Solving Pulley and Platform Homework w/ Acceleration

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Homework Help Overview

The discussion revolves around a physics problem involving two construction workers using pulleys to raise themselves on a platform. The problem includes variables such as the mass of the workers, the mass of the platform, and the forces they exert while pulling on the ropes. Participants are exploring the dynamics of the system, including the forces acting on both the workers and the platform.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss various methods of analyzing the forces acting on the workers and the platform, including treating them as a single system. There are attempts to clarify the roles of normal forces, tension, and the forces applied by the workers. Questions arise regarding the differences between the forces exerted by the workers and the tension in the ropes.

Discussion Status

The discussion is active, with participants providing insights and questioning each other's reasoning. Some participants express confusion about the forces involved, while others attempt to clarify the relationships between the forces and the resulting acceleration. There are ongoing explorations of the equations governing the system, with no clear consensus yet reached.

Contextual Notes

Participants are navigating assumptions about the system, such as the mass of the platform and the nature of the forces applied by the workers. There is mention of potential confusion regarding the normal force and the application of forces in the context of the problem.

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Homework Statement



Two construction workers each of mass m raise themselves on a hanging platform using pulleys as shown above. If the platform has a mass of 0.5m, the initial distance between the pulleys and the platform is d, and the workers each pull with a force f on the ropes, what is the acceleration a of the workers? Assume the pulleys and ropes are massless.

Homework Equations





The Attempt at a Solution



Forces on one Man:
FME (weight of the man),
The normal force NMP (contact of the man with the platform),
The tension force FMR (the response force that the rope applies to the man).
The force applied by 2 men pointing down

For the platform:
FPE (weight of the platform),
NPM (force on the platform due to the man, it is the pair of NMP) and remember you have two of this because you have 2 men,
T and T: The two tensions acting on the platform due to the ropes
 

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I would consider men+platform as one system. This reduces the number of forces involved.

What did you do after listing all those forces?
 
Eliminated Normal force. What answer do you get? I tried doing it by considering them as a one system too. Still stuck.
 
mfb said:
I would consider men+platform as one system. This reduces the number of forces involved.

What did you do after listing all those forces?

Considering them as one system:

2N downwards
2N upwards
2mg + .5 mg downwards
2f downwards (force applied by men)
2f upwards. (The tension in the strings attached to the platform)
2t upwards. (the tension in the strings held by men)

Where am I going wrong?
 
2f downwards (force applied by men)
They do not apply a downwards force on the pulley/men system.
2f upwards. (The tension in the strings attached to the platform)
2t upwards. (the tension in the strings held by men)
Where is the difference between f and t?

The other forces look good.
 
mfb said:
They do not apply a downwards force on the pulley/men system.
Where is the difference between f and t?

The other forces look good.

But the question says they pull down with a force f. What are the correct forces?
also, is there a difference between f and t in this case?
 
They pull the rope down, and themself up.
also, is there a difference between f and t in this case?
That was my question to you. Can the rope tension be different at both sides?
 
mfb said:
They pull the rope down, and themself up.
That was my question to you. Can the rope tension be different at both sides?

Okay, so tension and force are the same in this case. Can you please list out all the forces?
 
For the two men combined:
2N - 2F - 2Mg = 2Ma
For the platform
2f -.5Mg - 2N = .5Ma

Is that right?
 
  • #10
judas_priest said:
For the two men combined:
2N - 2F - 2Mg = 2Ma
Is that right?

Why the minus sign in -2F here?
 
  • #11
Because they're pulling it down.
Here's another way I did it in:
Forces on 2 men
tension upwards
normal upwards
weight downwards

Forces on platform:
normal downwards
weight downwards

I've ignored the forces applied by the two men to pull the string down, because they get balanced by the forces on platform pointing up
 
  • #12
F points upwards for the men, as I mentioned before. They are pulling the rope down, so they are pulling themself up.
Apart from that, post 9 is right.

I've ignored the forces applied by the two men to pull the string down, because they get balanced by the forces on platform pointing up
That does not work.
 
  • #13
But aren't they trying to pull it down? How would it point upwards?
And F gets balanced by the force on platform. Don't they?
 
  • #14
mfb said:
They are pulling the rope down, so they are pulling themself up.
Apart from that, post 9 is right.

That does not work.

Why not? They are reaction forces after all? How does F point upwards for men when they're applying force downwards.
Also, Isn't tension upwards?
 
  • #15
2N + 2F - 2Mg = 2Ma

2F -.5Mg - 2N = .5Ma

Solving them, this is what I get:

4F - 25m = 2.5Ma

Is that correct?
 
  • #16
judas_priest said:
But aren't they trying to pull it down?
Did you ever try to push a rope? Did it work?
You can only pull it, and get a positive tension in it.
They are "trying" to pull at the anchor on the top of the building. As this anchor feels a force downwards, they feel a force upwards (action/reaction).

Also, Isn't tension upwards?
Exactly, tension pulls them upwards.

Is that correct?
If you add the missing decimal point, yes.
 
  • #17
mfb said:
Did you ever try to push a rope? Did it work?
You can only pull it, and get a positive tension in it.
They are "trying" to pull at the anchor on the top of the building. As this anchor feels a force downwards, they feel a force upwards (action/reaction).

Exactly, tension pulls them upwards.

If you add the missing decimal point, yes.

I'm confused. If I'm pulling the rope, am I not applying the force downwards? Of course, it's upwards on the platform. And isn't the equation incomplete without tension?
Also, I did add the decimal points.
 
  • #18
I got the answer, but can someone explain to me the working of this problem? I haven't understood it.
I just got the answer by blindly following the posts
 
  • #19
mfb said:
If you add the missing decimal point, yes.
I think judas_priest has substituted g = 10.
judas_priest said:
If I'm pulling the rope, am I not applying the force downwards?
Yes, downwards on the rope, so the rope is applying the same force upwards on you.
The quick way to solve this problem is to consider the forces pulling upwards the pulleys. What do these add up to?
 
  • #20
haruspex said:
I think judas_priest has substituted g = 10.

Yes, downwards on the rope, so the rope is applying the same force upwards on you.
The quick way to solve this problem is to consider the forces pulling upwards the pulleys. What do these add up to?

Okay, got it. Thanks a lot!
 
  • #21
im trying to figure this out but I am really confused where did the 2N come from?
 
  • #22
lollikey said:
im trying to figure this out but I am really confused where did the 2N come from?
As in post #10? N is the normal force from the platform on each man. Total = 2N.
 
  • #23
ohh okay ! thank you ! I thought it was that but I got confused because on a later post it was said there is no normal force
 
  • #24
My problem :Two construction workers each of mass m raise themselves on a hanging platform using pulleys as shown above. If the platform has a mass of 5m and the workers each pull with a force f on the ropes, what is the acceleration a of the workers? Assume the pulleys and ropes are massless.

so would the forces be

2N up
2N down
7mg down
4f up

?

and you would add all of them up and equal to mass*acceleration (ma)?
 
Last edited:
  • #25
lollikey said:
My problem :Two construction workers each of mass m raise themselves on a hanging platform using pulleys as shown above. If the platform has a mass of 5m and the workers each pull with a force f on the ropes, what is the acceleration a of the workers? Assume the pulleys and ropes are massless.

so would the forces be

2N up
2N down
7mg down
4f up

?

and you would add all of them up and equal to mass*acceleration (ma)?
Yes, if the m in the last line represents the total mass, not the given m.
 
  • #26
so the equation would be 2N-2N+2f+2f+7mg=7mga
to find acceleration it would be (4f+7mg)/7mg= a ?
 
  • #27
it marked it wrong. I am soo confused! what am I doing wrong?
 
  • #28
lollikey said:
so the equation would be 2N-2N+2f+2f+7mg=7mga
7mga? That's multiplying two accelerations together.
 
  • #29
but I thought that since both men have a mass of m and the platforms mass is 5m the total mass would be 7m
so would the formula just be 2N-2N+2f+2f+7mga = 7ma ??
 
  • #30
I got it ! the answer is (4f-7mg) / 7m
 

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