Sum of magnitude of two forces acting at a point is $16$ N. If their resultant is normal to the smaller force, and has a magnitude $8$ N, then the forces are:

6 N

10 N

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8 N

8 N

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4 N

12 N

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2 N

16 N

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Solution

The correct option is A $6 N$ , $10 N$
$X + Y = 16$
$X < Y$ and $R$ is the resultant force perpendicular to $X$ .
$R^{2} = Y^{2} - X^{2}$
$= (Y - X) (Y + X) = 16 (Y - X)$
$\Rightarrow (Y - X) = \frac{R^{2}}{Y + X} = \frac{8^{2}}{16} = 4$
$Y - X = 4$ and $Y + X = 16$
On solving the two equations, we get $X = 6 N$ and $Y = 10 N$

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