Question

The moment of a force of 25 N about a point is 2.5 N m. Find the perpendicular distance of force from that point.

Answer 1

Moment of force = 2.5 N m
Force applied = 25
⊥ distance from the point of rotation =?
Moment of force = $Force\times$ $\perp distance$
2.5 = $25\times$ $x$

Therefore,

x = $\frac{2.5}{25}$ = $\frac{1}{10}$

= $\frac{1}{10}$ $\times$ 100

= 10cm