Respuesta :
Answer
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
[tex]F = \dfrac{GMm}{R^2}[/tex]
now,
[tex]F = - \dfrac{6.67 \times 10^{-11}\times 130 \times 30}{0.15^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{0.15^2}[/tex]
[tex]F = 3.12 \times 10^{-5}\ N[/tex]
Direction of force will be toward the mass of 430 Kg
b) position where force will be zero
[tex]F =-\dfrac{6.67 \times 10^{-11}\times 130 \times 30}{(0.3-x)^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{x^2} = 0[/tex]
[tex] -\dfrac{130}{(0.3-x)^2} +\dfrac{430}{x^2} = 0[/tex]
[tex] x^2- 0.6 x + 0.09=\dfrac{130}{430}x^2[/tex]
[tex] 0.7 x^2- 0.6 x + 0.09 =0[/tex]
solving the above equation
x = 0.1936 m
the distance of third mass will be at x = 0.1936 m from 430 Kg mass.
