Question

The sum of two forces acting at a point is $16N$. If the resultant force is $8N$ and its direction is perpendicular to smaller force then the forces are

• A. $6N$ and $10N$
• B. $8N$ and $7N$
• C. $4N$ and $12N$
• D. $2N$ and $14N$

Answer 1

The correct option is A

$6N$ and $10N$

The explanation for the correct option

Step 1: Given data

The sum of the two forces is $16N$

Let the force be $P$ and $Q$

$\therefore P+Q=16N....\left(i\right)$

Step 2: Formula used

$\tan \left(\alpha \right)=\frac{Q\sin \left(\theta \right)}{P+Q\cos \left(\theta \right)}$

Step 3: Calculating forces

The direction of the resultant force is perpendicular to the smaller force

$$\begin{gathered} \tan \left(\alpha \right)=\tan \left(90°\right) \\ =\frac{Q\sin \left(\theta \right)}{P+Q\cos \left(\theta \right)} \end{gathered}$$

Clearly $P+Q\cos \left(\theta \right)=0$

$\Rightarrow \cos \left(\theta \right)=-\frac{P}{Q}.....\left(ii\right)$

Given that the resultant force is $8N$

$$\begin{gathered} \therefore \sqrt{P^2+Q^2+2PQ\cos \left(\theta \right)}=8 \\ \Rightarrow P^2+Q^2+2PQ\left(-\frac{P}{Q}\right)=64 \\ \Rightarrow Q^2-P^2=64 \\ \Rightarrow \left(Q+P\right)\left(Q-P\right)=64 \\ \Rightarrow 16\left(Q-P\right)=64\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow Q-P=4........\left(\mathrm{iii}\right) \end{gathered}$$

Solving adding $\left(i\right)$ and $\left(iii\right)$ we get

$P+Q=16-P+Q=042Q=20\Rightarrow Q=10$

$$\begin{gathered} \because P+Q=16 \\ \Rightarrow P=16-10 \\ \Rightarrow P=6 \end{gathered}$$

Hence, option A is the correct answer.