Atwood Machine Related Question

[Jimmy25]

Homework Statement

A 59-kg window-washer stands on a 17-kg platform. The platform is fixed to a rope that passes over a pulley attached to the ceiling, which allows the window-washer to raise himself and the platform. (a) To accelerate himself and the platform at a rate of 0.84 m/s², with what force must he pull on the rope? (b) When his velocity reaches 2.3 m/s, he pulls so he and the platform go up at a constant speed. What force is he exerting on the rope?

Homework Equations

The Attempt at a Solution

I thought since there was only one pulley the ratio of the force of the window washer pulling on the rope to the net force due to tension on him and the platform would be 1:1. Evidently this is not the case. Is the force applied by the washer on the rope halfed by the pulley. If so why is this the case?

 

[G01]

Think of it this way:

The tension in the rope pulls upward on the platform and on the hands of the man pulling on the rope. So, the tension pulls upward at two points, meaning you only need half the force you would need if the rope only pulled at one point.

 

[Jimmy25]

So what would my FBD look like in this case?

I have T₁ pulling up on the platform and F_g pulling down on the platform.

Now, what I had before was F_a pulling down on the other side of the pulley and T₂ acting upwards canceling with T₁. But this is not correct.

Could you explain how the forces would look on the other side of the pulley?

 

[G01]

In your free body diagram, you have the rope pulling upward at two points, on the platform and the man.

This means that in your free body diagram, you will have two forces pulling upward, T_1 and T_2 with T_1=T_2.