22. As shown in the free-body diagram, the forces on the climber consist of TGfrom the rope, normal forceNFGon her feet, upward static frictional force sfGand downward gravitational force mgG. 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,0sin0cos.xNysFFTFTfmgφ==−+−¦¦In addition, the net torque about O (contact point between her feet and the wall) must also vanish: net0sinsin(180)OmgLTLτθθ−°−−¦From the torque equation, we obtain sin /sin(180).Tmgθθ=°−−Substituting the expression into the force equations, and noting that
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.