Question

Four identical masses of m are kept at the corners of a square of side 'a'. Find the gravitational force exerted on one of the masses by the other masses.

Answer 1

Hey mate,

● Answer-
F = (√2 + 1/2) Gm^2/a^2

● Explaination-
[Refer to the figure]
Let 4 identical masses be m1, m2, m3 and m4 such that
m1 = m2 = m3 = m4 = m

Now individual forces on mass m2 are-
F21 = Gm^2 / a^2
F23 = Gm^2 / a^2
F24 = Gm^2 / 2a^2

Resultant force on mass m2-
F = √(F21^2+F23^2) + F24
F = √[(Gm^2 / a^2)^2+(Gm^2 / a^2)^2] + Gm^2 / 2a^2
F = √2Gm^2 / a^2 + Gm^2 / 2a^2
F = (√2 + 1/2) Gm^2 / a^2

Resultant force on mass m2 by m1, m3 and m4 is (√2 + 1/2) Gm^2 / a^2.

Hope this is helpful...