Question

An 8.00 m long beam with a mass of 8.50 kg is suspended horizontally from a post, as shown. The left end of the beam is attached to a frictionless hinge on the vertical post that allows the beam to rotate up and down freely. The right end of the beam is supported by a wire that runs to the top of the post. The wire makes a 50° angle with the vertical post. The beam's shape is symmetric and it is made of a homogeneous material so that its center of mass is located at its geometric center. What tension in the wire is required to support the beam horizontally?

Answer 1

The torque equilibrium condition dives

$$mg\frac{L}{2}=T\cos 50^{\circ} L$$

So, the tension in a wire

$$T=\frac{mg}{2\cos 50^{\circ}}=\frac{8.50\times 9.81}{2\times \cos 50^{\circ}}=64.9\;\rm{N}$$