The net force on A is 1.7312 × 10⁻⁷ N
Newton's law of universal gravitation
It states that the force of attraction between two masses M and m is directly proportional to the product of their masses and inversely proportional to the square of their distance apart, R.
So, F = GMm/R² where G = universal gravitational constant = 6.67 × 10⁻¹¹ Nm²/kg²
Force of attraction between A and B
Now let
- m₁ = mass of object at A = 199.1 kg,
- m₂ = mass of object at B = 592.9 kg and
- R = distance between object at A and B = 7.5 m.
So, the force of attraction between them is
F = Gm₁m₂/R₁² (positive since it is directed towards B in the positive x direction)
Substituting the values of the variables into the equation, we have
F = Gm₁m₂/R₁²
F = 6.67 × 10⁻¹¹ Nm²/kg² × 199.1 kg × 592.9 kg/(7.5 m)²
F = 787369.4213 × 10⁻¹¹ Nm²/56.25 m²
F = 13997.68 × 10⁻¹¹ N
F = 1.399768 × 10⁻⁷ N
F ≅ 1.3998 × 10⁻⁷ N
Force of attraction between A and C
Also, let
- m₁ = mass of object at A = 199.1 kg,
- m₃ = mass of object at C = 330 kg,and
- R₂ = distance between object at A and C = 4 m + 7.5 m = 11.5 m
So, the force of attraction between them is
F' = Gm₁m₃/R₂² (positive since it is directed towards C in the positive x direction)
Substituting the values of the variables into the equation, we have
F' = Gm₁m₂/R₁²
F' = 6.67 × 10⁻¹¹ Nm²/kg² × 199.1 kg × 330 kg/(11.5 m)²
F' = 438239.01 × 10⁻¹¹ Nm²/132.25 m²
F' = 3313.72 × 10⁻¹¹ N
F' = 0.331372 × 10⁻⁷ N
F' ≅ 0.3314 × 10⁻⁷ N
Net force on A
So, the net force acting on object A is F" = F + F'
= 1.3998 × 10⁻⁷ N + 0.3314 × 10⁻⁷ N
= 1.7312 × 10⁻⁷ N
So, the net force on A is 1.7312 × 10⁻⁷ N
Learn more about Newton's law of universal gravitation here:
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