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 $0$ 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 $2kg$ is

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

Answer 1

Answer: (D)

Given $OP=OQ=OR=\sqrt{2}m$

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