Physics Solution Manual for 1100 and 2101

8 f k k fn in the y direction we have fy fn mg cos

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Unformatted text preview: er mg Mg 216 FORCES AND NEWTON'S LAWS OF MOTION In these applications we have taken upward as the positive direction, so that Robin’s acceleration is a, while the chandelier’s acceleration is –a. Solving the Robin-Hood equation for T gives T = mg + ma Substituting this expression for T into the Chandelier equation gives mg + ma – Mg = – Ma M –m a= g M +m or a. Robin’s acceleration is (195 kg ) – ( 77.0 kg ) M –m 2 2 a= 9.80 m/s = 4.25 m/s g = M +m 195 kg ) + ( 77.0 kg ) ( ( ) b. Substituting the value of a into the expression for T gives ( ) T = mg + ma = ( 77.0 kg ) 9.80 m/s 2 + 4.25 m/s2 = 1080 N ____________________________________________________________________________________________ 84. REASONING Newton’s second law, Equation 4.2a, can be used to find the tension in the coupling between the cars, since the mass and acceleration are known. The tension in the coupling between the 30th and 31st cars is responsible for providing the acceleration for the 20 cars from the 31st to the 50th car. The tension in the coupling between the 49th and 50th cars is responsible only for pulling one car, the 50th. SOLUTION a. The tension T between the 30th and 31st cars is Tx = (Mass of 20 cars)a x ( (4.2a) )( ) = ( 20 cars ) 6.8 × 103 kg/car 8.0 × 10−2 m / s 2 = 1.1 × 104 N b. The tension T between the 49th and 50th cars is Tx = (Mass of 1 car)a x ( (4.2a) )( ) = (1 car ) 6.8 × 103 kg/car 8.0 × 10−2 m / s 2 = 5.4 × 102 N ____________________________________________________________________________________________ Chapter 4 Problems 85. REASONING The box comes to a halt because the kinetic FN frictional force and the component of its weight parallel to the incline oppose the motion and cause the box to slow down. The distance that the box travels up the incline can be can be found mg sinθ 2 by solving Equation 2.9 ( v 2 = v0 + 2ax ) for x. Before we use this approach, however, we must first determine the acceleration fk of the box as it travels along the incline. 217 mg cos θ SOLUTION The figure above shows the free-body diagram for the box. It shows the resolved components of the forces that act on the box. If we take the direction up the incline as the positive x direction, then, Newton's second law gives ∑ Fx = – mg sin θ – f k = max – mg sin θ – µk FN = max or where we have used Equation 4.8, f k = µk FN . In the y direction we have ∑ Fy = FN – mg cos θ = 0 or FN = mg cos θ since there is no acceleration in the y direction. Therefore, the equation for the motion in the x direction becomes – mg sin θ – µ k mg cos θ = ma x or ax = – g (sin θ + µ k cos θ ) According to Equation 2.9, with this value for the acceleration and the fact that v = 0 m/s, the distance that the box slides up the incline is x=– 2 v0 2a = 2 v0 2 g (sin θ + µk cos θ ) = (1.50 m/s)2 = 0.265 m 2(9.80 m/s 2 )[sin 15.0° + (0.180)cos 15.0°] ____________________________________________________________________________________________ 86. REASONING Since we assume that there is no 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. The airplane (mass = M) undergoes a horizontal acceleration caused by the horizontal component Tx = T cos θ of the tension force, where θ is the angle that the cable makes with the horizontal. From Newton’s second law, the acceleration of the airplane is ax = ΣFx M = T cos θ M +y FN fsMAX Tx θ T +x Ty mg (1) Free-body diagram of the man 218 FORCES AND NEWTON'S LAWS OF MOTION The maximum tension in the cable is limited by the condition that the man’s feet must not slip. When the man pulls as hard as possible without slipping, the horizontal component of the tension acting on him matches the maximum static frictional force: Tx = T cos θ = fsMAX (see the free-body diagram of the man). The maximum static frictional force itself is given by fsMAX = µs FN (Equation 4.7). Together, these two relations yield T cos θ = µs FN (2) To evaluate the magnitude FN of the normal force that acts on the man, we must consider Newton’s third law. This law indicates that when the man (mass = m) pulls up on the cable, the cable pulls down on him (see the free-body diagram of the man). This additional downward force increases the upward normal force FN the runway exerts on him. Applying Newton’s second law to the vertical direction in this diagram, with zero acceleration, we see that ΣFy = FN − Ty − mg = FN − T sin θ − mg = 0 . Solving for FN yields FN = T sin θ + mg (3) Substituting Equation (3) into Equation (2) yields an expression in which the tension T is the only unknown quantity: T cos θ = µs (T sin θ + mg ) (4) We now solve Equation (4) for the tension in the cable: T cos θ = µsT sin θ + µs mg or T cos θ − µsT sin θ = µs mg T= or T ( cos θ − µs sin θ ) = µs mg µs mg cos θ − µs sin θ (5) Equation (5) may be substituted into Eq...
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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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