Boatswans Chair pulley Question

  • Thread starter Ryansf98
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  • #1
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Homework Statement


The question has been screenshotted, along with the given diagram. Basic stuff I know, but I can't seem to get my head around pulley systems. T
Screenshot (44).png


Screenshot (43).png

Homework Equations


F(net) = m ⋅ a(system)

The Attempt at a Solution


I understand the rope to be considered inextensible. Also, I understand that the tension must equal the weight of the man & chair, but beyond that I don't know how to break it down.

I labelled a reference frame with x being positive down.
 

Answers and Replies

  • #2
gneill
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Hi Ryansf98,

Welcome to Physics Forums!

I understand that the tension must equal the weight of the man & chair
What leads you to that conclusion?

Have you made a free body diagram of the man&chair?
 
  • #3
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Hi Ryansf98,

Welcome to Physics Forums!


What leads you to that conclusion?

Have you made a free body diagram of the man&chair?

I just read that over and now realise it doesn't make sense, as equal tension & weight would mean an acceleration of 0m/s^2 wouldn't it? What I meant was that tension must be equal throughout the rope.
 
  • #4
gneill
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I just read that over and now realise it doesn't make sense, as equal tension & weight would mean an acceleration of 0m/s^2 wouldn't it? What I meant was that tension must be equal throughout the rope.
Okay. Draw the resulting free body diagram for the man&chair.
 
  • #5
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Hi Ryansf98,

Welcome to Physics Forums!

Thank you by the way! I've used it for sometime looking at existing threads, but I couldn't find an existing problem like this one so thought it'd be a good time to start my own account!
 
  • #6
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Okay. Draw the resulting free body diagram for the man&chair.

I done that, labelling F(down) as 200N + mg, => F(down total) = 887N.

Regarding the tension, I labelled T moving up from the man to the smaller pulley. Would that then split into 1/2T either way up the pulley, or would it remain T?

I also labelled T pointing up on the rope which the man pulls down on.
 
  • #7
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Okay. Draw the resulting free body diagram for the man&chair.

I realise I jumped a gun a little there as you only suggested a free body diagram for the man and chair. For this, I labelled mg down and T up towards the smaller pulley. Would I in fact insert 200 N here also?
 
  • #8
gneill
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I realise I jumped a gun a little there as you only suggested a free body diagram for the man and chair. For this, I labelled mg down and T up towards the smaller pulley. Would I in fact insert 200 N here also?
How many ropes attach to the man&chair? If each has a tension T, what's the net force upwards?
 
  • #9
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How many ropes attach to the man&chair? If each has a tension T, what's the net force upwards?

One rope attached directly to the man and chair, therefore T up? Also, would 200N be T, due to Newtons 3rd Law?

Therefore, Net Force Up = 200N?
 
  • #10
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How many ropes attach to the man&chair? If each has a tension T, what's the net force upwards?

In fact, there are 3 ropes above the man and pulley, therefore F(up) = 3T.

T = 200N as the man exerts this force down the rope and so the rope exerts this force up on him.

That means that F(net of the system) = mg - 3T, => 686.7 - (3*200) = 86.7N downwards.

F = ma, a= 86.7/70, => a = 1.24 m/s^2?
 
  • #11
gneill
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One rope attached directly to the man and chair, therefore T up? Also, would 200N be T, due to Newtons 3rd Law?

Therefore, Net Force Up = 200N?
No, there are two ropes connected to the man and chair: One he holds directly, one connected to the chair.

The one he holds has a tension T = 200 N. What's the tension in the other rope? (consider a FBD of the small pulley).
 
  • #12
gneill
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In fact, there are 3 ropes above the man and pulley, therefore F(up) = 3T.

T = 200N as the man exerts this force down the rope and so the rope exerts this force up on him.

That means that F(net of the system) = mg - 3T, => 686.7 - (3*200) = 86.7N downwards.

F = ma, a= 86.7/70, => a = 1.24 m/2^2?
Yes! That's much better!
 
  • #13
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Yes! That's much better!

So generally no matter what, each rope has a tension T throughout the system, even if the rope reaches a pulley and looks to go into 2 separate ropes?

Thanks for the help!!
 
  • #14
gneill
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So generally no matter what, each rope has a tension T throughout the system, even if the rope reaches a pulley and looks to go into 2 separate ropes?
If the pulleys are massless and frictionless, then continuous ropes have the same tension. When separate ropes meet at a pulley, their tensions sum (vector sum).
 
  • #15
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If the pulleys are massless and frictionless, then continuous ropes have the same tension. When separate ropes meet at a pulley, their tensions sum (vector sum).

Makes sense. Again thanks for your help!
 

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