Problem_Set_8_Solutions

Problem_Set_8_Solutions - 18 Our system consists of the...

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18. Our system consists of the lower arm holding a bowling ball. As shown in the free-body diagram, the forces on the lower arm consist of T G from the biceps muscle, F G from the bone of the upper arm, and the gravitational forces, mg G and Mg G . Since the system is in static equilibrium, the net force acting on the system is zero: net, 0 ( ) y F T F m M g = = + ¦ In addition, the net torque about O must also vanish: net 0 ( )( ) (0) ( )( ) ( ) O d T F D mg L Mg τ = = + ¦ . (a) From the torque equation, we find the force on the lower arms by the biceps muscle to be 2 2 ( ) [(1.8 kg)(0.15 m) (7.2 kg)(0.33 m)](9.8 m/s ) 0.040 m 648 N 6.5 10 N. mD ML g T d + + = = = × (b) Substituting the above result into the force equation, we find F to be 2 2 ( ) 648 N (7.2 kg 1.8 kg)(9.8 m/s ) 560 N 5.6 10 N. F T M m g = + = + = = ×

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20. With pivot at the left end of the lower scaffold, Eq. 12-9 leads to m 2 g L 2 2 mgd + T R L 2 = 0 where m 2 is the lower scaffold’s mass (30 kg) and L 2 is the lower scaffold’s length (2.00 m). The mass of the package ( m = 20 kg) is a distance d = 0.50 m from the pivot, and T R is the tension in the rope connecting the right end of the lower scaffold to the larger scaffold above it. This equation yields T R = 196 N. Then Eq. 12-8 determines T L (the tension in the cable connecting the right end of the lower scaffold to the larger scaffold above it): T L = 294 N. Next, we analyze the larger scaffold (of length L 1 = L 2 + 2 d and mass m 1 , given in the problem statement) placing our pivot at its left end and using Eq. 12-9: m 1 g L 1 2 T L d T R ( L 1 d ) + T L 1 = 0 . This yields T = 457 N.
22. As shown in the free-body diagram, the forces on the climber consist of T G from the rope, normal force N F G on her feet, upward static frictional force s f G and downward gravitational force mg G . Since the climber is in static equilibrium, the net force acting on her is zero. Applying Newton’s second law to the vertical and horizontal directions, we have net, net, 0 sin 0 cos . x N y s F F T F T f mg φ φ = = = = + ¦ ¦ In addition, the net torque about O (contact point between her feet and the wall) must also vanish: net 0 sin sin(180 ) O mgL TL τ θ θ φ = = °− ¦ From the torque equation, we obtain sin /sin(180 ). T mg θ θ φ = ° − Substituting the expression into the force equations, and noting that s s N

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