Pulley Question - Please Help!

  • #1

Homework Statement


http://imgur.com/SE50jeP

Homework Equations


Fnet = m*a
FG = m*g

The Attempt at a Solution


I'm so confused about this problem and I still can't solve it after weeks.

The initial distance of the workers is irrelevant. If I treat the workers and the platform as a single system, then I don't have to worry about normal forces because they are internal forces. On my force body diagram, I have the tension force on the workers and platform due to the rope, f (two of them), and the weight force of the workers and the platform (together) as the external forces acting on the system:

(m + m + 0.5*m)*a = 2*f - (m + m + 0.5*m)*g

So the acceleration of the workers and the platform should be:

a = (2*f - (m + m + 0.5*m)*g)/(m + m + 0.5*m)

But the question doesn't allow you to enter "g", which suggests that the workers and the platform's weight is irrelevant. I don't understand why though. And what am I doing wrong? This is so frustrating. I have been trying to figure this problem out for weeks.

Thank you.
 

Answers and Replies

  • #2
haruspex
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I agree with you that g must feature in the answer, much as you have it, but you are missing something in the total external forces.
If you were to slice right across the diagram just above the workers' heads, what forces would you remove from the workers+platform system?
 
  • #3
I agree with you that g must feature in the answer, much as you have it, but you are missing something in the total external forces.
If you were to slice right across the diagram just above the workers' heads, what forces would you remove from the workers+platform system?

Since the workers are pulling with a force f on the rope, then due to Newton's Third Law, the rope would exert a force f on the workers in the opposite direction, so I am missing another 2 f forces, right?

So the acceleration should be:

a = (4*f - (m + m + 0.5*m)*g)/(m + m + 0.5*m)
 
  • #4
haruspex
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Since the workers are pulling with a force f on the rope, then due to Newton's Third Law, the rope would exert a force f on the workers in the opposite direction, so I am missing another 2 f forces, right?

So the acceleration should be:

a = (4*f - (m + m + 0.5*m)*g)/(m + m + 0.5*m)
Yes, or more simply ##\frac {8f}{5m}-g##
 
  • #5
Yes, or more simply ##\frac {8f}{5m}-g##

Thank you.
 

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