How Far Can Professor Fich Move Before Exceeding the Rope's Tension Limit?

[bluejay1]

Homework Statement

3B. A uniform wooden sign of mass 4.0 kg hangs beside a building wall. The sign is 2.00 m
high and 4.00 m wide. It is supported by a hinge at P, that is midway up one
edge, and by a light rope that is attached exactly three-quarters of the distance across the
upper edge. The rope makes an angle of 20.0 degrees with the horizontal.

(b) Write the conditions for static equilibrium of the sign and solve for the horizontal (H) and vertical (V) components of the force at P, and solve for the rope tension (T)

(c) Later, Prof. Fich of mass 90.0 kg climbs out of a window just below the sign, reaches up and grabs onto the bottom portion of the sign and, while hanging from the sign, begins to move away from the building. If the rope can only support a maximum tension of 1500 N, how far can Prof. Fich move away from the building ?

Homework Equations

net torque=0
Fnety=0
Fnetx=0

The Attempt at a Solution

I have the force equations set up.
Tsin20+Fpv-4g=0
Tcos20=Fph
I am having trouble setting up the torque. For the moment arm, can you go through the sign? I can't get the moment arms and angles set up.
So far I have
Tsin38.4(sqroot10)=mg(2)
Is this right?
(The three lines I've drawn on the sign are what I think the 2 moment arms are, and also the vector for Fg)

 

[Redbelly98]

I am having trouble setting up the torque. For the moment arm, can you go through the sign? I can't get the moment arms and angles set up.
So far I have
Tsin38.4(sqroot10)=mg(2)
Is this right?
(The three lines I've drawn on the sign are what I think the 2 moment arms are, and also the vector for Fg)

While I can't view your attachment yet, I think there is enough information in the worded description to see what is being asked.

For the torques, you just need to choose a reference point from which to calculate torques due to the forces that are present. Hint: it's generally useful to choose a point where 1 or more of the forces are acting. That way, those forces will have zero torque and the equation will be simpler to work with.

EDIT: Okay, I can view your attachment now. Your torque equation looks right.