Question

2 bodies of mass 5 kg and 20 kg are kept at the distance of 12 m . The third body of mass 3 kg is to be placed such that the force acting on this body is balanced . At what distance should it be kept from the body which has mass 5 kg ?

Answer 1

$$\begin{gathered} \mathrm{Condition}\mathrm{for}\mathrm{equilibrium}: \\ \frac{{\mathrm{Gm}}_1{\mathrm{m}}_3}{{\mathrm{x}}^2}=\frac{{\mathrm{Gm}}_2{\mathrm{m}}_3}{{\left(\mathrm{r}-\mathrm{x}\right)}^2} \\ \frac{5\times 3}{{\mathrm{x}}^2}=\frac{20\times 3}{{\left(12-\mathrm{x}\right)}^2} \\ \frac{{\left(12-\mathrm{x}\right)}^2}{{\mathrm{x}}^2}=4 \\ \frac{12-\mathrm{x}}{\mathrm{x}}=2 \\ 12-\mathrm{x}=2\mathrm{x} \\ 3\mathrm{x}=12 \\ \mathrm{x}=4\mathrm{m} \\ \mathrm{Ans}:4\mathrm{m}\mathrm{from}5\mathrm{kg} \end{gathered}$$