Question

Two forces are such that the sum of their magnitudes is $18 N$ and their resultant which has magnitude $12 N$, is perpendicular to the smaller force. Then the magnitudes of the forces are :

• A. $12 N , 6 N$
• B. $13 N , 5 N$
• C. $10 N , 8 N$
• D. $16 N , 2 N$

Answer 1

Answer: (B)

$A+B=18$...(1)
$ 12 =\sqrt{A^{2}+B^{2}+2 A B \cos \theta}$...(2)
$\tan \alpha =\frac{B \sin \theta}{A+B \cos 0}$
$\Rightarrow \tan 90^{\circ} =\frac{B \sin \theta}{A+B \cos \theta}$
$\Rightarrow \cos \theta=\frac{-A}{B}$...(3)
Solving Eqs. (1), (2) and (3), $A=5 N, B=13 N$