Why Does the Painter Only Need to Pull Down with 400N to Accelerate Upwards?

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Homework Help Overview

The problem involves a painter with a mass of 70kg sitting on a 10kg chair, connected to a pulley system. The question centers on the force the painter must exert downwards on the rope to achieve an upward acceleration of 0.20 m/s².

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between the forces acting on the painter and the chair, particularly the tension in the rope and how it relates to the forces required for upward acceleration.

Discussion Status

There are differing views on the correct force needed to pull down on the rope. Some participants suggest that the upward force from the tension in the rope is doubled due to the pulley setup, while others question the assumptions made about the forces involved.

Contextual Notes

Participants are exploring the implications of the pulley system and the distribution of forces, with some noting the complexity of the setup and the need for clearer visual aids to support their explanations.

ZanyCat
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This question seems incredibly straightforward, but my lecturer had a different answer to what I ended up with.

Essentially there's a painter with a mass of 70kg sitting on a chair of 10kg that is attached to a pulley on the roof via a rope (the pulley and rope are ideal,massless and frictionless). The painter is holding on to the rope that is hanging down from the other side of the pulley (I'll call this Part B of the rope, and the section attached to the chair -side of the pulley as Part A).

With what force must he pull down on the rope to accelerate upwards at 0.20 m/s^2?The net upward force acting on him/the chair must be 16N to accelerate at this rate. The FG is 80g (784) N downward, thus he needs a force of 800N from the rope pulling upwards.
By my thinking, if he pulls down with 800N on Part B of the rope, this will create a tension of 800N upwards on Part B, which will in turn yield a tension of 800N upwards on Part A of the rope.

My lecturer, however, is saying that he needs to pull downwards with only 400N. He was saying it had something to do with the fact that there are two forces acting upwards, namely the tension in Part B and Part A separately.

Which of us is going wrong, and where?
(sorry for the overly long explanation, a diagram would be a lot easier but my Paint skills leave a lot to be desired)
 
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By my thinking, if he pulls down with 800N on Part B of the rope, this will create a tension of 800N upwards on Part B, which will in turn yield a tension of 800N upwards on Part A of the rope.
800N of tension in the rope gives 800N upwards force both on the chair and him holding the rope, for a total force of 1600N upwards.

400N is the correct answer.

If you, standing next to the setup, would have to pull, you would need 800N (but have just half the rope velocity as you would not move upwards).
 
Was completely ignoring the contact point between the rope and his hand, thanks a lot :)
 
ZanyCat said:
Which of us is going wrong, and where?
I'm afraid it's you who are mistaken. If he pulls with a force of 800 N, that will make the tension in the rope 800 N. And since the rope pulls up on the person+chair system twice, that would make the total upward force equal to 1600 N, which is not what you want.
(sorry for the overly long explanation, a diagram would be a lot easier but my Paint skills leave a lot to be desired)
Here's a picture; it's called a Bosun's chair:
621ea107-a378-4787-91b3-c09daa95d090.gif
 

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