Question

The sum of the two forces is $18N$ and their resultant perpendicular to the lesser of the forces is $12N$, then the lesser force is

• A. $12N,6N$
• B. $13N,5N$
• C. $10N,8N$
• D. $16N,2N$

Answer 1

The correct option is B

$13N,5N$

Step 1: Given Data

Let the two forces be $A$ and $B$.

Sum of the two forces $=18N$

Let $\theta$ be the angle between the two forces.

Let $\alpha =90°$ be the angle between the lesser force and $12N$.

Step 2: Formula Used
As illustrated in the figure, the resultant $\overrightarrow{R}$ is the diagonal of which and $\overrightarrow{B}$ are adjacent sides according to the parallelogram law of vector addition.

$\overrightarrow{R}$ magnitude is given by,

Resultant force, $\sqrt{A^2+B^2+2AB\cos \theta }=12\dots ..\left(1\right)$
Also, $\frac{BSin\theta }{A+BCos\theta }=\tan \alpha$

Step 3: Calculate the Forces

We know that, $A+B=18...\left(2\right)$

Also, ​

$$\begin{gathered} \frac{Bsin\theta }{A+Bcos\theta }=\tan \alpha \\ \frac{Bsin\theta }{A+Bcos\theta }=\tan 90° \\ \frac{Bsin\theta }{A+Bcos\theta }=\frac{Sin90}{Cos90} \\ Cos\left(90\right)\left(BSin\theta \right)=Sin\left(90\right)(A+B\cos \theta ) \\ 0=A+B\cos \theta \\ Cos\theta =-\frac{A}{B} \end{gathered}$$

Squaring equation (1) and putting the value of $\cos \theta$, we get​
$$\begin{gathered} 144=A^2+B^2-2A^2 \\ or,B^2-A^2=144...\left(3\right) \end{gathered}$$
Solving equations $\left(1\right),\left(2\right)$ and$\left(3\right)$,

$A=5N,B=13N$

Hence, the correct answer is option (B).