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Chapter 4 Conceptual Questions: 15. (a) (b) N N F F m m The force F is the combination of the pushing by the seat back and the friction force by the seat surface. Multiple Choice Questions: 13. (d). Exercises: 7. x (sin 30) ŷ ] = (4.76 N) ^ x + (2.75 N) ŷ , F 1 = (5.5 N)[(cos 30) ^ x + (sin 37) ŷ ] = (2.80 N) ^ x + (2.11 N) ŷ . F 2 = (3.5 N)[(cos 37) ^ F = F 1 + F 2 + F 3 = 0, 25. F 3 = ( F 1 + F 2) = (–7.6 N) x̂ + (0.64 N) ŷ . (a) Since the object slides upward, there is no acceleration in the horizontal direction, so Fx = 0. Fx = N F cos 60 = 0, N = F cos 60 = (60 N) cos 60 = 30 N . (b) Fy = F sin 60 w = = F sin 60 mg = may, so ay = F sin 60 (60 N) sin 60 g= 9.80 m/s2 = 4.60 m/s2 . m 10.0 kg The negative sign means that the object is slowing down while it is sliding upward. 29. Applying Newton’s second law along the x- and y-axes gives Fx = max = -ma cos 30° = –(2.0 kg)(1.5 m/s2) cos 30° = –2.6 N Fy = may = ma sin 30° = (2.0 kg)(1.5 m/s2) sin 30° = 1.5 N F = –2.6 N xˆ + 1.5 N yˆ 40. (a) Fx = F cos 30 = (25 N) cos 30 = 21.65 N = max, y N so ax = = 0.72m/s2 . F 30 (b) Fy = N + F sin 30 w = may = 0, so x N = w F sin 30 = (30 kg)(9.80 m/s2) (25 N) sin 30 = w = mg 2.8 102 N . 41. (a) Fx = F cos = (30 N) cos 37 = 23.96 N = max, so y N 23.96 N ax = = 0.96m / s2 . 25 kg F x (b) Fy = N F sin w = may = 0, N = w + F sin 37 = (25 kg)(9.80 m/s2) + (30 N) sin 37 = 2.6 102 N 43. w = mg The acceleration is zero. The direction of the force F must be along and up the incline. Fx = mg sin F = max = 0, so y N F = mg sin = (25.0 kg)(9.80 m/s2 ) sin 30 = 123 N up the incline . F x 30 m g 44. (a) There are (2) two forces on the skier, the weight and the y normal force. N (b) The x-component of the weight is the side opposite to the angle shown, so sine is used. so Fx = mg sin = max, ax = g sin = (9.80 m/s2) sin 37 = 5.9 m/s2 . (c) From v2 = vo2 + 2ax, v = x m (5.0 m/s)2 + 2(5.9 m/s2)(35 m) = 21 m/s . 45. From Exercise 44(b), ax = g sin = (9.80 m/s2) sin 30 = 4.9 m/s2 down the slope. If up the slope is chosen as positive, then a = 4.9 m/s2. Taking xo = 0. v2 = vo2 + 2a(x xo), x= 0 2 – 25m / s 2 2 –4.9m / s2 = 64 m . (b) 48. (a) There are two rings. Fy = T + T w = 0. So 50. w mg (50 kg)(9.80 m/s2) T= 2 = 2 = = 2.7 102 N . 2 (b) In the vertical direction: Fy = T sin + T sin mg = 0. So T T= mg mg (55 kg)(9.80 m/s2) = = 3.8 102 N . 2 sin 2 sin 45 Isolate point A (or B). The vertical component of the tension T in the topmost cord supports half of the 10-kg weight. Therefore T sin 45° = (5 kg)(9.80 m/s2) = 69.3 N The horizontal forces at A balance, giving TAB = T cos 45° = (69.3 N) cos 45° = 49 N 51. The vertical component of the tension T in the lower cords attached at A and B supports the 10-kg weight, so T 2T cos 30° = mg = (10 kg)(9.80 m/s2) T = 57 N The tension in the vertical cord attached to the weight supports the full weight, so its tension is mg = 98 N . From Exercise 50, the other two tensions are 49 N and 69 N . 60. The acceleration is zero if both are at rest. For m1: Fx1 = T m1 g sin = m1ax = 0, T = m1 g sin. For m2: Fy2 = m2 g T = m2 ay = 0, So m2 g = T = m1 gsin. m2 = m1 sin = (2.0 kg) sin 37 = 1.2 kg . If both are moving at constant velocity, the answer is the same 1.2 kg because the acceleration is still zero and the forces must still balance out. 61. (a) For m1: Fx1 = T m1 g sin = m1ax. (1) For m2: Fy2 = m2 g T = m2 ay. (2) ax = ay = a in magnitude. Equation (1) + Equation (2) gives (m2 m1 sin)g = (m1 + m2)a. m2 m1 sin 2.50 kg (3.00 kg) sin 37 (9.80 m/s2) = 1.24 m/s2 a= g= m1 + m 2 2.50 kg + 3.00 kg =.1.2 m/s2 (m1 up and m2 down) 73. Yes , the coefficient of kinetic friction can be found. Fy = N mg cos = may = 0, For constant velocity, Or N = mg cos. ax = 0. So mg sin = k N = k (mg cos). Fx = mg sin fk = 0. Therefore k = sin = tan . cos 75. (a) Use the result from Exercise 66(a). s = tan, independent of m, size, etc. So it is still 30 . (b) s = tan 30 = 0.58 . 77. (a) If there were no friction, the minimum pushing force is equal to F = mg sin = (35.0 kg)(9.80 m/s2) sin 20 = 117 N. Since the person has to use a 150 N force, there is friction. So the answer is no , the incline is not frictionless. (b) fk = 150 N 117 N = 33 N . 79. (a) First assume that m1 has the tendency to move up the incline. For m1: Fy1 = N1 m1 g cos = 0, N1 = m1 g cos. Fx1 = T m1 g sin fsmax = T m1 g sin s N1 = T m1 g sin s m1 g cos = 0. So T = m1 g sin + s m1 g cos = m1 g(sin + s cos). For m2: Fx2 = m2 g T = 0, m2 g = T = m1 g(sin + s cos). Therefore y N m2 = m1 (sin + s cos) = (2.0 kg)(sin 37 + 0.30 cos 37) = F 1.7 kg. fk Next assume that m1 has the tendency to move down the incline. The static friction force will now be pointing up the incline, so the term s m1 g cos becomes negative. Repeating the calculation: m2 = m1 (sin s cos) = (2.0 kg)(sin 37 x m 0.30 cos 37) = 0.72 kg. Thus m2 can be anywhere between 0.72 kg and 1.7 kg . (b) When both are moving at constant velocity, the acceleration is still zero. However, s is replaced by k. Assume that m1 has the tendency to move up the incline. m2 = m1 (sin + k cos) = (2.0 kg)(sin 37 + 0.20 cos 37) = 1.5 kg. Assume that m1 has the tendency to move down the incline. m2 = m1 (sin k cos) = (2.0 kg)(sin 37 0.20 cos 37) = 0.88 kg. Therefore m2 can be anywhere between 0.88 kg and 1.5 kg . 84. (a) As the boat rises, the cables will go flatter (the angle decreases). The sum of the vertical components of the tensions in the cable must support the weight of the boat. The weight of the boat remains the same. As decreases, the tension must (1) increase . (b) Fy = 2T sin mg = ma = 0. At 45, T = mg (500 kg)(9.80 m/s2) = = 3.46 103 N . 2 sin 2 sin 45 At 30, T = (500 kg)(9.80 m/s2) = 4.90 103 N . 2 sin 30 As we see, the decrease in angle increases the tension. 88. (a) x y fk N mg Since the block is moving downward on the incline but slowing down (because it comes to rest), its acceleration must be up the incline; call this the +x-direction, and call the +y-direction perpendicular to this pointing upward from the surface of the incline. (It would also be possible to call the +x-axis pointing down the plane.) (b) Forces along the y-axis balance. F y N mg cos 0 Acceleration is along the x-axis. F x f k mg sin ma (c) Using kinematics to find the acceleration gives v2 = v02 + 2a(x – x0) 0 = (1.60 m/s)2 + 2a(1.10 m) a = 1.164 m/s2 up the incline fk = µkN = µkmg cos Putting this into the result from Newton’s second law gives µkmg cos –- mg sin = ma Solving for µk gives 1.164 m/s2 sin 30 2 a / g sin 9.80 m/s k 0.714 cos cos 30