Question

The resultant of two forces, one double then other in magnitude, is perpendicular to the smaller of the two forces. The angle between the two forces is

• A. 120${}^{\circ }$
• B. 60${}^{\circ }$
• C. 90${}^{\circ }$
• D. 150${}^{\circ }$

Answer 1

The correct option is A 120${}^{\circ }$

Let the two forces be $F_1$ and $F_2$ are given according to question $|F_1|=2|F_2|$

Resultant of two forces $F_1$ and $F_2$ are given by

$R=F_1+F_2$

$R$ is perpendicular upon smaller force

Here smaller forces $F_2$

So $R.F_2=0$

$\Rightarrow (F_1+F_2).F_2=0\Rightarrow F_1.F_2+F_2^2=0\Rightarrow |F_1||F_2|\cos \varphi +|F_2{|}^2=0$

Putting $|F_1|=2|F_2|$

$\Rightarrow 2|F_2||F_2|\cos \varphi +|F_2{|}^2=0\Rightarrow 2\cos \varphi +1=0\Rightarrow \cos \varphi =-\frac{1}{2}=\cos 120°$

Angle between two forces is $120°$