Question 7.3: A force of 40 N is exerted on a rod as shown. Find the momen...

In order to calculate the moment, the perpendicular distance between O and the line of action of the force must be found. This is shown on the diagram .

Here l = 1.5 × sin 50.

So the moment about O is

F × l = 40 × (1.5 × sin 50°)

= 46.0 Nm .

Alternatively you can resolve the 40 N force into components as in the next diagram.
The component of the force parallel to AO is 40 cos 50 N.
The component perpendicular to AO is 40 sin 50
(or 40 cos 40) N.

So the moment about O is
40 sin 50° × 1.5 = 60 sin 50°

= 46.0 Nm as before .