Tension & Torque Homework: 60.8 kg Man on 2.7m Scaffold

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SUMMARY

The problem involves calculating the tension in the right rope supporting a scaffold with a man standing on it. The man has a mass of 60.8 kg, and the scaffold has a mass of 17.2 kg, with a total length of 2.7 m. The man stands 0.45 m (one sixth of the scaffold's length) from the right end. The tension in the right rope is calculated using torque equilibrium, resulting in a value of 481.5 N after correcting initial miscalculations.

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Homework Statement



A man of mass 60.8 kg stands on a scaffold supported by a vertical rope at each end. The scaffold has a mass of 17.2 kg and is 2.7 m long. Suppose the man stands to the right from the middle of the scaffold that is a distance one sixth of the length of the scaffold. What is the tension in the right rope?

Homework Equations



Torque = F r sin(angle to horizontal)
F = m g

The Attempt at a Solution



clockwise torque = anticlockwise torque
(2.7)(17.2) + (60.8)(9.8)(1.8) = F (2.7)
...
F = 414.4 N

I even tried it without useing the mass of the board and then placing the force caused by the mass (Fg) at the center, but all these thing didn't work.
We are also asked for the tension in the left rope after the tension in the right rope. But I can't figure out the right one first. Thanks in advance...
 
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I realized my mistake, got it now, thanks anywas.. sorry I posted too soon.

clockwise torque = anticlockwise torque
(1.35)(17.2)(9.8) + (1.8)(60.8)(9.8) = F (2.7)
...
F = 481.5
 

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