Chapter 9, Problem 1Q

Figure 9-23 shows an overhead view of three particles on which external forces act. The magnitudes and directions of the forces on two of the particles are indicated. What are the magnitude and direction of the force acting on the third particle if the center of mass of the three-particle system is (a) stationary, (b) moving at a constant velocity rightward, and (c) accelerating rightward?

Figure 9-23 Question 1.

To determine

To find:

a) the force acting on third particle when center of mass is stationary.

b) the force acting on third particle when center of mass is moving rightward with constant velocity.

c) the force acting on third particle when center of mass is accelerating rightward.

Answer to Problem 1Q

Solution:

a) Force acting on third particle when center of mass is stationary, $F_3=2 N$ in the right direction.

b) Force acting on third particle when center of mass is moving rightward with constant velocity, $F=2 N$ in the right direction.

c) Force acting on third particle when center of mass is accelerating rightward, $F_3>2 N$ in the right direction

Explanation of Solution

1) Concept:

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 we can find the force acting on the third particle.

2) Formula:

$F_{Net}=\sum F$

3) Given:

i) $F_1=5 N$ acting on particle 1 in the left direction

ii) $F_2=3 N$ acting on particle 2 in the right direction

4) Calculation:

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

$F_{Net}=\sum F=0$

$F_{Net}=-5+3+F_3=0$

$F_3=2 N$

So, $F_3=2 N$ and the positive sign indicates that it should be in the right direction.

b) For the center of mass to move on the right with constant velocity, net force acting on the system should be zero.

$F_{Net}=\sum F=0$

$F_{Net}=-5+3+F_3=0$

$F_3=2 N$

So, $F_3=2 N$ and positive sign indicates that it should be in right direction.

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

$F_{Net}=\sum F>0$

$F_{Net}=-5+3+F_3>0$

$F_3>2 N$ in right direction.

So, for center of mass to accelerate rightward, we should have $F_3>2N$ in the right direction

Conclusion:

Considering the direction and magnitude of the acceleration of the center of mass, we can find the direction and magnitude of the net force acting on the third particle.