A 110-kg object and a 410-kg object are separated by 3.80 m.

(a) Find the magnitude of the net gravitational force exerted by these objects on a 41.0-kg object placed midway between them. N
(b) At what position (other than an infinitely remote one) can the 41.0-kg object be placed so as to experience a net force of zero from the other two objects?

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