Question

A 275 kg object and a 575 kg object are separated by 0.310 m. At what position (other than an infinitely remote one) can a 48.0 kg object be placed between the others so as to experience a net force of zero

Answer 1

x = 0.1267 m

Step-by-step explanation:

For this exercise we use the summation of the forces where the force is the gravitational force

F =
`G (m_1m_2)/( r^2)`

the equilibrium condition is

∑ F = -F₁₃ + F₂₃ = 0

F₁₃ = F₂₃

The gravitational force is attractive, so for the force to be zero the third body must be between the union line of the first two bodies

In this exercise they indicate the masses of the bodies, m₁ = 275 kg and m₂= 575 kg separated by a distance of r = 0.310 m, the third body of masses m₃ = 48.0 kg must be on the union line between them.

Let's set a reference system on body 1 with the x axis aligned with the two initial bodies

Let's look for the forces

F₁₃ =
`G (m_1 m_3)/(x^2)`

F₂₃ =
`G (m_2m_3)/((d-x)^2)`

we substitute

G \frac{m_1 m_3}{x^2} = G \frac{m_2m_3}{(d-x)^2}

`(m_1)/(x^2) = (m_2)/((d-x)^2)`

we solve

(d-x)² =
`(m_2)/(m_1)` x²

we substitute the values

(0.310 - x)² =
`(575)/(275)` x²

0.310² - 2 0.310 x + x² = 2.09 x²

1.09 x² + 0.620 x - 0.0961 = 0

let's solve the quadratic equation

x² + 0.5688x - 0.0881 = 0

x = [-0.5688 ±
`√(0.5688^2 + 4 0.0881 )`] / 2

x = [-0.5688 ± 0.82221] / 2

x₁ = 0.1267 m

x₂ = -0.6955 m

we see that the position between the two masses is x = 0.1267 m