USAPhO 2011 F=ma exam #12, (gravitational force, 3 body)

[Agrasin]

http://www.aapt.org/physicsteam/2012/upload/exam1-2011-1-3-answers_1.pdf

Homework Statement

You are given a large collection of identical heavy balls and lightweight rods. When two balls are placed at the ends of one rod and interact through their mutual gravitational attraction (as is shown on the left), the compressive force in the rod is F. Next, three balls and three rods are placed at the vertexes and edges of an equilateral triangle (as is shown on the right). What is the compressive force in each rod in the latter case?

Homework Equations

F = Gm² /r²

The Attempt at a Solution

In the first case (2 balls on one rod), the compressive force is 2*Gm² /r².

In the second case (3 balls, 3 rods, equilateral triangle), the compressive force can be measured by looking at one rod and the forces acting on it.

If you take the left-hand rod, there is a compressive force F on the rod, which would be there even without the triangle. Now, because of the triangle set-up, there are two more compressive forces working at angles of 30 to the horizontal. I need to add the vertical components of these forces to the original compressive force.

2* Fsin(30) = F

The additional compressive force due to the adjacent rods and ball is F. So, F(original) + F(due to triangle set up) = 2F.

2F is choice E. However, the answer is C, just F. How??

 

[haruspex]

In the first case (2 balls on one rod), the compressive force is 2*Gm² /r².

No. Consider the forces acting on one ball.

In the second case (3 balls, 3 rods, equilateral triangle), the compressive force can be measured by looking at one rod and the forces acting on it.

If you take the left-hand rod, there is a compressive force F on the rod, which would be there even without the triangle. Now, because of the triangle set-up, there are two more compressive forces working at angles of 30 to the horizontal. I need to add the vertical components of these forces to the original compressive force.

2* Fsin(30) = F

The additional compressive force due to the adjacent rods and ball is F. So, F(original) + F(due to triangle set up) = 2F.

I was not able to follow your logic. You would do better to consider the forces on one ball, as above.

 

[Agrasin]

The force on one ball? F = Gm2 /r2 pulling it along one rod, and again F = Gm2 /r2 pulling it along the other rod, 60 degrees from the first.

 

[haruspex]

The force on one ball? F = Gm2 /r2 pulling it along one rod, and again F = Gm2 /r2 pulling it along the other rod, 60 degrees from the first.

Not entirely sure how you are using F there. Is it the F in the statement of the problem, or generic?
Anyway, there are four forces on each ball in the three ball system: two gravitational and two mechanical.
Do you understand that the compressive force in the two ball system is just Gm²/r²?

 

[Agrasin]

Okay, big F is generic force.

Little f is the compressive force experienced by a rod in a two-ball system. Why would f be just Gm^2 / r^2? Wouldn't it be twice that because there are two balls squeezing, each of which experiences a force of Gm^2 / r^2 ?

Two gravitational and two mechanical forces on each ball in the 3-ball system? So each gravitation F is Gm^2 / r^2. Each mechanical (exerted in opposite direction by rod) must be the same because the ball is at rest.

...How do I get to the answer from there?

 

[Doc Al]

Wouldn't it be twice that because there are two balls squeezing, each of which experiences a force of Gm^2 / r^2 ?

Imagine a spring with a constant of 500 N/m. To compress it by one centimeter, what force must you exert on each end?

Hint to get the answer: Consider all the forces acting on each ball.

 

[haruspex]

Why would f be just Gm^2 / r^2? Wouldn't it be twice that because there are two balls squeezing, each of which experiences a force of Gm^2 / r^2 ?

Let's take the most complex situation: a massive object subject to a set of forces, undergoing acceleration. But to simplify it a little, it's a rod under compression. The compressive force will not be the same along its length. If you imagine cutting the rod at some point into two parts, the compressive force at that point is the force each part exerts on the other.
For a massless rod, the compressive force in it is therefore the force exerted on each end (the two forces are necessarily the same), not their sum.

Two gravitational and two mechanical forces on each ball in the 3-ball system? So each gravitation F is Gm^2 / r^2. Each mechanical (exerted in opposite direction by rod) must be the same because the ball is at rest.

That is indeed the right answer. The gravitational force ball A exerts on ball B is balanced by the force the rod joining them exerts on B. It's sort of obvious, but it doesn't hurt to go through the actual equations and see it emerge.

 

[Agrasin]

Thanks guys, that makes sense. The compressive force is just f = Gm^2 / r^2.