question_answer

Three equal masses of 1 kg each are placed at the vertices of an equilateral triangle PQ.R and a mass of 2 kg is placed at the centroid 0 of the triangle which is at a distance of V2 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

Correct Answer: D

Solution :

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