MCQ
The simple sum of two forces acting at a point is $16N$ and their sum is $8N$ and its direction is perpendicular to the smaller force, then the forces are:
  • $6N$ and $10N$
  • B
    $8N$ and $8N$
  • C
    $4N$ and $12N$
  • D
    $2N$ and $14N$

Answer

Correct option: A.
$6N$ and $10N$
Here $A + B = 16 ...(i)$
$\sqrt{\text{A}^2+\text{B}^2+2\text{AB}\cos\theta}=8\dots(\text{ii})$
and $\tan90^\circ=\frac{\text{B}\sin\theta}{\text{A}+\text{B}\cos\theta}$
or $\text{A}+\text{B}\cos\theta=\frac{\text{B}\sin\theta}{\tan90^\circ}=0$
or $\text{B}\cos\theta=-\text{A}$ or $\cos\theta=\frac{-\text{A}}{\text{B}}$
From $(ii), \text{A}^2+\text{B}^2+2\text{AB}\Big(\frac{-\text{A}}{\text{B}}\Big)=64$
or $B^2 - A^2 = 64$
Solving $(i)$ and $(iii),$ we get
$A = 6N$ and $B = 10N.$

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