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PHYS-207 Honors Section LL, HW4 Solutions 5.6. Since the tire remains stationary, by Newton’s second law, the net force must be zero: Fnet FA FB FC ma 0. From the free-body diagram shown on the right, we have 0 Fnet, x FC cos FA cos 0 Fnet, y FA sin FC sin FB FB , we first compute . With FA 220 N , FC 170 N , and 47, we get F cos (220 N) cos 47.0 cos A 0.883 28.0 FC 170 N To solve for Substituting the value into the second force equation, we find FB FA sin FC sin (220 N)sin 47.0 (170 N)sin 28.0 241 N. 5.17. The free-body diagram of the problem is shown to the right. Since the acceleration of the block is zero, the components of the Newton’s second law equation yield T – mg sin = 0 FN – mg cos = 0, where T is the tension in the cord, and FN is the normal force on the block. (a) Solving the first equation for the tension in the string, b gc h T mg sin 8.5 kg 9.8 m / s2 sin 30 42 N . (b) We solve the second equation in part (a) for the normal force FN: FN mg cos 8.5 kg 9.8 m/s 2 cos 30 72 N . (c) When the cord is cut, it no longer exerts a force on the block and the block accelerates. The xcomponent equation of Newton’s second law becomes –mgsin = ma, so the acceleration becomes a g sin (9.8 m/s2 )sin 30 4.9 m/s 2 . The negative sign indicates the acceleration is down the plane. The magnitude is 4.9 m/s 2. Note: The normal force FN on the block must be equal to mg cos so that the block is in contact with the surface of the incline at all time. When the cord is cut, the block has an acceleration a g sin , which in the limit 90 becomes g . 5.41. The mass of the bundle is m = (449 N)/(9.80 m/s2) = 45.8 kg and we choose +y upward. (a) Newton’s second law, applied to the bundle, leads to T mg ma a 387 N 449 N 45.8 kg which yields a = –1.4 m/s2 (or |a| = 1.4 m/s2) for the acceleration. The minus sign in the result indicates the acceleration vector points down. Any downward acceleration of magnitude greater than this is also acceptable (since that would lead to even smaller values of tension). (b) We use Eq. 2-16 (with x replaced by y = –6.1 m). We assume 0 = 0. v 2ay 2 1.35 m/s2 6.1 m 4.1 m/s. For downward accelerations greater than 1.4 m/s2, the speeds at impact will be larger than 4.1 m/s. 5.51. The free-body diagrams for m1 and m2 are shown in the figures below. The only forces on the blocks are the upward tension T and the downward gravitational forces F1 m1 g and F2 m2 g . Applying Newton’s second law, we obtain: T m1 g m1a m2 g T m2 a which can be solved to yield m m1 a 2 g m m 2 1 2m1m2 T g m m 1 2 (a) With m1 1.3 kg and m2 2.8 kg , the acceleration becomes Substituting the result back, we have: 2.80 kg 1.30 kg 2 2 2 a (9.80 m/s ) 3.59 m/s 3.6 m/s . 2.80 kg 1.30 kg (b) Similarly, the tension in the cord is T 2(1.30 kg)(2.80 kg) (9.80 m/s 2 ) 17.4 N 17 N. 1.30 kg 2.80 kg 5.87. From the reading when the elevator was at rest, we know the mass of the object is m = (65 N)/(9.8 m/s2) = 6.6 kg. We choose +y upward and note there are two forces on the object: mg downward and T upward (in the cord that connects it to the balance; T is the reading on the scale by Newton’s third law). (a) “Upward at constant speed” means constant velocity, which means no acceleration. Thus, the situation is just as it was at rest: T = 65 N. (b) The term “deceleration” is used when the acceleration vector points in the direction opposite to the velocity vector. We’re told the velocity is upward, so the acceleration vector points downward (a = –2.4 m/s2). Newton’s second law gives T mg ma T (6.6 kg)(9.8 m/s2 2.4 m/s2 ) 49 N. 6.16. (a) In this situation, we take f s to point uphill and to be equal to its maximum value, in which case fs, max = s FN applies, where s = 0.25. Applying Newton’s second law to the block of mass m = W/g = 8.2 kg, in the x and y directions, produces Fmin 1 mg sin f s , max ma 0 FN mg cos 0 which (with = 20°) leads to Fmin 1 mg sin s cos 8.6 N. (b) Now we take f s to point downhill and to be equal to its maximum value, in which case fs, max = sFN applies, where s = 0.25. Applying Newton’s second law to the block of mass m = W/g = 8.2 kg, in the x and y directions, produces Fmin 2 mg sin f s , max ma 0 FN mg cos 0 which (with = 20°) leads to Fmin 2 mg sin s cos 46 N. 6.48. We will start by assuming that the normal force (on the car from the rail) points up. Note that gravity points down, and the y axis is chosen positive upward. Also, the direction to the center of the circle (the direction of centripetal acceleration) is down. Thus, Newton’s second law leads to v2 FN mg m . r (a) When v = 11 m/s, we obtain FN = 3.7 103 N. (b) FN points upward. (c) When v = 14 m/s, we obtain FN = –1.3 103 N, or | FN | = 1.3 103 N. (d) The fact that this answer is negative means that FN points opposite to what we had assumed. Thus, r the magnitude of FN is | FN | 1.3 kN and its direction is down. 6.57. For the puck to remain at rest the magnitude of the tension force T of the cord must equal the gravitational force Mg on the cylinder. The tension force supplies the centripetal force that keeps the puck in its circular orbit, so T = mv2/r. Thus Mg = mv2/r. We solve for the speed: v Mgr (2.50 kg)(9.80 m/s 2 )(0.200 m) 1.81 m/s. m 1.50 kg