Uniform Rope Mechanics Problem: Finding Tension at a Distance x | Explained

[torquerotates]

Homework Statement

A uniform rope of mass m and length L is attached to a block of mass M. The rope is pulled with force F. Find the tension at a distance x from the end of the rope.

Homework Equations

T(total)=Ma, F-T(total)=ma & T(total)/L=T(x)/x

The Attempt at a Solution

T(total)=Ma, F-T(total)=ma => T(total)=(MF)/(1+mM)

This is the part I'm confused about. I assumed the following proportion. T(total)/L=T(x)/x
This implies, T(x)=(T(total)/L)x= (M/(L(mM+1)))x

Am I right in my reasoning?

 

[G01]

Hmm. I think my answer is different from yours, and I am not sure about your reasoning. I don't think you can assume that proportion.


Start the problem by considering the force at a point x. This force, T(x), is the total force, F, minus the force needed to accelerate the length of rope in front of it. What is this force needed to accelerate the segment of rope before point x?

HINT: You are going to need to use the mass per unit length of the rope, m/L.

Also, you will end up with a factor of "a', the acceleration? How can you express this in terms of F, m, and M?

HINT: Consider the motion of the box.

 

[tiny-tim]

Relevant equations[/b] T(total)=Ma, F-T(total)=ma & T(total)/L=T(x)/x

Hi torquerotates!

There's no such thing as total tension.

You can't add tension as if you were adding mass or forces.

The tension varies along the rope. It has an average, but not a total.

Hint: divide the problem into two sections:
i) the block plus part of the rope
ii) the rest of the rope.

Apply good ol' Newton's second law to i) and ii) combined, and to i) and ii) separately.