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

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

   $F = \dfrac{GMm}{R^2}$

now,

   $F = - \dfrac{6.67 \times 10^{-11}\times 130 \times 30}{0.15^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{0.15^2}$

   $F = 3.12 \times 10^{-5}\ N$

   Direction of force will be toward the mass of 430 Kg

b) position where force will be zero

 $F =-\dfrac{6.67 \times 10^{-11}\times 130 \times 30}{(0.3-x)^2} +\dfrac{6.67 \times 10^{-11}\times 430 \times 30}{x^2} = 0$

 $-\dfrac{130}{(0.3-x)^2} +\dfrac{430}{x^2} = 0$

 $x^2- 0.6 x + 0.09=\dfrac{130}{430}x^2$

 $0.7 x^2- 0.6 x + 0.09 =0$

solving the above equation

  x = 0.1936 m

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