Question

The two forces $2\sqrt{2}N$ and $xN$ are acting at a point their resultant is perpendicular to $xN$ and having magnitude of $\sqrt{6}N$. The angle between the two forces and magnitude of $x$ are:

• A. $\theta ={120}^o,x=\sqrt{2}N$
• B. $\theta ={30}^o,x=\sqrt{2}N$
• C. $\theta ={150}^o,x=\sqrt{3}N$
• D. $\theta ={150}^o,x=\sqrt{2}N$

Answer 1

The correct option is A $\theta ={120}^o,x=\sqrt{2}N$

$\text{Step 1: Magnitude of x [Refer Fig.]}$

Draw diagram using triangle law of vector addition.

Using Pythagoras theorem

$(2\sqrt{2}{)}^2=(\sqrt{6}{)}^2+x^2$

$\Rightarrow$ $x=\sqrt{2}$

$\text{Step 2: Angle calculation [Refer Fig.]}$

From triangle

$\tan \alpha =\frac{\sqrt{6}}{x}$ $=\frac{\sqrt{6}}{\sqrt{2}}$

$\Rightarrow$ $\alpha ={60}^{\circ }$

Now the angle between two forces is $\theta$

(For finding angle between two vectors we have to join tail of both vectors)

$\theta =180-\alpha$

$={120}^{\circ }$

Hence option $A$ is correct.