Two masses, `800 kg` and `600 kg` are at a distance `0.25 m` apart. The magnitude of total force experienced by a body of mass `1 kg` placed at a point distance `0.2 m` from the `800 kg` mass and `0.15 m` from the `600 kg` mass :

Explanation

Refer to Fig. The force on `I kg` at `P` due to body of mass `800 kg` at `A` is
`F_(A) = (G xx 800 xx 2)/((0.2)^(2)) = 40,000 G, along PA`
The force on `1 kg` at `P` due to body of mass `600 kg` at `B` is
`F_(B) = (G xx 600 xx 2)/((0.15)^(2)) = (160,000)/(3) G, along PB`

Since, `(0.25)^(2) = (0.2)^(2) + (0.15)^(2)` ,so
`AB^(2) = AP^(2) + PB^(2)`
Hence, `/_APB = 90^(@)`
`:.` Resultant force on `2 kg` body at `P` is
`F = sqrt(F_(A)^(2) + F_(B)^(2))`
`= [(40,000 G)^(2) + ((160,000G)/(3))^(2)]^(1//2)`
`= 10^(4)G [16 + (256)/(9)]^(1//2)`
`= 10^(4) xx 6.67 xx 10^(-11) xx (20)/(3) = 4.44 xx 10^(-6) N`