Question

The resultant of two forces having magnitude $3p$ and $2p$ is $R$. If the force having magnitude $3p$ is doubled keeping the same direction, then the resultant is also doubled. The angle between two forces is

• A. ${120}^o$
• B. ${150}^o$
• C. ${180}^o$
• D. ${125}^o$

Answer 1

The correct option is A ${120}^o$

Let $\theta$ be the angle between two forces. We know resultant of two forces $F_1$ and $F_2$ inclined at an angle θ is given by
$R=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \theta }$
Or $\begin{array}{}{R}^2=F_1^2+F_2^2+2F_1{F}_2\cos \theta & \text{(1)}\end{array}$
Here $F_1=3p$ and $F_2=2p$
$\therefore R^2=(3p{)}^2+(2p{)}^2+2\times 3p\times 2p\cos \theta$
or $R^2=9p^2+4p^2+12p^2\cos \theta$
$\begin{array}{}{R}^2=13p^2+12p^2\cos \theta & \text{(2)}\end{array}$
Now $F_1=2\times 3p=6p,R=2R,F_2=2p$
$\therefore$ From equation $\left(1\right)$,
$(2R{)}^2=(6p{)}^2+(2p{)}^2+2\times 6p\times 2p\cos \theta$
$4R^2=36p^2+4p^2+24p^2\cos \theta$
$4R^2=40p^2+24p^2\cos \theta$
or $\begin{array}{}{R}^2=10p^2+6p^2\cos \theta & \text{(3)}\end{array}$
equating $\left(2\right)$ and $\left(3\right)$, we get,
$13p^2+12p^{2}cos\theta =10p^2+6p^2\cos \theta$
Or $3p^2=-6p^2\cos \theta \cos \theta =-\frac{1}{2}$
$\therefore \theta =co{s}^{-1}(-\frac{1}{2})={120}^{\circ }$
$\therefore \theta ={120}^{\circ }$