# Tension on a rope with mass

• Solidmozza
In summary, at a distance x from the top of the rope, the tension in the rope is equal to the mass of the block multiplied by the inverse of the distance from the top of the rope.

## Homework Statement

A block with mass 'M' is attached to the lower end of a vertical, uniform rope with mass 'm' and length 'L'. A constant upward force 'F' is applied to the top of the rope, causing the rope and block to accelerate upward. Find the tension in the rope at a dstance 'x' from the top end of the rope, where 'x' can have any value from 0 to 'L'.

## Homework Equations

Newton's Second and Third Laws.

## The Attempt at a Solution

I'm a bit confused on this question. I've tried breaking the problem up into three parts - one for the block mass 'M', one for the top of the rope and one for a point 'x' on the rope - but I can't seem to get it to work. The actual constant force there is annoying too - for the top of the rope I have a force acting downwards of (m+M)g, and an upwards force that is greater than that of 'F', but I don't know how I can equate etc. The answer is F[M+m(1-x/L)]/(M+m) but I want to know why.

Try to apply Newton's 2nd law to the whole system in order to find the acceleration. Then try to look at an element of the rope at a distance x from the top end of the rope, and try to express the mass of that element somehow. Then apply Newton's 2nd law again. I don't have the time to write it down and check if it's right, though.

Ah sweet! That seems so obvious now - Silly me!
~~~~~
First, we consider the point on the highest point of the rope (ie where x=0). There are only 2 forces acting here: the combined weight force of the rope and the block, and the upward force. Since the whole thing is accelerating, we use Newtons second equation of motion... F=ma, now m = M+m (mass of rope+block), so we get a = F/(M+m). Now we take the point 'x' on the rope. Am I right in saying that there are only two forces here - the tension force due to action-reaction pairs which acts upwards, and the weight force? (It seems to work mathematically...). So we know that the mass of the section of the rope + block =.. M+M[(L-x)/L]. Thus again we use F=ma (although its tension force this time!), giving F= M+M[(L-x)/L] x F/(M+m), which is= F[M+m(1-x/L)]/(M+m) as required.

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