In Fig. 13-26, three particles are fixed in place. The mass of Bis greater than the mass of C. Can a fourth particle (particle D) be placed somewhere so that the net gravitational force on particle Afrom particles B, C,and Dis zero? If so, in which quadrant should it be placed and which axis should it be near?

The figure for particle-shell system

The force of attraction between any two bodies is directly proportional to the product of their masses and is inversely proportional to the square of the distance between them, according to Newton's universal law of gravitation.We can find the resultant force acting on a particle A due to particles B and C using vector addition. We can also find the direction of the resultant force by comparing the masses of particles B and A. Hence, from this, we can find the force vector at which the net gravitational force acting on particle A becomes zero.

Formula:

The Gravitational force of attraction between two bodies of masses M and m separated by distance R is, $F=\frac{GMm}{R^2}$ …(i)

Free body diagram:

The force is the resultant of the gravitational force exerted on particle A from particle C ($F_{AC}$) and the gravitational force exerted on particle A from particle B $\left(F_{AB}\right)$.

From equation (i), we can say that the gravitational force is directly proportional to the masses of the bodies and the mass of the particle B is greater than the mass of the particle A, the resultant force is closer towards the force$F_{AB}$_.

The net gravitational force acting on A will be zero if particle D will apply equal and opposite force to the resultant force F.

Therefore, particle D should be placed in the second quadrant near the positive y axis.