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Question:

A 115 kg scaffold is 7 m long. It is hanging with two wires, one from each end?

A 520 kg box sits 1.4 m from the left end. What is the tension in the right hand side wire? (g = 9.8 m/s2) ____________ Newtons

Answer:

Because the scaffold is not moving, we can assume that the net Torque acting on the scaffold equals zero. Net Torque = 0 torque = Force x distance (d) Use the left end of the scaffold as the center point. This will eliminate the Torque caused by the left wire (d = 0, so torque = 0) There are now three forces creating torque on the scaffold. There is the weight (m x g) of the scaffold, at 3.5m from the left end. m1 x g x 3.5m There is the weight of the crate, located 1.4m from the left end: m2 x g x 1.4m These two torques are countered by the torque caused by the Tension (T) of the right wire, located 7m from the left side. T x 7m The sum of these torques is zero. (T x 7m) - (m1 x g x 3.5m) - (m2 x g x 1.4m) = zero solve for T T = ((m1 x g x 3.5m) + (m2 x g x 1.4m)) / 7m T = ((115kg x 9.8m/s^2 x 3.5m) + (520kg x 9.8m/s^2 x 1.4m)) = 1.58 x 10^3 N Hope this helps.

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