Pulley Free Body Diagram and rope tension

[armoredhalo00]

Homework Statement

The figure above shows an 80 kilogram person standing on a 20 kilogram platform suspended by a rope passing over a stationary pulley that is free to rotate. The other end of the rope is held by the person. The masses of the rope and pulley are negligible. You may use g = 10 m/ s2. Assume that friction is negligible, and the parts of the rope shown remain vertical.
a. If the platform and the person are at rest, what is the tension in the rope?



My problem is that I'm not identifying the direction of the forces in the free body diagram.

On the left side of the pulley i have this: The Mg of the platform and the person doing down and the Tension of the rope going up. On the other side of the pulley I have the force of the person holding the rope down and the Tension of the rope up. (This is where I feel I'm wrong I just need an explanation if I am)

As result I get the T=1000N.

Homework Equations

The Attempt at a Solution

I know the answer is T= 500 N because the equation is supposed to be 2T=1000N

 

[PhanthomJay]

You are not drawing your FBD's correctly. In a FBD, you isolate all or part of the system, and note the forces acting on that isolated system. Forces that may be internal to the entire system become external to the isolated system. Try drawing a FBD that isolates the person/platform from the pulley, a diagram which cuts through both ropes and encircles the man and platform. Identify all forces acting, and use Newton 1.

 

[armoredhalo00]

Is this right Phantom?

I have on the left side of the pulley the mg of the platform and person countered by the Tension up.

On the right side i have a force down. So i suppose that that Force down which is given by the person is also Tension.

Thus 2T=1000N

 

[phever]

The forces in their respective components should be balanced out if the system of particles is at equilibrium.

 

[PhanthomJay]

Is this right Phantom?

I have on the left side of the pulley the mg of the platform and person countered by the Tension up.

On the right side i have a force down. So i suppose that that Force down which is given by the person is also Tension.

Thus 2T=1000N

Are you drawing a FBD of the pulley? Your wording is not clear to me. When you draw a FBD, draw a circle around the part of the system you want to isolate. Wherever that circle 'cuts' through a contact point, there is a force there. Also, the weight force, if any, is always external to that isolated system. So if you draw a FBD of the pulley, you have 2 equal tension forces acting down, and the tension in the ceiling cord acting up. That tells you that the tension in the ceiling cord is twice the tension in each of the lower cords (per Newton 1). Now by drawing a FBD that encircles the person and platform and upper ceiling cord, you can solve for the upper ceiling cord tension and then solve for the lower rope tension. But it is easier to draw a FBD that encircles the man and platform and 2 cords to solve for the tension in the cords supporting the man and platform.

 

[SammyS]

...

My problem is that I'm not identifying the direction of the forces in the free body diagram.

Can you scan your FBD and post it? That would be a big help.