# Pulleys and springs

1. Apr 24, 2013

### Saitama

1. The problem statement, all variables and given/known data
(see attachment)

2. Relevant equations

3. The attempt at a solution
At equilibrium conditions, the forces acting on $m_2$ are the tension due to the strings and its weight. $T=m_2g+T'$ where T is the tension in the string above $m_2$ and T' is the tension in the string below it. Equating forces for $m_1$, $T/2=m_1g \Rightarrow T=2m_1g$. I am not sure what to do next? How would I go about finding the initial acceleration of B?

Any help is appreciated. Thanks!

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2. Apr 24, 2013

### PhanthomJay

That is correct for the given equilibrium condition.
Do the free body diagrams again, with the string cut, but this time, after identifying forces, use Newton 2 (the accelerations of the 2 masses may differ).

3. Apr 24, 2013

### consciousness

For the FBD of m1 shouldnt it be

$2T=m_1g$

As two strings pass over?

My approach to this question is that due to the spring the Tension in the string(which isn't cut) before and after cutting will remain the same. This is because the elongation of the spring will not change JUST after cutting. Next after making FBD of m2 we can get the answer. But I am not 100% sure.

4. Apr 24, 2013

### PhanthomJay

Oh sure, thanks for the catch, my apologies, I glanced over that too quickly.
It will have shortened an infinitesimal distance just after the cut in an infinitesimal time, but still have an initial acceleration dv/dt.

5. Apr 24, 2013

### Saitama

When the string is cut, the forces acting on $m_2$ are the tension and its weight. For $m_2$, $m_2g-T=m_2a_2$. The forces acting on $m_1$ is the tension and its weight. $m_1g-2T=m_1a_1$. Have I got this right?

Thanks for the catch!

6. Apr 24, 2013

### PhanthomJay

Yes, provided you have the correct relation and signage between a1 and a2.

7. Apr 24, 2013

### Saitama

But I am still not sure what to do next? Can I have some hints? :)

8. Apr 24, 2013

### PhanthomJay

You have 3 unknowns T, a1, and a2, but so far you have listed just 2 equations. You need another one that describes the relationship between a1 and a2.

a1 and a2 are not the same.

9. Apr 24, 2013

### Saitama

$a_1=2a_2$?

10. Apr 24, 2013

### PhanthomJay

Which mass moves the furthest and fastest over an initial time, 1 or 2? Sometimes it is best to draw stop motion sketches and visualize the movement.

11. Apr 24, 2013

### Saitama

I can't figure this out. I don't know what to do with the spring. :(

12. Apr 24, 2013

### PhanthomJay

i suppose if you don't know what to do with the spring, try ignoring it and replace it with a cord! The problem asks for the initial acceleration of m2, that is, it's instantaneous acceleration at the instant the lower string is cut. The spring might be very stiff with very little elongation under load, or it might be very stretchy like that wonderful slinky toy, with a lot of elongation under load. But in either case, the spring will compress when the cut is made, but initially, it doesn't matter, what matters is that the mass m2 will rise up, and the mass m1 and the pulley supporting m1 will fall downward. So pretend you have a stiff spring, make it an inextensible cord if you will, and you must then answer the question about the a2 and a1 relationship.

13. Apr 25, 2013

### Saitama

Here's how I look at it. If $m_2$ moves up a distance x, the pulley to which $m_1$ is attached should move 2x. Right?

14. Apr 25, 2013

### PhanthomJay

If m2 moves up x, then pulley of m1 moves down x/2. Right?

15. Apr 28, 2013

### Saitama

Sorry for such a late reply, I had some problems with my internet connection.

Ugh..that's the biggest problem I face in these type of problems. If the acceleration of m_2 is a, then acceleration of m_1 is a/2.
For m_2, $T-m_2g=m_2a$ and for m_1, $m_1g-2T=m_1a/2$. Solving the two equations, I get
$$a=\frac{m_1-2m_2}{2m_2+m_1/2}g$$
which is wrong.

Last edited: Apr 28, 2013
16. Apr 28, 2013

### rl.bhat

Even though m1 is greater than m2, because of the stretched spring m2 moves downwards. Taking up positive down negative sign, rewrite your two equations and solve for a2.

17. Apr 28, 2013

### consciousness

Using the approach I gave in the 3rd post I am getting the answer

$$a=g\frac{m_1-2m_2}{2m_2}$$ (upwards)

Can you please tell me if this is the correct answer? I have also used another method given by my teacher to get the same answer.

18. Apr 29, 2013

### consciousness

I think i understood why the constraint relation approach doesn't work. It might have something to do with the fact that objects at rest can also have accelerations(like a vertically thrown ball at its highest point). For an object at rest its acceleration can only be found by newtons second law.

19. Apr 29, 2013

### Saitama

That's the correct answer.

What's the other method have you used? One of my friend also suggested me the same way as you did in your post #3.

20. Apr 29, 2013

### consciousness

The other method is that when the string is burnt the NET force on m2 will be the force it applied on m2 before it was burnt but in opposite direction (ie upwards). Now we can get the answer in one equation very quickly. This result is very useful and can be easily derived also.

Oh and its adivisible to ignore my post #18 i feel that the logic is flawed now.

Last edited: Apr 29, 2013