Gravitational Force on three spheres

[TmrK]

Homework Statement

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?

Homework Equations

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

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.

 

[Doc Al]

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

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

Just call the mass m₃ and keep going.

 

[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=[itex]\frac{6.67x10^{-11 N}*2.85 kg*2.85 kg}{0.550m²}[/itex]

$\Sigma$F=[itex]\frac{5.4177075x10^-40}{.3025}[/itex]

$\Sigma$F= 1.79x10^-39

Then for solving for a,

a=[itex]\frac{1.79x10^-39}{2.85}[/itex]

a=6.28x10^-40

 

[vela]

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=[itex]\frac{6.67x10^{-11 N}*2.85 kg*2.85 kg}{0.550m²}[/itex]

This line is incorrect. Newton's second law says$$\sum_i {\vec{F}_3}_i = m_3 \vec{a}_3$$where F_3i 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.

 

[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?

 

[vela]

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'm not sure what that's supposed to mean.

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

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.

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?

Write down the two equations for the sums in the x and y directions.

 

[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.

 

[vela]

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.

 

[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.

 

[vela]

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:

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.

Which forces? There are six forces in this problem, two forces on each of the three masses.

Come to think of it, when it comes to the direction of the third sphere on the top point of the triangle,

What do you mean by "direction of the third sphere"? Do you mean the direction of the forces on the third sphere?

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.

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 F_1x, etc. One equation will let you solve for a_x, and the other will let you solve for a_y.

 

[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.

 

[vela]

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.

 

[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.

 

[Doc Al]

Exactly how does m₃ cancel out?

The gravitational force on m₃ is proportional to m₃. Thus when you compute the acceleration of m₃, the mass cancels.

And also, I realized that the force all along was G itself.

Not sure what you mean by that, since G is a constant.

 

[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?

 

[Doc Al]

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

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?

 

[TmrK]

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?

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.

 

[Doc Al]

Then if I really cannot use G again,

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?

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.

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.