Question

The boom in the figure below weighs 2600 N and is attached to a frictionless pivot at its lower end. It is not uniform; the distance of its center of gravity from thepivot is 35.0 % of its length.

http://session.masteringphysics.com/problemAsset/1263413/2/yg.10.60.jpg

Part A: Find the tension in the guy wire.
Part B: Find the horizontal component of the force exerted on the boom at its lower end.
Part C: Find the vertical component of the force exerted on the boom at its lower end.

Answer 1

2) Let L be the length of the boom.
Then net torque on the boom is:
2600N*L/2*cos50 + 6000N*L*cos50 - T*L*sin50 = 0
We get T = 6125N.

3) The horizontal net force is: Fx - T = 0, Fx = T = 6125N.
The vertical net force is: Fy - 2600N - 6000N = 0, Fy = 8600N.