Sum of magnitudes of two forces acting at a point is $16 \mathrm{~N}$. If their resultant is normal to smaller force, and has a magnitude $8 \mathrm{~N}$, then forces are

• Correct answer
  A

  $6 \mathrm{~N}, 10 \mathrm{~N}$
• B

  $8 \mathrm{~N}, 8 \mathrm{~N}$
• C

  $4 \mathrm{~N}, 12 \mathrm{~N}$
• D

  $2 \mathrm{~N}, 14 \mathrm{~N}$

Solution

$(16-x)^2=8^2+x^2$
$\begin{aligned} 256+x^2-32 x & =64+x^2 \\ 32 x & =192 \\ x & =\frac{192}{32}=6\end{aligned}$ $\therefore$ Forces $6 \mathrm{~N}, 10 \mathrm{~N}$

Hence, the correct answer is option: 'A'.