• question_answer When forces ${{F}_{1}},{{F}_{2}},{{F}_{3}}$ are acting on a particle of mass m such that ${{F}_{1}}$ and ${{F}_{2}}$ are mutually perpendicular, then the particle remains stationary. If the force ${{F}_{1}}$ is now removed, then the acceleration of the particle is   [AIEEE 2002] A) ${{F}_{1}}/m$ B) ${{F}_{2}}{{F}_{3}}/m{{F}_{1}}$ C) $({{F}_{2}}-{{F}_{3}})/m$ D) ${{F}_{2}}/m$

 [a] In this question, we have to think about that "a body or system to be in equilibrium, net force acting on it should be zero. The particle is remains stationary under the action of three forces ${{F}_{1}},{{F}_{2}}$ and ${{F}_{3}}$, it means resultant force is zero. ${{F}_{1}}=-({{F}_{2}}+{{F}_{3}})$ Since, in second case, ${{F}_{1}}$ is removed (in terms of magnitude we are talking now), the forces acting are ${{F}_{2}}$ and ${{F}_{3}}$ the resultant of which has the magnitude as ${{F}_{1}}$, so acceleration of particle is $\frac{{{F}_{1}}}{m}$ in the direction opposite to that of ${{F}_{1}}$.