A uniform metallic rod PQ of length 2 m is acted upon by two forces A and B along the directions as shown in the figure. Find the magnitude and position of the resultant normal force that acts on the rod.

Explanation

The forces that act normally to the rod are
` A cos 60^(@) = 90 xx (1)/(2)=45N`
` B cos 60^(@) = 30 xx (1)/(2) = 15 N`.
As the two forces are unlike parallel forces, the resultant of these two forces is outside the rod at a point 'O' and in the direction of greater force, i.e. , A.

Magnitude of R = 45 - 15 = 30 N
Position of R: Let 'R' is at 'O' and at a distance of x m from 'P'. Then
` 45 xx x=15xx(2+x)`
`3x=2+x`
`2x=2`
` x = 1 m `