Uniform rope of length L and mass M, work to pull it up.

[timnswede]

Homework Statement

A uniform rope of length L and mass M is held on a frictionless table by a force
F. One fourth of its length (¼L) is hanging over the edge. Find the work required
to pull the hanging part back onto the table.

Homework Equations

W=Fd

The Attempt at a Solution

I have no idea how to solve it, but what I've written down is mass per unit length is lambda=M/L and the part on the table has length .75L. The part hanging off weighs (lambda).25L and the part on the table weighs (lambda).75L. I'm not really sure where to go from here.

 

[Dr.D]

Where is the center of mass at the beginning time, and where is it at the final time? That should get you thinking.

 

[timnswede]

I haven't learned how to calculate center of mass yet :/ This question is from a sample test, and my test is this week so I may not have been taught how to do this yet. Could you tell me what I would need to know besides center of mass to solve the problem, to see if I could do those parts?

 

[Dr.D]

You could then do a little bit of work-energy thinking, and have the answer.

 

[timnswede]

Like work-KE theorem? Then wouldn't I need to know the final velocity? We haven't talked about potential energy yet.

 

[Dr.D]

The way the problem is worded suggests (but does not say explicitly) that this is a quasi-static process, so velocity at all times is zero.

 

[Simon Bridge]

There is no need to worry about quasi-static processes here.

The energy used to pull the rope onto the table goes to two (main) places:
1. the work needed to get the rope on the table - change in gravitational potential energy.
2. the work done accelerating the rope - change in kinetic energy.
You are asked to figure out 1 - so the acceleration part does not matter - so you can pick any acceleration you like.
It is best to choose the acceleration that makes the calculations simple ... i.e. a=0 would be a good choice. This means the finial velocity is the same as the initial velocity... so pick a nice number.

Note: in the problem statement it talks about a "force" ... how big is that force at the start?

Since you have not done potential energy yet - so I wonder how the examiner expects you to solve the problem.
Perhaps you have to use force times distance or the question is too advanced for what you currently know?
Have you done some examples using calculus maybe?

 

[timnswede]

There is no need to worry about quasi-static processes here.

The energy used to pull the rope onto the table goes to two (main) places:
1. the work needed to get the rope on the table - change in gravitational potential energy.
2. the work done accelerating the rope - change in kinetic energy.
You are asked to figure out 1 - so the acceleration part does not matter - so you can pick any acceleration you like.
It is best to choose the acceleration that makes the calculations simple ... i.e. a=0 would be a good choice. This means the finial velocity is the same as the initial velocity... so pick a nice number.

Note: in the problem statement it talks about a "force" ... how big is that force at the start?

Since you have not done potential energy yet - so I wonder how the examiner expects you to solve the problem.
Perhaps you have to use force times distance or the question is too advanced for what you currently know?
Have you done some examples using calculus maybe?

Well the force would be equal to λ(¼L)g I believe, since it's just holding the rope in place. I have done problems involving ropes in my calculus class, but that just involved pulling it straight up, would this be similar to that? We still have a couple lectures before the exam, so I believe the professor will be going over potential energy before the test, I just wanted a head start.

 

[Simon Bridge]

Well the force would be equal to λ(¼L)g I believe, since it's just holding the rope in place.

Only if $\lambda = M/L$ - it is best practice to write equations as much as you can in terms of the things you are told. So $F_0=\frac{1}{4}Mg$ ... in fact, if the length hanging over the end is $xL: x<1$, then $F(x)=xMg$ will maintain a constant velocity.

I have done problems involving ropes in my calculus class, but that just involved pulling it straight up, would this be similar to that?

Yes. The important lesson is about how you set up problems. But you should also have done some problems like this in physics... OTOH:

We still have a couple lectures before the exam, so I believe the professor will be going over potential energy before the test, I just wanted a head start.

... it may be that the lesson's you need to be able to do this sort of thing have yet to appear.

 

[timnswede]

Only if $\lambda = M/L$ - it is best practice to write equations as much as you can in terms of the things you are told. So $F_0=\frac{1}{4}Mg$ ... in fact, if the length hanging over the end is $xL: x<1$, then $F(x)=xMg$ will maintain a constant velocity.Yes. The important lesson is about how you set up problems. But you should also have done some problems like this in physics... OTOH:... it may be that the lesson's you need to be able to do this sort of thing have yet to appear.

Maybe I'm thinking of this wrong, but to pull the rope up wouldn't the force have to be greater than the force for just holding it up? Otherwise the integral that I'm thinking of right now would be ¼Mg∫(0,¼L)dx. the (0,¼L) are the limits.
We didn't really do any complicated work problems in calculus, just problems where it gave us the length and weight of the rope and how far to pull it up.

 

[Simon Bridge]

Maybe I'm thinking of this wrong, but to pull the rope up wouldn't the force have to be greater than the force for just holding it up?

If you pull the rope at a constant speed, the net force must be zero.

Otherwise the integral that I'm thinking of right now would be ¼Mg∫(0,¼L)dx. the (0,¼L) are the limits.

$$W=\int F(x)\;dx$$ ... all right, but you have to remember what x means here when working out the limits, and where did that 1/4 come from in your version?

Another way to set up the integral is to figure the work to lift the mass element dm = (M/L)dy that lies between y and y+dy from the edge of the table.

The main thing is to think about the physics involved.

 

[timnswede]

If you pull the rope at a constant speed, the net force must be zero.

$$W=\int F(x)\;dx$$ ... all right, but you have to remember what x means here when working out the limits, and where did that 1/4 come from in your version?

Another way to set up the integral is to figure the work to lift the mass element dm = (M/L)dy that lies between y and y+dy from the edge of the table.

The main thing is to think about the physics involved.

Well for my integral I did the force which is ¼Mg times the distance dx. The ¼L is the part of the rope that is being lifted. Is that wrong? The part about dm=(M/L)dy is not something I have done before, so I can't say I really understand that. But basically the way I'm thinking of it is I'm just finding the work to lift the part that is over the edge of the table, since that is the same force as on the edge of the rope that is on the table.

 

[Simon Bridge]

But the force is not $\frac{1}{4}Mg$ all the time - that's just at the beginning.

 

[timnswede]

Oh that's true, forgot about that. If I brought back lambda, would it be ∫λ(¼L)gdL? With the same limits.
I really need to get some sleep now, maybe that'll help me think about this problem too.

 

[OldEngr63]

Simon, not to quibble too much, but if there is no acceleration work, then the system is in equilibrium all the time and that is a static process. Thus, it seems to me that the introduction of the idea of a quasi-static process is quite appropriate. It simply means that the system is always in equilibrium.

 

[timnswede]

OK, I finally got it. I just called the part hanging off the table "y" and λ is M/L so, ∫(0,¼L)λygdy, so I end up with MgL/32 after substituting M/L for Lambda. Checked with my professor and he said that was correct.

 

[Simon Bridge]

Simon, not to quibble too much, but if there is no acceleration work, then the system is in equilibrium all the time and that is a static process. Thus, it seems to me that the introduction of the idea of a quasi-static process is quite appropriate. It simply means that the system is always in equilibrium.

I agree ... it is reasonable to introduce the subject, but it is not necessary to do so. Please don't take my comment as a criticism.
It is best practice in education to build on what the student already knows as the path to understanding, rather than introduce new concepts ... however, it is also fair in these forums to give students an idea of more advanced concepts as they come up. Finding the balance can be tricky, so it is just as well there are so many people with different approaches prepared to answer the questions. You bringing up quasi-static means I don't have to ;)

 

[Simon Bridge]

OK, I finally got it. I just called the part hanging off the table "y" and λ is M/L so, ∫(0,¼L)λygdy, so I end up with MgL/32 after substituting M/L for Lambda. Checked with my professor and he said that was correct.

Well done. It is very useful to make connections between different disciplines. Do you see what I mean about concentrating on physics instead of equations?

btw... the formal way to set this up:
Assign "down" to the +y axis, then the rope hangs in range $0\leq y\leq L/4$.
The part of the rope between $y$ and $y+dy$ has mass $dm = (M/L)dy$.
The work to lift that mass to the table level is $dW = g y dm = (gM/L)y\;dy$[*]
... now it's just a matter of adding up all the $dW$'s; vis: integrate both sides.
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[*] Work is change of energy: change in gravitational potential energy of a mass $m$, close to the surface of the Earth, is $\Delta U_g = mg\Delta y$ ... where $y$ is the vertical component of position.