# Parallel Forces Problem

1. Nov 8, 2004

### DethRose

Hey...not sure how to solve this problem...i tried to do it like a normal one eg)2 painters standing on each end but have no idea how to account for the extra sign underneath the original...i cant use equilibrium cause i have 2 unknowns

anyways here is the question

A sign is 3.05 m long, and weighs 360 N. It is made from material which has uniform consistancy. A weight of 60 N hangs from the sign 1m from the end. Calculate the tension in each support rod. a=1.25m, b=0.80m, c=1m, d=1m.

a,b,and c are the lengths from left to right of the sign, and d is the length of the weight from the right side of the sign.

Ive been working on this question all day and cannot figure it out...help would be greatly appreciated

thanks

2. Nov 9, 2004

### Leong

Moment/Torque in static equilibrium

Moment counter clock-wise : Positive; Moment clock-wise : Negative
Moment at point b,
$$T_{c}(1-0.8)+T_{a}(1.25-0.8)+T_{d}(1.525+0.525-0.8)-360(1.525-0.8)-60(1.525+0.525-0.8)=0$$
Moment at point c,
$$T_{a}(1.25-1)+T_{d}(1.525+0.525-1)-T_{b}(1-0.8)-360(1.525-1)-60(2.05-1)=0$$
Moment at point a,
$$T_{d}(2.05-1.25)+T_{b}(1.25-0.8)-T_{c}(1.25-1)-360(1.525-1.25)-60(2.05-1.25)=0$$
Moment at point d,
$$360(0.525)-T_{b}(3.05-1-0.8)-T_{c}(3.05-1-1)-T_{a}(3.05-1-1.25)=0$$
I assume that there are 4 rods at point a,b,c and d.