With what magnitude must the man pull on the rope

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Homework Help Overview

The problem involves a man pulling on a rope connected to a pulley, with the goal of determining the magnitude of the force required for him to rise with a constant velocity or with a specified upward acceleration. The combined weight of the man and chair is given as 95 kilograms.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the importance of drawing a Free Body Diagram (FBD) to analyze the forces acting on the system. Questions arise regarding the frictionless nature of the pulley and the configuration of the rope. There is exploration of how many ropes are involved and how they contribute to the net tension acting on the man and chair.

Discussion Status

The discussion is actively exploring the forces involved, with participants revising their understanding of the Free Body Diagram based on clarifications about the pulley and the setup. There is an acknowledgment of multiple tensions acting on the system, and participants are questioning the signs of the forces involved.

Contextual Notes

Participants are working under the assumption that the pulley is frictionless and are clarifying the setup of the rope and its connection to the chair. There is a focus on understanding the net forces and their directions without reaching a final conclusion.

passphysicsC
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Homework Statement


A man is strapped to a chair which is connected to a pulley (combined weight of chair and person = 95 kilograms). With what magnitude must the man pull on the rope if he is to rise

a) with a constant velocity

b) with an upwards velocity of 1.3 meters per seconds squared.


tension = mass X acceleration
 
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first draw a FBD labelling forces, that will tell you much about the problem
 
this is what I drew as a Free Body Diagram with the information given:
Tension
|
|
|
person/chair
|
|
|
Weight
 
Last edited:
Is the pully frictionless?
 
passphysicsC said:
A man is strapped to a chair which is connected to a pulley (combined weight of chair and person = 95 kilograms). With what magnitude must the man pull on the rope if he is to rise
Is the man pulling himself and his chair up and does the rope loop around a pulley attached to the ceiling? (A diagram will help.)

If so, you'll need to revise your free body diagram.
 
oh sorry for not being more specific :(. the pulley is frictionless, and the man is pulling himself and the chair up via the rope (connected to the chair) that goes through the pulley connected to the ceiling and then falls back down again, allowing the man to grab it and pull himself and the chair up
 
Good. Now revise your free body diagram accordingly. Hint: How many times does the rope pull up on the 'man + chair'?
 
well in the positive y-direction (up) there is tension and in the negative y- direction there is weight and the force which is caused by the man pulling on the rope in order to life himself up. so would it be the same free body diagram, but with a negative force added to the bottom?
 
Doc Al said:
Hint: How many times does the rope pull up on the 'man + chair'?
I'll rephrase my question. Draw an imaginary box around the 'man + chair' contraption. How many ropes pass through the boundary of that box?
 
  • #10
ohhhh okayyy

then two ropes, one rope connected to the chair, and one rope which the man uses to pull himself and the chair up...would there be two different tensions then?
 
  • #11
passphysicsC said:
then two ropes, one rope connected to the chair, and one rope which the man uses to pull himself and the chair up...would there be two different tensions then?
Good! There would be two separate tensions acting. (Not necessarily different though.) One end of the rope pulls up on his hands; the other pulls up on his chair. They both count towards the net force on the 'man + chair'.

(Compare this situation to one where the rope attaches to the ceiling without a pulley and thus is not doubled over.)
 
  • #12
alright, so then the Net Tension = ((mass x acceleration) + weight) + (mass x acceleration)?
 
  • #13
passphysicsC said:
alright, so then the Net Tension = ((mass x acceleration) + weight) + (mass x acceleration)?
I find that very confusing.

Instead, just write ΣF = ma. Call the tension in the rope 'T'. What's the net force (ΣF)?
 
  • #14
Σf = (t + w) + (t)
 
  • #15
sorry, for some reason when I posted it, none of the letter remained capitalized
 
  • #16
passphysicsC said:
Σf = (t + w) + (t)
What direction do these force have? (Do they all have the same sign?)
 
  • #17
well the first tension is positive since it is holding the chair and man up. the weight (w) is negative since it the force working against the tension. I believe the last tension would be negative since it is caused by the man pulling on the rope to pull himself and the chair up.
 
  • #18
passphysicsC said:
well the first tension is positive since it is holding the chair and man up.
OK. The rope pulls up on the chair, thus is a positive force.
the weight (w) is negative since it the force working against the tension.
The weight acts down, so is negative.
I believe the last tension would be negative since it is caused by the man pulling on the rope to pull himself and the chair up.
All we care about are forces on the man/chair. The rope pulls up on the man and up on the chair. Both tension forces are positive.
 

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