Particles of masses m, 2m, 3m and 4m are plac | vedclass

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(C) Let the center of the square be $P$. The distance from each corner to the center $P$ is $x = \frac{a}{\sqrt{2}}$.The gravitational forces exerted

Vedclass · Accepted Answer

$\frac{4\sqrt{2}Gm^2}{a^2}$

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(C) Let the center of the square be $P$. The distance from each corner to the center $P$ is $x = \frac{a}{\sqrt{2}}$.The gravitational forces exerted by the masses at corners $A, B, C, D$ on the mass $m$ at $P$ are:$F_{PA} = \frac{G(m)(m)}{x^2} = \frac{Gm^2}{x^2} = F$$F_{PB} = \frac{G(2m)(m)}{x^2} = 2F$$F_{PC} = \frac{G(3m)(m)}{x^2} = 3F$$F_{PD} = \frac{G(4m)(m)}{x^2} = 4F$These forces act along the diagonals. The net force along the diagonal $AC$ is $F_{net, AC} = F_{PC} - F_{PA} = 3F - F = 2F$ (directed towards $C$).The net force along the diagonal $BD$ is $F_{net, BD} = F_{PD} - F_{PB} = 4F - 2F = 2F$ (directed towards $D$).The resultant force $F_{net}$ is the vector sum of these two perpendicular forces:$F_{net} = \sqrt{(2F)^2 + (2F)^2} = 2\sqrt{2}F$Substituting $F = \frac{Gm^2}{x^2} = \frac{Gm^2}{(a/\sqrt{2})^2} = \frac{2Gm^2}{a^2}$:$F_{net} = 2\sqrt{2} \left( \frac{2Gm^2}{a^2} \right) = \frac{4\sqrt{2}Gm^2}{a^2}$

Particles of masses $m, 2m, 3m$ and $4m$ are placed at the corners of a square of side $a$. What is the net gravitational force on a particle of mass $m$ placed at the center?

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