CHAPTER 14 STATIC EQUILIBRIUM ActivPhysics can help with these Problems: Activities 7.1–7.6 Section 14-1: Conditions for Equilibrium Problem 1. Five forces act on a rod, as shown in Fig. 14-25. Write the torque equations that must be satisfied for the rod to be in static equilibrium taking the torques (a) about the top of the rod and (b) about the center of the rod. Problem 2. A body is subject to three forces: F1 = 2ı̂ + 2̂ N, applied at the point x = 2 m, y = 0 m; F2 = −2ı̂ − 3̂ N, applied at x = −1 m, y = 0; and F3 = 1̂ N, applied at x = −7 m, y = 1 m. (a) Show explicitly that the net force on the body is zero. (b) Show explicitly that the net torque about the origin is zero. (c) To confirm the assertion following Equation 14-4 that the net torque must be zero about any other point, evaluate the net torque about the point (3 m, 2 m), the point (−7 m, 1 m), and about any other point of your choosing. Solution P P (a) Fi = (2ı̂ + 2̂ − 2ı̂ − 3̂ + ̂)N = 0. (b) ( τ i )0 = [2ı̂ × (2ı̂ + 2̂) + (−ı̂) × (−2ı̂ − 3̂) + (−7ı̂ + ̂) × ̂]N·m = (4 + 3 − 7)P k̂ N·m = 0. (c) For any point r0 = (x0 ı̂ + y0 ̂)m, (ri − r0 ) × Fı̂ = {[(2 − x0 )ı̂ − y0 ̂] × (2ı̂ + 2̂) + [(−1 − x0 )ı̂ − y0 ̂] × (−2ı̂ − 3̂) + [(−7 − x0 )ı̂ + (1 − y0 )̂] × ̂} N·m = [(4 + 3 − 7) + x0 (−2 + 3 − 1) + y0 (2 − 2 + 0)]k̂ N·m = 0. Problem figure 14-25 Problem 1. 3. Suppose the force F3 in the preceding problem is doubled so the forces no longer balance and the body is therefore accelerating. Show that (a) the torque about the point (−7 m, 1 m) is still zero, but that (b) the torque about the origin is no longer zero. What is the torque about the origin? Solution Solution All of the forces lie in the plane of Fig. 14-25, so all of the torques about any point on the rod are into or out of the page. Suppose the latter direction, out of the page or counterclockwise, is positive. Moreover, all of the forces are perpendicular to the rod, so their lever arms about any point on the rod (recall that the magnitude of the torque is force times lever arm) can easily be read-off from Fig. 14-25. (a) About the top of the rod, F4 and F5 contribute zero torque, and Equation 14-2 becomes 0 = 31 ℓF3 − 32 ℓF2 + ℓF1 . (b) About the center of the rod, the perpendicular distances to F2 and F3 are 61 ℓ, and to F1 , F4 and F5 are 21 ℓ, so 0 = 12 ℓ(F1 + F4 − F5 ) − 61 ℓ(F2 + F3 ). (a) Since r3 = (−7ı̂ + ̂) m is the point of application of F3P , the total torque about r3 is just due to F1 and F2 : ( τ i )3 = (r1 − r3 ) × F1 + (r2 − r3 ) × F2 = [(2ı̂ + 7ı̂ − ̂) × (2ı̂ + 2̂) + (−ı̂ + 7ı̂ − ̂) × (−2ı̂ − 3̂)]N·m = [(9×2)k̂ − (1×2)(− P k̂) + 6 × P(−3) k̂ + (1×2)(−k̂)] N·m = 0. (b) ( τ i )0 = (ri × Fi ) = [2ı̂ × (2ı̂ + 2̂) + (−ı̂) × (2ı̂ − 3̂) + (−7ı̂ + ̂) × (2̂)] N·m = [4k̂ + 3k̂ − 14k̂] N·m = −7k̂ N·m. Problem 4. A rod of mass m and length ℓ is falling freely in a horizontal orientation, with no torque about its center of mass. Find the magnitude of the torque about either end. Why does your answer not violate the point made in Problem 8? 208 CHAPTER 14 Solution The magnitude of the torque on a body of mass M in a constant gravity field is |τ grav | = |rcm × M g| = rcm M g sin θ. With origin at either end, rcm = 12 ℓ, and sin θ = 1 for horizontal orientation, hence τ grav = 1 2 M gℓ. With origin at the CM (rcm = 0) the torque is zero, but this does P not contradict the result of Problem 8 since F = M g 6= 0 in free fall. Fı̂ = (1 m)F k̂. Thus, −x3 + y3 = 1 m, or the line of action of F3 passes through the point of application of F2 (the point (0, 1 m)). Any point on this line is a suitable point of application for F3 (e.g. the point (0, 1 m)). (b) In this case, F1 = −F2 so F3 = 0, but r1 × F1 + r2 × F2 = (r2 − r1 ) × F2 6= 0 so r3 × F3 6= 0. Thus there is no single force that can be added to produce static equilibrium. Problem Problem 5. In Fig. 14-26 the forces shown all have the same magnitude F. For each of the cases shown, is it possible to place a third force so the three forces meet both conditions for static equilibrium? If so, specify the force and a suitable application point; if not, why not? 6. Are there any other application points for the force F3 in Problem 2 that will ensure that both static equilibrium conditions are met? Solution Equation 14-1 does not involve the points of application of the forces. Equation 14-2 can be satisfied for any point of application r′3 , for F3 , provided r′3 × F3 = r3 × F3 , or (x′3 ı̂ + y3′ ̂) × ̂ = (−7ı̂ + ̂) × ̂, which implies x′3 = −7. Thus, any point along the line x = −7 is a possible point of application for F3 , satisfying the equilibrium conditions. Problem 7. Four forces act on a body, as shown in Fig. 14-27. Write the set of scalar equations that must hold for the body to be in equilibrium, evaluating the torques (a) about point O and (b) about point P. F4 φ 2 P F3 1 φ F1 F2 O figure 14-27 Problem 7. figure 14-26 Problem 5. Solution Solution The conditions for static equilibrium, under the action of three forces, can be written as: F3 = −(F1 + F2 ) and r3 × F3 = −(r1 × F1 + r2 × F2 ). (a) In this case, F1 = F̂, r1 = (2 m)̂, F2 = Fı̂, and r2 = (1 m)̂. Thus, F3 = −F (ı̂ + ̂), which is a force of magnitude √ 2F , 45◦ down into the third quadrant (θx = 225◦ or −135◦ CCW from the x axis). The point of application, r3 , can be found from the second condition, r3 × F3 = (x3 ı̂ + y3 ̂) × (−Fı̂ − F̂) = (−x3 + y3 )F k̂ = −r1 × F1 − r2 × F2 = 0 − (1 m)̂× All of the forces lie in the same plane, which includes the points O and P, so there are two independent components of the force condition (Equation 14-1) and one component of the torque condition (Equation 14-2). Taking the x axis to the right, the y axis up and P the z axis out of the page in Figure P 14-27, we have: Fx =P 0 = −F1 + F2 sin φ + F3 , Fy = 0 = −F2 cos φ + F4 , (P τz )0 = 0 = −ℓ1 F2 − ℓ2 F3 sin φ + ℓ2 F4 cos φ, and ( τz )P = −ℓ2 F1 sin φ + (ℓ2 − ℓ1 )× F2 = 0. (The lever arms of all the forces about either O or P should be evident from Fig. 14-27). CHAPTER 14 209 Problem 8. In this problem you prove the statement in Section 14-1 that the choice of pivot point does not matter when applying the conditions for static equilibrium. Figure 14-28 shows an object on which the net force is assumed to be zero. Also, the net torque about the point O shown is zero. You’re to show that the net torque about any other point P is also zero. P To do so, write the net torque about P as τP = rP i × Fi , where the vectors rP are from P to the force application points, and the index i labels the different forces. Note in Fig. 14-28 that rP i = rOi + R, where R is a vector from P to O. Use this result in your expression for τ P and apply the distributive law to get two separate sums. Use the assumptions that Fnet = 0 and τ O = 0 to argue that both terms are zero. This completes the proof. figure 14-29(a) Problem 9 Solution. colinear with the weight, so τ B = 0. (c) τ C = 1 1 ◦ 2 ℓ mg sin 90 = 2 mg ℓ (but note that τ C = −τA ). (We also assumed that B and C are at the centers of their respective sides. Alternatively, the torques can be found from the lever arms shown.) Problem figure 14-28 Problem 8. 10. Figure 14-29b shows a thin, uniform plate of mass m in the shape of an equilateral triangle of side ℓ. The plate is in a vertical plane. Find the magnitude of the gravitational torque on the plate about each of the three points shown. Solution Solution With referenceP to Fig. 14-28, we can P P write τ p = rPP × F = (r + R) × F = rOi × Fi + i i Oi i R × Fi = τ O + R × Fnet = τ O . If Fnet = 0, the total torque about any two points is the same. The center of gravity is at the√center of the triangle, a perpendicular distance of ℓ/2 3 from any side. (a) The lever arm of the weight about point A is ℓ/2 so τ A = 21 mgℓ. (b) The lever arm about point B is zero, and τ B = 0. (c) The lever arm about point C is ℓ/4 (C is halfway up from the base) so τ C = 41 mgℓ. (Of course, τ A and τ C are in opposite directions. The torque could also be found from Equation 12-12.) Section 14-2: Center of Gravity Problem 9. Figure 14-29a shows a thin, uniform square plate of mass m and side ℓ. The plate is in a vertical plane. Find the magnitude of the gravitational torque on the plate about each of the three points shown. Solution The center of gravity is at the center of a uniform plate. In calculating the gravitational torque, one may consider the entire√weight as acting at the center of gravity. (a) rA =√ 2ℓ/2 at 135◦ from the weight of the plate, so τ A = ( 2ℓ/2)mg sin 135◦ = 12 mg ℓ. (b) rB is Problem 11. Three identical books of length L are stacked over the edge of a table as shown in Fig. 14-30. The top book overhangs the middle one by 21 L, so it just barely avoids falling. The middle book overhangs the bottom one by 14 L. How much of the bottom book can be allowed to overhang the edge of the table without the books falling? 210 CHAPTER 14 Problem 12. A 60-kg uniform tabletop 2.4 m long is supported by a pivot 80 cm from the left end, and by a scale at the right end (Fig. 14-31). How far from the left end should a 40-kg child sit if the scale is to read zero? Solution See the solution to the next problem, when Fs = 0. figure 14-29(b) Problem 10 Solution. figure 14-31 Problem 12 Solution. Solution In equilibrium, the farthest right the center of mass of the combination of three books can lie is directly above the edge of the table. (This is unstable equilibrium, since the slightest disturbance to the right would cause the books to fall.) The center of mass of each book is at its center, so if we take the origin at the edge with positive to the right, this condition becomes 0 = xcm 1 1 1 1 mx1 + m(x1 + L) + m(x1 + L + L) , = 3m 4 4 2 where x1 is the horizontal position of the center of the bottom book, and the centers of the other books are displaced as given. Therefore, 3x1 + L = 0, or x1 = − 31 L. If the center of the bottom book is 13 L to the left of the edge, then only 12 L − 13 L = 61 L can overhang on the right. (An argument based on torques is equivalent, since at the farthest right position, the normal contact force on the books acts essentially just at the table’s edge.) Section 14-3: Examples of Static Equilibrium Problem 13. Where should the child in Fig. 14-31 sit if the scale is to read (a) 100 N and (b) 300 N? Solution If we consider torques about the pivot point (so that the force exerted by the pivot does not contribute) then Equation 14-2 is sufficient to determine the position of the child. As shown on Fig. 14-31, the weight of the tabletop (acting at its center of gravity), the weight of the child (acting a distance x from the left end), and the scale force, Fs , produce zero torque about the pivot: P (1/g) ( τ )P = 0 = (Fs /g)(160 cm) − (60 kg)(40 cm) + (40 kg)(80 cm − x). Therefore, x = 20 cm + (Fs /9.8 N)4 cm. If (a) Fs = 100 N, then x = 20 cm + (400/9.8) cm = 60.8 cm, and if (b) Fs = 300 N, x = 142 cm. (Note that the child is on opposite sides of the pivot in parts (a) and (b), since without the child, Fs = 147 N.) Problem figure 14-30 Problem 11 Solution. 14. A 4.2-m-long beam is supported by a cable at its center. A 65-kg steelworker stands at one end of the beam. Where should a 190-kg bucket of concrete be suspended if the beam is to be in static equilibrium? Solution The sum of the torques on the beam (taken about its center, C, so that the cable’s tension and beam’s CHAPTER 14 211 Problem 16. A 23-m-long log of irregular cross section is lying horizontally, supported by a wall at one end and a cable attached 4.0 m from the other end, as shown in Fig. 14-33. The log weighs 7.5×103 N, and the tension in the cable is 6.2×103 N. Where is the log’s center of gravity? Problem 14 Solution. weight do not enter the equation) is equal to zero. Therefore, (190 kg)g(x) = (65 kg)g(2.1 m), or x = 71.8 cm, on the opposite side of C from the worker. (Note: Since sin θ = sin(π − θ) will cancel from the torque equation, the beam need not be horizontal to be in equilibrium; the steelworker’s mental equilibrium is greatest when the beam is horizontal.) Solution The log is in equilibrium under the torques exerted by the cable, gravity, and the wall. Calculating the torques about the point of contact with the wall (because the other two forces are given), we find (6.2 kN)(23 − 4) m = (7.5 kN)xCG, or xCG = 15.7 m, from the end on the wall. Problem 15. Two pulleys are mounted on a horizontal axle, as shown in Fig. 14-32. The inner pulley has a diameter of 6.0 cm, the outer a diameter of 20 cm. Cords are wrapped around both pulleys so they don’t slip. In the configuration shown, with what force must you pull on the outer rope in order to support the 40-kg mass? Solution Since each cord is tangent to its respective pulley, the lever arms are just the radii, as shown on the figure. The two torques are equal in magnitude, R1 F1 = R2 F2 , so that figure 14-33 Problem 16 Solution. Problem 17. Figure 14-34 shows a traffic signal, with masses and positions of its various members indicated. The structure is mounted with two bolts, located symmetrically about the vertical member’s centerline, as indicated. What tension force must the left-hand bolt be capable of withstanding? F1 = (40 kg)(9.8 m/s2 )(6 cm) = (20 cm) = 118 N. figure 14-34 Problem 17 Solution. figure 14-32 Problem 15 Solution. Solution The forces on the traffic signal structure, and their lever arms about point 0 (on the vertical member’s 212 CHAPTER 14 centerline between the bolts) are shown on Fig. 14-34. The normal forces exerted by the bolts and the ground on the vertical member are designated by Nℓ and Nr , measured positive upward. (Of course, the ground can only make a positive contribution, and the bolts only a negative contribution, to these normal forces.) The two conditions of static equilibrium needed to P determine Nℓ and Nr are: Fy = 0 = Nℓ + Nr − (9.8)(320 + 170 + 65) N (the vertical P component of Equation 14-1, positive up) and ( τz )0 = 0 = (Nr − Nℓ )(0.38 m) − (170 × 3.5 + 65 × 8)(9.8 N·m) (the out-of-the-page-component of Equation 14-2, positive CCW). These can be written as Nr + Nℓ = 5.44 kN, and Nr − Nℓ = 28.8 kN. Thus Nℓ = −11.7 kN, which is downward and must be exerted by the bolt. The reaction force on the bolt is upward and is a tensile force. (Really, Nℓ is the difference between the downward force exerted by the bolt and the upward force exerted by the ground. Tightening the bolt increases the tensile force it must withstand beyond the minimum value calculated above, under the assumption that the ground exerts no force.) Problem 18. Figure 14-35 shows how a scale with a capacity of only 250 N can be used to weigh a heavier person. The board is 3.0 m long, has a mass of 3.4 kg, and is of uniform density. It is free to pivot about the end farthest from the scale. What is the weight of a person standing 1.2 m from the pivot end, if the scale reads 210 N? Assume that the beam remains nearly horizontal. Problem 19. Figure 14-36a shows an outstretched arm with a mass of 4.2 kg The arm is 56 cm long, and its center of gravity is 21 cm from the shoulder. The hand at the end of the arm holds a 6.0-kg mass. (a) What is the torque about the shoulder due to the weights of the arm and the 6.0-kg mass? (b) If the arm is held in equilibrium by the deltoid muscle, whose force on the arm acts 5.0◦ below the horizontal at a point 18 cm from the shoulder joint (Fig. 14-36b), what is the force exerted by the muscle? Solution (a) The magnitude of the (external) torque on the arm is τ0 = [(4.2 kg)(0.21 m) + (6 kg)(0.56 m)] × 2 (9.8 m/s ) sin 105◦ = 40.2 N·m. The direction is clockwise (into the page) about the shoulder joint. (b) The deltoid muscle exerts a counterclockwise torque of magnitude F r sin θ = F (0.18 m) sin 170◦, which, under equilibrium conditions, equals the magnitude of the torque in part (a). Thus, F = 40.2 N·m/(0.18 m) sin 170◦ = 1.28 kN, underscoring the comment at the end of Example 14-5. The skeleto-muscular structure of the human extremities evolved for speed and range of motion, not mechanical advantage. Solution Since the board is at rest, the sum of the torques (positive CCW in Fig. 14-35) is zero about the pivot (due to the weight of the person, the weight of the board acting at its CG, and the scale force, as shown). P Thus, ( τz )P = 0 = (210 N)(3 m) − (3.4×9.8 N)× (1.5 m) − W (1.2 m), or W = 483 N. figure 14-36 Problem 19 Solution. Problem figure 14-35 Problem 18 Solution. 20. Figure 14-37 shows a portable infant seat that is supported by the edge of a table. The mass of the seat is 1.5 kg, and its center of mass is located 16 cm from the table edge. A 12-kg baby is sitting in the seat with her center of mass over the seat’s CHAPTER 14 center of mass. Find the forces FA and FB that the seat exerts on the table. Solution The infant seat is in equilibrium under the reaction forces to FA and FB , and the weight of the seat and infant. Thus, P Fy = 0 213 164FAx (−k̂) − 37.5M g(k̂), we find FAx = −(37.5)× 2 (15 kg)(9.8 m/s )/(164) = −33.6 N. From the x equation, FBx = −FAx = 33.6 N, and by assumption, FAy = FBy = 12 M g = 73.5 N. Of course, the forces exerted by the door on the hinges (by Newton’s third law) are the reactions to the forces, F A and F B , just calculated. = FB − FA − (1.5 kg + 12 kg)(9.8 m/s2 ) (positive up) ( P τ )A = 0 = (22 cm)FB − (16 + 22) cm(13.5)(9.8) N (positive CCW) Solving for FB from the torque equation and FA from the force equation, we find FB = 229 N and FA = 96.2 N, in the directions shown in the figure. Problem 21 Solution. Problem 940 figure 14-37 Problem 20 Solution. Problem 21. A 15.0-kg door measures 2.00 m high by 75.0 cm wide. It hangs from hinges mounted 18.0 cm from top and bottom. Assuming that each hinge carries half the door’s weight, determine the horizontal and vertical forces that the door exerts on each hinge. 22. Figure 14-38 shows a popular system for mounting bookshelves. An aluminum bracket is mounted on a vertical aluminum support by small tabs inserted into vertical slots. If each bracket in a shelf system supports 32 kg of books, with the center of gravity 12 cm out from the vertical support, what is the horizontal component of the force exerted on the upper of the two bracket tabs? Assume contact between the bracket and support occurs only at the upper tab and at the bottom of the bracket, 4.5 cm below the upper tab. Solution Solution If the door is properly hung, all the forces on the door are coplanar. We assume that the CM is at the geometrical center of the door. The conditions for equilibrium are: P Fx = 0 = FAx + FBx , P Fy = 0 = FAy + FBy − M g, P ( τ )B = 0 = rA × FA + rcm × M g = (164 cm̂) × (FAx ı̂ + FAy ̂) + (37.5 cmı̂ + 82 cm̂) × (−M ĝ), where we chose to calculate torques about the lower hinge at B. Expanding the cross products, 0 = The forces on the bracket are shown superposed on Fig. 14-38, assuming only a normal force on the P bottom. Equilibrium implies Fx = 0 = FAx + FB , 214 CHAPTER 14 P and ( τz )A = 0 = (4.5 cm)FB − (12 cm)W. (We chose to evaluate the z component (out of the page) of the torques about the upper tab, at A, and note that P Fy = 0 is not needed.) The solution for FAx is immediate: FAx = −FB = −(12/4.5)W = −(2.67)× (32 kg)(9.8 m/s2 ) = −836 N. The negative x component means that a tensile force is exerted on the upper tab, as expected. (Note that the upper tab must also support the weight of the books.) 5.00 kN. The force on the tie beam (the reaction to F B is a tension. (The function of the tie beam is precisely to relieve any horizontal force that the roof may exert on the walls.) figure 14-39 Problem 23 Solution. Problem figure 14-38 Problem 22 Solution. 24. Repeat Example 14-4, now assuming that the coefficient of friction at the floor is µ1 and that at the wall µ2 . Show that the minimum angle at which the board will not slip is now given by 1 − µ1 µ2 . φ = tan−1 2µ1 Problem 23. Figure 14-39 shows a house designed to have high “cathedral” ceilings. Following a heavy snow, the total mass supported by each diagonal roof rafter is 170 kg, including building materials as well as snow. Under these conditions, what is the force in the horizontal tie beam near the roof peak? Is this force a compression or a tension? Neglect any horizontal component of force due to the vertical walls below the roof. Ignore the widths of the various structural components, treating contact forces as though they were concentrated at the roof peak and at the outside edge of the rafter/wall junction. Solution The forces on one of the diagonal rafters are drawn on Fig. 14-39. If the rafters are symmetrical (without internal stress), and we neglect the weight of the tie beam, FB and FC will be horizontal. W and FA are vertical, the latter by assumption, and we suppose W acts at the center of the rafter. The Pequilibrium conditionsP needed to find FB are: Fx = 0 = FB − FC , and ( τ z )A = 0, or 0 = (4 m)FC − (3.2 m)FB − (2.4 m)W. (Point A was chosen to eliminate F A from the equation.) The solution is FB = 2.4W/(4 − 3.2) = figure 14-11(b) Problem 24 Solution. Solution The addition of a frictional force on the board where it contacts the wall is shown in the sketch (see Fig. 14-11b). We assume that f2 = µ2 N2 is the maximum frictional force and that it acts to oppose the fall of the board. (If f2 is not proportional to N2 , there is insufficient data to solve this Problem.) The equations of static equilibrium in Example 14-4 CHAPTER 14 become: f1 − N2 = 0 (horizontal), N1 + µ2 N2 − mg = 0 (vertical), and N2 L sin φ + µ2 N2 L cos φ − 1 2 mgL cos φ = 0 (torque about bottom of board). The minimum angle φ can be found from the horizontal equation and the requirement that N2 = f1 ≤ µ1 N1 . N1 can be eliminated from the vertical equation, N2 ≤ µ1 (mg − µ2 N2 ) or (1 + µ1 µ2 )× N2 ≤ µ1 mg. N2 can be eliminated from the torque equation, N2 (tan φ + µ2 ) = 12 mg, so this condition becomes 12 mg(1 + µ1 µ2 )/(µ2 + tan φ) ≤ µ1 mg, or tan φ ≥ (1 − µ1 µ2 )/2µ1 . Problem 215 Problem 26. Show that if the wall in the previous problem is frictionless, then a continuation of the rope line must pass through the center of the sphere. Solution From the preceding solution, it is clear that, in the absence of friction, the condition that the sum of the torques about the center of the sphere is zero implies that the lever arm of the tension is zero, or that the rope must be in line with the center. Problem 25. A uniform sphere of radius R is supported by a rope attached to a vertical wall, as shown in Fig. 14-40. The point where the rope is attached to the sphere is located so a continuation of the rope would intersect a horizontal line through the sphere’s center a distance R/2 beyond the center, as shown. What is the smallest possible value for the coefficient of friction between wall and sphere? 27. A garden cart loaded with firewood is being pushed horizontally when it encounters a step 8.0 cm high, as shown in Fig. 14-41. The mass of the cart and its load is 55 kg, and the cart is balanced so that its center of mass is directly over the axle. The wheel diameter is 60 cm. What is the minimum horizontal force that will get the cart up the step? Solution In equilibrium, the sum of the torques about the center of the sphere must be zero, so the frictional force is up, as shown. The √ lever arm of the tension in the rope is 12 R cos 30◦ = 3R = 4, and the weight and normal force exert no torque about the center. Thus, √ f R = T 3R/4. The sum of the horizontal components ◦ of the forces is zero also, so √ 0 = N − T sin 30 , or 1 T = 2N. Therefore f = 2 3 N. Since f ≤ µs N, this √ implies µs ≥ f /N = 12 3 = 0.87. F 60 cm 8.0 cm figure 14-41 Problem 27. figure 14-41 Problem 27 Solution. Solution figure 14-40 Problem 25 Solution. We assume that a horizontal push on the cart results in a horizontal force exerted on the wheels by the axle, as shown. (We also suppose both wheels share the forces equally, so they can be treated together.) Also shown are the weight of the cart and the normal force of the ground, both acting through the center of the wheels, and the force of the step, Fs . If we consider the sum of the torques (positive CCW) about the step, 216 CHAPTER 14 the latter does not contribute, and the wheels (and P cart) will remain stationary as long as ( τ )step = M gR sin θ − N R sin θ − F R cos θ = 0. When N = 0, however, the wheels begin to lose contact with the ground and go over the step. This occurs when F = M g tan θ. From the geometry of the situation, R(1 − cos θ) = h, the height of the step, so θ = cos−1 (1 − h/R) = cos−1 (1 − 8/30). Then F = (55× 9.8 N) tan(cos−1 (11/15)) = 500 N is the minimum force. Problem 28. Figure 14-42 shows the foot and lower leg of a person standing on the ball of one foot. Three forces act on the foot to maintain this equilibrium: the tension force T in the Achilles tendon, contact force F C at the ankle joint, and the normal force N of the ground that supports the person’s weight. The person’s mass is 70 kg, and the force application points are as indicated in Fig. 14-42. Find the magnitude of (a) the tension in the Achilles tendon and (b) the contact force at the ankle joint. Treat the tower as a uniform cylinder 7.0 m in diameter and 55 m high, and assume the ground supports the tower’s weight but does not provide any torque. Solution The center of mass of the tower must be somewhere over its footprint on the ground, or it will topple. This will be so, for a uniform cylindrical model of the tower, if the angle of tilt φ is less than the angle α that a diagonal makes with the length, as indicated on Fig. 14-43. Therefore φ ≤ α = tan−1 (7m/55m) = 7.25◦ . figure 14-43 Problem 29 Solution. Problem figure 14-42 Problem 28 Solution. Solution Solution If we approximate the bones in the foot as a massless, planar, rigid body, the equilibrium conditions for the P situation depicted in P Fig. 14-42 are: Fx = ◦ ◦ T sin 25 P − FC,x = 0, Fy = T cos 25 + N − FC,y ◦= 0, and ( τ )ankle joint = N (12 cm) − T (7 cm) sin 115 = 0. Then with N = 70 × 9.8 N, one finds (a) T = N (12 cm)/(7 cm) sin 115◦ = 1.30 kN, and (b) FC,x = T sin 25◦ =q 0.548 kN, FC,y = T cos 25◦ + N = 1.86 kN, and FC = 30. A uniform 5.0-kg ladder is leaning against a frictionless vertical wall, with which it makes a 15◦ angle. The coefficient of friction between ladder and ground is 0.26. Can a 65-kg person climb to the top of the ladder without it slipping? If not, how high can the person climb? If so, how massive a person would make the ladder slip? 2 + F 2 = 1.94 kN. FC,x C,y Problem 29. The leaning Tower of Pisa (Fig. 14-43) currently leans at a 4.7◦ angle to the vertical. Treating the tower as a solid cylinder, what is the maximum angle at which it can lean before falling over? It is shown in the solution to Problem 45, that the condition for a person of mass m to climb up a fraction of length of ladder without the ladder slipping is α ≤ µs cot θ + (mℓ /m)(µs cot θ − 21 ), where µs is the coefficient of friction with the floor, θ the angle with the vertical frictionless wall, and mℓ the mass of the ladder. For the situation in this problem, µs cot θ = 0.26 cot 15◦ = 0.970 and α ≤ 0.970 + (5/65)(0.470) = 1.01. Therefore (since α ≤ 1, by definition) a 65-kg person can climb all the way to the top. However, the right-hand side of the condition is less than 1 for m > mℓ (µs cot θ − 12 )/(1−µs cot θ) = 5 kg(0.470)/(0.0297) = 79.3 kg, so a person with mass greater than 79.3 kg causes the ladder to slip before reaching the top. CHAPTER 14 Problem 217 Therefore: 31. The boom in the crane of Fig. 14-44 is free to pivot about point P and is supported by the cable that joins halfway along its 18-m total length. The cable passes over a pulley and is anchored at the back of the crane. The boom has mass 1700 kg, distributed uniformly along its length, and the mass hanging from the end of the boom is 2200 kg. The boom makes a 50◦ angle with the horizontal. What is the tension in the cable that supports the boom? T2 = 0.5W sin 99.2◦/ sin 20.8◦ = 1.39W, T1 = T2 sin 60◦ / sin 35◦ = 2.10W, and w = T1 cos 35◦ + T2 cos 60◦ − W = 1.41W. figure 14-45 Problem 32 Solution. Problem figure 14-44 Problem 31 Solution. Solution The forces on the boom are shown superposed on the figure. By assumption, T is horizontal and acts at the CM of theP boom. To find T, we compute the torques about P, ( τ )P = 0, obtaining: T 12 ℓ sin 50◦ − mb g 21 ℓ cos 50◦ − mgℓ cos 50◦ = 0, or T = (2m + mb )g cot 50◦ = (4400 + 1700)(9.8 N) cot 50◦ = 50.2 kN. Problem 33. Figure 14-46 shows a 1250-kg car that has slipped over the edge of an embankment. A group of people are trying to hold the car in place by pulling on a horizontal rope, as shown. The car’s bottom is pivoted on the edge of the embankment, and its center of mass lies further back, as shown. If the car makes a 34◦ angle with the horizontal, what force must the group apply to hold it in place? Solution Three forces act on the car, as shown added to Fig. 14-46. The unknown force, FP , exerted by the edge of the embankment, does not contribute to Equation 14-2 (positive torques CCW) if evaluated about point P, so the tension necessaryPto keep the car in equilibrium can be found directly. ( τ )P = 0 = M g(2.4 m − 1.8 m) cos 34◦ − T (1.8 m) sin 34◦, or T = (1250×9.8 N)/3 tan 34◦ = 6.05 kN. 32. A uniform board of length ℓ and weight W is suspended between two vertical walls by ropes of length ℓ/2 each. When a weight w is placed on the left end of the board, it assumes the configuration shown in Fig. 14-45. Find the weight w in terms of the board weight W. Solution The conditions for equilibrium (about the origin drawn on the figure) are: P Fx = 0 = T2 sin 60◦ − T1 sin 35◦ , P Fy = 0 = T2 cos 60◦ + T1 cos 35◦ − w − W, P ( τz )0 = 0 = T2 ℓ sin 20.8◦ − W 12 ℓ sin 99.2◦ . figure 14-46 Problem 33 Solution. Problem 34. A uniform board of length ℓ is dangling over a frictionless edge, secured by a horizontal rope, as 218 CHAPTER 14 shown in Fig. 14-47. Show that the angle p it makes with the horizontal must be θ = sin−1 2d/ℓ, where d is the distance from the edge to the center of the board. Solution mg(0.94x − 0.01x2 ). (a) Equation 14-3, the condition for equilibrium, gives dU/dx = 0 = mg(0.94 − 0.02x), or x = 47 m (b) Since d2 U/dx2 = −0.02mg < 0, Equation 14-5 implies this is an unstable equilibrium. (c) h(47) = 0.94(47) − 0.01(47)2 = 22.1 m. See the solution to the next Problem. Problem 35. Figure 14-47 shows a uniform board dangling over a frictionless edge, secured by a horizontal rope. If the angle θ in Fig. 14-47 were 30◦ , what fraction would the distance d shown in the figure be of the board length ℓ? (a) (b) figure 14-48 Problem 37. Problem figure 14-47 Problems 34, 35. Solution The three forces acting on the board are in the same configuration as those in Problem 33, so Equation 14-2 about the edge gives M gd cos θ = T ( 21 ℓ − d) sin θ. If the edge is frictionless, then FP is perpendicular to the board, so Equation 14-1 requires T = FP sin θ and M g = FP cos θ. Substituting above, we find FP d cos2 θ = FP ( 12 ℓ − d) sin2 θ, or d(cos2 θ + sin2 θ) = p 2 −1 1 2d/ℓ. For θ = 30◦ , d/ℓ = 2 ℓ sin θ and θ = sin 2 1 1 ◦ 2 sin 30 = 8 . Section 14-4: Stability of Equilibria Problem 36. A portion of a roller coaster track is described by h = 0.94x − 0.010x2 , where h and x are the height and horizontal position in meters. (a) Find a point where the roller coaster car could be in static equilibrium on this track. (b) Is the equilibrium stable or unstable? (c) What is the height of the track at the equilibrium point? Solution The potential energy of the roller coaster car, in the equivalent one-dimensional problem, is U (x) = mgh = 37. A roly-poly toy clown is made from part of a sphere topped by a cone. The sphere is truncated at just the right point so that there is no discontinuity in angle as the surface changes from sphere to cone (Fig. 14-48a). If the clown always returns to an upright position, what is the maximum possible height for its center of mass? Would your answer change if the continuity-ofangle condition were not met, as in Fig. 14-48b? Solution Suppose the roly-poly rests on a flat horizontal surface. Its spherical surface is always tangent to the horizontal if the continuity-of-angle condition holds, as in sketch (a) (except when upside-down on the point of the cone). Gravity will always exert a restoring torque if the CM lies to the left of the vertical through the left-most point of contact, B, as shown. Since this vertical passes through O, the center of the sphere, the CM should lie “below” O, as measured from the bottom point, A, on the axis. If the continuity-of-angle condition is not met, the cone, in general, intersects the sphere in one of two circles (through BB ′ or CC ′ , as in sketch (b)). If CC ′ is the actual boundary, the reasoning in the first paragraph still applies. However, if BB ′ is the boundary, the maximum distance of the CM from point A should be < AQ (considerably “lower” than O). CHAPTER 14 219 Problem 39. The potential energy as a function of position for a certain particle is given by 3 x x2 x U (x) = U0 , + a + 4 x30 x20 x0 where U0 , x0 , and a are constants. For what values of a will there be two static equilibria? Comment on the stability of these equilibria. Solution Problem 37 Solution. Problem 38. A uniform rectangular block is twice as long as it is wide. Letting θ be the angle that the long dimension makes with the horizontal, determine the angular positions of any static equilibria, and comment on their stability. Solution Consideration of the block, tilted in a plane perpendicular to its thickness (the unspecified dimension in this question) reveals that θ = 0◦ is a stable equilibrium position, θ = 90◦ a metastable one, and θ + α = 90◦ an unstable one (α is the angle a diagonal makes with the longer side, as shown). Since tan α = ℓ/2ℓ = 12 , the unstable equilibrium is at θ = 90◦ − tan−1 ( 12 ) = 63.4◦. The equilibrium condition, dU/dx = 0 (Equation 14-3), requires 3(x/x0 )2 + 2a(x/x0 ) + 4 = 0. This quadratic has two real roots if the discriminant is √ 3. The roots are positive, i.e., a2 − 12 > 0, or |a| > 2 √ (x/x0 )± = 13 (−a ± a2 − 12). The second derivative of the potential energy, evaluated at these roots, is " # 2 d U U0 x = 2 6 + 2a dx2 ± x0 x0 ± p U0 2 . = ±2 a − 12 x20 Thus, the “plus” root is a position of metastable equilibrium (Equation 14-4), while the “minus” root represents unstable equilibrium, (Equation 14-5). A plot of the potential energy, which √is a cubic, will clarify these remarks. For |a| ≤ 2 3, U (x)has no wiggles, as shown. (U passes through the origin, but its position depends on the value of “a”, and is not shown.) 750 Problem 39 Solution. Problem 40. A cubical block rests on an inclined board with two sides parallel to the direction of the incline. The coefficient of static friction between block and board is 0.95. If the inclination angle of the board is increased, will the block first slide or first tip over? Solution Problem 38 Solution. Reference to the solution of Problem 57 shows that the cube will tip over when θ > tan−1 (w/h) = 45◦ , but 220 CHAPTER 14 will slide when θ > tan−1 0.95 = 43.5◦ . It thus slides before tipping. horizontal cable that supports the boom. Solution Paired Problems Problem 41. Figure 14-49 shows a 66-kg sign hung centered from a uniform rod of mass 8.2 kg and length 2.3 m. At one end the rod is attached to the wall by a pivot; at the other end it’s supported by a cable that can withstand a maximum tension of 800 N. What is the minimum height h above the pivot for anchoring the cable to the wall? As in the previous problem, the equilibrium condition for torques about the pivot does not contain the unknown pivot force, and thus allows the tension to be directly determined without use of the force equations. Thus, T ℓ sin 50◦ = M gℓ cos 50◦ + mg 31 ℓ cos 50◦, or T = (M + 31 m)g cot 50◦ = (2500 + 13 × 830)(9.8 N)× cot 50◦ = 22.8 kN. Solution Suppose that the sign is centered on the rod, so that its CM lies under the center of the rod. Then the total weight may be considered to act through the center of the rod, as shown. In equilibrium, Equation 14-2 calculated about the pivot (which does not contain the unknown pivot force) yields 0 = T ℓ sin θ − M g 21 ℓ, or T = M g/2 sin θ. But, tan θ = h/ℓ, so T = 21 M g× p 1 + ℓ2 /h2 (use the identity 1 + cot2 θ = csc2 θ). Therefore, the condition T ≤ p Tmax implies 1 + ℓ2 ÷ 2 (2Tmax /M g)2 − 1 = h2 ≤ (2T /M g) , or h ≥ ℓ/ pmax 2 2.3 m/ (1600/74.2×9.8) − 1 = 1.17 m. figure 14-50 Problem 42 Solution. Problem 43. A 4.2-kg plant hangs from the bracket shown in Fig. 14-51. The bracket has a mass of 0.85 kg, and its center of mass lies 9.0 cm from the wall. A single screw holds the bracket to the wall, as shown. Find the horizontal tension force in the screw. Hint: Imagine that the bracket is slightly loose and pivoting about its bottom end. Assume the wall is frictionless. Solution figure 14-49 Problem 41 Solution. Problem 42. A crane in a marble quarry is mounted on the rock walls of the quarry and is supporting a 2500-kg slab of marble as shown in Fig. 14-50. The center of mass of the 830-kg boom is located one-third of the way from the pivot end of its 15 m length, as shown. Find the tension in the We assume that the screw provides the total support for the bracket, exerting a force with horizontal component Fx (the reaction to which is a tensile force on the screw) and vertical component Fy (the reaction to which is a shearing force on the screw equal to the total weight) as shown. A normal contact force exerted by the wall could be distributed along the bracket (e.g., by tightening the screw), but if we only wish to estimate the minimum Fx , we may consider all the normal force to act at the lowest point of contact of the bracket, point O. Then Equation 14-2 about O gives Fx (7.2 cm) = [(4.2 kg)(28 cm) + (0.85 kg)× 2 (9.0 cm)](9.8 m/s ) or Fx = 170 N. CHAPTER 14 221 of the heaviest person who can safely ascend to the top of the ladder? (The center of mass of the ladder is at its center.) Solution figure 14-51 Problem 43 Solution. The forces on the uniform ladder are shown in the sketch, with the force exerted by the (frictionless) wall horizontal. The person is up the ladder a fraction α of its length. Equilibrium conditions require: P Fx = 0 = f − Fwall , P Fy = 0 = N − (mℓ + m)g, P ( τ )A = 0 = Fwall ℓ cos θ − mℓ g 21 ℓ sin θ − mgαℓ sin θ. The ladder will not slip if f ≤ µs N, which can be written as Problem 44. A 160-kg highway sign of uniform density is 2.3 m wide and 1.4 m high. At one side it is secured to a pole with a single bolt, mounted a distance d from the top of the sign. The only other place where the sign contacts the pole is at its bottom corner. If the bolt can sustain a horizontal tension of 2100 N, what is the maximum permissible value for the distance d? Solution The forces on the sign are in a similar configuration to those in the preceding problem, as shown. (The weight of the sign is shown acting at its center and the force at the bottom corner could have a vertical component.) To apply the limit on the tensile force Fx , we need only consider Equation 14-2 about point O: Fx (1.4 m − d) = (160×9.8 N)(1.15 m). Then Fx = (1.80 kN·m)/(1.4 m − d) < 2.1 kN implies that d < 1.4 m − 1.80 kN·m/2.1 kN = 54.1 cm. Problem 44 Solution. Problem 45. A 5.0-m-long ladder has mass 9.5 kg and is leaning against a frictionless wall, making a 66◦ angle with the horizontal. If the coefficient of friction between the ladder and ground is 0.42, what is the mass f = Fwall = ( 12 mℓ + αm)g tan θ ≤ µs N = µs (mℓ + m)g, or α ≤ [µs (mℓ + m) cot θ − 21 mℓ ]/m = µs cot θ + (mℓ /m)(µs cot θ − 21 ). (Here, we used the horizontal force equation to find f, the torque equation to find Fwall , and the vertical 222 CHAPTER 14 force equation to find N.) For a person at the top of the ladder, α = 1 and the condition for no slipping becomes m ≤ mℓ (µs cot θ − 12 )/(1 − µs cot θ). With the data given for the ladder (note that cot θ = tan 66◦ )m ≤ (9.5 kg)(0.42 tan 66◦ − 0.5)/(1 − 0.42 tan 66◦ ) = 74.3 kg. Problem 48. An isosceles triangular block of mass m and height h is in stable equilibrium, resting on its base on a horizontal surface. How much energy does it take to bring it to unstable equilibrium, resting on its apex? Hint: Consult Example 10-3. Solution Example 10-3 shows that the CM of an isosceles triangle is 31 the height from the base, or 23 the height from the apex. The difference in energy between the unstable and stable equilibrium mentioned is ∆U = mg ∆ycm = mg( 32 h − 13 h) = 31 mgh. Supplementary Problems Problem Problem 45 Solution. 49. A uniform pole of mass M is at rest on an incline of angle θ, secured by a horizontal rope as shown in Fig. 14-52. What is the minimum coefficient of friction that will keep the pole from slipping? Problem 46. To what vertical height on the ladder in the preceding problem could a 95-kg person reach before the ladder starts to slip? Solution For a 95-kg person, the condition that the ladder not slip gives α ≤ 0.42 tan 66◦ + (9/95)(0.42 tan 66◦ − 0.5) = 98.8% as the maximum fraction up along the length of the ladder. (See previous problem.) The maximum height above the ground is just αℓ cos θ = (0.988)(5 m) sin 66◦ = 4.51 m. Problem 47. A uniform, solid cube of mass m and side s is in stable equilibrium when sitting on a level tabletop. How much energy is required to bring it to an unstable equilibrium where it’s resting on its corner? Solution When balancing on a corner, the CM of a uniform cube (i.e., its center) √ is a distance p (s/2)2 + (s/2)2 + (s/2)2 = 3s/2 above the corner resting on the tabletop. When in stable equilibrium, the CM is s/2 above the tabletop. Thus, the√potential energy difference is ∆U = mg ∆ycm = mgs( 3 − 1)/2. figure 14-52 Problem 49 Solution. Solution The forces acting on the pole are the tension in the rope, gravity (acting at the CM at its center) and the contact force of the incline (perpendicular component N and parallel component f ) as shown. Consideration of Equation 14-2 about the CM shows that a frictional force f must be acting up the plane if the rod is to remain in static equilibrium. (Since the weight of the rod, mg, and the normal force, N, contribute no torques about the CM, there must be a force to oppose the torque of the tension, T.) The equations for static equilibrium (parallel and perpendicular components of Equation 14-1, and P CCW-positive component of EquationP14-2) are: F|| = 0 = f + T cos θ − mg sin θ, F = 0 = N − T sin θ − ⊥ P mg cos θ, and ( τ )cm = 0 = T 12 ℓ cos θ − f 21 ℓ. The solutions for the forces are f = 21 mg sin θ, T = 2 1 1 2 mg tan θ, and N = 2 mg(2 cos θ + sin θ/ cos θ), subject to the condition that f ≤ µN. Therefore sin θ ≤ µ(2 cos θ + sin2 θ/ cos θ) or µ ≥ tan θ ÷ (2 + tan2 θ). (By use of the identities sin 2θ = CHAPTER 14 2 sin θ cos θ, cos 2θ = cos2 θ − sin2 θ, and sin2 θ = 1 − cos2 θ, this may be rewritten as µ ≥ sin 2θ ÷ (3 + cos 2θ).) Problem 50. For what angle does the situation of Problem 49 require the greatest coefficient of friction? Solution The minimum coefficient of friction found in the previous problem, µmin (θ) = tan θ/(2 + tan2 θ), is a positive function which is zero at θ = 0◦ and 90◦ (the limits of its domain). Therefore it has a maximum 2 2 when dµmin √/dθ = 0, ◦or 2 + tan θ − 2 tan θ = 0, or −1 θ = tan 2 = 54.7 . Problem 51. One end of a board of negligible mass is attached to a spring of spring constant k, while its other end rests on a frictionless surface, as shown in Fig. 14-53. If a mass m is placed on the middle of the board, by how much will the spring compress? Solution If the frictionless surface is horizontal, the three forces acting on the board are vertical,P as shown. For equilibrium, N + Fs = mg and ( τ )cm = 0. The latter implies N = Fs , so Fs = 21 mg = k ∆x, and the compression of the spring is ∆x = mg/2k. 223 to show that anyone can climb to the top if µ ≥ tan θ, but that no one can if µ < 12 tan θ. Solution The condition for a person of mass m to be able to reach the top of the ladder, which was found in the solution to Problem 45, can be written as m ≤ mℓ (2µs − tan θ)/2(tan θ − µs ). Since m is positive, this condition cannot be fulfilled if µs ≤ 12 tan θ, i.e., no one can climb to the top without causing the ladder to slip, whereas if µs = tan θ, the limit is ∞, so anyone can climb to the top. (Note: the original expression, α ≤ µs cot θ + (mℓ /m)(µs cos θ − 12 ), shows that α = 1 is allowed for any m, provided µs cot θ > 1.) Problem 53. Figure 14-54 shows a wheel on a slope with inclination angle θ = 20◦ , where the coefficient of friction is adequate to prevent the wheel from slipping; however, it might still roll. The wheel is a uniform disk of mass 1.5 kg, and it is weighted at one point on the rim with an additional 0.95-kg mass m. Find the angle φ shown in the figure such that the wheel will be in static equilibrium. Solution P ThePwheel doesn’t slide if F|| = 0 and it doesn’t roll if ( τ )center = 0. (“||” means parallel to the incline, and “center” is the center and CM of the wheel. These are the only equilibrium conditions needed for the solution of this problem.) With reference to the forces shown added to Fig. 14-52, these conditions are f = (M + m)g sin θ and f R = mgR cos φ. Together, they imply f = mg cos φ = (M + m)g sin θ or φ = cos−1 [(1 + M/m) sin θ] = cos−1 [(1 + 1.5/0.95)× sin 20◦ ] = 28.1◦ . figure 14-53 Problem 51 Solution. figure 14-54 Problem 53 Solution. Problem 52. A uniform ladder of mass m is leaning against a frictionless vertical wall with which it makes an angle θ. The coefficient of static friction at the floor is µ. Find an expression for the maximum mass for a person who is able to climb to the top of the ladder without its slipping. Use your result Problem 54. The wheel in Fig. 14-54 has mass M and is weighted with an additional mass m. The slope angle is θ. Show that static equilibrium is possible only if m > M sin θ/(1 − sin θ). 224 CHAPTER 14 Solution Static equilibrium is possible only if the CM of the weighted wheel lies to the left of the point of contact with the incline (or the y axis in the sketch). The smallest additional mass would have to be placed horizontally to the left of the center, as shown. Then (M + m)xcm = M R sin θ − m(R − R sin θ) < 0, implies m > M sin θ/(1 − sin θ). (1 − xcm /ℓ). Equation 10-5 gives Z ℓ Z ℓ xcm = λx dx ÷ λ dx 0 = Z 0 0 ℓ (ax + bx2 ) dx ÷ Z ℓ (a + bx) dx 0 = (a 21 ℓ2 + b 31 ℓ3 )/(aℓ + b 12 ℓ2 ) = ℓ(3a + 2bℓ)/(6a + 3bℓ). 7 and note that For the values given, xcm /ℓ = 12 1 2 M = aℓ + 2 bℓ = 4 kg. Thus, Fsr = M gxcm /ℓ = 22.9 N and Fsℓ = 16.3 N. Problem 56. What horizontal force applied at its highest point is necessary to keep a wheel of mass M from rolling down a slope inclined at angle θ to the horizontal? Solution Problem 54 Solution. Problem 55. A 2.0-m-long rod has a density described by λ = a + bx, where λ is the density in kilograms per 2 meter of length, a = 1.0 kg/m, b = 1.0 kg/m , and x is the distance in meters from the left end of the rod. The rod rests horizontally with its ends each supported by a scale. What do the two scales read? Consider the conditions for static equilibrium of the wheel, under the action of the forces shown (Fapp is the applied horizontal force, Fc is the contact force of the incline, normal plus friction, and we assumed that the CM is at the center). The torques about the point of contact sum to zero, or Fapp R(1 + cos θ) = M gR sin θ. Therefore Fapp = M g sin θ/(1 + cos θ) = M g tan 21 θ. Problem 56 Solution. Problem 55 Solution. Solution The rod is in static equilibrium under P the three vertical forces shown in the sketch, so Fy = 0 P implies Fsr + Fsr = M g, and ( τ )cm = 0 implies Fsℓ xcm = Fsr (ℓ − xcm ). The solution for the left and right scale forces is Fsℓ = M g − Fsr = M g× Problem 57. A rectangular block twice as high as it is wide is resting on a board. The coefficient of static friction between board and block is 0.63. If the board is tilted as shown in Fig. 14-55, will the block first tip over or first begin sliding? Solution We suppose that the block is oriented with two sides parallel to the direction of the incline, and that its CM is at the center. The condition for sliding is that CHAPTER 14 mg sin θ > fsmax = µs N = µs mg cos θ, or tan θ > µs . For µs = 0.63, this condition is θ > tan−1 0.63 = 32.2◦ . The condition for tipping over is that the CM lie to the left of the lower corner of the block (see sketch). Thus θ > α, where α = tan−1 (w/h) is the diagonal angle of the block. For h = 2w, α = tan−1 0.5 = 26.6◦. Therefore, this block tips over before sliding. 225 Since tan α = ( 61 h)/( 14 h) = 32 > 0.63 = µs , this cone will slide before tipping over. Problem 59 Solution. Problem figure 14-55 Problem 57 Solution. Problem 58. What condition on the coefficient of friction in the preceding problem will cause the block to slide before it tips? Solution If µs < tan α = 0.5, the block in the previous problem will slide before tipping. Problem 59. A uniform solid cone of height h and base diameter 13 h is placed on the board of Fig. 14-55. The coefficient of static friction between the cone and incline is 0.63. As the slope of the board is increased, will the cone first tip over or begin sliding? Hint: Begin with an integration to find the center of mass. 60. In Fig. 14-56 a uniform boom of mass 350 kg is attached to a vertical wall by a pivot, and its far end is supported by a cable as shown. If the cable can withstand a maximum tension of 4.0 kN, what is the maximum value for the angle φ? Solution The situation is like that in Problem 41, but with different angles. In equilibrium, the net torque about the pivot point is zero, so M g 2ℓ sin 40◦ = T ℓ sin(40◦ − φ). (The exterior angle of a triangle equals the sum of the two alternate interior angles.) The condition T < Tmax means that sin(40◦ − φ) ≥ (M g/2 Tmax) sin 40◦ = 0.276, for the data given, so φ ≤ 40◦ − sin−1 (0.276) = 24.0◦ is required. Solution The analysis for Problem 57 applies to the cone, where α is the angle between the symmetry axis and a line from the CM to the edge of the base. The integration to find the CM is fastest when the cone is oriented like the aircraft wing in Example 10-3, for then, the equation of the sloping side is simple, as shown in the sketch. For mass elements, take thin disks parallel to the base, so dm = ρπy 2 dx = (3M/h3 )x2 dx, where ρ = M/ 13 πR2 h is the density (assumed constant) and M is the mass of the cone. Then xcm = M −1 ∫ x dm = (3/h3 ) ∫0h x3 dx = 34 h, or the CM is 14 h above the base. figure 14-56 Problem 60 Solution. Problem 61. An interstellar spacecraft from an advanced civilization is hovering above Earth, as shown in 226 CHAPTER 14 Fig. 14-57. The ship consists of two pods of mass m separated by a rigid shaft of negligible mass that is one Earth radius (RE ) long. Find (a) the magnitude and direction of the net gravitational force on the ship and (b) the net torque about the center of mass. (c) Show that the ship’s center of gravity is displaced approximately 0.083RE from its center of mass. Solution (a) The force on each pod (pod #1 over the North pole) can be calculated from the law of universal gravitation, Equation 9-1 (vectors in the x-y Earth-centered frame shown on Fig. 14-57): 1 GME m GME m 2 − √ ı̂ − √ ̂ . F1 = (−̂), F2 = √ (2RE )2 ( 5RE )2 5 5 2 When we replace GME by gRE , the net force becomes: ı̂ 2 1 F1 + F2 = −mg √ + ̂ + √ 4 5 5 5 5 = −mg(0.894ı̂ + 4.29̂) × 10−1 . This has magnitude (4.38×10−1 )mg and is directed 11.8◦ clockwise from the negative y axis. (b) The positions of the pods, relative to their CM midway between them, are r′1 = − 12 RE ı̂ and r′2 = 12 RE ı̂. Therefore, the net torque about the CM is: r′1 × F1 + r′2 × F2 1 mg mg 1 = − RE ı̂ × − ̂ + RE ı̂ × √ (−ı̂ − 2̂) 2 4 2 5 5 mgRE mgRE = k̂ − √ k̂ 8 5 5 = (3.56 × 10−2 )mgRE k̂ (out of page). (c) The CG is positioned between the pods such that the net gravitational torque about it is zero. If the positions of the pods relative to the CG are −α1 RE ı̂ and α2 RE ı̂, respectively, where α1 + α2 = 1, then mg mg ̂ + (α2 RE ı̂) × √ (−ı̂ − 2̂) O = (−α1 RE ı̂) × − 4 5 5 α1 2α2 = mgRE k̂. − √ 4 5 5 Solving for α1 (or α2 ), we √ √find α1 = 1 − α2 = 1 − 5 5α1 /8, or α1 = (1 + 5 5/8)−1 = 0.417 (and α2 = 0.583). Thus, the CG is (0.5 − α1 )RE = 0.0829RE closer to pod #1 than the CM. lfson0 20261 4A rt : 14 −figure 57.eps 14-57 Problem 61 Solution. PART 1 CUMULATIVE PROBLEMS CHAPTER 14 chapters in earlier parts, or they present special challenges. 227 Fig. 1. Where does the apple hit the ground? Neglect the effect of air resistance on either object as well as any friction between apple and post. Problem 1. A 170-g apple sits atop a 2.8-m-high post. A 45-g arrow moving horizontally at 130 m/s passes horizontally through the apple and strikes the ground 36 m from the base of the post, as shown in Solution We can assume that momentum is conserved during the inelastic collision (in a brief interval at t = 0) between the arrow (m1 ) and the apple (m2 ). The 228 CHAPTER 14 v = 130 m/s 2.8 m 3° figure 2 Cumulative Problem 2. figure 1 Cumulative Problem 1. velocities of the arrow before and after are specified to be horizontal (in the x direction), therefore the velocity of the apple (which was at rest before) is also horizontal after the collision. Thus, m1 v1i,x = m1 v1f,x + m2 v2f,x . Since both are moving horizontally after the collision, the arrow and the apple will each fall to the ground through the same vertical distance y (equal p to the height of the post), in the same time t = 2y/g. However, they strike the ground at different horizontal positions, which (in the absence of air resistance) are x1 = v1f,x t and x2 = v2f,x t, relative to the base of the post. Since x1 = 36 m, y = 2.8 m, and v1i,x = 130 m/s are given, v1f,x and v2f,x can be eliminated and a solution for x2 obtained: px2 = (m1 /m2 )(v1i,x − v1f,x )t =q (m1 /m2 )(v1i,x 2y/g − 2 x1 ) = (45/170)[(130 m/s) 2(2.8 m)/(9.8 m/s ) − 36 m] = 16.5 m. (Alternatively, since external horizontal forces are neglected, the center of mass of the arrow/apple system moves horizontally at constant speed until it reaches ground level, vcm,x = constant = m1 v1i,x /(m1 + m2 ) (its value before the collision). Then at ground level, m2 x2 = (m1 + m2 )xcm − m1 x1 = (m1 + m2 )vcm,x t − m1 x1 = m1 (v1i,x t − x1 ), as before.) Refer to relevant material in Chapters 4, 10, and 11 if necessary. Problem 2. A fire department’s tanker truck has a total mass of 21×103 kg, including 15×103 kg of water. Its brakes fail at the top of a long 3◦ slope and it begins to roll downward, starting from rest. In an attempt to stop the truck, firefighters direct a stream of water parallel to the slope, as shown in Fig. 2, beginning as soon as the truck starts to roll. The water leaves the 6.0-cm-diameter hose nozzle at 50 m/s. Will the truck stop before it runs out of water? If so, when? If not, what is the minimum speed reached? Solution The fire truck operates like a rocket, with the stream of water acting as expelled fuel. The rocket equation (Equation 10-10a) can be modified, as follows, to include the additional change in momentum due to the external force of gravity. In the text’s derivation, we now set ∆P = Fgrav ∆t (instead of ∆P = 0), so M (v + ∆v) + ∆m(v − vex ) − (M + ∆m)v = M ∆v − vex ∆m = Fgrav ∆t, and Equation 10-10a becomes M (∆v/∆t) = −vex (∆M ÷ ∆t) + Fgrav . This is a one-dimensional equation with positive components in the direction of the thrust, which is up the slope in this problem. The gravitational force is opposite to the thrust, and in the limit ∆m → 0, is just the downslope component of the fire truck’s weight, or −M g sin θ. Therefore, the equation of motion of the fire truck (which replaces Equation 10-10b) is M (dv/dt) = −vex (dM/dt) − M g sin θ. This equation can be integrated (after multiplication by dt/M ) to find the speed vf as a function of time tf , Z vf Z Mf Z tf dM dv = −vex vf − vi = dt − g sin θ M vi Mi ti Mi = vex ln − g sin θ(tf − ti ) Mf (this replaces Equation 10-11). For this fire truck, Mi = 21×103 kg and vi = 0, at ti = 0, and Mf ≥ (21 − 15)×103 kg = 6×103 kg, which is the truck’s mass without any water. Since vex is constant, so is the mass rate of flow of water (see Equation 18-4b), 3 thus dM/dt = −ρw vex πR2 = −(103 kg/m )× (50 m/s)π(3 cm)2 = −45π kg/s, where R = 3 cm is the radius of the nozzle, and ρw the density of water. Then the mass of the fire truck (at any time until the water runs out) is Mf = Mi − (45π kg/s)tf . At the start, the thrust, vex |dM/dt| = (50 m/s)× (45π kg/s) = 7.07 kN, is less than the magnitude of the downslope component of the weight, 2 Mi g sin θ = (21×103 kg)(9.8 m/s ) sin 3◦ = 10.8 kN, so the truck begins to roll downslope. When the water runs out, at tf = (15×103 kg)/(45π kg/s) = 106 s, vf = vex ln(21/6) − g sin θ(106 s) = 8.22 m/s, which is positive, upslope. Therefore, the firefighters succeed in stopping the truck (only instantaneously, unless they also stop the water and block the wheels) when vf = 0 = vex ln[Mi /(Mi − ρw vex πR2 t)] − (g sin θ)t, or, with the above numbers, ln[1 − (6.732×10−3 s−1 )t] = −(1.026×10−2 s−1 )t. A numerical solution for t, CHAPTER 14 obtained with three iterations of Newton’s method or widely available PC software, yields t = 88.8 s. Problem 3. A block of mass m1 is attached to the axle of a uniform solid cylinder of mass m2 and radius R by massless strings. The two accelerate down a slope that makes an angle θ with the horizontal, as shown in Fig. 3. The cylinder rolls without slipping and the block slides with coefficient of kinetic friction µ between block and slope. The strings are attached to the cylinder’s axle with frictionless loops so that the cylinder can roll freely without any torque from the string. Find an expression for the acceleration of the pair, assuming that the string remains taut. θ 229 motion of the cylinder is τP = (m2 g sin θ − T )R = IP α = ( 32 m2 R2 )(a|| /R), or m2 g sin θ − T = 32 m2 a|| (see the parallel-axis theorem in Chapter 12 for IP ). If T is eliminated by adding the two equations of motion, one finds a|| = 2g[(m1 + m2 ) sin θ − µm1 cos θ]÷ (2m1 + 3m2 ). Problem 4. A missile is launched from point A in Fig. 4, heading for target C. The launch angle is 45.0, and the launch speed is 1.90 km/s. An antimissile defense system is located at point B, 310 km downrange from the launch site. It fires an interceptor rocket at a 65.0◦ launch angle, with the intention of hitting the attacking missile when the latter is 270 km downrange of its launch site. (a) What should be the interceptor’s launch speed? (b) How long after the launch of the attacking missile should the interceptor be launched? (c) At what altitude does the interception take place? figure 3 Cumulative Problem 3. C 270 km 310 km A B figure 4 Cumulative Problem 4. Solution Cumulative Problem 3 Solution. Solution One must consider the forces on the block and the cylinder, exerted by gravity, the inclined surface, and the strings, as sketched (since the strings are assumed massless and other forces are neglected). If the strings remain taut, then the downslope acceleration, a|| , of the block and the cylinder’s center of mass are the same. If the string tension is parallel to the slope, the normal force on the block is N1 = m1 g cos θ, and the parallel component of its equation of motion is Fnet,|| = m1 g sin θ + T − µm1 g cos θ = m1 a|| . When rolling without slipping, the point of contact, P, of the cylinder with the slope, is instantaneously at rest, so the acceleration of its center of mass is a|| = αR (this follows from v = ωR and ω = ωc = ωcm ; see Section 12-5). Since only the string tension and gravity exert torques about point P, the equation of It is simplest to answer each part of this question using a different coordinate system, so that equations in the text can be used without modification. We are given enough information to specify the location of the point of interception; it is 270 km from point A, the missile launch site, and 310 − 270 = 40 km from point B, the interceptor rocket launch site. (c) The trajectory of the missile (Equation 4-9 with origin at point A in Fig. 4) gives the altitude of the point of interception when x = 270 m and other data for the missile launch are substituted: 2 y = (270 km) tan 45◦ − (0.0098 km/s )(270 km)2 2(1.90 km/s)2 cos2 45◦ = 72.1 km The time of flight of the missile to the point of interception is t = x/v0x = (270 km)÷ (190 cos 45◦ km/s) = 201 s. (a) With altitude from part (c) above and the other initial data for the interceptor rocket, its trajectory equation (Equation 4-9 with origin at point B and x axis leftward in Fig. 4) can be solved for the 230 CHAPTER 14 launch speed, as in Example 4-4. (The point of interception has coordinates x = 40 km and y = 72.1 km relative to point B.) Thus, " #1/2 2 (0.0098 km/s ) 40 km v0 = 2(40 tan 65◦ − 72.1) km cos 65◦ = 1.79 km/s. (b) The time of flight for the interceptor rocket is t = x/v0x = (40 km)/(1.79 cos 65◦ km/s) = 52.8 s. Therefore, it should be launched 201 − 52.8 = 148 s after the missile. Problem 5. A solid ball of radius R is set spinning with angular speed ω about a horizontal axis. The ball is then lowered vertically with negligible speed until it just touches a horizontal surface and is released (Fig. 5). If the coefficient of kinetic friction between the ball and the surface is µ, find (a) the linear speed of the ball once it achieves pure rolling motion and (b) the distance it travels before its motion is pure rolling. ω v figure 5 Cumulative Problem 5. Solution (a) While there is relative motion at the point of contact between the ball and the horizontal surface, the force of sliding friction (f = µN = µmg) slows the ball’s rotation and accelerates its center of mass. The equation for the latter is f = µmg = macm , or acm = µg (positive to the right in the sketch and Fig. 5). The equation for the former is τ = −f R = −µmgR = Iα = (2mR2 /5)α, where α = −5µg/2R is the angular acceleration about the horizontal axis through the center of the ball (positive clockwise in the sketch and Fig. 5). The accelerations are constant, so the velocities are given by Equations 2-17 and 12-9 as vcm = acm t = µgt and ω = ω0 + αt = ω0 − 5µgt/2R, where the ball is released at t = 0 and the initial velocities, v0 = 0 and ω0 , are given. The accelerated motion continues until the point of contact is instantaneously at rest (no more sliding friction). The ball rolls without slipping thereafter, at a constant velocity given by vcm = ωR. This occurs at a time t, when (ω0 − 5µgt/2R)R = µgt, or t = 2ω0 R/7µg. Thus, the final velocity is vcm = µgt = 2ω0 R/7. (b) The distance traveled during this time is ∆x = 12 acm t2 = 1 2 2 2 2 2 (µg)(2ω0 R/7µg) = 2ω0 R /49µg (or ∆x = vcm /2acm with the same result).

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