The discussion revolves around a physics problem involving two masses, m and M, connected by a string over a pulley. The original poster attempts to determine the time it takes for mass m to land on mass M, initially calculating it as 0.4 seconds but later noting the expected answer is 0.9 seconds. Participants explore the dynamics of the system, including the motion of M and the tension in the strings.
The discussion is active, with participants questioning each other's reasoning and clarifying concepts related to forces and motion. Some guidance has been offered regarding the need for separate free body diagrams and the importance of considering the geometry of the system. Multiple interpretations of the problem are being explored, particularly concerning the motion of the masses and the effects of tension.
Participants note the constraints of the problem, including the assumption that the pulleys are frictionless and the string is massless. There is also mention of the complexity introduced by the geometry of the system and the changing angles as m falls.
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 said: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?
Think about the forces on the pulley at the top of M. Why doesn't the pulley go flying off to the right?Ithilrandir said:(M+m) ax = T
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 said:Think about the forces on the pulley at the top of M. Why doesn't the pulley go flying off to the right?
It's held in place by a rigid bracket at the top of M.Ithilrandir said: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 said:It's held in place by a rigid bracket at the top of M.
What does that say about the forces on M?
Yes! And, of course, the normal force from m.Ithilrandir said: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?
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.haruspex said:Yes! And, of course, the normal force from m.
If the normal force of M on m is zero, how will m accelerate to the right?Ithilrandir said:I'm assuming that the normal force from M will be 0 when they are stationary
Talk me through this. In terms of Newton’s laws, what subsystem are you considering the forces on and acceleration of?Ithilrandir said:2T - N = (m+M) ax
The system of M I presume. With the 2 T pulling on M, then M push against m so they move together.haruspex said: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?
If only M, why (M+m)a?Ithilrandir said:The system of M I presume. With the 2 T pulling on M, then M push against m so they move together.
So a correction would beharuspex said:If only M, why (M+m)a?
All good.Ithilrandir said:So a correction would be
mg - T = may
2T - N = Max
N = max
2ax= ay
?
You should have got $$a_y = g \big ( \frac{4m}{M + 5m} \big ) = g/4$$Ithilrandir said:Thanks for all the helps in the thread, I've solved the questions. I need a lot more work.
Yes I didPeroK said:You should have got $$a_y = g \big ( \frac{4m}{M + 5m} \big ) = g/4$$