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. 6 N and 10 N
B. 8 N and 8N
C. 4 N and 12N
D. 2 N and 14 N

Answer

A

Explanation

As shown in the figure , F. is the resultant of `vecF_(1) and F_(2) and F_(1)` is the smaller force . Now
and `F_(1) +F_(2) + F^(2) = F_(1)^(2) +(8)^(2) " "...(i)`
and `F_(1) + F_(2) = 16 " or " F_(2) = 16 - F_(1) " "...(ii)`
Using (ii) and (i) , we have
which gives `F_(1) = 6N`
Therefore, `F_(2) = 16 - 6 = 10 N`