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

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SUMMARY

The discussion centers on the mechanics of a painter with a mass of 70 kg and a chair of 10 kg, using a pulley system to accelerate upwards at 0.20 m/s². The correct force required for the painter to pull down on the rope is 400 N, not 800 N as initially thought. The lecturer clarifies that the tension in the rope acts on both segments, effectively doubling the upward force when the painter pulls down. This results in a total upward force of 800 N from the tension created by the 400 N pull, allowing for the desired acceleration.

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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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