A physics professor of mass 58kg and height 1.68m is rappelling down a vertical
ID: 1349248 • Letter: A
Question
A physics professor of mass 58kg and height 1.68m is rappelling down a vertical cliff when she pauses for a moment. Her feet are touching the cliff, and she is leaning back so that her body makes an angle =40° with the vertical. She is tied into a harness that is connected to a rope that makes an angle of = 20° with the cliff face. The tension in the rope has a line of action that goes through her center of mass which is 1.0 m from her feet. Her hands are not exerting a force on the rope (it is fed through a figure eight shaped braking device that exerts the necessary tension)
a) Find the minimum coefficient of static friction between her feet and the cliff so that her feet do not slip on the cliff wall. (Hint: begin with the sum of torques about the center of mass).
b) Find the tension in the rope if her feet are about to slip and the coefficient of static friction has the value you obtained in part a).
Explanation / Answer
(a)
net torque about the feet = 0
T*cos20 = m*g*1*cos40
T*cos20 = 58*9.8*1*cos40
T = 463.4 N
normal force N = T*sin20
frictional force f = u*N
Fnet = 0
T*cos20 + f = m*g
463.4*cos20 + u*463.4*sin20 = 58*9.8
u = 0.84
-------
(b)
T*cos20 - m*g - u*T*sin20 = 0
T*cos20 - 58*9.8 - 0.84*T*sin20 = 0
T = 871.2 N <<---answer
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