Torque Equilibrium to find center of mass.

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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