Question

The resultant of two forces has magnitude of $20N$ . if one of the forces is of magnitude $20\sqrt{3}N$ and makes an angle of ${30}^{\circ }$ with the resultant then, the other forces must be of magnitude

• A. $10\sqrt{3}N$
• B. $20N$
• C. $20\sqrt{3}N$
• D. $10N$

Answer 1

The correct option is B $20N$

$\text{Step 1: Resolving forces along Resultant [Refer Fig.]}$

Let the first force be ${\overrightarrow{F}}_1=20\sqrt{3}N$ which makes ${30}^o$ degree with the resultant.

Let the other force be ${\overrightarrow{F}}_2$ which makes $\theta$ degree with the resultant.

Components of ${\overrightarrow{F}}_1$ and ${\overrightarrow{F}}_2$ along $\overrightarrow{R}$

$F_1\cos {30}^o+F_2\cos \theta =R$

$\Rightarrow$ $20\sqrt{3}\cos {30}^o+F_2\cos \theta =20$

$\Rightarrow$ $F_2\cos \theta =-10$ $....\left(1\right)$

$\text{Step 2: Resolving forces perpendicular to the resultant}.$

$F_1\sin {30}^o+F_2\sin \theta =0$

$\Rightarrow$ $20\sqrt{3}\times \frac{1}{2}+F_2\sin \theta =0$

$\Rightarrow$ $F_2\sin \theta =-10\sqrt{3}$ $....\left(2\right)$

$\text{Step 3: Solving equations}$

From equation $\left(1\right)$ and $\left(2\right)$

$F_2^2{\cos }^2\theta +F_2^2{\sin }^2\theta =100+300$

$\Rightarrow$ $F_2^2=400$

$\Rightarrow$ $F_2=20N$

Hence the magnitude of other force is $20N$

Option $B$ is correct.