A desperate hiker has to think fast to help his friend who has fallen below him.

Question

A desperate hiker has to think fast to help his friend who has fallen below him. Quickly, he ties a rope to a rock of mr = 396 kg and makes his way over the ledge (see the figure below). If the coefficient of static friction between the rock and the ground is ?s = 0.348, and the mass of the hiker is mh = 70.1 kg, what is the maximum mass of the friend, mf, that the rock can hold so the hikers can then make there way up over the ledge? Assume the rope is parallel to the ground and the point where the rope passes over the ledge is frictionless.

Explanation / Answer

I cannot see the diagram, so am not sure if I am doing this right, but here is by best answer with the given information.
We know that the two forces acting on the hikers are gravity and the tension in the rope, so the tension in the rope must be equal to the force of gravity of both the friends in order to keep them from falling. So:
T = mg
T = (70.1 + mf)(9.81)

T = 688 + 9.81mf

Now that we have T we look at the rock. Four forces act on the rock, the force of gravity on itself, the normal foce of the ground on the rock, the tension in the rope, and the friction that works to oppose the pull. Now, since the rock is parallel to the ground the normal force and gravity should cancel each other out, meaning you wont have to worry about it except to find the normal force.

N = mrg

N = 396(9.81)

N = 3885 N

Since the static force of friction and tension act in opposite direction and we want the boulder not to move, they must be equal.

T = Ffr          Ffr = sN

688 + 9.81mf = 0.348(3885)

mf = 67.68 kg

So the rock will only stay steady if the friend weighs 67.68 kg or less.

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