91Ó°ÊÓ

The figure shows an overhead view of three particles on which external forces act.

${\mathrm{F}}_1=5\mathrm{N}$acting on particle 1 in the left direction

${\mathrm{F}}_{2}3\mathrm{N}$acting on particle 2 in the right direction

We have to calculate the net force acting on the system. As the system contains three particles and the center of mass is to remain stationary, the net force acting on the system should be zero. Using this concept, we can find the force acting on the third particle.

Formulae:

The net forces acting on a body,${\mathrm{F}}_{\mathrm{Net}}=\underset{}{∑}\mathrm{F}$ (1)

For the center of mass to be stationary, the net force acting on the system should be equal to zero.

So, using equation (1), we can get the value of the force of the third particle as:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}=0 \\ -5\mathrm{N}+3\mathrm{N}+{\mathrm{F}}_3=0 \\ {\mathrm{F}}_3=2\mathrm{N} \end{gathered}$$

Hence, the value of the force is 2 N and the positive sign indicates that it should be in the right direction.

For the center of mass to move on the right with constant velocity, the net force acting on the system should be zero due to acceleration being zero. Now, using equation (i), the force on the third particle is given as:

$$\begin{gathered} {\mathrm{F}}_{\mathrm{Net}}=0 \\ -5+3\mathrm{N}+{\mathrm{F}}_3=0 \\ {\mathrm{F}}_3=2\mathrm{N} \end{gathered}$$

Hence, the value of the force is 2N and the positive sign indicates that it should be in the right direction.

For the center of mass to move in the right-hand direction with some acceleration

$$\begin{gathered} {\mathrm{F}}_{\mathrm{net}}>0 \\ -5\mathrm{N}+3\mathrm{N}+{\mathrm{F}}_3>0 \\ {\mathrm{F}}_3>2\mathrm{N} \end{gathered}$$

So, for the center of mass to accelerate rightward, we should have the value of the force >2N in the right direction