Answer:
a. the net force F = 1.684 × 10⁻⁵ N in the direction of the 555 kg mass. b. x = 0.1576 m or -0.8205 m
Explanation:
a. The force of gravitational attraction between two masses m₁ and m₂ separated by a distance, r is given by F = Gm₁m₂/r².
Let m₁, m₂ and m₃ be the masses and m₂ is midway between m₁ and m₃.
The force of attraction between m₁ and m₂ is F₁ = Gm₁m₂/r² and the force of attraction between m₂ and m₃ is F₂ = Gm₂m₃/r². The net force is F₁ - F₂ = Gm₁m₂/r² - Gm₂m₃/r²= Gm₂/r²(m₁ - m₃).
m₁ = 255 kg, m₂ = 555 kg and r = 0.390m/2 = 0.195 m
F = 6.67 × 10⁻¹¹ Nm²/kg² × 32 kg ( 255 - 555) kg/ (0.195 m)²
= 213.44 × 10⁻¹¹ (- 300)/(0.038025)N
= -1683944.77 × 10⁻¹¹ N
= -1.684 × 10⁻⁵ N.
So, the net force F = 1.684 × 10⁻⁵ N in the direction of the 555 kg mass.
b. Let x be the distance from the 255 kg mass and d - x be the distance from the 555 kg mass at zero net force, then
F = Gm₁m₂/x² = Gm₂m₃/(d - x)²
Simplifying,
(d - x)²/x² = Gm₂m₃/Gm₁m₂ = m₃/m₁.
Taking square root of both sides, we have
(d - x)/x = ± √ (m₃/m₁)
d/x - 1 = ± √ (m₃/m₁)
d/x = 1 ± √ (m₃/m₁)
x = d/(1 ± √ (m₃/m₁)) = 0.390/(1 ± √ (555/255)) = 0.390/(1 ± √ (2.1765)) = 0.390/(1 ± 1.4753)
x = 0.390/2.4753 or 0.390/-0.4753 = 0.1576 m or -0.8205 m.
x = 0.1576 m ≅ 0.158 m and -0.8205 m ≅ -0.821 m
x = 0.158 m or -0.821 m