Question

The sum of the magnitudes of two forces acting at a point is $16N$. The resultant of these forces is perpendicular to the smaller force and has a magnitude of $8N$. If the smaller force is of magnitude $x$, then the value of $x$ is

• A. $2N$
• B. $4N$
• C. $6N$
• D. $7N$

Answer 1

Answer: (C)

$6N$

Let two forces vector be ${\overrightarrow{F}}_1$ & ${\overrightarrow{F}}_2$.

$\left|{\overrightarrow{F}}_1\right|+\left|{\overrightarrow{F}}_2\right|=16N8N=\left|\overrightarrow{R}\right|$

smaller force $=x$

$\left|\overrightarrow{R}\right|$ is perpendicular for $K$ (smaller force)

Let $F_1$ be smaller force with magnitude $x$.

$\left|{\overrightarrow{F}}_2\right|=16-x⟶\left(i\right)$

${\left|{\overrightarrow{F}}_2\right|}^2=R^2+x^2$

$R^2={\left|F_2\right|}^2-x^2$

$8^2={(16-x)}^2-x^2$ [using eqn (i)]

$64=256+x^2-32x-x^2$

$x=\frac{192}{32}=6$

$x=6unit$