Question

The sum of the magnitude of two forces at a point is 16 N. If their resultant is normal to the smaller force and has a magnitude of 8 N. Then two forces are

• A. 6 N, 10 N
• B. 8 N, 8 N
• C. 4 N, 12 N
• D. 2 N, 14 N

Answer 1

The correct option is A 6 N, 10 N

Let the bigger force be $\overrightarrow{F_1}$and the smaller force be $\overrightarrow{F_2}$.

So, $F_1+F_2=16$ ....(a)

Also $|\overrightarrow{F_1}+\overrightarrow{F_2}|=8$

$⟹F_1^2+F_2^2+2F_1{F}_2=256-\left(1\right)$

and, $F_1^2+F_2^2+2\overrightarrow{F_1}.\overrightarrow{F_2}=64-\left(2\right)$

As, $\overrightarrow{F_1}+\overrightarrow{F_2}$ and $\overrightarrow{F_2}$ are normal, their dot product is zero.

i.e., $(\overrightarrow{F_1}+\overrightarrow{F_2}).\overrightarrow{F_2}=0$

$⟹\overrightarrow{F_1}.\overrightarrow{F_2}+F_2^2=0$

$⟹\overrightarrow{F_1}.\overrightarrow{F_2}=-F_2^2$

So from (2), $F_1^2-F_2^2=64⟹(F_1+F_2)(F_1-F_2)=64$

$⟹16(F_1-F_2)=64$

$⟹F_1-F_2=4$ ....(b)

Adding (a) and (b), we get $2F_A=20$ $⟹F_1=10N$

Thus from (a), $10+F_2=16$ $⟹F_2=6N$

$\therefore F_1=10N$, and $F_2=6N$