Question

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

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

Answer 1

The correct option is B $13N,5N$

Let the two forces be $F_1$ and $F_2$ such that ($F_1<F_2$)
Let $R$ be the resultant of these two forces.

According to question:
$F_1+F_2=18N$...(i)
Angle between $F_1$ and resultant ($R$) is ${90}^{\circ }$
$\therefore \tan {90}^{\circ }=\frac{F_2\sin \theta }{F_2\cos \theta +F_1}=\infty$
$⟹F_1+F_2\cos \theta =0$
$⟹\cos \theta =-\frac{F_1}{F_2}$....(ii)
$R^2=F_1^2+F_2^2+2F_1{F}_2\cos \theta$
$⟹144=F_1^2+F_2^2+2F_1{F}_2\times \frac{-F_1}{F_2}$
$⟹144=F_2^2-F_1^2$...(iii)
Using (i) and (iii), we get
$F_1=5N$ and $F_2=13N$