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 O of the triangle which is at a distance of $\sqrt2$ m from each of the vertices of the triangle. The force, in newton, acting on the mass of 2 kg is

• A. 2
• B. 1
• C. 3
• D. zero

Answer 1

Answer: (D)

zero

Answer 2

Given, OP =OQ =OR =$\sqrt2 \, m$ The gravitational force on the mass 2 kg due to the 1 kg mass at P is $F_{OP} = G \frac{2 \times 1}{(\sqrt2)^2} = G$ along OP Similarly, $\, \, \, \, F_{OQ} = G_1 \frac{2 \times 1}{(\sqrt2)^2} = G_1$ along OQ and $F_{OR} = G_1 \frac{2 \times 1}{(\sqrt2)^2} = G_1$ 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 2 kg mass at O, is \hspace15mm F = F_{OP} - (F_{OQ} sin \, 30^\circ + F_{OR} sin \, 30^\circ) \hspace25mm =G_1 - \big(\frac{G_1}{2} + \frac{G_1}{2}\big) \hspace25mm = 0