Answer:
The position of the small mass on the x- axis = 11.87 cm
Explanation:
From Newton's law of universal gravitation,
F₁ = Gm₁X/r₁² ................... Equation 1
Where F₁ = Force exerted on the on the small mass by the first sphere, m₁ = mass of the first sphere, X = mass of the small mass, r₁ = distance between the first sphere and the small mass.
Also,
F₂ = Gm₂X/r₂².................... Equation 2
Where F₂ = Force exerted by the second sphere on the small mass, m₂ = mass of the second sphere, X = mass of the small mass, r₂ = distance between the second sphere and the small mass.
Note: F₁ = F₂ ( Net force on the small mass due to the sphere is zero)
Therefore,
Gm₁X/r₁² = Gm₂X/r₂²
Equating the similar terms from both side of the equation,
m₁/r₁² = m₂/r₂².......................... Equation 3
Given: m₁ = 15.0 kg, m₂ = 7.00 kg, let r₁ = y cm, then r₂ = (20-y) cm
Substituting these values into equation 3
15/y² = 7/(20-y)²
15(20-y)² = 7y²
√{15(20-y)² = √7y²
√15×√(20-y)² = √7×√y² ( from the law of surd)
3.87(20-y) = 2.65y
77.4 - 3.87y = 2.65y
Collecting like terms and reordering the equation
2.65y+3.87y = 77.4
6.52y = 77.4
y = 77.4/6.52
y = 11.87 cm
Thus the position of the small mass on the x- axis = 11.87 cm
