Three identical particles A, B,and C of mass | vedclass

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(B) Let the mass of each particle be $M = 100 \, kg$. The distance $AB = BC = r = 13 \, m$. Particle $P$ is at distance $r = 13 \, m$ from $B$ on the

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(B) Let the mass of each particle be $M = 100 \, kg$. The distance $AB = BC = r = 13 \, m$. Particle $P$ is at distance $r = 13 \, m$ from $B$ on the perpendicular bisector of $AC$.The distance of $P$ from $B$ is $r_B = 13 \, m$.The distance of $P$ from $A$ and $C$ is $r_A = r_C = \sqrt{13^2 + 13^2} = 13\sqrt{2} \, m$.The gravitational force exerted by $B$ on $P$ is $F_B = \frac{GM^2}{r^2}$ directed towards $B$.The gravitational forces exerted by $A$ and $C$ on $P$ are $F_A = F_C = \frac{GM^2}{(13\sqrt{2})^2} = \frac{GM^2}{2r^2}$.The horizontal components of $F_A$ and $F_C$ cancel out. The vertical components add up:$F_{net} = F_B + 2 	imes F_A \cos(	heta)$,where $\cos(	heta) = \frac{13}{13\sqrt{2}} = \frac{1}{\sqrt{2}}$.$F_{net} = \frac{GM^2}{r^2} + 2 	imes \frac{GM^2}{2r^2} 	imes \frac{1}{\sqrt{2}} = \frac{GM^2}{r^2} (1 + \frac{1}{\sqrt{2}})$.Substituting $M = 100 \, kg$ and $r = 13 \, m$:$F_{net} = \frac{G 	imes 100^2}{13^2} (1 + 0.707) = \frac{10000}{169} 	imes 1.707 	imes G \approx 59.18 	imes 100 	imes G \approx 100 G$ (considering the magnitude of mass squared).Given the options,the closest value is $100 G$.

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