Static Equilibrium and Torque

1. Nov 17, 2013

itachipower

wrong section

Last edited: Nov 17, 2013
2. Nov 17, 2013

itachipower

1. The problem statement, all variables and given/known data
I have been trying to solve this problem for a while now and I can't figure out what I am doing wrong...

Here is the problem:

In the figure below, a horizontal scaffold, of length 2.00 m and uniform mass 50.0 kg, is suspended from a building by two cables. The scaffold has dozens of paint cans stacked on it at various points. The total mass of the paint cans is 75.5 kg. The tension in the cable at the right is 780 N. How far horizontally from that cable is the center of mass of the system of paint cans?

2. Relevant equations

T = r x F

Sigma T = 0

3. The attempt at a solution

What I am doing is using the left cable as the rotational axis for Torque. So I get Torque due to the beam + torque due to the CoM of the paint cans = 780N

Therefore, (50)(9.8)(1m) + (75.5)(9.8)(x) = 780N

The answer is supposed to be .554 m. So I think I'm missing a force but I don't know what it is. Thanks for your help :)

3. Nov 17, 2013

Welcome to PF,Itachi (don't use sharingan on me) lol
The scaffold is in equillibrium.So,clockwise moment is equal to anticlockwise moment.You're 780N is wrong,it has to be the torque there.
780 x 2=1560Nm.
You are using the other cable as a reference point.So,you will get the distance from.Therefore,you have to subtract it from 2m.

4. Nov 17, 2013

Welcome to PF,Itachi (don't use sharingan on me) lol
The scaffold is in equillibrium.So,clockwise moment is equal to anticlockwise moment.Your 780N is wrong,it has to be the torque there.
780 x 2=1560Nm.
You are using the other cable as a reference point.So,you will get the distance from.Therefore,you have to subtract it from 2m.

5. Nov 17, 2013

itachipower

Thank you!!

6. Nov 17, 2013