Physics Solution Manual for 1100 and 2101

# 7 as where i is the moment of inertia of the cd and is

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Unformatted text preview: n the end of the boom prevents the boom from falling down. The horizontal force H that also acts at the hinge pin prevents the boom from sliding to the left. The weight WL of the wrecking ball (the "load") acts at the end of the boom. Chapter 9 Problems 453 θ = 32° T φ–θ WB V θ = 32° WL P φ = 48° H By applying the equilibrium conditions to the boom, we can determine the desired forces. SOLUTION The directions upward and to the right will be taken as the positive directions. In the x direction we have (1) ∑ Fx = H − T cos θ = 0 while in the y direction we have ∑ Fy = V − T sin θ − WL − WB = 0 (2) Equations (1) and (2) give us two equations in three unknown. We must, therefore, find a third equation that can be used to determine one of the unknowns. We can get the third equation from the torque equation. In order to write the torque equation, we must first pick an axis of rotation and determine the lever arms for the forces involved. Since both V and H are unknown, we can eliminate them from the torque equation by picking the rotation axis through the point P (then both V and H have zero lever arms). If we let the boom have a length L, then the lever arm for WL is L cos φ , while the lever arm for WB is ( L / 2) cos φ . From the figure, we see that the lever arm for T is L sin(φ – θ ) . If we take counterclockwise torques as positive, then the torque equation is L cos φ ∑ τ = −WB − WL L cos φ + TL sin (φ − θ ) = 0 2 Solving for T, we have 1 W +W L (3) cos φ T=2 B sin(φ –θ ) a. From Equation (3) the tension in the support cable is T= 1 (3600 2 N) + 4800 N sin(48° – 32°) cos 48° = 1.6 × 104 N 454 ROTATIONAL DYNAMICS b. The force exerted on the lower end of the hinge at the point P is the vector sum of the forces H and V. According to Equation (1), ( ) H = T cos θ = 1.6 × 104 N cos 32° = 1.4 × 104 N and, from Equation (2) ( ) V = WL + WB + T sin θ = 4800 N + 3600 N + 1.6 × 104 N sin 32° = 1.7 × 104 N Since the forces H and V are at right angles to each other, the magnitude of their vector sum can be found from the Pythagorean theorem: FP = H 2 + V 2 = (1.4 ×104 N)2 + (1.7 ×104 N)2 = 2.2 ×104 N 27. SSM REASONING Since the man holds the ball motionless, the ball and the arm are in equilibrium. Therefore, the net force, as well as the net torque about any axis, must be zero. SOLUTION Using Equation 9.1, the net torque about an axis through the elbow joint is Στ = M(0.0510 m) – (22.0 N)(0.140 m) – (178 N)(0.330 m) = 0 Solving this expression for M gives M = 1.21× 103 N . The net torque about an axis through the center of gravity is Στ = – (1210 N)(0.0890 m) + F(0.140 m) – (178 N)(0.190 m) = 0 Solving this expression for F gives F = 1.01× 103 N . Since the forces must add to give a net force of zero, we know that the direction of F is downward . 28. REASONING Although the crate is in translational motion, it undergoes no angular acceleration. Therefore, the net torque acting on the crate must be zero: Στ = 0 (Equation 9.2). The four forces acting on the crate appear in the free-body diagram: its weight W, the kinetic friction force fk, the normal force FN, and the tension T in the strap. We will take the edge of the crate sliding along the floor as the rotation axis for applying Equation 9.2. Both the friction force and the normal force act at this point. These two forces, therefore, generate no torque about the axis of rotation, so the clockwise torque of the crate’s weight W must balance the counterclockwise torque of the tension T in the strap: Chapter 9 Problems 455 T L/2 H/2 θ d d θ W fk 25° W lw FN Lever arm of the crate’s weight Free-body diagram of the crate T lT = W l W or T= WlW lT = mg l W lT (1) We will apply trigonometry to determine the lever arms l W and l T for the weight and the tension, respectively, and then calculate the magnitude T of the tension in the strap. SOLUTION The lever arm l W of the crate’s weight is shown in the right-hand diagram ( ) above, and is given by l W = d cos θ + 25o , where d is the distance between the axis of rotation (lower edge of the crate) and the crate’s center of gravity, and θ is the angle between that line and the bottom of the crate. The right triangle in the free-body diagram of the crate (drawing on the left) shows how we can use trigonometry to determine both the length d and the angle θ. The length d is the hypotenuse of that right triangle, and the other sides are the half-height (H/2 = 0.20 m) and half-length (L/2 = 0.45 m) of the crate, so by the Pythagorean theorem (Equation 1.7) we find that 2 2 H L d = + = 2 2 ( 0.20 m )2 + ( 0.45 m )2 = 0.49 m We can find the angle θ from the inverse tangent function: H 2 −1 0.40 m o = tan = 24 0.90 m L2 θ = tan −1 456 ROTATIONAL DYNAMICS The lever arm l T of the tension force is illustrated in the drawing below, where we see that ( ) l T = L sin 61o − 25o = L sin 36o . Therefore, from Equation (1), the magnitude of the tension in the strap is T= mg l W lT = ( mgd cos 24o + 25o L sin 36o ) = ( 72 kg ) ( 9....
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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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