Eighth Edition Vector Mechanics for Engineers: Statics CE 102 Statics Chapter 4 Equilibrium of Rigid Bodies © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-1 Eighth Edition Vector Mechanics for Engineers: Statics Contents Introduction Free-Body Diagram Equilibrium of a Rigid Body in Three Dimensions Reactions at Supports and Connections for a Two-Dimensional Structure Reactions at Supports and Connections for a Three-Dimensional Structure Equilibrium of a Rigid Body in Two Dimensions Sample Problem 4.5 Statically Indeterminate Reactions Sample Problem 4.1 Sample Problem 4.2 Sample Problem 4.3 Equilibrium of a Two-Force Body Equilibrium of a Three-Force Body Sample Problem 4.4 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-2 Eighth Edition Vector Mechanics for Engineers: Statics Introduction • For a rigid body in static equilibrium, the external forces and moments are balanced and will impart no translational or rotational motion to the body. • The necessary and sufficient condition for the static equilibrium of a body are that the resultant force and couple from all external forces form a system equivalent to zero, F 0 M O r F 0 • Resolving each force and moment into its rectangular components leads to 6 scalar equations which also express the conditions for static equilibrium, Fx 0 Fy 0 Fz 0 Mx 0 My 0 Mz 0 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-3 Eighth Edition Vector Mechanics for Engineers: Statics Free-Body Diagram First step in the static equilibrium analysis of a rigid body is identification of all forces acting on the body with a free-body diagram. • Select the extent of the free-body and detach it from the ground and all other bodies. • Indicate point of application, magnitude, and direction of external forces, including the rigid body weight. • Indicate point of application and assumed direction of unknown applied forces. These usually consist of reactions through which the ground and other bodies oppose the possible motion of the rigid body. • Include the dimensions necessary to compute the moments of the forces. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-4 Eighth Edition Vector Mechanics for Engineers: Statics Reactions at Supports and Connections for a TwoDimensional Structure • Reactions equivalent to a force with known line of action. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-5 Eighth Edition Vector Mechanics for Engineers: Statics Reactions at Supports and Connections for a TwoDimensional Structure • Reactions equivalent to a force of unknown direction and magnitude. • Reactions equivalent to a force of unknown direction and magnitude and a couple.of unknown magnitude © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-6 Eighth Edition Vector Mechanics for Engineers: Statics Equilibrium of a Rigid Body in Two Dimensions • For all forces and moments acting on a twodimensional structure, Fz 0 M x M y 0 M z M O • Equations of equilibrium become Fx 0 Fy 0 M A 0 where A is any point in the plane of the structure. • The 3 equations can be solved for no more than 3 unknowns. • The 3 equations can not be augmented with additional equations, but they can be replaced Fx 0 M A 0 M B 0 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-7 Eighth Edition Vector Mechanics for Engineers: Statics Statically Indeterminate Reactions • More unknowns than equations • Fewer unknowns than • Equal number unknowns and equations but equations, partially improperly constrained constrained © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-8 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.1 SOLUTION: • Create a free-body diagram for the crane. • Determine B by solving the equation for the sum of the moments of all forces about A. Note there will be no contribution from the unknown reactions at A. A fixed crane has a mass of 1000 kg and is used to lift a 2400 kg crate. It is held in place by a pin at A and a rocker at B. The center of gravity of the crane is located at G. Determine the components of the reactions at A and B. • Determine the reactions at A by solving the equations for the sum of all horizontal force components and all vertical force components. • Check the values obtained for the reactions by verifying that the sum of the moments about B of all forces is zero. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4-9 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.1 • Determine B by solving the equation for the sum of the moments of all forces about A. M A 0 : B1.5m 9.81 kN2m 23.5 kN6m 0 B 107.1 kN • Create the free-body diagram. • Determine the reactions at A by solving the equations for the sum of all horizontal forces and all vertical forces. Fx 0 : Ax B 0 Ax 107.1kN Fy 0 : Ay 9.81kN 23.5 kN 0 Ay 33.3 kN • Check the values obtained. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 10 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.2 SOLUTION: • Create a free-body diagram for the car with the coordinate system aligned with the track. • Determine the reactions at the wheels by solving equations for the sum of moments about points above each axle. • Determine the cable tension by A loading car is at rest on an inclined solving the equation for the sum of track. The gross weight of the car and force components parallel to the track. its load is 5500 lb, and it is applied at at G. The cart is held in position by • Check the values obtained by verifying the cable. that the sum of force components perpendicular to the track are zero. Determine the tension in the cable and the reaction at each pair of wheels. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 11 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.2 • Determine the reactions at the wheels. M A 0 : 2320 lb 25in. 4980 lb 6in. R2 50in. 0 R2 1758 lb M B 0 : 2320 lb 25in. 4980 lb 6in. R1 50in. 0 R1 562 lb • Create a free-body diagram W x 5500 lb cos 25 4980 lb W y 5500 lb sin 25 • Determine the cable tension. Fx 0 : 4980 lb T 0 T 4980 lb 2320 lb © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 12 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.3 SOLUTION: • Create a free-body diagram for the frame and cable. • Solve 3 equilibrium equations for the reaction force components and couple at E. The frame supports part of the roof of a small building. The tension in the cable is 150 kN. Determine the reaction at the fixed end E. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 13 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.3 • Solve 3 equilibrium equations for the reaction force components and couple. 4.5 150 kN 0 F 0 : E x x 7.5 E x 90.0 kN Fy 0 : E y 420 kN 6 150 kN 0 7.5 E y 200 kN • Create a free-body diagram for the frame and cable. M E 0 : 20 kN 7.2 m 20 kN 5.4 m 20 kN 3.6 m 20 kN 1.8 m 6 150 kN 4.5 m M E 0 7.5 M E 180.0 kN m © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 14 Eighth Edition Vector Mechanics for Engineers: Statics Equilibrium of a Two-Force Body • Consider a plate subjected to two forces F1 and F2 • For static equilibrium, the sum of moments about A must be zero. The moment of F2 must be zero. It follows that the line of action of F2 must pass through A. • Similarly, the line of action of F1 must pass through B for the sum of moments about B to be zero. • Requiring that the sum of forces in any direction be zero leads to the conclusion that F1 and F2 must have equal magnitude but opposite sense. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 15 Eighth Edition Vector Mechanics for Engineers: Statics Equilibrium of a Three-Force Body • Consider a rigid body subjected to forces acting at only 3 points. • Assuming that their lines of action intersect, the moment of F1 and F2 about the point of intersection represented by D is zero. • Since the rigid body is in equilibrium, the sum of the moments of F1, F2, and F3 about any axis must be zero. It follows that the moment of F3 about D must be zero as well and that the line of action of F3 must pass through D. • The lines of action of the three forces must be concurrent or parallel. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 16 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.4 SOLUTION: • Create a free-body diagram of the joist. Note that the joist is a 3 force body acted upon by the rope, its weight, and the reaction at A. A man raises a 10 kg joist, of length 4 m, by pulling on a rope. Find the tension in the rope and the reaction at A. • The three forces must be concurrent for static equilibrium. Therefore, the reaction R must pass through the intersection of the lines of action of the weight and rope forces. Determine the direction of the reaction force R. • Utilize a force triangle to determine the magnitude of the reaction force R. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 17 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.4 • Create a free-body diagram of the joist. • Determine the direction of the reaction force R. AF AB cos 45 4 m cos 45 2.828 m CD AE 12 AF 1.414 m BD CD cot(45 25) 1.414 m tan 20 0.515 m CE BF BD 2.828 0.515 m 2.313 m tan CE 2.313 1.636 AE 1.414 58.6 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 18 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.4 • Determine the magnitude of the reaction force R. T sin 31.4 R sin 110 98.1 N sin 38.6 T 81.9 N R 147.8 N © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 19 Eighth Edition Vector Mechanics for Engineers: Statics Equilibrium of a Rigid Body in Three Dimensions • Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three dimensional case. Fx 0 Fy 0 Fz 0 Mx 0 My 0 Mz 0 • These equations can be solved for no more than 6 unknowns which generally represent reactions at supports or connections. • The scalar equations are conveniently obtained by applying the vector forms of the conditions for equilibrium, F 0 M O r F 0 © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 20 Eighth Edition Vector Mechanics for Engineers: Statics Reactions at Supports and Connections for a ThreeDimensional Structure © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 21 Eighth Edition Vector Mechanics for Engineers: Statics Reactions at Supports and Connections for a ThreeDimensional Structure © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 22 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.5 SOLUTION: • Create a free-body diagram for the sign. • Apply the conditions for static equilibrium to develop equations for the unknown reactions. A sign of uniform density weighs 270 lb and is supported by a ball-andsocket joint at A and by two cables. Determine the tension in each cable and the reaction at A. © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 23 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.5 TBD TBD TBD TBD TEC TEC • Create a free-body diagram for the sign. TEC Since there are only 5 unknowns, the sign is partially constrain. It is free to rotate about the x axis. It is, however, in equilibrium for the given loading. TEC © 2007 The McGraw-Hill Companies, Inc. All rights reserved. rD rB rD rB 8i 4 j 8k 12 1 2 2 3i 3 j 3k rC rE rC rE 6i 3 j 2 k 7 3 2 6 7i 7 j 7k 4 - 24 Eighth Edition Vector Mechanics for Engineers: Statics Sample Problem 4.5 • Apply the conditions for static equilibrium to develop equations for the unknown reactions. F i: j: k: MA j: k: A TBD TEC 270 lb j 0 Ax 23 TBD 76 TEC 0 Ay 13 TBD 73 TEC 270 lb 0 Az 23 TBD 72 TEC 0 rB TBD rE TEC 4 ft i 270 lb j 0 5.333TBD 1.714TEC 0 2.667TBD 2.571TEC 1080 lb 0 Solve the 5 equations for the 5 unknowns, TBD 101.3 lb TEC 315 lb A 338 lbi 101.2 lb j 22.5 lbk © 2007 The McGraw-Hill Companies, Inc. All rights reserved. 4 - 25 Problem 4.6 P A 45o B O D 45o C The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that = 45o, determine the reactions at B , C , and D. 26 Problem 4.6 P A 45 B Solving Problems on Your Own O o D 45 o C The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that = 45o, determine the reactions at B , C , and D. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it. 27 Problem 4.6 P A 45 B Solving Problems on Your Own O o D 45 o C The semicircular rod ABCD is maintained in equilibrium by the small wheel at D and the rollers at B and C. Knowing that = 45o, determine the reactions at B , C , and D. 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = 0 SFy = 0 SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = 0 SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis or SMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line. 28 Problem 4.6 Solution P A 45 B O o D 45 o Draw a free-body diagram of the body. C P sin P A P cos O 45o B D 45o C R B/ 2 B D B/ 2 C/ 2 C C/ 2 29 Problem 4.6 Solution P sin P A P cos O 45o B Write three equilibrium equations and solve for the unknowns. D 45o C R B/ 2 B D C/ 2 C B/ 2 C/ 2 + S MO = 0: (P sin) R _ D (R) = 0 + S Fx = 0: P cos + B/ 2 + S Fy = 0: _ _ 1 D = P sin C/ 2=0 (2) P sin + B/ 2 + C / 2 _ P sin = 0 _ 2P sin + B/ 2 + C / 2 = 0 (3) 30 P sin P A P cos O 45o C R B/ 2 (2) + (3) D 45o B B Problem 4.6 Solution D B/ 2 C/ 2 C C/ 2 P(cos _ 2sin) + 2 B/ 2 = 0 2 B= (2sin _ cos) P 2 (2) _ (3) (4) P(cos + 2sin) _ 2 C/ 2 = 0 2 C= (2sin + cos) P 2 (5) 31 P sin P A P cos O 45o B B EQ. (4) : EQ. (5) : EQ. (1) : D 45o C R B/ 2 B/ 2 Problem 4.6 Solution D For = 45o C/ 2 C sin = cos = 1/ 2 C/ 2 2 2 B= ( 2 2 2 2 C= ( 2 2 D = P/ 2 _ _ 1 1 )P= P; 2 2 1 3 )P= P; 2 2 1 B= P 2 C= 3 P 2 45o 45o D = P/ 2 32 Problem 4.7 4 in 4 in 20 lb A C q 40 lb B 2 in 3 in D 3 in E The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. 33 4 in 4 in 20 lb A C q 40 lb B 2 in 3 in D 3 in E Problem 4.7 Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it. 34 4 in 4 in 20 lb A C q 40 lb B 2 in 3 in D 3 in E Problem 4.7 Solving Problems on Your Own The T-shaped bracket shown is supported by a small wheel at E and pegs at C and D. Neglecting the effect of friction, determine the reactions at C , D , and E when q = 30o. 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = 0 SFy = 0 SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = 0 SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis or SMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line. 35 Problem 4.7 Solution 4 in 4 in 20 lb A C q 40 lb B 2 in 3 in D 3 in E Draw a free-body diagram of the body. 4 in 4 in 20 lb A C 40 lb B 2 in C D D E 3 in 3 in E 30o 36 Problem 4.7 Solution 4 in 4 in 20 lb A C 40 lb B 2 in C D D E 3 in 3 in Write equilibrium equations and solve for the unknowns. E 30o S Fy = 0: E cos 30o _ 20 _ 40 = 0 60 lb E= o = 69.28 lb cos 30 E = 69.3 lb 60o 37 Problem 4.7 Solution 4 in 4 in 20 lb A C 40 lb B 2 in C D D E 3 in 3 in E 30o S MD = 0: ( 20 lb)( 4 in) _ ( 40 lb)( 4 in) _ C ( 3 in) + E sin 30o ( 3 in) = 0 _ 80 _ 3C + 69.28 ( 0.5 )( 3 ) = 0 C = 7.974 lb C = 7.97 lb 38 Problem 4.7 Solution 4 in 4 in 20 lb A C 40 lb B 2 in C D D E 3 in 3 in + S Fx= 0: E sin 30o + C _ D = 0 E ( 69.28 lb )( 0.5 ) + 7.974 lb _ D = 0 30o D = 42.6 lb 39 Problem 4.8 y 1.2 m E 1.2 m D 1.5 m z C A B 5 kN 2m f 1m x A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. 40 Problem 4.8 y 1.2 m Solving Problems on Your Own E 1.2 m D 1.5 m z C A B 5 kN 2m f 1m x A 3-m pole is supported by a ball-and-socket joint at A and by the cables CD and CE. Knowing that the line of action of the 5-kN force forms an angle f=30o with the vertical xy plane, determine (a) the tension in cables CD and CE, (b) the reaction at A. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it. 41 y 1.2 m Problem 4.8 E Solving Problems on Your Own 1.2 m A 3-m pole is supported by a balland-socket joint at A and by the cables CD and CE. Knowing that D C the line of actionof the 5-kN force 1.5 m x A B forms an angle f=30o with the 5 kN vertical xy plane, determine (a) z f the tension in cables CD and CE, 1m 2m (b) the reaction at A. 2. Write equilibrium equations and solve for the unknowns. For three-dimensional body the six scalar equations SFx = 0 SFy = 0 SFz = 0 SMx = 0 SMy = 0 SMz = 0 should be used and solved for six unknowns. These equations can also be written as SF = 0 SMO = S (r x F ) = 0 where F are the forces and r are position vectors. 42 Problem 4.8 Solution y 1.2 m E 1.2 m Draw a free-body diagram of the body. D 1.5 m z C A B 5 kN 2m x y 1.2 m f E 1.2 m 1m D TCD Ax i 1.5 m z Az k TCE C B A x 30o Ay j 5 kN 2m 1m 43 Problem 4.8 Solution y 1.2 m Write equilibrium equations and solve for the unknowns. E 1.2 m D TCD Ax i 1.5 m z Az k TCE C B A x 30o Ay j 5 kN 2m 5 unknowns and 6 equations of equilibrium, but equilibrium is maintained, S MAC = 0 . rB/A = 2 i rC/A = 3 i 1m Load at B, FB = _ ( 5 cos 30o ) j + ( 5 sin 30o ) k = _ 4.33 j + 2.5 k CD = _ 3 i+ 1.5 j + 1.2 k CD = 3.562 m T TCD = TCD CD = CD (_ 3 i + 1.5 j + 1.2 k) CD 3.562 TCE = TCE CE = TCD (_ 3 i + 1.5 j _ 1.2 k) CE 3.562 44 Problem 4.8 Solution y 1.2 m E 1.2 m D TCD Ax i 1.5 m z Az k TCE C B A x 30o Ay j 5 kN 2m 1m SMA = 0: rC/A x TCD + rC/A x TCE + rB/A x FB = 0 i j k i j k i j k TCD TCE 3 0 0 + 2 0 0 =0 + 3 0 0 3.562 3.562 _ _ _ 3 1.5 1.2 3 1.5 1.2 0 _4.33 2.5 45 Problem 4.8 Solution y 1.2 m E 1.2 m D TCD Ax i 1.5 m z Az k TCE B A TCD TCE _ j: 3.6 + 3.6 5=0 3.562 3.562 C _ _ 3.6 T +3.6 T (1) x CD CE 17.81 = 0 30o Ay j 5 kN 2m (2) + 1.25 (1): Eq. (1): Equate coefficients of unit vectors to zero. _ 1m 9TCE _ TCD TCE _ k: 4.5 + 4.5 8.66 = 0 3.562 3.562 4.5 TCD+4.5 TCE = 30.85 (2) _ 53.11 = 0 ; 3.6TCD + 3.6 (5.902) _ 17.81 = 0 TCE = 5.90 kN TCD = 0.954 kN 46 Problem 4.8 Solution y 1.2 m E 1.2 m D TCD Ax i 1.5 m z Az k SF = 0: A + TCD + TCE + FB = 0 TCE C B A i: Ax + x 0.954 _ 5.902 _ ( 3) + ( 3) = 0 3.562 3.562 Ax = 5.77 kN 30o Ay j 5 kN 2m 1m 0.954 5.902 j: Ay + (1.5) + (1.5) _ 4.33 = 0 3.562 3.562 0.954 5.902 _ k: Az + (1.2) + ( 1.2) + 2.5 = 0 3.562 3.562 Ay = 1.443 kN Az = _ 0.833 kN A = ( 5.77 kN) i + ( 1.443 kN ) j - ( 0.833 kN ) k 47 Problem 4.9 A a B C 60 lb 10 in 20 in Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. 48 Problem 4.9 A Solving Problems on Your Own a B C 60 lb 10 in 20 in Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it. 49 Problem 4.9 Solving Problems on Your Own A a B C 60 lb 10 in 20 in Rod AC is supported by a pin and bracket at A and rests against a peg at B. Neglecting the effect of friction, determine (a) the reactions at A and B when a = 8 in., (b) the distance a for which the reaction at A is horizontal and the corresponding magnitudes of the reactions at A and B. 2. For a three-force body, solution can be obtained by constructing a force triangle. The resultants of the three forces must be concurrent or parallel. To solve a problem involving a three-force body with concurrent forces, draw the free-body diagram showing that the three forces pass through the same point. Complete the solution by using a force triangle. 50 Problem 4.9 Solution A (a) a = 8 in a B Draw a free-body diagram of the body. 20 in 10 in C A 60 lb 10 in F = 26.57 A D G tan = 21 B o 2 C 60 lb g E 8 in B 12 in 1 10 in 51 Problem 4.9 Solution 10 in Construct a force triangle. A F A D G B 2 C 60 lb g E 3 - FORCE BODY 8 in B 12 in Reaction at A passes through D where B and 60-lb load intersect AE = 1 1 EB = (8) = 4 in. 2 2 1 EF = BG = 10 _ 4 = 6 in 10 in DG = 2 BG = 2 (6) = 3 in. 1 1 FD = FG _ DG = 8 _ 3 = 5 in. FD 5 Tan g = AF = ; 10 g = 26.57o 52 Problem 4.9 Solution 10 in A A D G 2 60 lb 8 in = 26.57o 30 lb 60 lb A B C g E F FORCE TRIANGLE 1 10 in B 12 in B = 26.57o 30 lb 30 lb A=B= o = 67.08 lb sin 26.57 A = 67.1 lb 26.6o B = 67.1 lb 26.6o 53 Problem 4.9 Solution A (b) For A horizontal a B 20 in Draw a free-body diagram of the body. 10 in C 60 lb 10 in A F G B 2 C a A B = 26.57o 1 60 lb 54 Problem 4.9 Solution 10 in A F G 2 C 60 lb 1 a A B Construct a force triangle. B = 26.57o D ABF : BF = AF cos D BFG : FG = BF sin a = FG = AF cos sin a = (10 in.) cos 26.57o sin 26.57o a = 4.00 in. 55 Problem 4.9 Solution 10 in FORCE TRIANGLE A F G B 2 C a A B B = 26.57o 60 lb = 26.57o 1 A 60 lb A = 60 lb = 120 lb tan B = 60 lb = 134.16 lb sin A = 120.0 lb B = 134.2 lb 26.6o 56 Problem 4.10 a a a A P B 30o 30o C D Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle = 30o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. 57 Problem 4.10 Solving Problems on Your Own a a a A P B 30o 30o C D Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on the rods shown. Knowing that the wire attached at D forms an angle = 30o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. 1. Draw a free-body diagram of the body. This diagram shows the body and all the forces acting on it. 58 Problem 4.10 Solving Problems on Your Own a a a Rod AD supports a vertical load P and is attached to collars B and C, which may slide freely on A B C D the rods shown. Knowing that o o 30 30 the wire attached at D forms an P angle = 30o with the vertical, determine (a) the tension in the wire, (b) the reactions at B and C. 2. Write equilibrium equations and solve for the unknowns. For two-dimensional structure the three equations might be: SFx = 0 SFy = 0 SMO = 0 where O is an arbitrary point in the plane of the structure or SFx = 0 SMA = 0 SMB = 0 where point B is such that line AB is not parallel to the y axis or SMA = 0 SMB = 0 SMC = 0 where the points A, B , and C do not lie in a straight line. 59 Problem 4.10 Solution a a a Draw a free-body diagram of the body. A P B 30o C D 30o 30o 30o o 30o 30o 30 C A B C T D 30o B P a a a 60 30o 30o 30 C A B C 30 o 30 o 30o T D o B P a a Write equilibrium equations and solve for the unknowns. a S F = 0: 30o Problem 4.10 Solution _ P cos 30o + T cos 60o = 0 cos 30o 3/2 T= P T= o = P 1/2 cos 60 + S MB = 0: P a _ (C sin 30o) a + T cos 30o (2a) = 0 Pa_( 1 C)a+ 3P( 3 ) 2a = 0 2 1 C + (1 + 3) P = 0; 2 C=8P 2 _ C=8P 3 P 30o61 30o 30o 30 C A B C 30 o 30 o Problem 4.10 Solution 30o T D o B P a a + S F = 0: a _ _ B cos 30o + C cos 30o _ T sin 30o = 0 3 3 B 2 +8P 2 B= 7P _ 1 3 P ( 2 ) = 0; B=7P 30o 62