Question

The sum of two forces acting at a point is $6N$. If the resultant force is $8N$ and its direction is perpendicular to minimum force, then the force are

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

Answer 1

The correct option is C $6N\text{and}10N$

Given:
Sum of two forces is $16N$
Therefore, $F_1+F_2=16\cdots \cdots \left(i\right)$
The direction of the resultant force is perpendicular to the smaller force
$\tan \alpha =\tan 90=\frac{F_2\sin \Theta }{F_1+F_2\cos \Theta }$
Therefore, $F_1+F_2\cos \Theta =0\cos \Theta =-\frac{F_1}{F_2}$
We have,
$8=\sqrt{F_1^2+F_2^2+2F_1{F}_2\cos \Theta }$
Square both sides, we get
$64=F_1^2+F_2^2+2F_1{F}_2\cos \Theta$
Put the value,
$\cos \Theta =-\frac{F_1}{F_2}$
$64=(F_2+F_1)(F_2-F_1)\therefore (F_2+F_1=16N)(F_2-F_1)=4N\cdots \cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$, we get
$F_1=6N,F_2=10N$