Question

Objects with masses of 130 kg and a 430 kg are separated by 0.300 m.(a) Find the net gravitational force exerted by these objects on a 35.0 kg object placed midway between them.(b) At what position (other than infinitely remote ones) can the 35.0 kg object be placed so as to experience a net force of zero?

Answer 1

given,

mass of object

m₁ = 130 Kg

m₂ = 430 Kg

distance between them = 0.3 m

a) net force when 35 kg is place in between them

`F = (GMm)/(R^2)`

now,

`F = - (6.67 * 10^(-11)* 130 * 30)/(0.15^2) +(6.67 * 10^(-11)* 430 * 30)/(0.15^2)`

`F = 3.12 * 10^(-5)\ N`

Direction of force will be toward the mass of 430 Kg

b) position where force will be zero

`F =-(6.67 * 10^(-11)* 130 * 30)/((0.3-x)^2) +(6.67 * 10^(-11)* 430 * 30)/(x^2) = 0`

`-(130)/((0.3-x)^2) +(430)/(x^2) = 0`

`x^2- 0.6 x + 0.09=(130)/(430)x^2`

`0.7 x^2- 0.6 x + 0.09 =0`

solving the above equation

x = 0.1936 m

the distance of third mass will be at x = 0.1936 m from 430 Kg mass.