Torque with sign attached to a rod

[Painguy]

Homework Statement

a 54.8 kg uniform square sign, 2.00 m on a side, is hung from a 3.00 m rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.00 m above the point where the rod is fixed to the wall.

Homework Equations

The Attempt at a Solution

Ʃτ=Trsin(θ1) -mg(r)(sin(θ2))=0
cable length=sqrt(4^2 +3^2)=5
r=3
arctan(4/3)=51.3*=θ1
180-(51.3+90)=36.86=θ2
T(3)(sin(53.1)-(54.6)(9.8)(3)sin(90)=0
T=1605/2.389=669N

That's wrong...what am I forgetting to do?

 

[SammyS]

Homework Statement

a 54.8 kg uniform square sign, 2.00 m on a side, is hung from a 3.00 m rod of negligible mass. A cable is attached to the end of the rod and to a point on the wall 4.00 m above the point where the rod is fixed to the wall.
[ IMG]http://i.imgur.com/6Apo0YU.png[/PLAIN]

Homework Equations

The Attempt at a Solution

Ʃτ=Trsin(θ1) -mg(r)(sin(θ2))=0
cable length=sqrt(4^2 +3^2)=5
r=3
arctan(4/3)=51.3*=θ1
180-(51.3+90)=36.86=θ2
T(3)(sin(53.1)-(54.6)(9.8)(3)sin(90)=0
T=1605/2.389=669N

That's wrong...what am I forgetting to do?

What is it you're trying to find ?

 

[haruspex]

T(3)(sin(53.1)-(54.6)(9.8)(3)sin(90)=0

How far from the wall is the centre of mass of the sign (which weighs 54.8kg, not 54.6)?