Time for a mass to land on another mass

  • #1
Ithilrandir
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Homework Statement:
Given the system shown in Fig. 5-9, consider all surfaces frictionless. If m = 150g is released when it is d = 1m above the base of M = 1650g, how long after release Δt, will m strike the base of M?
Relevant Equations:
...
Originally I had thought this was a normal question and simply did the normal s = (1/2)at2, which got the answer of about 0.4s. The answer is however 0.9s, so I double checked the diagram.

The right tail of M is tied to the string connecting m, and two strings connects the top of M to the pulley. I think I can do make M move in a circular fashion once m has traveled downwards a certain point, but I don't know how to formulate the in between parts.
 

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  • #2
Halc
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and two strings connects the top of M to the pulley.
Think of it as a little rigid crane holding the pulley out from M, not as strings holding it, which would just let the pulley fall.

Nothing is going to move in a circular direction. M will slide to the right, being pulled from both the bottom and top strings.

What's the tension on the string? That's probably most important, and not a trivial answer.

How far does M need to move in order for m to reach the bottom?
 
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  • #3
Ithilrandir
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The tension on the bottom of M will be mg, the tension at the top I'm not certain at all. Won't the top become slack once M starts moving right?
 
  • #4
Halc
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The pulleys are frictionless and the string massless, so the tension should be the same everywhere. The string will become slack only at the end once m comes to rest on M.

And no, the tension is not mg because if it was, m would be entirely supported and would not begin moving at all.
 
  • #5
Ithilrandir
82
2
Letting the tension in the string pulling m be T1, and the tension in the string pulling the top of M be T2,

mg-T1 = (m+M) a1

Horizontal component of 2T2-T1 = (M+m) a2

Horizontal acceleration of M = a1 + a2

Vertical acceleration of m = a1

Does these seem right so far?
 
  • #6
haruspex
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Letting the tension in the string pulling m be T1, and the tension in the string pulling the top of M be T2,
Not sure what you mean there. There is only one string.
At the top of M, there is a rigid bracket supporting the pulley.
mg-T1 = (m+M) a1
How do you arrive at that? I.e. which free body are you considering and in what direction are you evaluating the forces on it?
 
  • #7
Ithilrandir
82
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Not sure what you mean there. There is only one string.
At the top of M, there is a rigid bracket supporting the pulley.
So the top part of M is rigid?

How do you arrive at that? I.e. which free body are you considering and in what direction are you evaluating the forces on it?

For m to fall, mg must be greater than T1, so resultant force F = mg - T1. The acceleration due to resultant force will be F/(m+M) as the string is pulling m and M at the same time.
 
  • #8
haruspex
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So the top part of M is rigid?
Yes.
For m to fall, mg must be greater than T1, so resultant force F = mg - T1. The acceleration due to resultant force will be F/(m+M) as the string is pulling m and M at the same time.
This is one of the reasons it is so encouraged to draw separate free body diagrams for each body that can move separately.
The net force on the hanging mass is mg - T1. Its mass is m. Its acceleration follows from those two facts. It doesn't 'know' anything about other forces or masses in the system.
 
  • #9
Ithilrandir
82
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The net force on the hanging mass is mg - T1. Its mass is m. Its acceleration follows from those two facts. It doesn't 'know' anything about other forces or masses in the system.
Hmm... Even with separate free body diagrams, is it not still necessary to take into account of the m+M as they're connected? Additionally, once M starts moving right, the downward velocity of m should also be affected.
 
  • #10
PeroK
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Hmm... Even with separate free body diagrams, is it not still necessary to take into account of the m+M as they're connected? Additionally, once M starts moving right, the downward velocity of m should also be affected.
The motion of the two masses is constrained by the geometry of the pulleys. To understand the problem, I would imagine letting ##m## fall a short distance and redrawing the diagram with the new positions of ##m## and ##M##.
 
  • #11
haruspex
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is it not still necessary to take into account of the m+M as they're connected?
Yes, but only via the forces that act directly on each.
For m, there are three forces: gravity, the tension in the cable, and the normal force from M. Only two of those act vertically, so only those affect the vertical acceleration of m.
M's involvement in that is taken care of by T1. You do not need to bring it into that equation in any other way.

Then you can move on to considering horizontal forces and accelerations.
 
  • #12
Halc
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So the top part of M is rigid?
All of M is rigid, as are all the brackets holding the pulleys. That's what I meant in my first reply.
There is only one string, and thus only one tension T on it that is the same everywhere, so no need for a T1 through T3.
mg-T1 = (m+M) a1
For m to fall, mg must be greater than T1, so resultant force F = mg - T1.
Agree with that part, but not the right side of the equation. You got the (m+M) part correct. The mass of m adds to M since both of them are going to accelerate to the right. But the a1 part is wrong since you're ignoring the pulley arrangement. Remember my first post? I asked
Halc said:
How far does M need to move in order for m to reach the bottom?
You didn't answer that.
The acceleration due to resultant force will be F/(m+M) as the string is pulling m and M at the same time.
Yes, but given string tension T, what is the force F to the right on (M+m)?

BTW, while I've not actually done the arithmetic, I'd have guessed considerably greater than 0.9 seconds for m to make the 1 meter trip.
 
  • #13
PeroK
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BTW, while I've not actually done the arithmetic, I'd have guessed considerably greater than 0.9 seconds for m to make the 1 meter trip.
The ##0.9s## checks out!
 
  • #14
Halc
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The ##0.9s## checks out!
So much for gut feel...
 
  • #15
Ithilrandir
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The motion of the two masses is constrained by the geometry of the pulleys. To understand the problem, I would imagine letting ##m## fall a short distance and redrawing the diagram with the new positions of ##m## and ##M##.
I did a drawing in paint to visualize the new positions, and when M moves to the right by Δs, m moves diagonally by Δs, where Δs2 = Δx2 + Δy2. If the top of M is fixed then M should rotate anti-clockwise too, but the drawing would be extremely complex.

Yes, but only via the forces that act directly on each.
For m, there are three forces: gravity, the tension in the cable, and the normal force from M. Only two of those act vertically, so only those affect the vertical acceleration of m.
M's involvement in that is taken care of by T1. You do not need to bring it into that equation in any other way.

Then you can move on to considering horizontal forces and accelerations.

Once m moves moves away from its original position, the downward component will become mgcosθ - T1, whereby Θ is the angle the string makes with M, and it changes over time. The horizontal acceleration of M should remain the same. I'm thinking of making Θ = ωt, but without a fixed speed I find it not very feasible.
 
  • #17
Halc
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If the top of M is fixed then M should rotate anti-clockwise too, but the drawing would be extremely complex.
M is rigid, not fixed. M does not rotate. It slides freely on the base. But how far if m moves down 1 meter?
 
  • #18
PeroK
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I did a drawing in paint to visualize the new positions, and when M moves to the right by Δs, m moves diagonally by Δs, where Δs2 = Δx2 + Δy2. If the top of M is fixed then M should rotate anti-clockwise too, but the drawing would be extremely complex.



Once m moves moves away from its original position, the downward component will become mgcosθ - T1, whereby Θ is the angle the string makes with M, and it changes over time. The horizontal acceleration of M should remain the same. I'm thinking of making Θ = ωt, but without a fixed speed I find it not very feasible.
Okay, so you've got the wrong idea here. When ##m## has fallen a short distance, the system looks like it does at the start, except ##m## is lower, ##M## and ##m## have moved to the right, and the string has run through all the pulleys.

For this problem, you simply decompose the motion into ##x## and ##y## components: ##\Delta x, \Delta y, v_x, v_y## and ##a_x, a_y##, with the y-component applying to ##m## and the x-component applying to both ##m## and ##M##.
 
  • #19
Ithilrandir
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Okay, so you've got the wrong idea here. When ##m## has fallen a short distance, the system looks like it does at the start, except ##m## is lower, ##M## and ##m## have moved to the right, and the string has run through all the pulleys.

For this problem, you simply decompose the motion into ##x## and ##y## components: ##\Delta x, \Delta y, v_x, v_y## and ##a_x, a_y##, with the y-component applying to ##m## and the x-component applying to both ##m## and ##M##.
In this case, vertically, may = mg - T,
horizontally, (m+M) ax = T?

If so, how does the horizontal acceleration affect the time taken to hit M?
 
  • #20
PeroK
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In this case, vertically, may = mg - T,
horizontally, (m+M) ax = T?

If so, how does the horizontal acceleration affect the time taken to hit M?
You need to:

a) Fix the mistake you made. Look more closely at the system.

b) Find a relationship between ##a_x## and ##a_y##.
 
  • #21
Ithilrandir
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You need to:

a) Fix the mistake you made. Look more closely at the system.

b) Find a relationship between ##a_x## and ##a_y##.

I can see the angle of the string changing, which probably changes the vertical component of the tension. If the tension is constant then ax will be constant too,

So this will hold,
(m+M) ax = T

may = mg - T,
Becomes
may = mg - T cos Θ, the angle of the string with the vertical.

This is what I've seen so far from it.
 
  • #22
PeroK
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So this will hold,
(m+M) ax = T

This is not right. Look more closely!

may = mg - T cos Θ, the angle of the string with the vertical.

This is what I've seen so far from it.
There is no need to involve the angle ##\theta## here. You have the x-y components. That is enough.
 
  • #23
Ithilrandir
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This is not right. Look more closely!
One thing which I am not entirely certain is this: do a Δx of M = a Δy of m? I'm confused as to whether this is the case from looking at the diagram in various ways.
 
  • #24
PeroK
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One thing which I am not entirely certain is this: do a Δx of M = a Δy of m? I'm confused as to whether this is the case from looking at the diagram in various ways.
If you draw a large accurate diagram, there can only be one answer. The string is of fixed length.
 
  • #25
Ithilrandir
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If you draw an accurate diagram, there can only be one answer. The string is of fixed length.
Then Δx of M ≠ Δy of m. Admittedly my diagram drawing have always been rather bad so I've mostly relied on imagining it in my head, but my drawing of the diagram in paint is honestly not that helpful either.

the system looks like it does at the start, except ##m## is lower, ##M## and ##m## have moved to the right, and the string has run through all the pulleys.
When M moves right by Δx, the length of string that hangs m will increase by Δx too. So the x of m increases by Δx first and the y decreases by some amount, and then with the increase in length by Δx drops down further.
 
  • #26
PeroK
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Then Δx of M ≠ Δy of m. Admittedly my diagram drawing have always been rather bad so I've mostly relied on imagining it in my head, but my drawing of the diagram in paint is honestly not that helpful either.


When M moves right by Δx, the length of string that hangs m will increase by Δx too. So the x of m increases by Δx first and the y decreases by some amount, and then with the increase in length by Δx drops down further.
I get ##\Delta y = 2 \Delta x##.
 
  • #27
haruspex
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When M moves right by Δx, the length of string that hangs m will increase by Δx too.
The total length of string remains constant. If M moves to the right by Δx, what happens to the two horizontal sections of string? What about the two vertical sections?
 
  • #28
Ithilrandir
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I get ##\Delta y = 2 \Delta x##.
I... have no idea how you got that. I had imagined Δy to be slightly less than Δx.

The total length of string remains constant. If M moves to the right by Δx, what happens to the two horizontal sections of string? What about the two vertical sections?
Which two are you referring to?
 
  • #29
haruspex
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Which two are you referring to?
There are four straight sections of the string, two vertical (one near the right wall and one from which m hangs) and two horizontal (one near the floor, one at the top).
Are there any others??
If M moves to the right by Δx, what happens to the lengths of the two horizontal sections of string?
What happens to the lengths of the two vertical sections?
 
  • #30
Ithilrandir
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There are four straight sections of the string, two vertical (one near the right wall and one from which m hangs) and two horizontal (one near the floor, one at the top).
Are there any others??
If M moves to the right by Δx, what happens to the lengths of the two horizontal sections of string?
What happens to the lengths of the two vertical sections?
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?
 
  • #31
haruspex
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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.
 
  • #32
Ithilrandir
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OK I've corrected my workings based on 2Δx = Δy, but I think there's still some errors.

mg - T = F = may

ax = (1/2)ay

(M+m) ax = T
 
  • #33
haruspex
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(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?
 
  • #34
Ithilrandir
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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.
 
  • #35
haruspex
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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?
 

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