**1. The problem statement, all variables and given/known data**

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?

**2. Relevant equations**

T= d x F

Trigonometry (sin=o/h, cos=a/h, tan=o/a)

Fg= m x g

**3. 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.

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