Three particles each of mass m_{1} are placed | vedclass

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(A) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let the mass $m_{2}$ be placed at point $D$,which is the midpoint of side $BC$.Th

Vedclass · Accepted Answer

$\frac{12 G m_{1} m_{2}}{L^{2}}$

Vedclass · Answer

(A) Let the vertices of the equilateral triangle be $A, B,$ and $C$. Let the mass $m_{2}$ be placed at point $D$,which is the midpoint of side $BC$.The gravitational forces on $m_{2}$ due to the masses $m_{1}$ at $B$ and $C$ are equal in magnitude and opposite in direction,so they cancel each other out.The distance of point $D$ from vertex $A$ is the altitude $h$ of the equilateral triangle.In the right-angled triangle $ABD$,the hypotenuse $AB = \frac{L}{3}$ and $\angle BAD = 30^{\circ}$.Thus,$h = AB \cos 30^{\circ} = \frac{L}{3} 	imes \frac{\sqrt{3}}{2} = \frac{L}{2\sqrt{3}}$.The net force on $m_{2}$ is the force due to the mass $m_{1}$ at $A$:$F = G \frac{m_{1} m_{2}}{h^{2}} = G \frac{m_{1} m_{2}}{\left(\frac{L}{2\sqrt{3}}ight)^{2}} = G \frac{m_{1} m_{2}}{\frac{L^{2}}{12}} = \frac{12 G m_{1} m_{2}}{L^{2}}$.

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