JEE MAIN - Physics (2002 - No. 58)
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
$${F_1}/m$$
$${F_2}{F_3}/m{F_1}$$
$$\left( {F{}_2 - {F_3}} \right)/m$$
$${F_2}/m$$
Explanation
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 then particle will start moving due to the force $${F_2}$$ and $${F_3}$$ in the resultant of $${F_2}$$ and $${F_3}$$ and it should be equal and opposite to $${F_1}.$$
$$i.e.$$ $$\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right| = \left| {{{\overrightarrow F }_1}} \right|$$
$$\therefore$$ $$\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right|} \over m} \Rightarrow a = {{{F_1}} \over m}$$
$$i.e.$$ $$\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right| = \left| {{{\overrightarrow F }_1}} \right|$$
$$\therefore$$ $$\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} \right|} \over m} \Rightarrow a = {{{F_1}} \over m}$$
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