Physics
posted by Anonymous .
A man of mass 69.5 kg stands on a scaffold supported by a vertical rope at each end. The scaffold has a mass of 22.3 kg and is 2.4 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?

You have to write equations about arbritary points, but it is very convenient.
I will do one for you, about the right end.
69.5g*2/6*2.422.3g*1.2 + Tl*2.4 =0
You do the left equation.
Dont forget the sum of vertical forces..
Tl+Tr 22.3g69.5g=0 
Let T1 be the tension in the left rope and T2 be the tension in the right rope. A vertical force balance tells you that
T1 + T2 = (M1+M2)g = (69.5 + 22.3)* (9.8) N
You don't need that equation to solve for T2.
You can use the fact that the torque about the point where the left rope meets the scaffold is zero. The weight of the scaffold acts at the midpoint, which is 1.6 m from the left end. Therefore
M1*g*1.6 m = T2* 2.4 m
Solve for T2 (in Newtons) 
i still have a few questions
1) wouldn't the 1.2 be 1.6 instead?
2) and how come the first equation equals zero. isn't it somewhat of a rearrangement of the second equation, so wouldn't it equal Tr? 
using drwls method i got 454N and it was wrong

Whoops. I forgot the torque due to the scaffold's weight, on the left side of the equation, which is 22.3 kg*1.2 m * 9.8 m/s^2. That's why my equation gave you the wrong answer.

thanx :)
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