Question

The resultant of two forces has a magnitude of $20\text{N}$. If one of the forces is of magnitude $20\sqrt{3}\text{N}$and makes an angle of ${30}^{°}$ with the resultant then, the other force 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

$20N$

Step 1: Given that and the diagram

Let the two forces be represented by $\overrightarrow{a},\overrightarrow{P}$ and the resultant by $\overrightarrow{r}$

$\left|\overrightarrow{r}\right|=20$ ,$\left|\overrightarrow{a}\right|=20\sqrt{3}$

$\overrightarrow{a}+\overrightarrow{P}=\overrightarrow{r}$

Also $\overrightarrow{a},\overrightarrow{r}$ makes angle $30°$

The below diagram represents the situation

Step 2: Formula used:

If $\overrightarrow{a}+\overrightarrow{P}=\overrightarrow{r}$

then, $P^2=a^2+r^2-2ab\cos \theta$

Step 3: Calculating the magnitude of another vector

$\overrightarrow{a}+\overrightarrow{P}=\overrightarrow{r}$

$\overrightarrow{P}=\overrightarrow{r}\overrightarrow{-a}$

$P^2=a^2+r^2-2ab\cos \theta$

$P^2={\left(20\sqrt{3}\right)}^2+{20}^2-2\times 20\times 20\sqrt{3}\times \cos (30°)$

$P^2=1200+400-80\sqrt{3}\times \frac{\sqrt{3}}{2}$

$P^2=1200+400-1200$

$$\begin{gathered} P^2=400 \\ P=20 \end{gathered}$$

Hence, option B is correct.