Question

Three uniform spheres are located at the corners of an equilateral triangle. each side of the triangle has a length of 1.20 m. two of the spheres have a mass of 2.80 kg each. the third sphere (mass unknown) is released from rest. considering only the gravitational forces that the spheres exert on each other, what is the magnitude and direction of the initial acceleration of the third sphere?

Answer 1

To calculate the initial acceleration of the third sphere in an equilateral triangle, we need to consider the gravitational forces between the spheres. Using Newton's law of gravitation and Newton's second law, we can calculate the magnitude and direction of the acceleration. The magnitude of the initial acceleration is approximately 0.035 m/s^2 and the direction is towards the center of the triangle.

Step-by-step explanation:

To calculate the initial acceleration of the third sphere, we need to consider the gravitational forces between the spheres. Each sphere exerts a gravitational force on the other spheres according to Newton's law of gravitation: F = G * (m1 * m2) / r^2, where F is the force, G is the gravitational constant, m1 and m2 are the masses of the two spheres, and r is the distance between them.

In this case, the two anchored spheres have a mass of 2.80 kg each, while the third sphere has a mass of unknown (let's call it m).

Using the equation for gravitational force, we can calculate the force between the two anchored spheres and the third sphere, and then use Newton's second law (F = ma) to find the acceleration:

a = F / m = (G * (m1 * m2) / r^2) / m = G * m1 * m2 / (r^2 * m)

Now, we can substitute the given values: G = 6.67 * 10^-11 Nm^2/kg^2, m1 = m2 = 2.80 kg, and r = 1.20 m.

After calculating the result, we find that the magnitude of the initial acceleration of the third sphere is approximately 0.035 m/s^2. The direction of the acceleration is towards the center of the equilateral triangle.

Answer 2

Refer to the diagram shown below.

Because of symmetry, equal forces, F, exist between the sphere of mass m and each of the other two spheres.
The acceleration of the sphere with mass m will be vertical as shown.

The gravitational constant is G = 6.67408 x 10⁻¹¹ m³/(kg-s²)

Calculate F.
F = [ (6.67408 x 10⁻¹¹ m³/(kg-s²))*(m kg)*(2.8 kg)]/(1.2 m)²
= 1.2977 x 10⁻¹⁰ m N

The resultant force acting on mass m is
2Fcos(30°) = 2*(1.2977 x 10⁻¹⁰m N)*cos(30°) = 2.2477 x 10⁻¹⁰m N

If the initial acceleration of mass m is a m/s², then
(m kg)(a m/s²) = (2.2477 x 10⁻¹⁰m N)
a = 2.2477 x 10⁻¹⁰ m/s²

Answer:
The magnitude of the acceleration on mass m is 2.25 x 10⁻¹⁰ m/s².
The direction of the acceleration is on a line that joins mass m to the midpoint of the line joining the known masses.