Question

The square of the resultant of two equal forces acting at a point is equal to three times their product. Angle between them is

• A. $30^{\circ}$
• B. $45^{\circ}$
• C. $60^{\circ}$
• D. $90^{\circ}$

Answer 1

Answer: (C)

$60^{\circ}$

Answer 2

Let two forces be $F_1$ and $F_2$ acting at a point and angle between them is $\theta$.
Resultant of $F_1$ and $F_2$ is given by
$F = \sqrt{F_1^2 + F_2^2 + 2F_1 F_2 \cos \theta}$ ....(i)
According to question,
$F^2 = 3 F_1 F_2 = 3 F_{1^2} \, \, \, \, [\because \, \, \, \, F_1 = F_2]$
$F^{2} =F_{1^{2}} + F_{2^{2}} + 2F_{1 }F_{2} \cos\theta$
or $3F_{1^{2} } + F_{1^{2} }+2F_{1^{2}} \cos \theta$
or, $3 = 1+ 1+2 \cos \theta$
or, $\cos \theta =\frac{1}{2}$
or, $\theta =\cos^{-1} \left(\frac{1}{2} \right) = 60^{\circ}$