# Tough physics question

1. Mar 2, 2006

### Finns14

Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 1.48 m. Two of the spheres have a mass of 2.18 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 of the initial acceleration of the third sphere?

Thats the question I know that some how I have to use the equations

F=G(m1m2/d^2) but besides for that I don't know. I know I probably the force towards one sphere planet and then separate that vector into x and y and then double that to include both sphere but I am still very confused. Thanks in advanced.

2. Mar 2, 2006

### Hootenanny

Staff Emeritus
$$F = \frac{GMm}{d^2}$$
and equate with newtons law ($F =ma$)
$$ma = \frac{GMm}{d^2}$$
The masses of the unknown sphere (m) will cancel giving:
$$a = \frac{GM}{d^2}$$
Remeber this is the only the acceleration of one sphere. As you say you will then consider the vectors (draw a diagram). You can double the force, but you will first have to find the centre of mass of the two known balls, then assume the force acts from that point.

Hope this helps

3. Mar 2, 2006

### Finns14

Ok question in the second equations m cancels on both sides but what is M?

4. Mar 2, 2006

### Finns14

Also I am confused on the angle you would use if you need to consider a point between 2 spheres? Can any one clarify that for me a little better?

5. Mar 2, 2006

### Hootenanny

Staff Emeritus
If you find the centre of mass of the two spheres of know mass, it will be the mid point between the two masses. You can then consider the two masses as a single mass of $2 \times 2.18 = 4.36kg$, this is the $M$, the combined mass of the two spheres. $d$ is then the distance from this point to the unknown mass.

6. Mar 2, 2006

### topsquark

Ummmm...There's nothing wrong with Hootenanny's post, but I think it might be more instructive to do it this way:

Let's think of it this way. Force is a vector. So what you need to do is to consider the force from EACH mass separately and add them vectorally to get the net force on your moving mass. (Then of course, netF=ma will cancel out the unknown mass in the equations.)

-Dan