Physics Math

Question

Consider the 62.0 kg mountain climber in the figure. https://imgur.com/a/TJIqs A.) Find the tension in the rope (in N) and the force that the mountain climber must exert with her feet (in N) on the vertical rock face to remain stationary. Assume that the force is exerted parallel to her legs. Also, assume negligible force exerted by her arms. B.)What is the minimum coefficient of friction between her shoes and the cliff?

Answer 1

There are 3 unknowns: T, F (exerted by rock on the climber), and friction f

So we need 3 equations.

 

Equilibrium of forces in vertical direction:

 

T cos31° + F sin15° + f = mg = (62.0)(9.8) N = 607.6 N

 

(E1)   

 

0.8572T + 0.2588F + f = 607.6

 

Equilibrium of forces in horizontal direction:

 

T sin31° - F cos15° = 0

 

(E2)   

 

0.5150T - 0.9659F = 0

 

Moment equilibrium about contact point between feet and rock, around axis vertical to plane of the diagram:

 

Assuming an unknown horizontal moment arm of length r (which will drop out of the equation anyway),

 

rT cos31° = rmg

 

(E3)   

 

0.8572T = 607.6

 

Solve equations (E1), (E2), and (E3) for T, F, and f.  I get

 

T = 708.8 N

F = 377.9 N

f = -97.8 N (the minus sign means f is pointing down)

 

|f| = μF cos15° ⇒

 

μ = |f| / (F cos15°) = 97.8 / (377.9 cos15°) = 0.2679

 

That is what I get.  Check each step very carefully!