Question

A uniform 2.0 kg horizontal beam 50 cm long is
bolted to a brick wall and supports a 5.0 kg lighting
fixture. Calculate the torque produced by the
combined weight of the beam and the light about
the point where the beam meets the wall

Answer 1

QUESTION:

A uniform $2.0\mathrm{kg}$ horizontal beam $50~\mathrm{cm}$ long is bolted to a brick wall and supports a $5.0\mathrm{kg}$ lighting fixture. Calculate the torque produced by the combined weight of the beam and the light about the point where the beam meets the wall

SOLUTION:

Let's draw a sketch. As beam is uniform, its center of mass is in the middle.

There are two forces that have torque about the point where the beam meets the wall: $m\vec{g}$ and $M\vec{g}$ . Hence, the torque is

$$\tau = m g l + M g \frac {l}{2}$$

Here $l = 50 \, \text{cm}$ is the length of the beam.

$$\tau = 5 \cdot 9.8 \cdot 0.5 + 2 \cdot 9.8 \cdot 0.25 = 29.4 \, \text{N} \cdot \text{m}$$

ANSWER:

29.4 N·m