Question

When forces $F_1,F_2,F_3$ are acting on a particle of mass $m$ such that $F_2$ and $F_3$ are mutually perpendicular, then the particle remains stationary. If the force $F_1$ is now removed then the magnitude of acceleration of the particle is

• A. $\frac{F_1}{m}$
• B. $\frac{F_2{F}_3}{m{F}_1}$
• C. $\frac{(F_2-F_3)}{m}$
• D. $\frac{F_2}{m}$

Answer 1

The correct option is A $\frac{F_1}{m}$

Given,

$\overrightarrow{F_1}+\overrightarrow{F_2}+\overrightarrow{F_3}=0=F_{\text{net}}$

Now if $F_1$ is removed, let acceleration be ${}^{\prime }{a}^{\prime }$.

Then, $\overrightarrow{F_2}+\overrightarrow{F_3}=m\overrightarrow{a}$

$\Rightarrow -\overrightarrow{F_1}=m\overrightarrow{a}$

$\Rightarrow \overrightarrow{a}=-\frac{\overrightarrow{F_1}}{m}$

$\therefore \left|\overrightarrow{a}\right|=\left|\frac{\overrightarrow{F_1}}{m}\right|$

Hence, $\left(A\right)$ is the correct answer.