Question

A 40.0 kg uniform beam is attached to a wall with a hinge while its far end is supported by a cable such that the beam is horizontal.

If the angle between the beam and the cable is θ = 59.0° what is the tension in the cable?

Please show all work, and answer in sig-figures.

Thank you for all your help.

Answer 1

This is a torque question. The sum of all torque = 0. Torque is rXF. or rFsinθ

The torque DOWN or clockwise is die to the force of gravity, mg, but it is located at the center of the plank. We dont know the length of the plank, so call it L. The force is located at 1/2 L. So

torque DOWN = rXF yields (1/2)L(mg)

The torque UP is the perpendicular component of the tension, or sin 59, so

rXF yields LTsin(59)

since it is in equilibrium, the torques are equal

LTsin(59) = (1/2)L(mg)

simplify to T = mg/(2sin(59))

so T = (40 kg)(9.8) / (2sin(59)) = 228.66 N