# Homework Help: Gravitational Force on three spheres

1. Sep 13, 2011

### TmrK

1. The problem statement, all variables and given/known data
Three uniform spheres are located at the corners of an equilateral triangle. Each side of the triangle has a length of 0.550 m. Two of the spheres have a mass of 2.85 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?

2. Relevant equations
F= G*m1m2/r^2
G=6.67x10^-11

3. The attempt at a solution
I took the first formula & wrote down two separate ones as

F= G*m1m3/r^2
F= G*m2m3/r^2

where m3 = the unknown mass.

Basically, I do not know how I would find the third "unknown mass" if the Force is not known.

Last edited: Sep 13, 2011
2. Sep 13, 2011

### Staff: Mentor

Hint: You are asked to find the acceleration, not the mass or force.

Just call the mass m3 and keep going.

3. Sep 13, 2011

### TmrK

Before I reply, would I use the formula

$\Sigma$F=ma

to solve for a, which would give me

a=$\Sigma$F/m?

Here is my work:

$\Sigma$F=$\frac{6.67x10-11 N*2.85 kg*2.85 kg}{0.550m2}$

$\Sigma$F=$\frac{5.4177075x10-40}{.3025}$

$\Sigma$F= 1.79x10-39

Then for solving for a,

a=$\frac{1.79x10-39}{2.85}$

a=6.28x10-40

Last edited: Sep 13, 2011
4. Sep 13, 2011

### vela

Staff Emeritus
This line is incorrect. Newton's second law says$$\sum_i {\vec{F}_3}_i = m_3 \vec{a}_3$$where F3i are the individual forces acting on mass 3 and a3 is the acceleration of mass 3. In your original post, you have expressions for the magnitude of the two forces acting on mass 3. You need to use the geometry of the problem to figure out their directions, so you can calculate the vector sum of those two forces.

A good idea would be to draw a picture. Place the two known masses on the x-axis, centered around the origin, and place mass 3 at the appropriate spot on the y-axis.

5. Sep 13, 2011

### TmrK

Okay, from what I understand, being that this is an equilateral triangle, each angle would be 60$\circ$. Therefore, each sphere on the x-line is Fcos(60$\circ$), and the third sphere on the y-line is Fsin(60$\circ$).

I could then use F=Gm1m/r2 or F=Gm2m/r2 to find out the mass of the third sphere.

From there, I can use the force found out from Fsin(60$\circ$) and the other two forces from cosine to find out the acceleration. Correct or did I go into another wrong route?

6. Sep 13, 2011

### vela

Staff Emeritus
I'm not sure what that's supposed to mean.
No, you can't solve for m3. It turns out you don't need to know it because it will cancel out. So, as Doc Al suggested, just call it m3, write down your equations, and it will divide out.
Write down the two equations for the sums in the x and y directions.

7. Sep 13, 2011

### TmrK

I mean 60 degrees. And the x & y equations would refer to the displacement equations, yes?

And to sum it all up, although I listed all of the equations, I would not know what to start out with and which equation to end with.

8. Sep 13, 2011

### vela

Staff Emeritus
I know what the 60∘ stood for. I don't understand what "each sphere on the x-line is Fcos(60∘), and the third sphere on the y-line is Fsin(60∘)" is supposed to mean.

There are two forces on the third sphere. Describe in words which way they point.

9. Sep 13, 2011

### TmrK

What I mean is that since each of these spheres are in a position that makes up an equilateral triangle, the direction of these forces are 60 degrees north of east AND north of west. Come to think of it, when it comes to the direction of the third sphere on the top point of the triangle, if split in half one direction would be 30 degrees east of north and 30 degrees west of north (assuming the top of the triangle is in the positive y-coordinate range. The total force delivered by the third sphere on top of the triangle would be the total of the two separate force as mentioned earlier.

10. Sep 13, 2011

### vela

Staff Emeritus
Sorry if it seems I'm being pedantic here, but it's really hard to figure out what you mean when you use language so imprecisely. For example:
Which forces? There are six forces in this problem, two forces on each of the three masses.
What do you mean by "direction of the third sphere"? Do you mean the direction of the forces on the third sphere?
You mean the total or net force exerted on the third sphere, right? When you write "delivered by," it sounds like you're talking about the gravitational force on the other spheres due to their attraction to the third sphere.

Back to the problem, so I think what you've said is that there are two forces on the third sphere. One force — let's call it F1 — points downward and to the left, making a 30-degree angle with the y-axis, and the other force — we'll label this on F2 — points downward and to the right, making a 30-degree angle with the y-axis. So far, so good.

As you wrote in your first post, the magnitudes of the forces are
\begin{align*}
|\vec{F}_1| &= \frac{Gm_1m_3}{r_{13}^2} \\
|\vec{F}_2| &= \frac{Gm_2m_3}{r_{23}^2}
\end{align*}
where r13 = r23 = 0.550 m and m1=m2=2.85 kg.

Now you want to break these forces into horizontal and vertical components. Then using F=ma, you'll get the two equations:
\begin{align*}
\sum F_x &= F_{1_x} + F_{2_x} = m_3a_x \\
\sum F_y &= F_{1_y} + F_{2_y} = m_3a_y
\end{align*}
You need to plug in what you get for F1x, etc. One equation will let you solve for ax, and the other will let you solve for ay.

11. Sep 13, 2011

### TmrK

So in short, I just need to find out what exactly is the magnitude of the forces before I can do anything else. But here's the problem: there is NOTHING given as to exactly on how much force there is between the spheres. That part is not given.

12. Sep 13, 2011

### vela

Staff Emeritus
No, that's not correct. I've essentially told you how to do the whole problem, including addressing the point you're bringing up yet again. I'll let you think about it for a while.

13. Sep 13, 2011

### TmrK

Exactly how does m3 cancel out? And also, I realized that the force all along was G itself. Please tell me I've gone one step forward in the positive direction.

14. Sep 14, 2011

### Staff: Mentor

The gravitational force on m3 is proportional to m3. Thus when you compute the acceleration of m3, the mass cancels.
Not sure what you mean by that, since G is a constant.

15. Sep 14, 2011

### TmrK

To find F, all I have to do is use G to replace as F, so really I'll be using G twice, yes?

16. Sep 14, 2011

### Staff: Mentor

What do you mean by 'use G'? G is a constant that appears in the Newtonian formula for gravitational force. You cannot replace F by G. Or do you mean something else?

17. Sep 14, 2011

### TmrK

Then if I really cannot use G again, then I would have to use all of the components to mind m3. However I can't being that there is no magnitude of the force mentioned at all. The problem alone I posted does not give me of what the magnitude of the forces are. Not one time is it even mentioned. So really, I cannot do anything if I don't have the magnitude of the force to work with.

18. Sep 14, 2011

### Staff: Mentor

What are you talking about? G is just a constant. Whenever you calculate the gravitational force between two spheres you'll use G. So what?
You have everything needed to find the acceleration of m3. You have the mass of m1 and m2, and the distance between the spheres. You don't have a value for m3, but you don't need one.

To find the force between m1 and m3, use Newton's law of gravity. (The expression will be in terms of m3.)

To find the force between m2 and m3, use Newton's law of gravity. (The expression will be in terms of m3.)

Add up those force vectors to find the net force on m3. (The expression will be in terms of m3.)

Use Newton's 2nd law to find the acceleration of m3.