Question

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

• A. $\overrightarrow{F_1}/m$
• B. $2\overrightarrow{F_1}/m$
• C. $-\overrightarrow{F_1}/m$
• D. $-2\overrightarrow{F_1}/m$

Answer 1

The correct option is D $-2\overrightarrow{F_1}/m$

Since $\overrightarrow{F1},\overrightarrow{F2},\overrightarrow{F3}$ are acting on the same point,
$\overrightarrow{F1}+\overrightarrow{F2}+\overrightarrow{F3}=0$
Therefore $\overrightarrow{F1}=-\overrightarrow{F2}-\overrightarrow{F3}$
Total force on the particle $=-\overrightarrow{F1}+\overrightarrow{F2}+\overrightarrow{F3}=-2\overrightarrow{F1}$
$\therefore$ Acceleration of the particle $=\frac{Totalforce}{Mass}=\frac{-2\overrightarrow{F1}}{m}$