# Homework Help: Constrained motion problem

1. Aug 1, 2013

### Saitama

1. The problem statement, all variables and given/known data
If acceleration of A is 2m/s2 to the left and acceleration of B is 1m/s2 to left, then acceleration of C is (see attachment 1)
A)1m/s2 upwards
B)1m/s2 downwards
C)2m/s2 downwards
D)2m/s2 upwards

Ans: A

2. Relevant equations

3. The attempt at a solution
(see attachment 2, I hope the labels are clear)
Let the radius of smaller pulleys be r and that of bigger pulleys be R. I have marked the lengths of different portions of string. L is the length of total string.
$$L=x_1+2 \times \pi r+x_2+x_3+2\times \pi R/4+2x_4$$
Differentiating twice w.r.t time
$$0=\ddot{x_1}+\ddot{x_2}+\ddot{x_3}+2\ddot{x_4}$$
Now I am confused as to what should be the value of $\ddot{x_1}$.

Any help is appreciated. Thanks!

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2. Aug 1, 2013

### verty

I suggest expressing $x_1$ as a function of time, then differentiating twice.

Without calculus, it is not as easy. I can't think of a useful hint to give in this case except to say, use your intuition.

Last edited: Aug 1, 2013
3. Aug 1, 2013

### tiny-tim

Hi Pranav-Arora!

Isn't x1 a function of x2 and x3 ?

4. Aug 1, 2013

### Saitama

Yes but then how do I write $x_1$ in terms of $x_2$ and $x_3$? Something like this: $x_2+x_3+\text{some constant}$?

5. Aug 1, 2013

### tiny-tim

yes!!!

(what's wrong with that?? )

6. Aug 1, 2013

### Saitama

Okay so I get,
$$0=2\ddot{x_2}+2\ddot{x_3}+2\ddot{x_4}$$
$$\ddot{x_4}=-(\ddot{x_2}+\ddot{x_3})$$
$\because \ddot{x_2}=2 m/s^2$ and $x_2=\ddot{x_2}=1 m/s^2$
$$\ddot{x_4}=-3 m/s^2$$

This is definitely wrong.

7. Aug 1, 2013

8. Aug 1, 2013

### Saitama

Can you please explain a bit more? I am still not able to understand what to do with that info. :(

9. Aug 1, 2013

### tiny-tim

so x2' = 2, and x3' = … ?

10. Aug 1, 2013

### haruspex

I assume you meant $\ddot{x_3}=1 m/s^2$, but that's still wrong. Which way is B moving?

11. Aug 1, 2013

### PeterO

If A moves left while B is stationary, C would move up.

If B moved left while A was stationary, C would move down.

So the effects of leftward motion of A & B have opposite effects on C.

12. Aug 2, 2013

### Saitama

Is that a typo or are you actually asking me to calculate the first derivative? :uhh:

13. Aug 2, 2013

### tiny-tim

sorry, i meant '' not '

14. Aug 2, 2013

### Saitama

But that still doesn't help. Both the accelerations are in the same direction. Should I work on the signs of first derivatives?

15. Aug 2, 2013

### tiny-tim

But x2 is measured to the left, and x3 is measured to the right.

16. Aug 2, 2013

### Saitama

So $\ddot{x_2}=2 m/s^2$ and $\ddot{x_3}=-1 m/s^2$?

But is it wrong to think about the signs of first derivative? I never specified any directions, like you said that x2 is measured to the right.

17. Aug 2, 2013

### tiny-tim

i don't see how it helps
yes you did!

it's clear from your diagram that x2 and x3 are always positive, and so must be measured from the fixed end, which is in the middle

18. Aug 2, 2013

### Saitama

Okay, so in general, I have to specify the directions and measure distances from a fixed end. I guess I have to do some more practice on these problems. -.-'

Anyways, I get $\ddot{x_4}=-1 m/s^2$. Do I have to put my reasoning for the negative sign like this: Since $x_4$ is positive, the negative sign indicates that the block C moves upwards. Would that be enough?

Thanks a lot tiny-tim!

19. Aug 2, 2013

### tiny-tim

It probably doesn't need any reasoning.

If you want to provide a reason, just say that x4 is decreasing, and x4 is the depth below the main level

20. Aug 2, 2013

### Saitama

Okay. Thanks again! :)