A 52.2 kg mountain climber is rappelling down a steep mountain face. Her foot ma

Question

A 52.2 kg mountain climber is rappelling down a steep mountain face. Her foot makes contact with the rock at point A. Her rope attaches to her harness at point B, which is located at x1 = 0.88 m and y1 = 0.34 m relative to point A. Her centre of mass is found at point C, which has an x coordinate x2 = 1.06 m relative to point A.

a) Calculate to 4 sig.figs. the tension in the rope if the rope makes an angle of ? = 50° with respect to the horizontal.

b) What is the force of friction between her foot and the rock at point A? Include the appropriate sign in your answer to indicate the direction according to the usual sign conventions. Make sure to use 4 sig.figs for the tension calculated above.

c) What is the minimum coefficient of static friction between her foot and the rock to prevent her from slipping?

Explanation / Answer

taking moments about point A

under equlibrium the net torque is zero

T cos theta y1 + T sin theta x1 = mg x2

T = mg x2/cos theta y1 + sin theta x1

=52.2 ( 9.8) (1.06)/ cos 50 ( 0.34 ) + sin 50 (0.88)

=607.45 N

the net vertical force is zero

Fy = 0

T sin theta + f = mg

f = mg - T sinthea

= 52.2 ( 9.8)-607.45 N sin 50

=46.22 N

normal force is

N = T cos theta = 607.45 cos 50 = 390.46 N

(c)

the minimum coefficient of static friction between her foot and the rock to prevent her from slipping is

m_u = f/ N = 46.22 N/ 390.46 N = 0.118

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