Question

At what angle should the.two force vectors $2F$ and $\sqrt{2}F$ act so that the resultant force is $\sqrt{10}$ F?

Answer 1

$\left|{\overrightarrow{F}}_1\right|=2F$ Let $0$ be angle between

$\left|{\overrightarrow{F}}_2\right|=\sqrt{2F}$ $F_1$ & $F_2$

$\left|\overrightarrow{R}\right|=\sqrt{10}F$

$\left|\overrightarrow{R}\right|=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$

${10F}^2={4F}^2+{2F}^2+2\times 2\times \sqrt{2}{F}^2\cos \theta$

${4F}^2=4\sqrt{2}{F}^2\cos \theta$

$\frac{1}{\sqrt{2}}=\cos \theta \Rightarrow \theta ={\cos }^{-1}\left(\frac{1}{\sqrt{2}}\right)$