Coefficient of friction problem

Nov 2016
9
0
Hi everyone. I'm having trouble with a question and I was wondering if I can have some help? The question is:

"The diagram shows a man of mass 70 kg at rest while abseiling down a vertical cliff. Assume that the rope is attached to the man at his centre of mass. You should model the man as a uniform rod and assume that he is not holding the rope. Find the minimum value of the coefficient of friction between his feet and the cliff if he remains at rest."

The diagram essentially has a vertical line representing the cliff. Then at the bottom of the cliff, there is a diagonally up-right line representing the man's body, and this is at 60 degrees to the vertical. Then mid-way up his body, there is another line perpendicular to his body that represents the rope, so this goes back towards the cliff.



My solution attempt:

The total forces: friction (F) going up from his feet, reaction (R) going right from his feet, weight (70g) at mid-point of him going down, tension (T) going back towards the cliff along the rope.

Moments around his feet: 70g * x sin 60 = T * x (I assume that the weight and rope are at the same point since it says the rope is tied to his centre of mass, so let's just call the distance x) therefore T = 70g sin 60

Resolving horizontally: R = Tcos60 = (70g sin 60) (cos60)

Resolving vertically, F + Tsin60 = 70g

F + (70g sin 60 )sin60 = 70g

F = 70g - (70g sin 60) sin60

F = 70g(1 - sin60) (sin 60)

F = MR (where M is the coefficient of friction)

M((70g sin 60) (cos60)) = 70g(1 - sin60)(sin 60)

M = 0.268

But the answer is 0.577
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
I think your error is going from this:

F = 70g - (70g sin 60) sin60

to this:

F = 70g(1 - sin60) (sin 60)

This second equation should be F = 70g(1-sin60 sin60).

See if that fixes it for you.
 
Last edited:
Nov 2016
9
0
By changing the equations to

R = T sin60

F + T cos60 = 70g

I have a value of 1.32, which can't be right because it can't be greater than 1...
 

ChipB

PHF Helper
Jun 2010
2,367
292
Morristown, NJ USA
Sorry, I think I was probably editing my previous response when you responded. The error lay elsewhere than getting sines and cosines reversed - please go back and try the correction I suggested.
 
Last edited:
Nov 2016
9
0
All works now - I must have responded during your edit because it was a different equation there.

Thanks for the help though. :)