Mass on string landing on moving platform with pulley system

[haruspex]

Oh I see. So when M moves right by Δx, the top Horizontal string moves left by Δx, m moves by 2 Δx because of the combined effects of the vertical at the right moving up by Δx then the mass m moving down by Δx. Is that right?

Yes, m descends by 2 Δx, but I would explain it this way: the two horizontal sections each shrink by Δx, but the right hand vertical section stays the same length, so the left hand vertical section must extend by 2 Δx.

 

[Ithilrandir]

OK I've corrected my workings based on 2Δx = Δy, but I think there's still some errors.

mg - T = F = ma_y

a_x = (1/2)a_y

(M+m) a_x = T

 

[haruspex]

(M+m) ax = T

Think about the forces on the pulley at the top of M. Why doesn't the pulley go flying off to the right?

 

[Ithilrandir]

Think about the forces on the pulley at the top of M. Why doesn't the pulley go flying off to the right?

I had thought it was nailed to the wall, but now that I look at it, it's held in place to M by the ropes at the top.

 

[haruspex]

I had thought it was nailed to the wall, but now that I look at it, it's held in place to M by the ropes at the top.

It's held in place by a rigid bracket at the top of M.
What does that say about the forces on M?

 

[Ithilrandir]

It's held in place by a rigid bracket at the top of M.
What does that say about the forces on M?

So the two lines are rigid brackets and not ropes? I had been thinking they were ropes. SO M, aside from its own weight, experiences two T, one from the top and one from the bottom?

 

[haruspex]

So the two lines are rigid brackets and not ropes? I had been thinking they were ropes. SO M, aside from its own weight, experiences two T, one from the top and one from the bottom?

Yes! And, of course, the normal force from m.

 

[Ithilrandir]

Yes! And, of course, the normal force from m.

In that case I'm assuming that the normal force from M will be 0 when they are stationary , as there do not seem to be any Horizontal force from m.

mg - T = ma_y

2T - N = (m+M) a_x

2a_x= a_y

These seem right so far?

 

[haruspex]

I'm assuming that the normal force from M will be 0 when they are stationary

If the normal force of M on m is zero, how will m accelerate to the right?

2T - N = (m+M) ax

Talk me through this. In terms of Newton’s laws, what subsystem are you considering the forces on and acceleration of?

 

[Ithilrandir]

If the normal force of M on m is zero, how will m accelerate to the right?

Talk me through this. In terms of Newton’s laws, what subsystem are you considering the forces on and acceleration of?

The system of M I presume. With the 2 T pulling on M, then M push against m so they move together.

 

[haruspex]

The system of M I presume. With the 2 T pulling on M, then M push against m so they move together.

If only M, why (M+m)a?

 

[Ithilrandir]

If only M, why (M+m)a?

So a correction would be

mg - T = ma_y

2T - N = Ma_x
N = ma_x
2a_x= a_y
?

 

[haruspex]

So a correction would be

mg - T = ma_y

2T - N = Ma_x
N = ma_x
2a_x= a_y
?

All good.

 

[Ithilrandir]

Thanks for all the helps in the thread, I've solved the questions. I need a lot more work.

 

[PeroK]

Thanks for all the helps in the thread, I've solved the questions. I need a lot more work.

You should have got $$a_y = g \big ( \frac{4m}{M + 5m} \big ) = g/4$$

 

[Ithilrandir]

You should have got $$a_y = g \big ( \frac{4m}{M + 5m} \big ) = g/4$$

Yes I did