91Ó°ÊÓ

At equilibrium, the net torque is zero. For this problem, the clockwise torque due to the weight of the beam is equal to the torque due to the tension in the cable.

Let m be the mass of the beam, T be the tension in the cable, and L be the length of the beam.

You can assume that the beam is uniform, and the weight of the beam is acting at the middle of the beam.

Also, assume that the cable makes \(\theta \) angle with the beam.

Now, at equilibrium, the net torque is zero. Then,

\(\begin{aligned}{c}T\sin \theta \times L = mg \times \frac{L}{2}\\\sin \theta = \frac{{\frac{{mg}}{2}}}{T}\end{aligned}\).

Now, you know that

\(\begin{aligned}{c}\sin \theta \le 1\\\frac{{\frac{{mg}}{2}}}{T} \le 1\\T \ge \frac{{mg}}{2}\end{aligned}\).

Hence, the tension in the cable must be at least half the weight of the beam.