Tension on a Rope with Variable Mass Distribution

[Solidmozza]

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.

Thanks in advance.

 

[radou]

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.

 

[Solidmozza]

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.

Thanks for your help :D