Question

The sum of two forces acting at a point is 16 N. If the resultant force is 8 N and its direction is perpendicular to minimum force then the forces are

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

Answer 1

Answer: (A)

6 N and 10 N

Answer 2

$A + B = 16$ (given) …(i)

$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} = \tan 90{^\circ}$

∴ $A + B\cos\theta = 0$⇒ $\cos\theta = \frac{- A}{B}$ …(ii)

$8 = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}$ …(iii)

By solving eq. (i), (ii) and (iii) we get $A = 6N,$ $B = 10N$

$A = 6NB = 10N$