Question

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

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

Answer 1

The correct option is A

$6N$ and $10N$

Step 1: Given Data

Let the two forces be $P$ and $Q$.

Sum of the two forces $P+Q=16N$

The angle between the lesser force and $8N$ is $90°$

Step 2: Formula Used

Resultant vector,$Q=\sqrt{R^2+P^2+2PR\cos 90°}$

Step 3: Calculate the Forces,

$$\begin{gathered} Q=\sqrt{R^2+P^2+2PR\cos 90°} \\ {Q}^2=R^2+P^2 \\ Putp=16-Q \\ {Q}^2=64+{(16-Q)}^2 \\ {Q}^2=64+256-32Q+Q^2 \\ 32Q=320 \\ Q=10N \end{gathered}$$

​$$\begin{gathered} P+Q=16N \\ P+10N=16N \\ P=6N \end{gathered}$$
Solving these we get, $P=6N$ and $Q=10N$

Hence, the correct option is A.