# Time for a mass to land on another mass

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?

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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 = may

2T - N = (m+M) ax

2ax= ay

These seem right so far?

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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.

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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 = may

2T - N = Max
N = max
2ax= ay
?

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So a correction would be

mg - T = may

2T - N = Max
N = max
2ax= ay
?
All good.

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

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