A 30 kg hanging mass is supported by three cables and is in static equilibrium. The tension in the center of the vertical cable is 126 N. A free body diagram is given with the one cable making an angle of 15 deg and another cable making an an angle of 40 deg. Determine the tension in the other two cables.
I'm not sure.
The Attempt at a Solution
Tension of the first cable = mg/(sin theta1 + cos theta1 + tan theta2)
So t1=30kg*9.8m/s^2/(sin 15 + cos 15 + tan 40) = 142.45 N
Tension of the second cable = T1cos theta1/cos theta2
So t2=142.45Ncos15/cos40 = 179.62 N
Does this look correct?