Rock Climbing Energy Conservation (further help)

[JJB11]

Homework Statement

I know that this question has been dealt with on this forum before, however it is part (b) that is new and that I need help with. I have re-written part (a) for clarity.

Question: The greatest instantaneous acceleration a person can survive is 25g, where g is the acceleration of free fall. A climber's rope should be selected such that, if the climber falls when the rope is attached to a fixed point on a vertical rock, the fall will be survived.

A climber of mass m is attached to a rope which is attached firmly to a rock face at B. When at a point A, a distance L above B, the climber falls.

a) Assuming the rope obeys Hooke's Law up to breaking, use the principle of conservation of energy and the condition for greatest instantaneous acceleration to show the the part AB of the rope (of unstretched length L) must be able to stretch by more than without breaking for the climber to survive.

b) A particular rope has a breaking strength of 25 times the weight of the climber, and obeys Hooke's law until breaking, when it has stretched by 20% of its length. Determine whether this rope is suitable.

Homework Equations

The Attempt at a Solution

(a) I've got the answer to this question.

I said that for the climber to survive:

1/2Fx > (2L+x)mg

as I figured the max distance the climber could fall before the rope became taut would be 2L (ie L below B) and then it would extend, thus losing more GPE. As F=ma+mg=26mg, substitute and get x must be greater than or equal to L/6.

(b) I know that the given breaking strength is insufficient to decelerate the climber beyond 25g so this is okay. However I can't work out whether the string will break or not as I'm not sure which energy balance equation I can use to find the rope extension.

 

[KataKoniK]

Hello there,

Thank you for your question. In order to determine whether the given rope is suitable, we will need to use the equation for Hooke's Law, which is F = kx, where F is the force applied to the rope, k is the spring constant, and x is the displacement of the rope from its original length. We can also use the principle of conservation of energy, which states that the total energy of a system (in this case, the climber and the rope) remains constant.

Let's start by finding the maximum force that the rope can withstand before breaking. We know that the breaking strength of the rope is 25 times the weight of the climber, so we can write this as F_break = 25mg. Now, let's use the equation for Hooke's Law to determine the maximum displacement of the rope before it reaches its breaking strength. We can write this as:

F_break = kx_max

Substituting in the value for F_break, we get:

25mg = kx_max

Now, let's consider the energy balance equation. We know that the total energy of the system must remain constant, so we can write:

Initial energy = Final energy

The initial energy of the system is the gravitational potential energy of the climber at point A, which is given by mgh, where m is the mass of the climber, g is the acceleration due to gravity, and h is the height of point A above point B. The final energy of the system is the energy stored in the rope, which is given by the equation for elastic potential energy, 1/2kx^2, where k is the spring constant and x is the displacement of the rope.

We can write this equation as:

mgh = 1/2kx^2

Now, let's substitute in the value for x_max that we found earlier:

mgh = 1/2k(25mg)^2

Simplifying, we get:

h = 1/2k(625g^2)

Now, we know that the rope obeys Hooke's Law until breaking, when it has stretched by 20% of its length. This means that the final length of the rope is 1.2 times its original length. We can write this as:

Final length = 1.2L

Substituting this into our equation, we get:

h = 1/