How to find the equilibrium position of three masses (two of them fixed)?

[gnitsuk]

Homework Statement

How to find the equilibrium position of three masses (two of them fixed)

Relevant Equations

F=Gmm'/r^2

Question: Suppose that two bodies of masses m1 and m2 are a fixed distance d apart, and that both of them act on a third body of mass m. Find the position of the third body such that the two forces are in equilibrium.

I have solved this question and obtained the answer listed in the back of the book from which this question comes. The given answer is:

Equilibrium point is on the line joining the two bodies, at a distance d/(1 + sqrt(y)) from the body of mass m1. Where y = m2/m1.

The solving of this requires the solution of a quadratic equation.

In my case I let D be the distance of m from m1 and the quadratic I needed to solve was d^2*(y-1) + 2dD - d^2 = 0.

I used the quadratic formula to solve this for D, and to get the book answer I took the positive sign in that formula.

My question is, if I take the negative sign I and up with the answer: d/(1 - sqrt(y))

In what way is this answer not valid (I can see that if y > 1 then m2 > m1 and so this would give a negative distance, and that m cannot be to the left of m1 for equilibrium, but if m2 < m1 then sqrt(y) < 1 and so d/(1 - sqrt(y)) would be positive).

{Aside: I appreciate that I can do it this way: (d-D)^2=D^2*y therefore (d-D)^2=D^2=(sqrt(y))^2 and then taking the root of each side I get to the answer in the book, but I was trying to understand the negative root in the approach where I solve the quadratic with the quadratic formula).

Thanks for any help.

 

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[haruspex]

The equation you write down to start with, $m_1/x^2=m_2/(d-x)^2$, is only correct if $0<x<d$.
Outside that range, the equation becomes $m_1/x^2=-m_2/(d-x)^2$.

 

[Sai prince]

In the situation you described where two bodies of mass m1 and m2 act on a third body of mass m, you obtained the quadratic equation d^2*(y-1) + 2dD - d^2 = 0 well, where D is the distance of the third body from the body of mass m1. Solving this quadratic equation will yield two possible solutions: D = d/(1 + sqrt(y)) and D = d/(1 - sqrt(y)).

You note correctly that if y > 1 (implying m2 > m1), then solving D = d/(1 - sqrt(y)) will yield a negative path, which does not make physical sense in this case in addition that the equilibrium position cannot move to the left m1, e.g. You have searched right. Thus, in this particular case, only the positive solution D = d/(1 + sqrt(y)) is valid and represents the equilibrium position of the third body

It is important to consider the physical constraints and to interpret the solution in the context of the problem to prove it. In this respect, negative radicalization is incompatible with a balanced position of material reason.

 

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[haruspex]

In the situation you described where two bodies of mass m1 and m2 act on a third body of mass m, you obtained the quadratic equation d^2*(y-1) + 2dD - d^2 = 0 well, where D is the distance of the third body from the body of mass m1. Solving this quadratic equation will yield two possible solutions: D = d/(1 + sqrt(y)) and D = d/(1 - sqrt(y)).

You note correctly that if y > 1 (implying m2 > m1), then solving D = d/(1 - sqrt(y)) will yield a negative path, which does not make physical sense in this case in addition that the equilibrium position cannot move to the left m1, e.g. You have searched right. Thus, in this particular case, only the positive solution D = d/(1 + sqrt(y)) is valid and represents the equilibrium position of the third body

It is important to consider the physical constraints and to interpret the solution in the context of the problem to prove it. In this respect, negative radicalization is incompatible with a balanced position of material reason.

As I read it, @gnitsuk understood all that, but was puzzled as to how the extraneous solution had arisen. I explained that in post #2.

Note that the analogous question can be asked in electrostatics, but in that case the 'mass' can be negative. A correct procedure is to look for solutions between them and solutions outside them separately, writing the appropriate starting equation for each case, and filtering the answers according to the band in which that equation is valid.

 

[gnitsuk]

Thanks for all the replies, they have let me understand the issue and resolve my confusion.