Physics Solution Manual for 1100 and 2101

197 ssm reasoning and solution the free body diagram

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Unformatted text preview: s is ΣFy = −T sin16.0° − T′ sin 16.0° = −(21 N) sin 16.0° − (21 N) sin 16.0° = −11.6 N The magnitude F of the net force exerted on the tooth is F= ( ΣFx ) + ( ΣFy ) 2 2 2 2 = ( 0 N ) + ( −11.6 N ) = 11.6 N 196 FORCES AND NEWTON'S LAWS OF MOTION ______________________________________________________________________________ 56. REASONING At first glance there seems to be very little information given. However, it is enough. In part a of the drawing the bucket is hanging stationary and, therefore, is in equilibrium. The forces acting on it are its weight and the two tension forces from the rope. There are two tension forces from the rope, because the rope is attached to the bucket handle at two places. These three forces must balance, which will allow us to determine the weight of the bucket. In part b of the drawing, the bucket is again in equilibrium, since it is traveling at a constant velocity and, therefore, has no acceleration. The forces acting on the bucket now are its weight and a single tension force from the rope, and they again must balance. In part b, there is only a single tension force, because the rope is attached to the bucket handle only at one place. This will allow us to determine the tension in part b, since the weight is known. SOLUTION Let W be the weight of the bucket, and let T be the tension in the rope as the bucket is being pulled up at a constant velocity. The free-body diagrams for the bucket in parts a and b of the drawing are as follows: T 92.0 N 92.0 N W W Free-body diagram for part a Free-body diagram for part b Since the bucket in part a is in equilibrium, the net force acting on it is zero. Taking upward to be the positive direction, we have Σ F = 92.0 N + 92.0 N − W = 0 or W = 184 N Similarly, in part b we have ΣF = T − W = 0 or T = W = 184 N Chapter 4 Problems 57. 197 SSM REASONING AND SOLUTION The free body diagram for the plane is shown below to the left. The figure at the right shows the forces resolved into components parallel to and perpendicular to the line of motion of the plane. L L T R T R W θ W sin θ W cos θ If the plane is to continue at constant velocity, the resultant force must still be zero after the fuel is jettisoned. Therefore (using the directions of T and L to define the positive directions), T – R – W(sin θ) = 0 (1) L – W (cos θ) = 0 (2) From Example 13, before the fuel is jettisoned, the weight of the plane is 86 500 N, the thrust is 103 000 N, and the lift is 74 900 N. The force of air resistance is the same before and after the fuel is jettisoned and is given in Example 13 as R = 59 800 N. After the fuel is jettisoned, W = 86 500 N – 2800 N = 83 700 N From Equation (1) above, the thrust after the fuel is jettisoned is T = R + W (sin θ) = [(59 800 N) + (83 700 N)(sin 30.0°)] = 101 600 N From Equation (2), the lift after the fuel is jettisoned is L = W (cos θ) = (83 700 N)(cos 30.0°) = 72 500 N a. The pilot must, therefore, reduce the thrust by 103 000 N – 101 600 N = 1400 N b. The pilot must reduce the lift by 74 900 N – 72 500 N = 2400 N ____________________________________________________________________________________________ 198 FORCES AND NEWTON'S LAWS OF MOTION 58. REASONING The worker is standing still. Therefore, he is in equilibrium, and the net force acting on him is zero. The static frictional force fs that prevents him from slipping points upward along the roof, an angle of θ degrees above the horizontal; we choose this as the −x direction (see the free-body diagram). The normal force FN is perpendicular to the roof and thus has no x component. But the gravitational force mg of the earth on the worker points straight down, and thus has a component parallel to the roof. We will use this freebody diagram and find the worker’s mass m by applying Newton’s second law with the acceleration equal to zero. +y FN fs mg sin θ θ +x mg θ Free-body diagram of the worker SOLUTION The static frictional force fs points in the −x direction, and the x component of the worker’s weight mg points in the +x direction. Because there are no other forces with x components, and the worker’s acceleration is zero, these two forces must balance each other. The x component of the worker’s weight is mg sin θ, therefore fs = mg sin θ. Solving this relation for the worker’s mass, we obtain m= fs g sin θ = 390 N (9.80 m/s2 )(sin 36o ) = 68 kg 59. REASONING The sum of the angles the right and left surfaces +y make with the horizontal and the angle between the two surface F1 F2 must be 180.0°. Therefore, the angle that the left surface makes with respect to the horizontal is 180.0° − 90.0° − 45.0° = 45.0°. In 45.0° 45.0° the free-body diagram of the wine bottle, the x axis is the +x horizontal. The force each surface exerts on the bottle is perpendicular to the surfaces, so both forces are directed 45.0° above the horizontal. Letting the surface on the right be surface 1, W and the surface on the left be surface 2, the forces F1 and F2 a...
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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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