Bosun's chair problem - force on the seat?

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The discussion revolves around the mechanics of a boy using a bosun's chair, specifically the forces at play when he pulls down on the rope with a force T to lift himself. Initially, it was assumed that the force on the chair equals the boy's weight, but this was incorrect. The key insight is that when the boy pulls down with force T, he effectively shares the load with the chair, meaning the chair's upward force is half of his weight if the chair is considered weightless. If the boy is accelerating upwards, he must pull with a force greater than half his weight to account for the additional upward force needed. Understanding these dynamics clarifies the relationship between the forces acting on both the boy and the chair.
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A friend and I have been having some difficulties with a problem involving a bosun's chair. A boy lifts himself up by pulling down on the rope with a force T, which lifts him up the ground. The problem is asking what would the force on the chair be as he is lifting. Our initial thought would be that it is the same as his weight. He is pulling down with T, which at the same time pulls himself up with T, and also the chair with T. Since the same upward force is added to his weight and also to the force of the chair against him, we assumed the force on the chair is equal to his weight. However, we submitted this online and it said it was wrong. Any ideas?
 
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It might be easier to understand if you think of person + chair as a single object. At how many points does the rope pull up on that object? In terms of rope tension, what's the upward force on the object? And how does the force with which the man pulls on his end of the rope relate to the rope tension?
 
Stinky said:
A friend and I have been having some difficulties with a problem involving a bosun's chair. A boy lifts himself up by pulling down on the rope with a force T, which lifts him up the ground. The problem is asking what would the force on the chair be as he is lifting. Our initial thought would be that it is the same as his weight. He is pulling down with T, which at the same time pulls himself up with T, and also the chair with T. Since the same upward force is added to his weight and also to the force of the chair against him, we assumed the force on the chair is equal to his weight. However, we submitted this online and it said it was wrong. Any ideas?

your question is not clear enough. Do you mean the net force on the chair? Do you mean the normal force on the chair? Are you neglecting the weight of the chair? Is the boy moving up at constant speed?
 
Stinky said:
A friend and I have been having some difficulties with a problem involving a bosun's chair. A boy lifts himself up by pulling down on the rope with a force T, which lifts him up the ground. The problem is asking what would the force on the chair be as he is lifting. Our initial thought would be that it is the same as his weight. He is pulling down with T, which at the same time pulls himself up with T, and also the chair with T. Since the same upward force is added to his weight and also to the force of the chair against him, we assumed the force on the chair is equal to his weight. However, we submitted this online and it said it was wrong. Any ideas?

I am going to assume that you are looking for the nromal force on the chair and that the speed is constant and the weight of the chair is negligible.

A FBD of the chair will then contain two forces: the normal of the boy on the chair and the tension upward.

The FBD of the boy shows three forces: gravity down, a normal force up and a tension up (he pulls down on the rope so the rope pulls up on him).

If you now combine the two as a single object, what FBD do you get? Then the answer will be obvious.
 
Sorry, I will be clearer. He is accelerating upward at a constant rate. I am trying to figure out the force that the boy exerts onto the chair as he and the chair are accelerating upwards.
 
nrqed gave you a systematic approach to follow that can easily be modified to incorporate acceleration: Draw a free body diagram and apply Newton's 2nd law.

But I'd like you to first understand the key idea of the bosun's chair. Forget acceleration for a moment and assume the chair itself weighs nothing. How hard does he have to pull to support his weight? (Read what I wrote above.)
 
He will have to pull with a force equal to half of his weight to support his weight, correct?
 
Stinky said:
He will have to pull with a force equal to half of his weight to support his weight, correct?
Exactly. So what provides the other force (half of his weight) needed to support him?
 
He provides half of it by pulling down on the rope, and therefore upwards on himself, and the chair pushes up with the other half.
 
  • #10
Stinky said:
He provides half of it by pulling down on the rope, and therefore upwards on himself, and the chair pushes up with the other half.
Exactly. Whatever total upward force is needed on the man, half is provided by the rope and half by the normal force of the seat. So, what upward force is needed if he is accelerating?
 
  • #11
A total force greater than his weight, or he must pull with a force greater than half his weight.
 
  • #12
Stinky said:
A total force greater than his weight, or he must pull with a force greater than half his weight.
Right. Now answer the question in terms of the information given in the problem statement (from your first post). It says: He pulls down on the rope with a force T.
 
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