How to find tension using integration

[PhizKid]

Homework Statement

A block of mass M is attached to a rope of mass m and length l, and the rope is being pulled with a force F on a frictionless surface and there is no gravity. Find the tension in the rope.

Homework Equations

F = ma

The Attempt at a Solution

A friend has been trying to help me with this, but I don't understand his explanations. As he explained it, I need to take a small segment of the rope to find a differential equation to integrate. I don't really see the logic or reasoning behind this, but I guess if I want to integrate an equation, the equation has to be a differential and I need to somehow obtain that equation with the information I'm given. I can't use a point on the rope because a derivative is a slope and you can't find a slope with a point, so I need to take a very small segment of the rope.

So here is my force diagram of the forces acting on the rope:

So we have: T_M(x + \Delta x) - T_F(x + \Delta x) = \Delta m \cdot a

There are 2 different tensions so I'm not sure how to combine them into a single tension..

 

[Chestermiller]

If the mass m of the rope is distributed uniformly over its length l, how much mass is contained between locations x and x + Δx?

 

[PhizKid]

If the mass m of the rope is distributed uniformly over its length l, how much mass is contained between locations x and x + Δx?

<br /> \frac{\Delta m}{m} = \frac{\Delta x}{l} \\\\<br /> \Delta m \cdot l = m \cdot \Delta x \\\\<br /> \Delta m = \frac{m \cdot \Delta x}{l}<br />

 

[Chestermiller]

<br /> \frac{\Delta m}{m} = \frac{\Delta x}{l} \\\\<br /> \Delta m \cdot l = m \cdot \Delta x \\\\<br /> \Delta m = \frac{m \cdot \Delta x}{l}<br />

OK. So

m\frac{\Delta x}{l}a=T(x+\Delta x)-T(x)

Next step: Divide both sides by Δx, and let Δx approach zero.

 

[PhizKid]

OK. So

m\frac{\Delta x}{l}a=T(x+\Delta x)-T(x)

Next step: Divide both sides by Δx, and let Δx approach zero.

I don't understand where T(x+\Delta x)-T(x) comes from. What is T? What happened to the Tension due to the Force, and the Tension from the block on the rope? (The 2 Tensions in the force diagram I drew)

 

[Chestermiller]

I don't understand where T(x+\Delta x)-T(x) comes from. What is T? What happened to the Tension due to the Force, and the Tension from the block on the rope?

Excellent questions. Since the rope has mass distributed along its length, the tension in the rope is not a constant. It varies with position x along the rope.

If we consider the cross section at location x, then the portion of the rope to the right of this cross section is pulling on the portion of the rope to the left of this cross section with a force T directed to the right. The portion of the rope to the left of this cross section is pulling back on the portion to the right of this cross section with the same force T directed to the left. This is how tension in a rope works.

Now for the forces at the ends. The force F pulls on the rope at x = l, and so the tension in the rope at x = l is

T(l) = F

The force that the end of the rope at x = 0 exerts on the block is T(0). This, of course, will be less than T(L). You need to integrate the differential equation for the rope to get how T(0) is related to T(l).

 

[PhizKid]

Thank you Chestermiller. So it looks mathematical from here:

<br /> m\frac{\Delta x}{l}a=T(x+\Delta x)-T(x) \\\\<br /> \frac{ma}{l}=\frac{dT}{dx} \\\\<br /> \int_{?}^{?}\frac{dT}{dx} = \frac{ma}{l}<br />

Is this correct so far? I'm not sure what to make the limits for the integration. And do I need to substitute 'a' for 'F / (m + M)' ?

 

[Chestermiller]

Thank you Chestermiller. So it looks mathematical from here:

<br /> m\frac{\Delta x}{l}a=T(x+\Delta x)-T(x) \\\\<br /> \frac{ma}{l}=\frac{dT}{dx} \\\\<br /> \int_{?}^{?}\frac{dT}{dx} = \frac{ma}{l}<br />

Is this correct so far? I'm not sure what to make the limits for the integration. And do I need to substitute 'a' for 'F / (m + M)' ?

dT =\frac{ma}{l}dx
T-T(0) =\frac{ma}{l}x
T(l)-T(0) =ma
T(0) =T(l)-ma=F-ma
Ma=T(0)=F-ma
(M+m)a=F
T=F-ma(1-\frac{x}{l})=F-F\frac{m}{M+m}(1-\frac{x}{l})
T=F(\frac{M}{M+m}+\frac{m}{M+m}\frac{x}{l})

 

[PhizKid]

I'm not entirely sure what you did. It looks like you did:

<br /> \int_{0}^{l} dT = \int_{0}^{l} \frac{ma}{l} dx \\\\<br /> T = \frac{ma}{l}(l) - \frac{ma}{l}(0) \\\\<br /> T = ma<br />

But I'm not sure. I see what we are integrating with respect to x, but if x is 0 it looks like T(0) = 0 but according to your calculations, it doesn't seem to be true.

 

[Chestermiller]

You integrated the left hand side incorrectly. It should be T(l) - T(0). T(0) is not zero. It is equal to the force of the rope on the mass M. Anyway, I didn't integrate from 0 to l, I integrated from x = 0 to x = x (i.e., arbitrary x).

 

[PhizKid]

Oh, I see: T(x) - T(0) = \frac{ma}{l}x
Is the same thing you wrote.

Then what I don't get is why you chose to integrate from 0 to x? Why those 2 limits, and not, say, x to l or 0 to l?

 

[Chestermiller]

Oh, I see: T(x) - T(0) = \frac{ma}{l}x
Is the same thing you wrote.

Then what I don't get is why you chose to integrate from 0 to x? Why those 2 limits, and not, say, x to l or 0 to l?

Well, you wanted to know the tension as a function of x (as implied in the problem statement), so x has to be in there somewhere. The results wouldn't have been any different if I had chosen to integrate between x and l. Try it an see. The third equation of my earlier post is the result of substituting x = l into the previous relationship. It is also the result of directly integrating from 0 to l.

Chet