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(B) Let the equilateral triangle be $PQR$ with side length $l$. The masses $M$ are at $P, Q,$ and $R$. The mass $m$ is placed at the midpoint $L$ of s

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$\frac{4GMm}{3l^2}$

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(B) Let the equilateral triangle be $PQR$ with side length $l$. The masses $M$ are at $P, Q,$ and $R$. The mass $m$ is placed at the midpoint $L$ of side $QR$.The gravitational force exerted by the mass $M$ at $Q$ on mass $m$ at $L$ is $F_Q = \frac{GMm}{(l/2)^2}$ directed towards $Q$.The gravitational force exerted by the mass $M$ at $R$ on mass $m$ at $L$ is $F_R = \frac{GMm}{(l/2)^2}$ directed towards $R$.Since $F_Q$ and $F_R$ are equal in magnitude and opposite in direction,they cancel each other out $(F_Q + F_R = 0)$.The net force on $m$ is therefore only due to the mass $M$ at vertex $P$.The distance $PL$ is the altitude of the equilateral triangle,given by $PL = l \sin 60^{\circ} = l \frac{\sqrt{3}}{2}$.The magnitude of the force is $F = \frac{GMm}{(PL)^2} = \frac{GMm}{(l\sqrt{3}/2)^2} = \frac{GMm}{3l^2/4} = \frac{4GMm}{3l^2}$.

Three particles each of mass $M$ are placed at the three corners of an equilateral triangle of side $l$. What is the gravitational force due to this system of particles on another particle of mass $m$ placed at the midpoint of any side?

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