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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