Torque with sign attached to a rod

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SUMMARY

The discussion centers on calculating the tension in a cable supporting a 54.8 kg uniform square sign attached to a 3.00 m rod. The sign's dimensions are 2.00 m on each side, and it is anchored to a wall 4.00 m above the rod's fixed point. The participant used the torque equation Ʃτ=Trsin(θ1) - mg(r)(sin(θ2))=0 and calculated the tension T to be 669 N, but identified an error in the mass of the sign, which should be 54.8 kg instead of 54.6 kg. The correct center of mass distance from the wall is also a point of confusion.

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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.
6Apo0YU.png


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?
 
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Painguy said:

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 ?
 

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Painguy said:
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)?
 

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