# Static rock climber problem

1. Jul 15, 2006

### mikejones2000

In Figure 12-38, a 50 kg rock climber is in a lie-back climb along a fissure, with hands pulling on one side of the fissure and feet pressed against the opposite side. The fissure has width w = 0.20 m, and the center of mass of the climber is a horizontal distance d = 0.40 m from the fissure. The coefficient of static friction between hands and rock is µ1 = 0.49, and between boots and rock it is µ2 = 1.6.

The figure shoes a rock climber in this position and I presume only his hands and feet are touching the rock. it wants to know:
a) What is the least horizontal pull by the hands and push by the feet that will keep the climber stable?
N
(b) For the horizontal pull of (a), what must be the vertical distance h between hands and feet?
m
(c,d) If the climber encounters wet rock, so that µ1 and µ2 are reduced, what happens to the answers to (a) and (b)? (Select all that apply.)
The force in part (a) increases.
The force in part (a) decreases.
The force in part (a) doesn't change.
h increases.
h decreases.
h does not change.

I think I approached the first question right: I wrote out all the forces acting on the climber in the x direction and y direction and found that
µ1F1+µ2F2=mg, I figured that F1=F2 and solved for F but think I should be approaching this from a torque perspective but am rather clueless because there is the force of weight which is perpindicular to the pulling and pushing forces. As for as the last two questions are concerned, im rather stumpted, any help would be appreciated.

2. Jul 16, 2006

### andrevdh

You have no justification for saying that F1 = F2. That is why you need to set up more equations (at least one more) to solve for the two unknown forces. You need to set up torque equations about some chosen point for rotational equilibrium. Simplify the problem to a stick experiencing two upwards and two inwards forces at its ends and its weight being in both rotational and translational equilibrium.