Question

A uniform door weighs 54.0 N and is 1.30 m wide and 2.60 m high. What is the magnitude of the torque due to the door’s own weight about a horizontal axis perpendicular to the door and passing through a corner?

Answer 1

ANSWER:

35.1 J

STEP-BY-STEP EXPLANATION:

The point of application of the force of gravity is at the geometrical center of the door, so

$\begin{gathered} r=\frac{\text{width of the door}}{2} \\ \text{ replacing} \\ r=\frac{1.3}{2} \\ r=065\text{ m} \end{gathered}$

The net force is equal to its weight that acts at the center of the door. Now above equation changes as:

$\begin{gathered} \tau=F\cdot d \\ \text{ replacing} \\ \tau=54\cdot0.65 \\ \tau=35.1\text{ J} \end{gathered}$

The torque acting on the door is 35.1 J