Two fixed point masses of 10 kg and 20 kg are located at (0 m, 0 m) and (5 m, 0 m). Where should a third point mass of 30 kg be kept so that the net force of attraction on 30 kg mass due to point masses 10 kg and 20 kg is zero?

Force on 30 kg mass due to 10 kg mass. F1=6.67×10−11×10×30x2 towards left Force on 30 kg mass due to 20 kg mass F2=6.67×10−11×30×20(5−x)2towards right Since the net force of 30 kg mass is zero, ∴ F1=F2 ∴6.67×10−11×10×30x2             =6.67×10−11×30×20(5−x)2 Or (5−x)2=2x2 or 25+x2−10x=2x2 Or x2+10x−25=0 x=2.05m