Question

Three particles, each of mass m, are kept at the corners of an equilateral triangle of side 'l'. Force exerted by this system on another particle of mass m placed at the midpoint of any side will be:

• A. $\frac{5G{m}^2}{l^2}$
• B. $\frac{G{m}^2}{2l^2}$
• C. $\frac{4G{m}^2}{3l^2}$
• D. $\frac{5G{m}^2}{3l^2}$

Answer 1

The correct option is C $\frac{4G{m}^2}{3l^2}$

From the diagram, it can be seen that the force exerted by either particle at the end of the side on the particle at the midpoint of the side will be equal in magnitude and opposite in direction.

Thus, these two forces will cancel out each other.

Thus, only the particle at the opposite corner of the equilateral triangle will be significant in this case.

The gravitational force between this particle and the particle at midpoint of the side is given by:

$\because r=l\frac{\sqrt{3}}{2}$ is the distance between these two particles

$\therefore F=\frac{Gmm}{r^2}$

$\therefore F=\frac{G{m}^2}{{\left(l\frac{\sqrt{3}}{2}\right)}^2}$

$\therefore F=\frac{4G{m}^2}{3l^2}$

Hence, the force exerted by a system of particles of equal masses on an equilateral triangle on another particle at a midpoint of the side of the triangle is $\frac{4G{m}^2}{3l^2}$