Newton's laws for masses connected by a string over a pulley

[Saitama]

Homework Statement

(see attachment)

Homework Equations

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!

 

[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

 

[Saitama]

Does not the tension act on B?

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

The pulley is fixed to B... And can be also normal force between the blocks.

ehild

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.

 

[ehild]

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

 

[Saitama]

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


ehild

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?

 

[ehild]

Add the pulley to the big block.

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

ehild

 

[Saitama]

Add the pulley to the big block.

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

 

[ehild]

I see... So what are your equations?

ehild

 

[Saitama]

I see... So what are your equations?

ehild

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?

 

[haruspex]

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?

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.)

 

[Saitama]

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.)

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.

 

[ehild]

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


ehild

 

[Saitama]

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


ehild

Is it T in both the directions?

 

[ehild]

Why should it be different?

 

[Saitama]

Why should it be different?

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.

 

[ehild]

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

ehild

 

[Saitama]

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

ehild

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?

 

[ehild]

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

 

[Saitama]

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

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!

 

[ehild]

Congratulation! I never dreamt that would be the result

ehild

 

[haruspex]

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!

Except, that's the ratio of magnitudes . Presumably that's what was wanted. The two accelerations are in different directions, so asking for the ratio of the accelerations is a bit poor.