# Homework Help: Pulley question

1. Jul 2, 2007

### Mr Virtual

1. The problem statement, all variables and given/known data

In my book, this question is accompanied with a diagram. But I will try to explain the diagram as best as I can.
There are two pulleys A and B connected with a single thread. One end of thread, P, is along the pulley A, and the other end, Q, along B. Now let us take a mass M, tie it to a thread, and tie this thread at O, which is the centre of the distance AB. Naturally the thread AB will sag downwards.
The angle AOB, created by the sag is twice-theta. Let us now start pulling both the free ends P and Q downwards with velocity u. What is the speed with which the mass M will rise up?
The tension of the strings, friction etc are not to be taken into account here.

2. Relevant equations

Sorry, I don't have any.

3. The attempt at a solution

The angle AOB is twice-theta. Let us draw OX, such that each of AOX and BOX is theta.
Now, downward speed at pulley A = upward speed at pully A = u
Therefore, speed with which OA is rising = u
Similarly, speed with which OB is rising = u

The speed of M = cos component of AO + cos component of BO
= u cos theta + u cos theta
= 2u cos theta

Am I right?

regards
Mr V

2. Jul 3, 2007

### Mr Virtual

Err... it would be so nice if somebody gives any hint.

Mr V

3. Jul 3, 2007

### HallsofIvy

I think you need to clarify more. The two pulleys are attached by a thread. Is the thread running horizontally or vertically? What is supporting the two pulleys?

My best guess is that the two pulleys are supported by wires, say, attached to their centers so that the line between them is horizontal. Then a thread is run through the two pulleys, hanging down on either end. Finally, a mass is attached to the thread equidistant from the two pulleys. The two ends are pulled downward with velocity u. You do not say what the distance between the two pulleys is and that is important. I'm going to assume a distance L, so that the initial length of thread between the two pulleys is $L/ sin(\theta)$ and the distance of the mass below the two line between the two pulleys is $L cot(\theta)$. Note that $\theta$ is not constant but will increase as the two ends are pulled downward.

If each end is pulled downward with velocity u m/s, then the length of the thread between the pulleys decreases by 2u m/s: $Lcos(\theta)- 2ut$. If y is the distance the mass is below the horizontal line then, by the Pythagorean theorem, $y^2= (L cos(\theta)- 2ut)^2- L^2/4$. Differentiating both sides of the [itex] 2y dy/dt= 2(L(cos(\theta)- 2ut)(-Lsin(
theta)- 2u).

4. Jul 3, 2007

### Mr Virtual

The pulleys are running horizontally. You just imagine two pulleys A and B situated horizontally in air (but fixed rigidly in place), without any support. A single thread PQ runs horizontally on these pulleys. One end of thread, P, is along A (that is if we pull P downwards, pulley A and B will start rolling anti-clockwise with the velocity with which P is pulled) and the other end of the thread, Q, is along B (same here; if we pull Q, A and B will roll clockwise with velocity with which Q is pulled, and P (along A)will rise up with the same velocity).

Now let us take a mass M, tie the end X of a thread OX to it, and tie the end O to the centre of the thread running between A and B. That is, the thread PQ has its centre at a point at the centre of length AB. This is where we tie OX.
Thread PQ sags down at the point O (where OX is tied to PQ), making an angle twice-theta, like a clock in 10:10 position.
If we now pull both the ends P and Q downwards simultaneously with the same velocity, then what is the velocity with which M rises up.

I hope I am clear now?

Mr V

Last edited: Jul 3, 2007
5. Jul 3, 2007

### PhanthomJay

Is it as clear as (theta + theta + 2(theta)) = sum of angles in a triangle = 180 degrees, therfore , 4(theta) =180, theta (initial) = 45 degrees?

6. Jul 3, 2007

### Mr Virtual

Er.. yes. If you consider the triangle AOB, then angle(AOB)=2(theta), and angle(OAB) = angle (OBA) = theta. So, theta=45 degrees. Correct. But please keep in mind that, in reality, side AB of triangle AOB does not exist, because it has sagged down to form angle AOB. I hope it make sense.

Now, when we pull PA and QB separately (but simultaneously) with velocity u, both AO and BO will rise with vel u. The horizontal component of AO=u cos theta, and of BO=u cos theta. The net velocity acting on M = horizontal comp of AO + horizontal comp of BO= ucos theta + u cos theta = 2 u cos theta. Right?

Mr V

Last edited: Jul 3, 2007
7. Jul 3, 2007

### PhanthomJay

I'm not sure why you're looking in the horizontal direction, since the mass does not move horizontally, only vertically (the horizontal components don't add, they cancel, since they are in opposite directions). From the physics of the problem, as the mass rises, theta gets smaller, and the tension in the thread gets higher, requiring a greater and greater downward force at P and Q to move the block up (approaching infinity as the thread approaches the horizontal, assuming the thread is infinitely strong and does not break). But this seems more of a calculus problem than a physics one, per HallsofIvy's attempt. I pass on this one. I just wanted to establish the initial condition of theta=45 degrees.

8. Jul 4, 2007

### Mr Virtual

I am talking about the horizontal "component" of u, not the horizontal direction. The horizontal component of u is u cos theta, and the vertical component is u sin theta. It is another thing that u cos theta (horizontal component) is working in vertical "direction".
u sin theta cancels u sin theta because they are working in opposite directions "horizontally".

This is a fairly simple problem. No calculus or hard maths is needed here. I can give you the options provided in my book. One of them is correct.

1. 2u cos theta
2. u/ cos theta
3. 2u/ cos theta
4. u cos theta

I think the first option is correct, because the horizontal components of each u (direction: vertical) add up to give a net upward velocity = 2 * u cos theta.

Mr V

Last edited: Jul 4, 2007
9. Jul 5, 2007

### Mr Virtual

Can anyone spend a little time here too, pleeeease?

Thanks
Mr V

10. Jul 6, 2007

### Staff: Mentor

OK. Point X is the midpoint between the top of the pulleys.
The rate at which both string ends are being pulled down is = u. Therefore, the rate at which OA (or OB) shrinks = u. Use the properties of a right triangle to deduce at what rate OX must shrink (XA is constant, of course).

No. Why are you adding speeds? You can analyze this using side A or side B, getting the same answer of course; these are not independent speeds--adding them makes no sense.

11. Jul 6, 2007

### Mr Virtual

Yeah, adding the speeds was a mistake. So, is the answer u cos theta ?

Mr V

12. Jul 6, 2007

### Staff: Mentor

Yes.........

13. Jul 6, 2007

### Mr Virtual

The rate at which OX is shrinking will definitely be smaller than the rate at which OA or OB is shrinking. At least that is what I think.

14. Jul 6, 2007

### Mr Virtual

Thanks a lot.

15. Jul 6, 2007

### PhanthomJay

And I'm wondering how the problem would be different if the angle AOB, indicated initially as "twice" theta, was, instead, "thrice" theta?

16. Jul 6, 2007

### Staff: Mentor

I think that that is just the definition of theta, not an initial condition: Theta is defined as half of the angle AOB.

17. Jul 6, 2007

### Mr Virtual

EDITED

Mr V

Last edited: Jul 6, 2007
18. Jul 6, 2007

### Mr Virtual

Yeah you are right. We are always going to consider half of the total angle. I will edit my previous post.

Last edited: Jul 6, 2007
19. Jul 6, 2007

### PhanthomJay

Oh, sure, now i get it; I had been assuming that theta was the angle BAO or ABO, instead of half AOB. Thanks.