The net gravitational force on mass A due to masses B and C is required.
The correct option is a. [tex]2.8\times 10^{-8}\ \text{N}[/tex] to the right.
G = Gravitational constant = [tex]6.674\times 10^{-11}\ \text{Nm}^2/\text{kg}^2[/tex]
[tex]r_{AB}[/tex] = Distance between A and B = 0.4+0.1 = 0.5 m
[tex]r_{AC}[/tex] = Distance between A and C = 0.1 m
[tex]m_A=m_B=m_c[/tex] = Mass of each particle = 2 kg
The required force is
[tex]F_A=F_{AB}+F_{AC}\\ =\dfrac{Gm_Am_B}{r_{AB}^2}+\dfrac{Gm_Am_C}{r_{AC}^2}\\ =Gm^2\left(\dfrac{1}{r_{AB}^2}+\dfrac{1}{r_{AC}^2}\right)\\ =6.674\times 10^{-11}\times 2^2\left(\dfrac{1}{0.5^2}+\dfrac{1}{0.1^2}\right)\\ =2.8\times 10^{-8}\ \text{N}[/tex]
The magnitude of force will be [tex]2.8\times 10^{-8}\ \text{N}[/tex]
The direction will be towards the right since C is closer to A.
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