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

Solution :
When $${F_1},{F_2}$$  and $${F_3}$$ are acting on a particle then the particle remains stationary. This means that the resultant of $${F_1},{F_2}$$  and $${F_3}$$ is zero, When $${F_1}$$ is removed, $${F_2}$$ and $${F_3}$$ will remain. But the resultant of $${F_2}$$ and $${F_3}$$ should be equal and opposite to $${F_1}.{\text{i}}{\text{.e}}{\text{.}}\left| {{{\vec F}_2} + {{\vec F}_3}} \right| = \left| {{{\vec F}_1}} \right|$$
$$\therefore a = \frac{{\left| {{{\vec F}_2} + {{\vec F}_3}} \right|}}{m} \Rightarrow a = \frac{{{F_1}}}{m}$$