Solve Mass Suspended from Uniform Boom

[NikkiNik]

Homework Statement

A mass M=171 kg is suspended from the end of a uniform boom as shown. The boom (mass=83.0 kg, length=2.40 m) is at an angle θ=69.0 deg from the vertical, and is supported at its mid-point by a horizontal cable and by a pivot at its base. Calculate the tension in the horizontal cable.

http://i32.photobucket.com/albums/d2/NikkiNik88/staticsboom.gif

Homework Equations

T=rsin(theta)

The Attempt at a Solution

Theta = 69 deg, alpha=21 deg

mg(L/2)cos(theta) + Mg(L)cos(theta) + T(L/2)sin(alpha) = 0

When I solved for T I got 4169.25N but that is incorrect please help

 

[ideasrule]

Why are you using cosine for mg(L/2) and sine for the others, instead of the other way round? Think more carefully about which functions you should use.

 

[NikkiNik]

Would I be using sine for the masses because they are vertical and cosine for the tension because it's horizontal? If that's the case, is the rest of my work correct if I switch sine and cosine?

 

[ideasrule]

Would I be using sine for the masses because they are vertical and cosine for the tension because it's horizontal? If that's the case, is the rest of my work correct if I switch sine and cosine?

Yes! So:

mg(L/2)sin(theta) + Mg(L)sin(theta) + T(L/2)cos(theta) = 0

The answer you get will be negative, but that's only because the tension acts to counteract the torque applied by gravity.