Two bodies of masses \(8 \mathrm{~kg}\) are placed at the vertices \(A\) and \(B\) of an equilateral triangle \(A B C\). A third body of mass \(2 \mathrm{~kg}\) is placed at the centroid \(G\) of the triangle. If \(A G=B G=C G=1 \mathrm{~m}\), where should a
fourth body of mass \(4 \mathrm{~kg}\) be placed, so that the resultant force on the \(2 \mathrm{~kg}\) body is zero?

According to the question, the arrangement of the masses is as shown below,

Gravitational force between two masses is given as
\(F=\frac{G m_{1} m_{2}}{r^{2}}\)
where, \(G=\) gravitational constant and \(r=\) distance between them.
From the given values, we can say that force between masses at \(A\) and \(G\). \(=\) Force between masses at \(B\) and \(G\).
\(\Rightarrow F_{A}=\frac{G m_{A} m_{G}}{A G^{2}} \text { and } F_{B}=\frac{G m_{B} m_{G}}{B G^{2}} \)
\( \text {or } F_{A}=F_{B}=\frac{G \times 8 \times 2}{1^{2}}=16 G...(i)\)
\(\left[\because\right.\) Given, \(m_{1}=m_{A}=m_{B}=8 \mathrm{~kg}, m_{2}=m_{G}=2 \mathrm{~kg}\),
\(A G=B G=r=1 \mathrm{~m} \text { ] }\)
From the figure, resultant of \(F_{A}\) and \(F_{B}\) is given as
\(F_{A B} =\sqrt{F_{A}^{2}+F_{B}^{2}+2 F_{A} F_{B} \cos \theta} \)
\( =\sqrt{2 F_{A}^{2}+2 F_{A}^{2} \cos 120^{\circ}} \)
\( =\sqrt{F_{A}^{2}}=\sqrt{(16 G)^{2}}=16 G \)
\( \Rightarrow F_{A B} =F_{A}=F_{B}=16 G \text { [from Eq. (i)] }\)
For resultant force on \(2 \mathrm{~kg}\) body to be zero, \(4 \mathrm{~kg}\) body should be placed at a certain distance from \(G\) such that
\(\mathbf{F}_{C G}=-\mathbf{F}_{A E}\)
\(\Rightarrow \left|\mathbf{F}_{C G}\right|=\left|\mathbf{F}_{A B}\right|\)
or \(\frac{G \times m_{C} m_{G}}{x^{2}}=G(\mathbf{1} 6)\)
where, \(x=\) distance between \(4 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) body.
Here, \(m_{C}=4 \mathrm{~kg}\)
\(\Rightarrow \frac{G \times 4 \times 2}{x^{2}} =G(16) \)
\( \Rightarrow x^{2} =\frac{1}{2} \)
\( \Rightarrow x =\frac{1}{\sqrt{2}} \mathrm{~m}\)
This means, \(4 \mathrm{~kg}\) body should be placed at point \(P\) on line \(C G\) such that \(P G=\frac{1}{\sqrt{2}} \mathrm{~m}\).