# Netwon law's - confused about the forces

1. Apr 28, 2013

### Saitama

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

2. Relevant equations

3. The attempt at a solution
I am not sure how to begin with this one. I understand how the motion takes place but I am really not sure how to begin making the equations.
Considering the first case, B moves down and since the string is inextensible, A moves to left. So the forces acting on B are its weight and the tension from the string. Right? But what are the forces acting on A except its weight and the normal reaction from ground? I can only see that the force arise from the pulley attached but how would I calculate that force?

Any help is appreciated. Thanks!

#### Attached Files:

• ###### nlm1.jpg
File size:
18.3 KB
Views:
78
2. Apr 28, 2013

### ehild

Does not the tension act on Edit: [STRIKE]B[/STRIKE] A? The pulley is fixed to it... And can be also normal force between the blocks.

ehild

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

### Saitama

Do you mean A? I already said that the tension acts on B.

Sorry, can't understand this. Are you talking about the normal reaction between the blocks? I don't see how this will come into play.

4. Apr 28, 2013

### ehild

Yes, I meant A. Sorry. And think about the normal force.

ehild

5. Apr 28, 2013

### Saitama

Please see the FBD of A and B attached. (I haven't drawn the normal reaction from the ground on A and its weight)

Do they look correct?

#### Attached Files:

• ###### fbd1.png
File size:
2.7 KB
Views:
53
6. Apr 28, 2013

### ehild

Add the pulley to the big block.

I have to leave now. Think, and solve :)

ehild

7. Apr 28, 2013

### Saitama

I already added it. The tension on A comes from the pulley. Right?

8. Apr 28, 2013

### ehild

I see... So what are your equations?

ehild

9. Apr 28, 2013

### Saitama

For horizontal motion:
$N=20a$ and $T-N=40a$
$\Rightarrow T=60a$

For vertical motion:
$20g-T=20a$
$\Rightarrow 20g=80a$
$\Rightarrow a=g/4$

Net acceleration of B: $\sqrt{2}a=g\sqrt{2}/4$

Does this look correct?

10. Apr 28, 2013

### haruspex

That looks right to me for case (i). What about case (ii)? (I take issue with one aspect of the question. It ought to specify ratio of magnitudes of accelerations.)

11. Apr 28, 2013

### Saitama

I am having trouble determining the tension acting on the block A in case ii). Should it be simply T in the left direction? The pulley is at some angle to horizontal which is confusing me.

12. Apr 28, 2013

### ehild

There is a horizontal and a vertical tension, both of the same magnitude.

ehild

13. Apr 28, 2013

### Saitama

Is it T in both the directions?

14. Apr 28, 2013

### ehild

Why should it be different?

15. Apr 28, 2013

### Saitama

Okay, thanks. Please check if my equations are correct.

Case ii)
For A, the forces acting on it in the horizontal direction are the normal reaction due to B and the tension. Both the forces are in left direction.
$T+N=40a$

For B, the forces acting in the horizontal direction is only the normal reaction due to A. But this force is in direction opposite to the motion of B and this doesn't look right to me.

16. Apr 28, 2013

### ehild

well, the normal force can not pull B.... So what can happen with B?

ehild

17. Apr 28, 2013

### Saitama

It can move vertically downwards, the normal reaction won't do anything to the motion of B? Does this mean the normal reaction will be zero?

18. Apr 28, 2013

### ehild

It looks so. The big block accelerates to the left and leaves the small one behind in the first instant.

19. Apr 28, 2013

### Saitama

Okay so the normal reaction is zero.
Hence $T=40a$ and $20g-T=20a$
$\Rightarrow 20g=60a \Rightarrow a=g/3$

Finding out the ratio of accelerations, it comes out to be $3/2\sqrt{2}$ which is correct.

Thanks a lot ehild!

20. Apr 28, 2013

### ehild

Congratulation! I never dreamt that would be the result

ehild