2(70 kg)(9.8 m/s )(0.20 m)17 N.44(2.0 m)hmgaFH==≈(b) From the y-component of the force equation, we obtain 22(70 kg)(9.8 m/s )1.7 10 N.44vmgF≈×34. As shown in the free-body diagram, the forces on the climber consist of the normal force from the wall, the vertical component vFand the horizontal component hFof the force acting on her four fingertips, and the downward gravitational force mg. Since the climber is in static equilibrium, the net force acting on her is zero.
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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.