Question

Two equal forces act at a point. The square of their resultant is three times their product. Find the angle between them?

Answer 1

Answer :-

Let the two equal forces be  $F_1\:\:,\:\:F_2$

$\implies F=F_1=F_2$

The square of their resultant is three times their product.

Let the resultant be R.

$\implies \:\:\:\;\;\;R^2=3*F_1*F_2\\\\\implies \;\;\;\;\;\; R^2=3F^2\;\;\;[Since , F_1=F_2=F]\\\\\implies (\sqrt{F_1^2+F_2^2+2F_1.F_2.cos\theta} )^2=3F^2\\\\\implies F^2+F^2+2F.F.cos\theta = 3F^2\:\;\;[Since , F_1=F_2=F]\\\\\implies 2F^2+2F^2.cos\theta =3F^2 \\\\\implies 3F^2-2F^2=2F^2.cos\theta\\\\\implies F^2=2F^2.cos\theta\\\\\implies 1=2cos\theta\\\\\implies cos\theta=\frac{1}{2}\\\\ \implies \theta =cos^{-1}(\frac{1}{2})\\\\\implies \theta = \frac{\pi }{3} \\\\\implies \theta =60^o$