Q. Two forces are such that the sum of their magnitudes is $18\ N$ and their resultant is perpendicular to the smaller force and magnitude of resultant is 12 N. Then the magnitudes of the forces are
A) $12\ N,\ 6\ N$
B) $13\ N,\ 5\ N$
C) $10\ N,\ 8\ N$
D) $16\ N,\ 2\ N$
Explanation
(b) $A + B = 18$ …(i)
$12 = \sqrt{A^{2} + B^{2} + 2AB\cos\theta}$ …(ii)
$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta} = \tan{90{^\circ}} \Rightarrow \cos\theta = - \frac{A}{B}$ …(iii)
By solving (i), (ii) and (iii), $A = 13\ N$ and $B = 5N$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.