Question 156: Laws of Motion

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

Accepted Answer

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 $${F_1}.$$
$$i.e.$$ $$\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} ight| = \left| {{{\overrightarrow F }_1}} ight|$$
$$	herefore$$ $$\,\,\,\,\,a = {{\left| {{{\overrightarrow F }_2} + {{\overrightarrow F }_3}} ight|} \over m} \Rightarrow a = {{{F_1}} \over m}$$

Answer

$${F_1}/m$$

Answer

$${F_2}{F_3}/m{F_1}$$

Answer

$$\left( {F{}_2 - {F_3}} \right)/m$$

Answer

$${F_2}/m$$

Laws of Motion

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

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