The sum of magnitude of two forces acting at a point 16N and their resultant 83 N is at 90∘ with the force of smaller magnitude. The two forces (in Ns) are

Let F1→ and F2→ be two forces and F→ be their resultant. Let F1→ be the smaller force.In ∆ABC, F22=F2+F12 (OR) F2=F22−F12(83)2=F22−F12192=(F2+F1)(F2−F1)∴F1+F2=16N (GIVEN)Put this value in (i), we get F2−F1=19616=12N.....(iii)Solving (ii) and (iii), we get F1=2n, F2=14N∴ forces are 2N and 14 N .

The sum of magnitude of two forces acting at a point 16N and their resultant 83 N is at 90∘ with the force of smaller magnitude. The two forces (in Ns) are

Let F1→ and F2→ be two forces and F→ be their resultant. Let F1→ be the smaller force.In ∆ABC, F22=F2+F12 (OR) F2=F22−F12(83)2=F22−F12192=(F2+F1)(F2−F1)∴F1+F2=16N (GIVEN)Put this value in (i), we get F2−F1=19616=12N.....(iii)Solving (ii) and (iii), we get F1=2n, F2=14N∴ forces are 2N and 14 N .

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The sum of magnitude of two forces acting at a point 16N and their resultant $8 \sqrt[]{3} N$ is at $90^{\circ}$ with the force of smaller magnitude. The two forces (in Ns) are

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$11, 5$

$9, 7$

$6, 10$

$2, 14$

answer is D.

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Detailed Solution

Let $\vec{F_{1}} a n d \vec{F_{2}}$ be two forces and $\vec{F}$ be their resultant. Let $\vec{F_{1}}$ be the smaller force.

$\begin{array}{l} I n ∆ A B C, F_{2}^{2} = F^{2} + F_{1}^{2} (O R) F^{2} = F_{2}^{2} - F_{1}^{2} \\ {(8 \sqrt{3})}^{2} = F_{2}^{2} - F_{1}^{2} \\ 192 = (F_{2} + F_{1}) (F_{2} - F_{1}) \\ ∴ F_{1} + F_{2} = 16 N (G I V E N) \end{array}$
Put this value in (i), we get $F_{2} - F_{1} = \frac{196}{16} = 12 N ..... (i i i)$
Solving (ii) and (iii), we get $F_{1} = 2 n, F_{2} = 14 N$
$∴$ forces are 2N and 14 N .