Easy. P5.3 For equilibrium: f = F and n = F g. Also, f = n, i.e., f n F F g. (a) 75.0 N N N N (b) ma y.

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1 Chapter 5 Homework Solutions Easy P5.3 For equilibrium: f = F and n = F g. Also, f = n, i.e., (a) f n F F g s k 75.0 N N N N ANS. FIG. P5.3 P5.4 F y ma y : n mg 0 f s s n s mg This maximum magnitude of static friction acts so long as the tires roll without skidding. F x ma x f s ma The maximum acceleration is a s g The initial and final conditions are: x i = 0, v i = 50.0 mi/h = 22.4 m/s, v f = 0. Then, v f 2 v i 2 2a(x f x i ) v i 2 2 s gx f (a) x f v 2 i 2 s g m 22.4 m s x f x f v 2 i 2 s g 9.80 m s m 22.4 m s x f m s 2

2 P5.10 (a) See the free-body diagram of the suitcase in the figure on the right. m suitcase = 20.0 kg, F = 35.0 N F x ma x : 20.0 N Fcos 0 ma y : n Fsin F g 0 F y Fcos 20.0 N cos 20.0 N 35.0 N ANS. FIG. P5.10(a) (c) With Fg = (20.0 kg)(9.80 m/s 2 ), n F g Fsin N n 167 N m s 2 P5.16 (a) F mv kg r m N inward a v 2 r m s m m s 2 inward P5.26 (a) The external forces acting on the water are the gravitational force and the contact force exerted on the water by the pail. (c) The contact force exerted by the pail is the most important in causing the water to move in a circle. If the gravitational force acted alone, the water would follow the parabolic path of a projectile. When the pail is inverted at the top of the circular path, it cannot hold the water up to prevent it from falling out. If the water is not to spill, the pail must be moving fast enough that the required centripetal force is at

3 least as large as the gravitational force. That is, we must have m v 2 r mg or v rg 1.00m 9.80m s m s (d) If the pail were to suddenly disappear when it is at the top of the circle and moving at 3.13 m/s, the water would follow the parabolic path of a projectile launched with initial velocity components of v xi = 3.13 m/s, v yi = 0. P5.37 F Gm 1m N m 2 / kg kg r m N 2.00 kg P5.38 For two 70 kg persons, modeled as spheres, F g Gm 1m N m 2 / kg 2 70 kg70 kg ~ 10 7 N r 2 2 m 2 Medium P5.7 Newton s second law for the 5.00-kg mass gives T f k = (5.00 kg)a Similarly, for the 9.00-kg mass, (9.00 kg)g T = (9.00 kg)a Adding these two equations gives: 9.80 m/ s kg 9.00 kg 9.80 m/ s kg Which yields a = 5.60 m/s 2. Plugging this into the first equation above gives a ANS. FIG. P5.7

4 5.60 m/ s 2 T 5.00 kg 9.80 m/ s kg 37.8 N P5.9 m = 3.00 kg, = 30.0, x = 2.00 m, t = 1.50 s (a) At constant acceleration, Solving, x f v i t 1 2 at 2 a 2 x f v i t t 2 2(2.00 m 0) (1.50 s) m/ s 2 ANS. FIG. P5.9 From the acceleration, we can calculate the friction force, answer (c), next. (c) Take the positive x axis down parallel to the incline, in the direction of the acceleration. We apply Newton s second law: F x mg sin f ma Solving, f = m(g sin a ) Substituting, f = (3.00 kg)[(9.80 m/s 2 )sin m/s 2 ] = 9.37 N Applying Newton s law in the y direction (perpendicular to the incline), we have no burrowing-in or taking-off motion. Then the y component of acceleration is zero: F y n mg cos 0 Thus Because we have n = mg cos f = k n µ k f mg cos 9.37 N (3.00 kg) 9.80 m/ s 2 cos 30.0 o (d) v f v i at so v f 0 (1.78 m/ s 2 )(1.50 s) 2.67 m/ s

5 P5.12 (a) To find the maximum possible value of P, imagine impending upward motion as case 1. Setting F x 0: Pcos50.0 n 0 with f s, max s n: Setting f s, max s Pcos P 0.161P F y 0: Psin P 3.00 kg9.80 m/ s 2 0 P max 48.6 N ANS. FIG. P5.12 To find the minimum possible value of P, consider impending downward motion. As in case 1, Setting F y 0: f s, max 0.161P Psin P 3.00 kg9.80 m/ s 2 0 P min 31.7 N (c) If P 48.6 N, the block slides up the wall. If P 31.7 N, the block slides down the wall. We repeat the calculation as in part (a) with the new angle. Consider impending upward motion as case 1. Setting Setting F x 0: Pcos13 n 0 f s, max s n: f s, max s Pcos P 0.244P F y 0: Psin P 3.00 kg9.80 m/ s 2 0 P max N The push cannot really be negative. However large or small it is, it cannot produce upward motion. To find the minimum possible value of

6 P, consider impending downward motion. As in case 1, Setting f s, max 0.244P F y 0: Psin P 3.00 kg9.80 m/ s 2 0 P min 62.7 N P 62.7 N. The block cannot slide up the wall. If P 62.7 N, the block slides down the wall. P5.13 (a) See the figure on the right. For Block #1, T m 1 g m 1 a For Block #2, 68.0 N T m 2 g m 2 a Adding these equations gives, 68.0 N k m 1 m 2 g m 1 m 2 a or a 68.0 N g 1.29 m/ s 2 m 1 m 2 ANS. FIG. P5.13 (c) T = k m 1 g + m 1 a = m 1 (g + a) = (12.0 kg)[(0.100)(9.80 m/s 2 ) T 27.2 N + (1.29 m/s 2 )]

7 P5.14 (a) The free-body diagrams for each object appear on the right. Let a represent the positive magnitude of the acceleration aĵ of m 1, of the acceleration aî of m 2, and of the acceleration aĵ of m 3. Call T 12 the tension in the left rope and T 23 the tension in the cord on the right. For m 1, F y ma y T 12 m 1 g m 1 a For m 2, F x ma x T 12 k n T 23 m 2 a and For m 3, F y ma y, giving n m 2 g 0 F y ma y, giving T 23 m 3 g m 3 a we have three simultaneous equations: T N 4.00 kga T N T kga T N 2.00 kga Add them up (this cancels out the tensions): 39.2 N 3.43 N 19.6 N 7.00 kga ANS. FIG. P5.14(a) (c) a 2.31 m s 2, down for m 1, left for m 2, and up for m 3 Now T N 4.00 kg 2.31 m s 2 T N and T N 2.00 kg T N 2.31 m s 2 (d) If the tabletop were smooth, friction disappears ( k = 0), and so the acceleration would become larger. For a larger acceleration, according to the equations above, the tensions change: T 12 m 1 g m 1 a T 12 decreases

8 T 23 m 3 g m 3 a T 23 increases P5.21 T cos 5.00 mg 80.0 kg (a) T = 787 N: 9.80 m s 2 r T 68.6 N î 784 Nĵ T sin 5.00 ma c : a c m s 2 toward the center of the circle. The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion. ANS. FIG. P5.21 P5.25 Let the tension at the lowest point be T. From Newton s second law, F ma and T mg ma c mv 2 T m g v 2 r r T 85.0 kg9.80 m s m s 10.0 m 1.38 kn N 2 ANS. FIG. P5.25 He doesn t make it across the river because the vine breaks.

9 Hard P5.44 (a) If the car is about to slip down the incline, f is directed up the incline. F y ncos f sin mg 0 where f s n. Substituting, n and f Then, F x mg cos 1 s tan s mg cos 1 s tan v min n sin f cos m v 2 min R Rg tan s 1 s tan yields ANS. FIG. P5.44 When the car is about to slip up the incline, f is directed down the incline. Then, F y This yields ncos f sin mg 0 with f s n mg n cos1 s tan and f s mg cos1 s tan In this case, v max F x n sin f cos m v 2 max R Rg tan s 1 s tan If v min Rg tan s 1 s tan 0, then s tan., which gives

10 P5.55 (a) First, draw a free-body diagram, (top figure) of the top block. Since ay = 0, n1 = 19.6 N, and f k k n N 5.88 N From F x ma T, 10.0 N 5.88 N 2.00 kga T or a T 2.06 m/ s 2 (for top block). Now draw a free-body diagram (middle figure) of the bottom block and observe that F x Ma B gives f 5.88 N 8.00 kga B or a B m/ s 2 (for the bottom block). In time t, the distance each block moves (starting from rest) is d T 1 2 a T t 2 and d B 1 2 a B t 2. For the top block to reach the right edge of the bottom block, (see bottom figure), it is necessary that d T d B L or ANS. FIG. P m/ s2t d B 1 2 which gives t 2.13 s. From above, m s m/ s2t m 2.13 s m.