Q: Three forces $ \displaystyle \vec{F_1}, $ $ \displaystyle \vec{F_2} $ and $ \displaystyle \vec{F_3} $ are simultaneously acting on a particle of mass ‘m’ and keep it in equilibrium. If $\displaystyle \vec{F_1} $ force is reversed in direction only, the acceleration of the particle will be.

(a)$ \displaystyle \frac{\vec{F_1}}{m} $

(b)$ \displaystyle \frac{\vec{2 F_1}}{m} $

(c)$\displaystyle -\frac{\vec{F_1}}{m} $

(d)$ \displaystyle -\frac{\vec{2 F_1}}{m} $

Ans: (d)

Solution :$ \displaystyle \vec{F_1}+\vec{F_2}+\vec{F_3} = 0 $

$ \displaystyle \vec{F_2}+\vec{F_3} = – \vec{F_1} $ …(i)

If $ \displaystyle \vec{F_1} $ force is reversed in direction ,

$ \displaystyle \vec{F_2}+\vec{F_3}-\vec{F_1} = m \vec{a} $

$ \displaystyle – \vec{F_1}-\vec{F_1} = m \vec{a} $

$ \displaystyle – 2\vec{F_1} = m \vec{a} $

$ \displaystyle \vec{a} = -\frac{\vec{2 F_1}}{m} $