Physics Solution Manual for 1100 and 2101

# 77 85 kg 980 ms a 59 103 ms 2 m m 109 000 kg 81

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Unformatted text preview: g an appropriate equation of kinematics from Chapter 3. SOLUTION Assume that the boat and trailer are moving in the +x direction. Newton’s second law is ΣFx = ma x (see Equation 4.2a), where the net force is just the tension +T in the hitch, so ΣFx = T . Thus, T = ma x (1) Since the initial and final velocities, v0x and vx, and the time t are known, we may use Equation 3.3a from the equations of kinematics to relate these variables to the acceleration: v x = v0 x + a x t (3.3a) Solving Equation (3.3a) for ax and substituting the result into Equation (1), we find that v −v 11 m/s − 0 m/s T = max = m x 0 x = ( 410 kg ) = 160 N t 28 s ____________________________________________________________________________________________ 80. REASONING Since we assume that there is no +y frictional force resisting the airplane’s motion, the only horizontal force acting on the airplane arises because of the tension (magnitude = T) in the cable. From Newton’s FN second law ΣF = Ma (Equation 4.1), we conclude that the airplane’s acceleration is given by a = ΣF/M = T/M, fsMAX T where M is the mass of the airplane. The harder the man +x pulls on the cable, the greater the tension T, and the greater the airplane’s acceleration. According to Newton’s third law, however, the cable also exerts an mg opposing horizontal force of magnitude T on the man. Thus, if he is to keep his footing, T cannot exceed the Free-body diagram maximum force of static friction fsMAX the runway of the man exerts on him. Therefore, the airplane’s acceleration is greatest when these two forces have equal magnitudes: T = fsMAX (see the free-body diagram of the man). The maximum static frictional force the runway can exert is determined by the relation fsMAX = µs FN (Equation 4.7). Because the man has no acceleration in the vertical direction, the normal force must balance the downward pull of gravity: FN = mg. 214 FORCES AND NEWTON'S LAWS OF MOTION SOLUTION Combining FN = mg and fsMAX = µs FN (Equation 4.7), we obtain the maximum tension in the cable: T = fsMAX = µs FN = µs mg (1) We can now substitute Equation (1) for the tension into Newton’s second law (a = T/M), and calculate the maximum possible acceleration of the airplane: ( ) 2 T µs mg ( 0.77 )( 85 kg ) 9.80 m/s a= = = = 5.9 ×10−3 m/s 2 M M 109 000 kg 81. REASONING AND SOLUTION If the +x axis is taken to be parallel to and up the ramp, then ΣFx = max gives T – fk – mg sin 30.0° = max where fk = µ kFN . Hence, Also, ΣFy = may gives T = max + µ kFN + mg sin 30.0° (1) FN – mg cos 30.0° = 0 since no acceleration occurs in this direction. Then FN = mg cos 30.0° (2) Substitution of Equation (2) into Equation (1) yields T = max + µ kmg cos 30.0° + mg sin 30.0° T = (205 kg)(0.800 m/s2) + (0.900)(205 kg)(9.80 m/s2)cos 30.0° + (205 kg)(9.80 m/s2)sin 30.0° = 2730 N ____________________________________________________________________________________________ Chapter 4 Problems 215 82. REASONING To determine the man’s upward acceleration by + means of Newton’s second law, we first need to identify all of the forces exerted on him and then construct a free-body diagram. The earth pulls down on the man with a gravitational force W = mg. Once T T he begins accelerating upward, he is no longer in contact with the ground, so there is no normal force acting on him. The pulling force P that he exerts on the rope does not appear in his free-body diagram because it is not a force exerted on him. Each end of the rope exerts a W tension force on him. If we assume that the rope is massless, and ignore friction between the rope and the branch, then the magnitude of the tension T is the same everywhere in the rope. Because the man Free-body diagram of the man pulls down on the free end intentionally and on the other end inadvertently (because it is tied around his waist), Newton’s third law predicts that both ends of the rope pull upward on him. The third law predicts that the free end of the rope pulls up on the man with a force exactly equal in magnitude to that of the 358-N pulling force. Thus, in addition to the downward gravitational force, there are two upward tension forces with magnitudes T = 358 N acting on the man, as illustrated in the free-body diagram. SOLUTION Taking up as the positive direction and applying Newton’s second law to the man’s free-body diagram yields ΣF = 2T − W = ma Solving for the acceleration a, we find a= 83. 2 ( 358 N ) 2T − W 2T − mg 2T = = −g = − 9.80 m/s 2 = 0.14 m/s 2 m m m 72.0 kg SSM REASONING The free-body diagrams for Robin (mass = m) and for the chandelier (mass = M) are given at the right. The tension T in the rope applies an upward force to both. Robin accelerates upward, while the chandelier accelerates downward, each acceleration having the same magnitude. Our solution is based on separate applications of Newton’s second law to Robin and the chandelier. T Robin SOLUTION Applying Newton’s second law, we find T – mg = ma 14 244 4 3 Robin Hood and T – Mg = – Ma 14 244 4 3 Chandelier T Chandeli...
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## This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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