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

[ZanyCat]

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 F_G 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)

 

[mfb]

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).

 

[ZanyCat]

Was completely ignoring the contact point between the rope and his hand, thanks a lot :)

 

[Doc Al]

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:

 

[Nickie]

I would approach this problem by first drawing a free body diagram of the situation. This will help us visualize the forces acting on the painter and chair system.

From the given information, we know that the total mass of the painter and chair is 80kg (70kg + 10kg). We also know that the acceleration of the system is 0.20 m/s^2. Using Newton's second law (F=ma), we can determine that the net upward force on the system must be 16N (80kg x 0.20 m/s^2).

Now, let's consider the forces acting on the system. The only external force acting on the system is the tension in the rope. This tension is acting upwards on both parts of the rope (Part A and Part B). According to Newton's third law, for every action, there is an equal and opposite reaction. This means that the tension in Part A of the rope will be equal and opposite to the tension in Part B of the rope.

Therefore, the total tension in the rope must be 16N. This means that the painter must pull down on the rope with a force of 16N in order to create a net upward force of 16N and accelerate at 0.20 m/s^2. This is equivalent to the force of 800N that you calculated for the tension in the rope.

Your lecturer's answer of 400N is incorrect because it does not take into account the equal and opposite tension in Part A of the rope. It is important to consider all the forces acting on a system and their interactions in order to accurately solve a physics problem.

In conclusion, the correct answer to the question is that the painter must pull down on the rope with a force of 16N (or 800N of tension in the rope) in order to accelerate upwards at 0.20 m/s^2. This aligns with your initial thinking and calculation.