Pulling force for Bosun's chair?

  • Thread starter Theorγ
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  • #1
Theorγ
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Homework Statement


A pulley is attached to the ceiling, with a rope going over that pulley and both objects are mass less. At one end of the rope, a Bosun's chair with a man sitting on it is attached, and the man is holding on to the other end of the rope. The man and chair's mass combined is 95 kg. What is the force the man must use to pull himself up at constant velocity?

Homework Equations


[tex]F_{net} = m a[/tex]

The Attempt at a Solution


I first drew all the forces on the system by using the following equations:

Forces on the man and chair
[tex]-F_{g} + T = m a[/tex]

Forces on the other end of rope
[tex]F_{pull} - T = m a[/tex]

Definition of Constant velocity
[tex]a = 0[/tex]

Subbing in the given values will give the notion that the man must pull his own weight to go up, but apparently this is not the answer, so what am I missing?
 
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Answers and Replies

  • #2
PhanthomJay
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Homework Statement


A pulley is attached to the ceiling, with a rope going over that pulley and both objects are mass less. At one end of the rope, a Bosun's chair with a man sitting on it is attached, and the man is holding on to the other end of the rope. The man and chair's mass combined is 95 kg. What is the force the man must use to pull himself up at constant velocity?

Homework Equations


[tex]F_{net} = m a[/tex]

The Attempt at a Solution


I first drew all the forces on the system by using the following equations:

Forces on the man and chair
[tex]-F_{g} + T = m a[/tex]

Forces on the other end of rope
[tex]F_{pull} - T = m a[/tex]

Definition of Constant velocity
[tex]a = 0[/tex]

Subbing in the given values will give the notion that the man must pull his own weight to go up, but apparently this is not the answer, so what am I missing?
You are missing the fact that if you draw a free body diagram of the chair with the man on it, there are two ropes pulling up on them...one on the man and one on the chair. The tension on each rope must be equal, and the sum of those 2 tensions must equal the weight of the man and chair.
 
  • #3
Theorγ
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There's a rope pulling on the man?
 
  • #4
Delphi51
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The man pulls down on the rope. According to Newton's third law, the rope will pull up on the man with an equal force.

If you imagine a man hanging on to a rope half way up a cliff, you will see this force must exist!
 
  • #5
Theorγ
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The man pulls down on the rope. According to Newton's third law, the rope will pull up on the man with an equal force.

If you imagine a man hanging on to a rope half way up a cliff, you will see this force must exist!

So do you mean:

[tex]-F_{g} + F_{n} + F_{pull} = m a[/tex]
[tex]F_{pull} - F_{n} - F_{g} = m a[/tex]
 
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  • #6
PhanthomJay
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You have already established that a = 0, so if you look at the forces on the man only, it's Ft + Fn - Fg = 0 ( the normal force acts in the same direction as the tension force). But you still cannot find Ft ,since Fn is also unknown. So instead, draw a FBD of the man-chair system, isolating them from the pulley. Then the normal force becomes internal to your system, and does not enter into the equation using Newton 1. Only the rope tensions and the weights act when looking at it this way. Solve for Ft.
 
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