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

Question

Dr. Abubacker · Accepted Answer

(d) [image:8_image53.emf]
Here, $OP = OQ = OR = \sqrt{2}\ m$
The gravitational force on mass $2\ kg$ at $O$ due to mass
$1\ kg$ at $P$ is $F_{OP} = \frac{G 	imes 2 	imes 1}{\left( \sqrt{2} ight)^{2}} = G$ along $OP$
The gravitational force on mass $2\ kg$ at $O$ due to mass
$1\ kg$ at $Q$ is $F_{OQ} = \frac{G 	imes 2 	imes 1}{\left( \sqrt{2} ight)^{2}} = G$ along $OQ$
The gravitational force on mass $2\ kg$ at $O$ due to mass
$1\ kg$ at $R$ is $F_{OR} = \frac{G 	imes 2 	imes 1}{\left( \sqrt{2} ight)^{2}} = G$ along $OR$
Resolve forces $F_{OQ}$ and $F_{OR}$ into two rectangular components
$F_{OQ}\cos 30{^\circ}$ and $F_{OR}\cos{30{^\circ}}$ are equal in magnitude of equal and opposite direction
$= F_{OP} - (F_{OQ}\sin{30{^\circ} + F_{OR}\sin{30{^\circ})}}$
$= G - \left( G 	imes \frac{1}{2} + G 	imes \frac{1}{2} ight) = G - G = Zero\ N$

Q. 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

Explanation

Reviewed by Dr. Abubacker

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