Question

A uniform door weighs 70.0 N and is 1.20 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

Given,

The weight of the door, W=70.0 N

The height of the door, h=2.60 m

The width of the door, l=1.20 m

The torque is the product of the force and the distance between the axis and the point where the force is being applied.

The center of mass of the rectangle will be at its geometric center.

The distance between the horizontal axis passing through the corner of the door and the center of mass of the door is,

`d=\sqrt[]{((h)/(2))^2+((l)/(2))^2}`

On substituting the known values,

`\begin{gathered} d=\sqrt[]{((2.60)/(2))^2+((1.20)/(2))^2} \\ =1.43\text{ m} \end{gathered}`

Thus the required torque is given by,

`\tau=Wd`

On substituting the known values,

`\begin{gathered} \tau=70.0*1.43 \\ =100.1\text{ Nm} \end{gathered}`

Thus the torque due to the door's own weight about a horizontal axis perpendicular to the door and passing through a corner is 100.1 Nm