# Static Equilibrium Weight Problem

1. Nov 16, 2009

### r_swayze

The mass of the sign shown is 28.5 kg. Find the weight supported by (a) the left support and (b) the right support.

How do I set this problem up properly?

Here is my set up even though I know its wrong.

m = 28.5kg, mg = 279.3N

Fl + Fr = mg

Fl = mg - Fr

t = torque, r = radius

tl + tr = mg

Fl*rl + Fr*rr = mg

(mg - Fr)*rl + Fr*rr = mg

mg*rl - Fr*rl + Fr*rr = mg

then I plug in the radius of the right and left posts using 1.2m for the left post and .3m for the right post, and 279.3 for mg and solve.

How do I properly set this problem up?

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2. Nov 16, 2009

### rl.bhat

Weight of the sign board act at the center of the board in the downward direction.
When you write Fl + Fr = mg, Fl and Fr are in the upward direction.
To find the torque, select a reference point, say left end of the sign board.
Then find the torques due to Fl, Fr and mg. Which one is CW and which one is CCW?
Equate them to find Fr. Then proceed further.

3. Nov 16, 2009

### r_swayze

ah ok, I see it now. I didnt account for the torque of the sign itself.

thank you!