Statics problem: man lowering himself from tree w/ friction

[bkw2694]

Homework Statement

The 180-lb tree surgeon lowers himself with the rope over a horizontal limb of the tree. If the coefficient of friction between the rope and the limb is 0.60, compute the force (P) which the man must exert on the rope to let himself down slowly.

Homework Equations

T1/T2 = e^(βμ)

Where β = the angle of the rope on the log, μ = 0.60, T1 = the bigger tension (180 lbs), and T2 = the smaller tension (P)

The Attempt at a Solution

So I looked at the picture and I believe β = 180° which = π rads.

Then I plugged those into my equation and got:

180/P = e^(.6π), this gave me P = 27.33 lbs, which is incorrect.

The correct answer should be 23.7 lbs, but I have no idea what I'm doing wrong. I'm guessing my β is the wrong number, but I'm not sure how to find the correct number if that's the case.

 

[haruspex]

Your mistake is in assuming the greater tension equals the man's weight. Draw a FBD for the man.

 

[bkw2694]

Your mistake is in assuming the greater tension equals the man's weight. Draw a FBD for the man.

Sorry, I'm a little confused so I don't know if my FBD is right. I made a diagram of just the guy and the two sides of the rope, with a downward force of 180 lbs due to his weight, a downward force of P on the piece of rope he's holding (since he's pulling down) and an upward force of 180 + P since that satisfies equilibrium. So then when I make a diagram of the branch with the two pieces of rope, I have (180 + P) going downward and a force of P going upward. Am I on the right track here?

 

[haruspex]

I made a diagram of just the guy and the two sides of the rope, with a downward force of 180 lbs due to his weight, a downward force of P on the piece of rope he's holding

The rope he's holding is pushing down on him with force P??

 

[bkw2694]

The rope he's holding is pushing down on him with force P??

Ah, so I was thinking of it as if he were pulling down, but it should be that the tension of P is upward, which would make the tension on the other side: T1 = (180 - P). Thanks so much!

And for anyone else who might find this problem, I'll finish the solution: (180-P)/P = e^(.6π) ⇒ P = 23.7 lbs