A 100 kg scaffold is 7 m long. It is hanging with two wires, one from each end. A 600 kg box sits 1.8 m from the left end. What is the tension in the left hand side wire?(g = 9.8 m/s2)answer in newtons
This question can be solved by first drawing a free-body diagram and then summing the moments (torques) around the right end of the scaffold. We'll designate the tension in the left hand wire with T. So after making our free-body diagram we find the sum of the moments around the right end of the scaffold (where each moment = force x distance from point we are taking it about), setting it to zero because the system experiences no angular motion 0 = 7m * -T + 600kg * 9.8m/s^2*(7m - 1.8m) + 100kg * 9.8m/s^2 * 3.5m (Notice that T is negative because it would produce a clockwise rotation about the right end of the scaffold, and the weights are positive because their rotation would be counter-clockwise) Now we just solve for T, which comes out to be T = 4858N
