 42 22 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.153 Determine JO and kO . for the area in Problem 8.152. Solution: For a semicircle: Iyy D Ixx D 1 R4 . 8 4 4 24 D 94.248 ft4 . 8 Ix D kx D 2Ix D 2.236 ft. 42 22 Also use the solution to Problem 8.152. JO D Ix C Iy D 294.248 D 188.5 ft4 kO D 2JO D 3.16 ft 42 22 Problem 8.154 Determine Ix and kx . Solution: Break the area into three parts: Part (1) The rectangle with base 2a and altitude h; Part (2) The triangle on the right with base b a and altitude h, and Part (3) The triangle on the left with base b a and altitude h. Part (1) The area is y 3 ft 3 ft 6 ft A1 D 2ah D 24 ft2 . The centroid is x x1 D 0 2 ft h and y1 D D 3 ft. 2 2 ft y b The area moment of inertia about the centroid is Ixc1 D 1 12 2ah3 D Part (2): A2 D x2 D a C y2 D ba D 2.3333 ft, 3 1 36 h 1 hb a D 3 ft2 , 2 2 h D 4 ft, 3 Ixc2 D 1 ah3 D 72 ft4 . 6 a x The composite moment of inertia Ix D y1 2 A1 C Ixc1 C y2 2 A2 C Ixc2 C y3 2 A3 C Ixc3 , Ix D 396 ft4 b ah3 D 6 ft4 . kx D Ix D A 396 D 3.633 ft 30 Part (3): A3 D A2 , x3 D x2 , y3 D y2 , Ixc3 D Ixc2 . The composite area is A D A1 C A2 C A2 D 30 ft2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 679 Problem 8.155 Determine Iy and ky for the area in Problem 8.154. Solution: Divide the area as in the solution to Problem 8.154. Part (1) The area is A1 D 2ah D 24 ft2 . The centroid is x1 D 0 and h y1 D D 3 ft. The area moment of inertia about the centroid is 2 Iyc1 D 1 12 h2a3 D Part (2): A2 D x2 D a C y2 D 2 ha3 D 32 ft4 3 1 hb a D 3 ft2 , 2 ba D 2.3333 ft, 3 2 h D 4 ft, 3 Iyc2 D 1 36 hb a3 D 0.1667 ft4 . Part (3): A3 D A2 , x3 D x2 , y3 D y2 , Iyc3 D Iyc2 . The composite area is A D A1 C A2 C A2 D 30 ft2 . The composite moment of inertia, Iy D x21 A1 C Iyc1 C x22 A2 C Iyc2 C x23 A3 C Iyc3 , Iy D 65 ft4 and ky D Iy D 1.472 ft A Problem 8.156 The moments of inertia of the area are Ix D 36 m4 , Iy D 145 m4 , and Ixy D 44.25 m4 . Determine a set of principal axes and the principal moment of inertia. Solution: The principal angle is D 1 2Ixy tan1 D 19.54° . 2 Iy Ix y 3m 4m 3m x The principal moments of inertia are IxP D Ix cos2 2Ixy sin cos C Iy sin2 D 20.298 D 20.3 m4 IyP D Ix sin2 C 2Ixy sin cos C Iy cos2 D 160.70 m4 680 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.157 The moment of inertia of the 31-oz bat about a perpendicular axis through point B is 0.093 slugft2 . What is the bat’s moment of inertia about a perpendicular axis through point A? (Point A is the bat’s “instantaneous center,” or center of rotation, at the instant shown.) 31 D 0.06023 slugs. 1632.17 Use the parallel axis theorem to obtain the moment of inertia about the center of mass C, and then use the parallel axis theorem to translate to the point A. Solution: The mass of the bat is m D IC D IA D 12 12 2 12 C 14 12 m C 0.093 D 0.0328 slug-ft2 2 m C 0.0328 D 0.3155 slug-ft2 C 12 in B 14 in A Problem 8.158 The mass of the thin homogenous plate is 4 kg. Determine its moment of inertia about the y axis. Solution: Divide the object into two parts: Part (1) is the semicircle of radius 100 mm, and Part (2) is the rectangle 200 mm by 280 mm. The area of Part (1) A1 D y 100 mm 140 mm x R2 D 15708 mm2 . 2 140 mm The area of Part (2) is 200 mm A2 D 280200 D 56000 mm2 . The composite area is A D A2 A1 D 40292 mm2 . The area mass density is D 4 D 9.9275 ð 105 kg/mm2 . A For Part (1) x1 D y1 D 0, Iy1 D 1 R4 D 3898.5 kg-mm2 . 8 For Part (2) x2 D 100 mm. Iy2 D x22 A2 C 1 12 2802003 D 74125.5 kg-mm2 . The composite: Iy D Iy2 Iy1 D 70226 kg-mm2 D 0.070226 kg-m2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 681 Problem 8.159 Determine the moment of inertia of the plate in Problem 8.158 about the z axis. Solution: Use the same division of the parts and the results of the solution to Problem 8.158. For Part (1), Ix1 D 1 R4 D 3898.5 kg-mm2 . 8 For Part (2) Ix2 D 1 12 2002803 D 36321.5 kg-mm2 . The composite: Ix D Ix2 Ix1 D 32423 kg-mm2 , from which, using the result of the solution to Problem 8.158 Iz D Ix C Iy D 32422 C 70226 D 102649 kg-mm2 D 0.10265 kg-m2 y Problem 8.160 The homogenous pyramid is of mass m. Determine its moment of inertia about the z axis. Solution: The mass density is D x 3m m D 2 . V w h h The differential mass is dm D ω2 dz. The moment of inertia of this element about the z axis is z 1 dIZ D ω2 dm. 6 Noting that ω D dIz D w4 6h4 z4 dz D y y wz , then h mw2 4 z dz. 2h5 Integrating: Iz axis D x w z w h 2 mw 2h5 h z4 dz D 0 1 mw2 10 Problem 8.161 Determine the moment of inertia of the homogenous pyramid in Problem 8.160 about the x and y axes. Solution: Use the results of the solution of Problem 8.160 for the Noting that ω D mass density. The elemental disk is dm D ω2 dz. The moment of inertia about an axis through its center of mass parallel to the x axis is dIX D 1 12 Ix axis D w4 12h4 ω2 dm. Ix 682 axis D 1 12 h z4 dz C 0 w2 h2 h z4 dz. 0 Integrating and collecting terms Use the parallel axis theorem: w z, the integral is h Ix ω2 dm C m axis Dm 1 2 3 2 w C h . 20 5 z2 dm. m By symmetry, Iy axis D Ix axis c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 8.162 The homogenous object weighs 400 lb. Determine its moment of inertia about the x axis. y y Solution: The volumes are Vcyl D 4692 z 6 in 9 in 3 D 11,706 in , Vcone D 13 62 36 D 1357in3 , 36 in x 36 in 46 in x 46 in so V D Vcyl Vcone D 10,348 in3 . The masses of the solid cylinder and the material that would occupy the conical hole are mcyl D mcone D Vcyl V Vcone V 400 32.2 400 32.2 D 14.052 slug, D 1.629 slug. Using results from Appendix C, Ix axis D 1 3 mcyl 92 mcone 62 2 10 D 551 slug-in2 D 3.83 slug-ft2 Problem 8.163 Determine the moments of inertia of the object in Problem 8.162 about the y and z axes. Solution: See the solution of Problem 8.162. The position of the center of mass of the material that would occupy the conical hole is 3 36 D 37 in. 4 x D 46 36 C 6 in x , x′ From Appendix C, X Iy 0 axiscone y′ y D mcone 3 3 362 C 62 80 20 36 in D 87.97 slug-in2 . The moment of inertia about the y axis for the composite object is Iy axis D mcyl )1 2 3 46 $ Iy 0 C 14 92 axiscone * C x2 mcone % D 7877 slug-in2 D 54.7 slug-ft2 . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 683 Problem 8.164 Determine the moment of inertia of the 14-kg flywheel about the axis L. 50 mm 70 mm 120 mm L 100 mm 440 mm 500 mm 150 mm Solution: The flywheel can be treated as a composite of the objects shown: The volumes are = – + – + 4 V1 D 1502502 D 294.5 ð 105 mm3 , 1 2 – 5 6 3 V2 D 1502202 D 228.08 ð 105 mm3 , V3 D 502202 D 76.03 ð 105 mm3 , V4 D 50602 D 5.65 ð 105 mm3 , V5 D 100602 D 11.31 ð 105 mm3 , V6 D 100352 D 3.85 ð 105 mm3 . The volume V D V1 V2 C V3 V4 C V5 V6 D 144.3 ð 105 mm3 , so the density is υD 14 D 9.704 ð 107 kg/mm3 . V The moment of inertia is IL D 12 υV1 2502 12 υV2 2202 C 12 υV3 2202 12 υV4 602 C 12 υV5 602 12 υV6 352 D 536,800 kg-mm2 D 0.5368 kg-m2 . 684 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 9.1 In Active Example 9.1, suppose that the coefficient of static friction between the 180-lb crate and the ramp is s D 0.3. What is the magnitude of the smallest horizontal force the rope must exert on the crate to prevent it from sliding down the ramp? 20⬚ Solution: The free-body diagram is shown. Assume that the crate is on the verge of slipping down the ramp. Then the friction force is f D s N and points up the ramp. The equilibrium equations are y x T 20⬚ Fx : s N C T cos 20° W sin 20° D 0, Fy : N T sin 20° W cos 20° D 0. Setting W D 180 lb and s D 0.3 and solving yields W f N N D 173 lb, T D 10.4 lb. 10.4 lb. Problem 9.2 A person places a 2-lb book on a table that is tilted at 5° relative to the horizontal. She finds that if she exerts a very small force on the book as shown, the book remains in equilibrium, but if she removes the force, the book slides down the table. What force would she need to exert on the book (in the direction parallel to the table) to cause it to slide up the table? 15⬚ Solution: If the person can hold the book in equilibrium with a very small force, but the book slips when she removes the force, the maximum friction force must be nearly equal to the value necessary to maintain the book in equilibrium. Assume that the book is in equilibrium and f D s N (Fig. a). The equilibrium equations are Fx : f 2 lb sin 15° D 0, Fy : N 2 lb cos 15° D 0. The coefficient of friction is then s D f D tan 15° D 0.268. N Now assume that the person exerts a force F on the book that is parallel to the table and slip up the table is impending (Fig. b). Then the friction force f D s N opposes the impending motion and the equilibrium equations are Fx : F f 2 lb sin 15° D 0, Fy : N 2 lb cos 15° D 0. Solving yields F D 1.04 lb . c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 685 Problem 9.3 A student pushes a 200-lb box of books across the floor. The coefficient of kinetic friction between the carpet and the box is k D 0.15. (a) (b) If he exerts the force F at angle ˛ D 25° , what is the magnitude of the force he must exert to slide the box across the floor? If he bends his knees more and exerts the force F at angle ˛ D 10° , what is the magnitude of the force he must exert to slide the box? F F Solution: a 200 lb Fx : F cos ˛ f D 0 α Fy : N 200 lb F sin ˛ D 0 f D 0.15 N (a) ˛ D 25° ) F D 35.6 lb f (b) ˛D 10° ) F D 31.3 lb N Problem 9.4 The 2975-lb car is parked on a sloped street. The brakes are applied to both its front and rear wheels. (a) (b) If the coefficient of static friction between the car’s tires and the road is s D 0.8, what is the steepest slope (in degrees relative to the horizontal) on which the car could remain in equilibrium? If the street were icy and the coefficient of static friction between the car’s tires and the road was s D 0.2, what is the steepest slope on which the car could remain in equilibrium? Solution: Let ˛ be the slope of the street in degrees. The equilibrium equations and impending slip friction equation are Fx : W sin ˛ f D 0, Fy : N W cos ˛ D 0, f D s N Solving, we find that f D W sin ˛, N D W cos ˛, s D tan ˛. ˛ D tan1 s . a ˛ D tan1 0.8 D 38.7° . b ˛ D tan1 0.2 D 11.3° . a ˛ D 38.7° , b ˛ D 11.3° . 686 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 9.5 The truck’s winch exerts a horizontal force on the 200-kg crate in an effort to pull it down the ramp. The coefficient of static friction between the crate and the ramp is s D 0.6. 20⬚ (a) If the winch exerts a 200-N horizontal force on the crate, what is the magnitude of the friction force exerted on the crate by the ramp? (b) What is the magnitude of the horizontal force the winch must exert on the crate to cause it to start moving down the ramp? Solution: Assume the crate doesn’t slip 1962 N F% : N 1962 N cos 20° C 200 N sin 20° D 0 F- : f 1962 N sin 20° 200 N cos 20° D 0 20° fmax D 0.6 N F = 200 N f (a) Solving N N D 1775 N, f D 859 N, fmax D 1065 N Sincef < fmax , (b) f D 859 N F% : N 1962 N cos 20° C F sin 20° D 0 F- : f 1962 N sin 20° F cos 20° D 0 ) F D 380 N f D 0.6 N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 687 Problem 9.6 The device shown is designed to position pieces of luggage on a ramp. It exerts a force parallel to the ramp. The suitcase weighs 40 lb. The coefficients of friction between the suitcase and ramp are s D 0.20 and k D 0.18. 20⬚ (a) Will the suitcase remain stationary on the ramp when the device exerts no force on it? (b) What force must the device exert to push the suitcase up the ramp at a constant speed? Solution: (a) Assume that the suitcase is in equilibrium with no external force exerted on it (Fig. a). From the equilibrium equations Fx : f W sin 20° D 0, Fy : N W cos 20° D 0, we obtain f D W sin 20° D 13.7 lb, N D W cos 20° D 37.6 lb. The maximum friction force fmax D s N D 0.237.6 lb D 7.25 lb is less than the friction force necessary for equilibrium, so the suitcase will not remain in equilibrium with no force exerted on it. (b) Now assume that the device exerts a force F on the suitcase and is pushing it up the ramp at a constant speed (Fig. b). Then the friction force f D k N opposes the motion and the equilibrium equations are Fx : F k N W sin 20° D 0, Fy : N W cos 20° D 0. Solving yields N D 37.6 lb, F D 20.4 lb. a No, b 20.4 lb. 688 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 9.7 The coefficient of static friction between the 50-kg crate and the ramp is s D 0.35. The unstretched length of the spring is 800 mm, and the spring constant is k D 660 N/m. What is the minimum value of x at which the crate can remain stationary on the ramp? x k 50⬚ Solution: 490.5 N Fs D 660 N/m0.8 m x F% : Fs 490.5 N sin 50° C f D 0 50° F- : N 490.5 N cos 50° D 0 N f D 0.35 N f Solving: x D 0.398 m D 398 mm Fs Problem 9.8 The coefficient of kinetic friction between the 40-kg crate and the slanting floor is k D 0.3. If the angle ˛ D 20° , what tension must the person exert on the rope to move the crate at constant speed? a 10⬚ Solution: 392.4 N T ˛ D 20° , k D 0.3 α F% : T cos ˛ f 392.4 N sin 10° D 0 F- : T sin ˛ C N 392.4 N cos 10° D 0 f f D 0.3 N 10° Solving T D 177 N N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 689 Problem 9.9 In Problem 9.8, for what angle ˛ is the tension necessary to move the crate at constant speed a minimum? What is the necessary tension? Solution: See figure for 9.8 k D 0.3 F% : T cos ˛ f 392.4 N sin 10° D 0 F- : T sin ˛ C N 392.4 N cos 10° D 0 f D 0.3 N ) TD 184.1 N cos ˛ C 0.3 sin ˛ To find the angle for minimum T 184.1 Nsin ˛ 0.3 cos ˛ dT D D 0 ) tan ˛ D 0.3 d˛ cos ˛ 0.3 sin ˛2 ˛ D 16.7° ) T D 176.3 N Problem 9.10 Box A weighs 100 lb, and box B weighs 30 lb. The coefficients of friction between box A and the ramp are s D 0.30 and k D 0.28. What is the magnitude of the friction force exerted on box A by the ramp? A B 30° Solution: The sum of the forces parallel to the inclined surface is B F D A sin ˛ C B C f D 0, α A from which f D A sin ˛ B D 100 sin 30° 30 D 20 lb f N 690 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 9.11 In Problem 9.10, box A weighs 100 lb, and the coefficients of friction between box A and the ramp are s D 0.30 and k D 0.28. For what range of the weights of the box B will the system remain stationary? Solution: The upper and lower limits on the range are determined by the weight required to move the box up the ramp, and the weight that will allow the box to slip down the ramp. Assume impending slip. The friction force opposes the impending motion. For impending motion up the ramp the sum of forces parallel to the ramp are BMAX α A N µ sN F D A sin ˛ BMAX C S A cos ˛ D 0, BMIN from which α A BMAX D Asin ˛ C s cos ˛ µ sN D 100sin 30° C 0.3 cos 30° D 75.98 lb N For impending motion down the ramp: F D A sin ˛ BMIN s A cos ˛ D 0, from which B D Asin ˛ s cos ˛ D 100sin 30° 0.3 cos 30° D 24.02 lb Problem 9.12 The mass of the box on the left is 30 kg, and the mass of the box on the right is 40 kg. The coefficient of static friction between each box and the inclined surface is s D 0.2. Determine the minimum angle ˛ for with the boxes will remain stationary. a Solution: If the boxes slip when ˛ is decreased, they will slip toward the right. Assume that slip toward the right impends, the free body diagrams are as shown. y y T α x T 30° (40)(9.81) (30)(9.81) The equilibrium equations are 30⬚ 0.2 NB Fx D T 0.2 NA 309.81 sin ˛ D 0, (1) Fy D NA 309.81 cos ˛ D 0, (2) Fx D T 0.2 NB C 409.81 sin 30° D 0, (3) Fy D NB 409.81 cos 30° D 0, (4) 0.2 NA NA NB x Summing Equations (1) and (3), we obtain 0.2 NA 0.2 NB 309.81 sin ˛ C 409.81 sin 30° D 0. Solving Equation (2) for NA and Equation (4) for NB and substituting the results into ° Equation (5) gives 15 sin ˛ C 3 cos ˛ D 10 4 cos 30 . (6) Using the identity cos ˛ D 1 sin2 ˛ and solving Equation (6) for sin ˛, we obtain sin ˛ D 0.242, so ˛ D 14.0° c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 691 Problem 9.13 The coefficient of kinetic friction between the 100-kg box and the inclined surface is 0.35. Determine the tension T necessary to pull the box up the surface at a constant rate. T 60⬚ 3T Solution: F- : 3T 981 N sin 60° f D 0 981 N F% : N 981 N cos 60° D 0 60° f D 0.35 N Solving: T D 340 N N f 692 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 9.14 The box is stationary on the inclined surface. The coefficient of static friction between the box and the surface is s . (a) (b) If the mass of the box is 10 kg, ˛ D 20° , ˇ D 30° , and s D 0.24, what force T is necessary to start the box sliding up the surface? Show that the force T necessary to start the box sliding up the surface is a minimum when tan ˇ D s . T β α Solution: T mg y α β α f x N ˛ D 20° s D 0.24 m D 10 kg g D 9.81 m/s2 (a) Fx : T cos ˇ C f C mg sin ˛ D 0 Fy : N C T sin ˇ mg cos ˛ D 0 ˇ D 30° , f D s N Substituting the known values and solving, we get T D 56.5 N, N D 64.0 N, f D 15.3 N Solving the 2nd equilibrium eqn for N and substituting for f f D s N in the first eqn, we get T cos ˇ C s mg cos ˛ s T sin ˇ C mg sin ˛ D 0 Differentiating with respect to ˇ, we get Tsin ˇ s cos ˇ dT D dˇ cos ˇ C s sin ˇ Setting dT D 0, we get dˇ tan ˇ D s c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 693 Problem 9.15 To explain observations of ship launchings at the port of Rochefort in 1779, Coulomb analyzed the system shown in Problem 9.14 to determine the minimum force T necessary to hold the box stationary on the inclined surface. Show that the result is TD sin ˛ s cos ˛mg . cos ˇ s sin ˇ Solution: mg y T β α α x N f = µsN Fx : T cos ˇ C mg sin ˛ s N D 0 Fy : N C T sin ˇ mg cos ˛ D 0 ˛ is fixed, ˇ is variable. Solve the second eqn for N and substitute into the first. We get 0 D Ts sin ˇ cos ˇ D mgsin ˛ s cos ˛ or T D mgsin ˛ s cos ˛ cos ˇ s sin ˇ To get the conditions for the minimum, set dT D0 dˇ Tsin ˇ C s cos ˇ dT D D0 dˇ cos ˇ s sin ˇ For the min. tan ˇ D s . Note ˇ is negative! 694 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 9.16 Two sheets of plywood A and B lie on the bed of a truck. They have the same weight W, and the coefficient of static friction between the two sheets of wood and between sheet B and the truck bed is s . (a) (b) If you apply a horizontal force to sheet A and apply no force to sheet B, can you slide sheet A off the truck without causing sheet B to move? What force is necessary to cause sheet A to start moving? If you prevent sheet A from moving by applying a horizontal force on it, what horizontal force on sheet B is necessary to start it moving? A B Solution: (a) The friction force exerted by sheet A on B at impending motion is fAB D s W. The friction force exerted by sheet B on the bed of the truck is fBT D s 2W, since the normal force is due to the weight of both sheets. Since fBT > fAB , the top sheet will begin moving before the bottom sheet. Yes fAB (a) fAB W fBT fAB The force required to start sheet A to move is (b) 2W W FB F D fAB D s W. (b) F W fBT 2W The force on B is the friction between A and B and the friction between B and the truck bed. Thus the force required to start B in motion is FB D fAB C fBT D 3s W. Problem 9.17 The weights of the two boxes are W1 D 100 lb and W2 D 50 lb. The coefficients of kinetic friction between the left box and the inclined surface are s D 0.12 and k D 0.10. Determine the tension the man must exert on the rope to pull the boxes upward at a constant rate. 30⬚ W1 30⬚ W2 Solution: F- : T 100 lb sin 30° f 50 lb D 0 100 lb T F% : N 100 lb cos 30° D 0 f D 0.10 N Solving: 30° 50 lb T D 109 lb f N c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 695 Problem 9.18 In Problem 9.17, for what range of tensions exerted on the rope by the man will the boxes remain stationary? Solution: See the figure in 9.17. First solve for the largest force Tmax F- : Tmax 100 lb sin 30° f 50 lb D 0 F% : N 100 lb cos 30° D 0 ) Tmax D 110.4 lb f D 0.12 N Next solve for the smallest force Tmin . We need to turn the friction force in the opposite direction. F- : Tmin 100 lb sin 30° C f 50 lb D 0 F% : N 100 lb cos 30° D 0 ) Tmin D 89.6 lb f D 0.12 N Thus for the boxes to remain stationary we must have 89.6 lb < T < 110.4 lb Problem 9.19 Each box weighs 10 lb. The coefficient of static friction between box A and box B is 0.24, and the coefficient of static friction between box B and the inclined surface is 0.3. What is the largest angle ˛ for which box B will not slip? A B Strategy: Draw individual free-body diagrams of the two boxes and write their equilibrium equations assuming that slip of box B is impending. a Solution: We have 6 unknowns, 4 equilibrium equations and 2 10 lb T friction equations FA% : T 10 lb sin ˛ f2 D 0 FA- : N2 10 lb cos ˛ D 0 A f2 FB% : f2 C f1 10 lb sin ˛ D 0 N2 10 lb FB- : N1 N2 10 lb cos ˛ D 0 f1 B f1 D 0.3N1 , f2 D 0.24N2 Solving we find ˛ D 40.0° α N1 696 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. Problem 9.20 The masses of the boxes are mA D 15 kg and mB D 60 kg. The coefficient of static friction between boxes A and B and between box B and the inclined surface is 0.12. What is the largest force F for which the boxes will not slip? A F B 20⬚ Solution: We have 6 unknowns, 4 equilibrium equations and 2 friction equations. 147.15 N T FA% : T F 147.15 N sin 20° C f2 D 0 FA- : N2 147.15 N cos 20° D 0 A f2 FB% : T 588.6 N sin 20° f1 f2 D 0 T F N2 FB- : N1 N2 588.6 N cos 20° D 0 B f1 D 0.12N1 , f2 D 0.12N2 Solving we find 588.6 N f1 F D 267 N 20° N1 Problem 9.21 In Problem 9.20, what is the smallest force F for which the boxes will not slip? Solution: See the solution for 9.20 — change the directions of all of the friction forces. FA% : T F 147.15 N sin 20° f2 D 0 FA- : N2 147.15 N cos 20° D 0 FB% : T 588.6 N sin 20° C f1 C f2 D 0 ) F D 34.8 N FB- : N1 N2 588.6 N cos 20° D 0 f1 D 0.12N1 , f2 D 0.12N2 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. 697 Problem 9.22 In Example 9.2, what clockwise couple M would need to be applied to the disk to cause it to rotate at a constant rate in the clockwise direction? M C 1 h 2 1 h 2 r E D B A b Solution: Assume that the disk is rotating in the clockwise direction. From the free-body diagram of the disk, MD : M C R sin k r D 0. From the free-body diagram of the brake, MA : F 12 h C R cos k h C R sin k b D 0. Solving these two equations yields MD 1 2 hrFk . h C bk Problem 9.23 The homogeneous horizontal bar AB weighs 20 lb. The homogeneous disk weighs 30 lb. The coefficient of kinetic friction between the disk and the sloping surface is k D 0.24. What is the magnitude of the couple that would need to be applied to the disk to cause it to rotate at a constant rate in the clockwise direction? 5 ft 1 ft A B 20⬚ Solution: From the free-body diagram of the bar, MB : 20 lb2.5 ft C Ay 5 ft D 0 ) Ay D 10 lb. From the free-body diagram of the disk. Fx : Ax C N sin 20° C k N cos 20° D 0, Fy : N cos 20° k N sin 20° 30 lb D 0, MA : M C k N1 ft D 0. Solving yields Ax D 26.5 lb, N D 46.6 lb, M D 11.2 ft-lb. M D 11.2 ft-lb. 698 c 2008 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced