Torque Equilibrium to find center of mass.

[MedChill]

Homework Statement

1. A non-uniform beam with a mass of 85kg has a length of 6.2m and is attached to a wall and supported at an angle of 28 degrees from the wall by a horizontal rope which is attached to the wall above the beam . The tension in the rope is found to be 310N. How far from the pivot point is the center of mass of the beam?

Homework Equations

T= d x F
Trigonometry (sin=o/h, cos=a/h, tan=o/a)
Fg= m x g

The Attempt at a Solution

Well I am pretty lost in the question but I gave it a shot anyways.
Fg= 85 x 9.8= 833 N
Fg(perpendicular)= 833sin28= 391.1 N
Ft= 310 N
Ft(perpendicular)= 310cos28= 273.7 N
∑= 391.1-273.7= 117.4 N
T= 117.4 x 6.2= 727.88

This is where I think I messed up but don't know where else to go from here.

 

[Simon Bridge]

Welcome to PF;
The trick with this sort of question is to sketch it out and draw in the forces, and label points.

So call the pivot O, the other end A and the point the rope joins the wall is B ... pick an arbitrary point off-center on the beam to represent the center of mass and call it C.

OC=x (which you want to find).
I don't see any reference to an unknown distance in your analysis.
Sketch in the forces at O, A, B, and C, don't forget the directions.

Now you should be in a position to write the expression for the total torque about O, which will include the unknown x. Solve for x.

Note - torque is usually given as a cross product as in T = d x F, but the weight is just a scalar product Fg=mg.
Formally: $\vec \tau = \vec r \times \vec F$ and $\vec F\!_g=m\vec g$

 

[paisiello2]

Maybe a free body diagram would help us follow what you are trying to do.

But you got the right equations, just need to execute them properly.