How Does Friction Affect Rock Climbing Stability?

[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, I am rather stumpted, any help would be appreciated.

 

[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.

 

[kyle8921]

I would approach this problem by first analyzing the forces acting on the climber in the x and y directions, as you have done. This allows us to determine the minimum horizontal pull and push forces required to keep the climber stable, which is the answer to part (a). However, as you mentioned, we also need to consider the torque acting on the climber in order to fully understand the stability of the climber.

To determine the torque, we can use the formula τ = r x F, where r is the distance from the pivot point (the fissure) to the force and F is the magnitude of the force. In this case, the pivot point is the center of mass of the climber, which is a horizontal distance of 0.40 m from the fissure. Thus, the torque from the weight of the climber (mg) is given by τ = (0.40 m)(50 kg)(9.8 m/s^2) = 196 Nm. This torque must be balanced by the torque from the horizontal forces applied by the hands and feet.

To determine the vertical distance h between the hands and feet (part b), we can use the fact that the torque from the horizontal forces must be equal to the torque from the weight of the climber. This gives us the equation τ = r x F = (0.40 m)(F) = 196 Nm. Solving for F, we get F = 490 N. This is the minimum horizontal pull and push forces required to keep the climber stable.

In the case of wet rock, the coefficient of static friction between the hands and rock (µ1) and between the feet and rock (µ2) are reduced. This means that the maximum force of friction that can be exerted by the hands and feet is also reduced. Therefore, the force in part (a) will decrease. However, the torque from the weight of the climber will remain the same. This means that the vertical distance h between the hands and feet will also decrease in order to balance the torque. This is because a smaller force will need to be applied at a shorter distance in order to produce the same torque.

In summary, as the coefficient of static friction decreases, the minimum horizontal pull and push forces required to keep the climber stable will decrease, and the vertical distance between the hands and feet will also decrease. This is because a smaller force can