# Homework Help: Confusing mechanics(Extracted from my university physics paper)

1. Apr 23, 2013

### aerograce

1. Initially the system of objects shown in the following figures is held motionless. All surfaces, pulley, and wheels are frictionless. At the instant after the system of objects is released, m2 moves vertically downward.

Caculate the acceleration of m1, m2 and M, and the tension T in the string.

I attached a photo of the figure, including my professor's solution as well.
So you can actually see how my professor approached this problem:

Assume:
acceleration of m2: a
acceleration of M: A
Then the acceleration of m1: a-A

So we have:

m2g-T=m2a
T=MA
T=m1(a-A)

What confused me is the expression of accelerations:

As for why acceleratoin of m1 is equal to a-A, I can fully understand. Because:
a(m1->ground)=a(m1->A)+a(A->ground)

But now, lets look at m2
a(m2->ground)=a(m2->A) + a(A->ground)

As we all know that, a(m2->A) is pointing downwards. However, there should be an acceleration horizontally put on m2 for the second item, a(A->ground). Thats according to the equation, right?

But now, I find that, there is no horizontal force exerted on m2.

That is what is confusing me. And I dont know whether my professor's solution is correct or not. I think he should be correct but I really need further explaination on the solution. Can anyone help me out? Thank you!

2. Apr 23, 2013

### haruspex

The solution is correct. As you say, there is no horizontal force on m2. M will accelerate away from it, leaving it to accelerate vertically downwards. As time goes by, the string m2 hangs from will no longer be vertical, and m2 will acquire horizontal acceleration, but that's beyond the time to be considered.

3. Apr 24, 2013

### aerograce

Thank you! But how to consider this equation: a(m2->ground)=a(m2->M)+a(M->ground).
And how come the acceleration on one string can not be the same? When I approached this problem, I capitalized on the same acceleration on the same string and I got the wrong answer.

Thank you so much!

4. Apr 24, 2013

### haruspex

The acceleration of m2 relative to M does have a horizontal component, equal and opposite to the horizontal acceleration of M relative to the ground. Relative to the ground, M accelerates horizontally only, while m2 accelerates vertically only.
The acceleration of the string has the same component along the string in all parts. It's not clear to me how you used this. Can you post the equations you based on it?

5. Apr 24, 2013

### aerograce

Er, can you look at the figure I attached? It shows my prof's method. When he was solving this problem, he thinks the acceleration of m1 is a-A,and the acceleration of m2 is a. This already shows that the acceleration of the string doesn't have the same component along the string.

6. Apr 24, 2013

### haruspex

Ok, I see. I made the same mistake you did. Consider the length of the string in the horizontal section. Because of m1's acceleration, it is shortening with an acceleration of a-A. but because the pulley is also accelerating, it is shortening with an additional acceleration of A. So in total the length is shortening with an acceleration of a. The vertical section must be lengthening with the same magnitude of acceleration.

7. Apr 24, 2013

### aerograce

This is the photo again.

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• ###### IMAG1854.jpg
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8. Apr 24, 2013

### aerograce

Thank you so much. So can I interpret your words as this:

The acceleration on the string is same for all parts, it is equal to a-A.
Due to the movement of the pulley, there is an extra acceleration for m2, which is equal to A.
So the acceleration for m2 is equal to a.

Er, but how to explain this, when you have this equation
a(m2->ground)=a(m2->M)+a(M->ground), it still shows that there should be a horizontal acceleration on m2. But here, only vertical acceleration is shown,

9. Apr 24, 2013

### haruspex

No, I conceded that was wrong. In an inertial reference frame, the horizontal section accelerates horizontally a-A. The vertical section accelerates in quite a complex way: it has a rotational acceleration. The top accelerates A horizontally and a vertically, while the bottom only accelerates a vertically. What's conserved, of course, is the length of the string.
No, that's the wrong argument. m2 has no A acceleration because there's no horizontal force on it. It is not 'due to the movement of the pulley', it's due to the fact that the string is massless and is free to rotate about the pulley.

10. Apr 28, 2013

### aerograce

I understand a bit, but how to clarify that a(m2->ground)=a(m2->M)+a(M->ground) is wrong? Since this equation show that a(m2->ground) has a horizontal acceleration.

11. Apr 28, 2013

### haruspex

It isn't wrong, and there's no contradiction:
a(m2->ground) = a(m2->M)+a(M->ground) = 0
a(m2->M) = -a(M->ground)

12. May 3, 2013

### aerograce

Er, my thought is a(m2->M) is vertical, which can be verified by fixing M. a(M->ground) is horizontal, since M is moving horizontally. So a(m2->ground) has both horizontal and vertical component.

13. May 3, 2013

### aerograce

Can you further explain, what you meant by a complex way? How do you deduce it? Why the top part has a horizontal acceleration but the bottom doesn't have ?

14. May 3, 2013

### haruspex

No, a(m2->ground) is vertical. There are only vertical forces acting on it. m2 will not stay touching M. M will accelerate away from m2.
Since the string remains under tension, and is massless, each section of the string remains straight. The top of the vertical section necessarily accelerates horizontally to stay with the pulley, while the bottom part must stay with m2. There being no horizontal forces on m2, the bottom part of the string accelerates vertically. An element of the string at the top part of that section must also accelerate downwards to match since the string does not stretch. Therefore the vertical section of the string has rotational acceleration.

15. Nov 25, 2015

### qzwxec

I understand everything in this problem except for one part. Why is the force on the cart equal to the tension? How does the tension from the rope get transferred on the cart, and if it does, then what is the effect on the system of the reaction force from the cart on the rope? I don't see any normal force exerted by the cart based on the diagram.

16. Nov 26, 2015

### haruspex

The pulley is part of the cart. It helps to consider that part of the rope that's in contact with the pulley as part of it too. So there's only one horizontal force on the cart, T. Similarly, the vertical part of the rope exerts a vertically downward force T on the cart. Or to combine those, the rope exerts a force $T\sqrt 2$ on the cart at 45 degrees. The reaction from the cart is equal and opposite to that.

17. Nov 26, 2015

### qzwxec

Thank you, I get it now, but this actually raises another question. I had a different problem very similar to this one, in fact exactly the same except that now there is a horizontal force acting on the cart that pushes it to the right. The problem asks for the minimum horizontal force on the cart necessary for both blocks to stay stationary relative to the cart. I did find the answer to this one, and verified it as correct, but I noticed that my solution did not take into account the fact that the rope pushes the cart backwards with a force T.

The exact problem is here: (same diagram)

What horizontal force must be applied to the cart shown in the figure so that the blocks remain stationary relative to the cart? Assume all surfaces, wheels, and pulleys are frictionless.

My solution (correct according to answer key)

m1a=T
m2a=m2g-T=0

Substituting the first equation into the second one gives
m1a=m2g

Therefore,

a=(m2/m1)*g

Since F net, ext = F = (m1+m2+M)a
F=(m1+m2+M)*(m2/m1)*g

As I previously stated, I didn't take tension into account, but my answer still turned out to be correct. Intuitively, I was thinking that the F net, ext should instead be F-T because the tension force pushes the cart to the left, but at the same time tension is an internal force, so I'm unsure about which is correct.

18. Nov 27, 2015

### haruspex

certainly if you consider the whole assembly as a unit then tension is internal, so cancels out. But then you have F responsible for accelerating all three masses.
If you consider the cart+pulley in isolation then tension acts to oppose F horizontally, reducing the net force on F. There is also a horizontal reaction force from the suspended mass. Both of these act to reduce the net horizontal force on the cart, but the cart has less mass than the whole system. These two effects balance out, leading to the same acceleration for the cart.