Question:

A horizontal beam is attached to a wall. The length and weight are 10.0 m and 200 N, respectively. The far end is supported by a wire and the angle between the beam and wire is 60\(^o\). A person who has 500-N weight stands 2.00 m from the wall. What is the tension in the wire?

Answer:

This system is at equilibrium, so the sum of all of the torques must be zero.

\[  \sum \tau = 0  \]

The torque is defined as

\[  \tau = rF\sin \theta  \]

where \(r\) and \(F\) are the lever arm and the force. The angle \(\theta\) is taken from the beam. The weight of the beam acts on the middle of the length by assuming that the mass is uniformly distributed. Plug in the numbers to calculate the net torque:

\[  500\cdot 2.00 \sin(270^o) + 200\cdot (5.00)\sin(270^o) + T\cdot 10.0 \sin(60^o) = 0    \]

\[  – 1000 – 1000 + 10.0T\sin(60^o)  = 0     \]

\[  T = \frac{2000}{10\sin(60^o)}       \]

\[  T = 231 \quad \mathrm{N}  \]