The sum of magnitudes of two forces acting at a point is `16 N`. If the resultant force is `8 N` and its direction is perpendicular to smaller force, then the forces are :-

Options

A. `6N & 10N`
B. `8N & 8N`
C. `4N & 12N`
D. `2N & 14N`

Answer

A

Explanation

`F_(1)+F_(2)=16` .....(1)
`8=sqrt(F_(1)^(2)+2F_(1)F_(2)cos theta)`
`rArrF_(1)^(2)+F_(2)^(2)+2F_(1)F_(2)+2F_(1)F_(2)costheta=64` ..........(2)

`tan alpha=(F_(2) sin theta)/(F_(1)+F_(2) cos theta)=tan 90^(@)=oo`
`F_(2)costheta=-F_(1)`
Solving (1),(2) and (3)
`F_(1)=6N,F_(2)=10N`