Question

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

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

Answer 1

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

$F_2>F_1$ (Because Resultant is more close to taue $F_2$)

$F_1+F_2=18N$

Now, $F_2\cos \theta =F$ resultant $=12\to \left(i\right)$

$F_1=F_2\sin \theta \to \left(ii\right)$

Divide equation $\left(ii\right)$ by $\left(i\right)$

$\tan \theta =F_1/12$

So, in equation $\left(ii\right)$

$F_1=F_2\times \frac{F_1}{\sqrt{(F_1{)}^2+144}}$

$(18-F_1{)}^2=F_1^2+144$

So, $F_1=65N$

So, $F_2=13N$

Option $\left(B\right)$