Answer:
a) Fₙ = 2,273 10⁻⁷ N and b) x₃ = 1,297 m
Explanation:
This problem can be solved using the law of universal gravitation and Newton's second law for the equilibrium case. The Universal Gravitation Equation is
F = G m₁ m₂ / r₁₂²
a) we write Newton's second law
Σ F = F₁₃ - F₃₂
Body 1 has mass of m₁ = 110 kg and we will place our reference system, body 2 has a mass of m₂ = 410 kg and is in the position x₂ = 3.80 m
Body 3 has a mass of m₃ = 41.0 kg and is in the middle of the other two bodies
x₃ = (x₂-x₁) / 2
x₃ = 3.80 / 2 = 1.9 m
Fₙ = -G m₁ m₃ / x₃² + G m₃ m₂ / x₃²
Fₙ = G m₃ / x₃² (-m₁ + m₂)
Calculate
Fₙ = 6.67 10⁻¹¹ 41.0 / 1.9² (- 110 + 410)
Fₙ = 2,273 10⁻⁷ N
Directed to the right
b) find the point where the force is zero
The distance is
x₁₃ = x₃ - 0
x₃₂ = x₂ -x₃= 3.8 -x₃
We write the park equation net force be zero
0 = - F₁₃ + F₃₂
F₁₃ = F₃₂
G m₁ m₃ / x₁₃² = G m₃ m₂ / x₃₂²
m₁ / x₁₃² = m₂ / x₃₂²
Let's look for the relationship between distances, substituting
m₁ / x₃² = m₂ / (3.8 - x₃)²
(3.8 - x₃) = x₃ √ (m₂ / m₁)
x₃ + x₃ √ (m₂ / m₁) = 3.8
x₃ (1 + √ m₂ / m₁) = 3.8
x₃ = 3.8 / (1 + √ (m₂ / m₁))
x₃ = 3.8 / (1 + √ (410/110))
x₃ = 1,297 m
When body 3 is in this position the net force on it is zero
