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1. The problem statement, all variables and given/known data
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?
2. Relevant equations
F = Gm2 /r2
3. 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??
1. The problem statement, all variables and given/known data
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?
2. Relevant equations
F = Gm2 /r2
3. 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??
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