Ch12-p024 - 24 As shown in the free-body diagram the forces on the climber consist of the normal forces FN 1 on his hands from the ground and FN 2

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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.

Ask a homework question - tutors are online