Question

Three equal masses $m$ are placed at the three corners of an equilateral triangle of side $a$. Find the force exerted by this system on another particle of mass $m$ placed at (a) the mid-point of a side, (b) at the centre of the triangle.

Answer 1

According to diagram

By Pythagoras theorem,

$a^2=\frac{a^2}{4}+h^2$

$h^2=\frac{3a^2}{4}$

$h=\frac{\sqrt{3}a}{2}$

Now, the force exerted by this system on another particle of mass m placed at the mid-point of a side

So,

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

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

Now, the force exerted by this system on another particle of mass m placed at the center of the triangle

According to figure the force from each particle is at an angle of ${120}^0$with respect to another. So, all forces cancel out and net force is zero.