Question

Three equal masses of $1\,kg$ each are placed at the vertices of an equilateral triangle $PQR$ and a mass of $2\,kg$ is placed at the centroid $0$ of the triangle which is at a distance of $\sqrt{2}$ m from each of the vertices of the triangle. The force, in newton, acting on the mass of $2\,kg$ is

• A. $2$
• B. $ \sqrt{2} $
• C. $1$
• D. zero

Answer 1

Answer: (D)

Given $O P=O Q=O R=\sqrt{2} m$

The gravitational force on mass $2\, kg$ due to mass $1\, kg$ at $P$.
$F_{O P}=G \frac{2 \times 1}{(\sqrt{2})^{2}}=G$ along $O P$
Similarly,
$F_{O Q}=G \frac{2 \times 1}{(\sqrt{2})^{2}}=G$ along $O Q$
and $F_{O R}=G \frac{2 \times 1}{(\sqrt{2})^{2}}=G$ along $O R$
$F_{O Q} \cos 30^{\circ}$ and $F_{O R} \cos 30^{\circ}$ are equal and acting in opposite directions, then cancel out each other.
Then the resultant force on the mass $2\, kg$ at $O$
$F=F_{O P}-\left(F_{O Q} \sin 30^{\circ}+F_{O R} \sin 30^{\circ}\right)$
$F=G-\left(\frac{G}{2}+\frac{G}{2}\right)$
$F=0$ (zero)