The iron door of a building is 3 m broad. It can be opened by applying a force of 100 N normally at the middle of the door. Calculate : (a) the torque needed to open the door. (b) the least force and it's point of application to open the door.

As we know,

Moment of force (torque) = force (f) x distance (r)

Given,

f = 100 N

As force is applied at the middle of the door, hence,

$\text{r} = \big(\dfrac{1}{2} \times 3 \big) \text{ m}$

Therefore, r = 1.5 m

Substituting the values in the formula above we get,

Torque = 100 N x 1.5 m = 150 Nm

Hence, the torque needed to open the door = 150 Nm.

(b) The force applied will be least if it is applied at the farthest point from the hinges. Therefore, the force should be applied at the free end of the door. i.e., at the distance of 3 m from the hinges.

Let the least force required be F.

r = 3 m (i.e., point of application of least force)

Moment of force (torque) = f x r

torque = 150 Nm

Substituting the values in the formula above we get,

$150 = \text{F} \times 3 \\[0.5em] \Rightarrow \text{F} = \dfrac{150}{3} \\[0.5em] \Rightarrow \text{F} = 50 \text { N}\\[0.5em]$

Hence, the least force = 50 N.