Question

Three equal masses of $1kg$ each are placed at the vertices of an equilateral triangle $PQR$ and a mass of $2kg$ is placed at the centroid O 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 $2kg$ is

• A. 2
• B. 1
• C. 3
• D. zero

Answer 1

Answer: (D)

Given, $OP=OQ=OR=\sqrt{2}m$
The gravitational force on the mass $2kg$ due to the $1kg$ mass at $P$ is $F_{OP}=G\frac{2\times 1}{(\sqrt{2}{)}^2}=G$ along $OP$
Similarly, $F_{OQ}=G_1\frac{2\times 1}{(\sqrt{2}{)}^2}=G_1$ along $OQ$ and
$F_{OR}=G_1\frac{2\times 1}{(\sqrt{2}{)}^2}=G_1$ along $OR$
$F_{OQ}\cos 30^{\circ }$ and $F_{OR}\cos 30^{\circ }$ are equal and acting in opposite directions, thus they cancel out. Hence, the resultant force on the $2kg$ mass at $O$, is
$F=F_{OP}-(F_{OQ}\sin 30^{\circ }+F_{OR}\sin 30^{\circ })$
$=G_1-\left(\frac{G_1}{2}+\frac{G_1}{2}\right)$
$=0$