Question

Forces $M$ and $N$ acting at a point $O$make an angle $150°$. Their resultant acts at $O$ has magnitude $2$ unit and is perpendicular to $M$. Then, in the same unit, the magnitudes of $M$ and $N$ are

• A. $2\surd 3,4$
• B. $\surd 3/2,2$
• C. $3,4$
• D. $4,5$

Answer 1

The correct option is A

$2\surd 3,4$

Step 1. Given data

Angle between force $M$ and $N$ is $\theta =150°$

Magnitude of resultant of two forces, $R=2unit$ and angle with respect to force $M$ is ${\theta }_R=90°$

Step 2. Finding magnitudes of $M$ and $N$

The resultant is the vector sum of two or more vectors. It is the result of adding two or more vectors together.

Resultant is given by

$$\begin{gathered} R^2=M^2+N^2+2M.N\cos \theta \\ \Rightarrow 2^2=M^2+N^2+2M.N\cos 150° \\ \Rightarrow 4=M^2+N^2+2M.N\left(-\frac{\sqrt{3}}{2}\right) \\ \Rightarrow 4=M^2+N^2-\sqrt{3}M.N\left(1\right) \end{gathered}$$

Also, we have

$$\begin{gathered} N\cos 60°=R \\ N\sin 60°=M \\ \Rightarrow \tan 90°=\frac{M}{R}=\frac{M}{2} \\ \Rightarrow M=2\sqrt{3} \end{gathered}$$

Putting in equation $\left(1\right)$, we have

$$\begin{gathered} \Rightarrow 4=4\times 3+N^2-\sqrt{3}\times 2\sqrt{3}.N \\ \Rightarrow N^2+8-6N=0 \\ \Rightarrow N=4 \end{gathered}$$

Hence, the correct option is A