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On the other hand, the frictional force can also be written as 1 s sN f F μ = , where s is the coefficient of static friction between his feet and the ground. From the above equation and the values given in the problem statement, we find s to be 22 2 2 0.914 m 0.940 m cot 0.216 2.10 m (2.10 m) (0.914 m) s da d LL La μθ == = = −− . 24. As shown in the free-body diagram, the forces on the climber consist of the normal forces 1 N F on his hands from the ground and 2 N F on his feet from the wall, static frictional force s f and downward gravitational force mg . Since the climber is in static equilibrium, the net force acting on him is zero. Applying Newton’s
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