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 acceleration of the particle is

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

Answer 1

The correct option is A $F_1/m$

Given, the particle remains stationary under forces $F_1,F_2,F_3$ and $F_2$ and ,$F_3$ are mutually perpendicular
For equilibrium of system,
$F_1=\sqrt{F_2^2+F_3^2,}$ as $\theta ={90}^{\circ }$
In the absence of force $F_1$,
Acceleration $=\frac{\text{Net force}}{\text{Mass}}$
Acceleration $=\frac{\sqrt{F_2^2+F_3^2}}{m}=\frac{F_1}{m}$
Final answer: (a)