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 TF mMg == + ¦ In addition, the net torque about O must also vanish: net ) ( ) ( 0 ) ( ) ( ) ( ) O dT F Dm g LM g 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 22 ( ) 648 N (7.2 kg 1.8 kg)(9.8 m/s ) 560 N 5.6 10 N. FT Mm 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 ) + TL 1 = 0 . This yields T = 457 N.
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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, 0s i n 0c o s . xN ys FF T FT f m g 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 τθ θ ° ¦ From the torque equation, we obtain sin /sin(180 ).
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This note was uploaded on 11/16/2009 for the course PHYSICS 11a taught by Professor Silvera during the Fall '07 term at Harvard.

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Problem_Set_8_Solutions - 18 Our system consists of the...

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