A 33kg traffic light hangs from a vertical beam by two wires. The wire on the left side of the light has an angle of 53 degrees to the beam and the right side wire has an angle of 37 degrees with the beam. Calculate the tension in each of the wires. Would the tension just be 33kg(sin53) + 33kg(cosin53) for the left side, and 33kg(sin37) + 33kg(cosine37) for the right side?
physics - drwls, Saturday, April 2, 2011 at 10:59am
Call the left side tension T1 and the right side tension T2. Vertical and horizontal equilibrium equations are:
Mg = T1 cos53 + T2 cos37
0 = T1 sin53 - T2 sin37
Mg = T1 cos 53 + T1*cos37*(sin53/sin37)
T1 = Mg/(0.6918 + 1.0598)
= 0.5709 M*g
You set up the equations or did the algebra incorrectly, and you also omitted the g factor that you need to get the tension in Newtons.
physics - Mark, Saturday, April 2, 2011 at 12:56pm
Thanks , I knew this just didn't seem right.