F = ma Exam 2011 # 12 (Gravitational Attraction of Rods and Balls)

[SignaturePF]

Homework Statement

12. 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?
(A) √1/3F
(B)√3 / 2 F
(C) F
(D)√3F
(E) 2F

The correct answer is C

Homework Equations

F_g = GmM / r^2

The Attempt at a Solution

Why is the force still F?
Is it because the only force that the rod has to equal in magnitude is the gravitational acceleration between two balls at a time?

 

[Simon Bridge]

Take a look at the top ball on the triangle - what is the total force on it?
What proportion of that force goes along each rod?

 

[SignaturePF]

The top ball has Force Gm^2 / R^2 (where R is length of triangle) on the left
same on the right. We'll call this force F, as done in the problem.
However, when these two forces are added the x components of each cancel out, leaving the net force as:
F_net = 2Fsin60
F_net = root(3)F
so each rod will face root(3)/2F, where did I go wrong?

 

[Simon Bridge]

You did the proportions back to front.
Draw the vectors on the diagram.

 

[SignaturePF]

I don't follow what you mean. When I draw the vectors on the diagram I get the same result.

 

[Simon Bridge]

Surely you get a vector of magnitude F pointing along one rod and another vector of magnitude F pointing along the other rod - the total force being the sum of these vectors?