Three equal masses of $1\text{kg}$ each are placed at the vertices of an equilateral triangle $\text{PQR}$ and a mass of $2\text{kg}$ is placed at the centroid $O$ of the triangle which is at a distance of $\sqrt{2}\text{m}$ from each of the vertices of the triangle. The net force (in Newton), acting on the mass of $2\text{kg}$ is

The correct option is D zero
Given:
$\text{OP=OQ=OR=}\sqrt{2}\text{m}$
$m=1\text{kg};M=2\text{kg}$


The gravitational force on mass $2\text{kg}$ due to mass $1\text{kg}$ at $\text{P}$.

$F_{OP}=G\frac{2\times 1}{(\sqrt{2}{)}^2}=G\text{along OP}$

Similarly,
$F_{OQ}=G\frac{2\times 1}{(\sqrt{2}{)}^2}=G\text{along OQ}$

$F_{OR}=G\frac{2\times 1}{(\sqrt{2}{)}^2}=G\text{along OR}$

Net force at $O$ in direction parallel to the base of triangle is due to force $F_{OQ}$ and $F_{OR}$ .

$F_{OQ}\cos {30}^{\circ }$ and $F_{OR}\cos {30}^{\circ }$ are equal and acting in opposite directions, thus balance each other.
Therefore net force at $O$ will be in vertical direction.

The resultant force on the mass $2\text{kg}$ at $O$ is given by

$F_{net}=F_{OP}-(F_{OQ}\sin {30}^{\circ }+F_{OR}\sin {30}^{\circ })$

$\Rightarrow F_{net}=G-(\frac{G}{2}+\frac{G}{2})=0$

Hence, option $\left(d\right)$ is correct.

Why this question: To help students understand the Principle of Superposition. Key Concept: Principle of Superposition - The gravitational force between two masses is independent of other masses in the vicinity and the net force on a mass is equal to the vector sum of forces by all the masses.