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Suppose that I have a situation like the figure below, with several circular objects in contact. I know:

  • $\vec{A},\vec{B},\vec{C}$: the positions of the points of contact between circles
  • $\vec{O}$: the center of one circle
  • $\vec{F}$: the force applied to that circle

Image of circles

I need to find the forces $\vec{N}_1,\vec{N}_2,\vec{N}_3$ which the main circle exerts on the other circles. How can I do that?

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Hi tvolodimir, and welcome to Physics Stack Exchange! This site is not a problem-solving service; it's for conceptual questions about physics. If you edit your question to explain exactly what the concept that is giving you trouble is and ask about that, rather than asking how to do the problem, I'll be happy to reopen this. For more information, see our FAQ on homework-like questions. –  David Z Dec 15 '11 at 9:03
    
@DavidZaslavsky - sorry. I write simple 2d physics engine. And I have difficulty with this problem. When balls have contacts i dont now how to calculate the forces acting when each ball to another which touched. In topic I describe some type of case. In fact I would like to learn how to solve the problem when the circle tangent with N circles. –  tvolodimir Dec 15 '11 at 10:15
    
I've edited your question to reflect what I think you meant. (The solid-state-physics tag threw me off for a while; this problem actually has nothing to do with solid state physics) Please read this over and see if it accurately represents what you wanted to ask. Even if this is what you meant to ask, we will need some more information to solve the problem. For example, are the three smaller circles fixed in place at their centers? –  David Z Dec 16 '11 at 20:57
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1 Answer 1

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Some assumptions:

  1. The masses of the circles are known
  2. There is no friction between the circles (no tangential forces)
  3. If the circles move the small ones do not loose contact with the big one
  4. There is no external forces but $\vec{F}$
  5. The small circles do not interact

The 2nd and the 3rd Newton's laws (with assumptions 2. and 4.) give for the big circle: $$ M\vec{a} = \vec{F} - \sum_i \vec{N}_i, \qquad (1) $$ where $M$ and $\vec{a}$ are the mass and the acceleration of the big circle.

Assumption 3. means that the surfaces of the circles at the contact point move with the same normal acceleration i.e.: $$ \frac{\vec{N}_i}{m_i} = \vec{a}_i = \frac{\vec{a} \cdot \vec{N}_i}{N_i}, \qquad (2) $$ where $m_i$ and $\vec{a}_i$ are the mass and the acceleration of the $i$-th small circle.

Let's direct $x$ coordinate axis along $\vec{F}$ and $y$ axis in proper direction to get right coordinate system.

Let's denote the angle between $OA$ and $\vec{F}$ as $\beta_1$, the angle between $OB$ and $\vec{F}$ as $\beta_2$ the angle between $OC$ and $\vec{F}$ as $\beta_3$ and so on.

Note that $\beta_1$ and $\beta_2$ are negative while $\beta_3$ is positive.

Now we can rewrite equations (1) and (2) as follows: $$ \begin{aligned} M a_x &= F - \sum_{j=1}^n N_j \cos\beta_j; \\ M a_y &= - \sum_{j=1}^n N_j \sin\beta_j; \\ \frac{N_i}{m_i} &= a_x \cos\beta_i + a_y \sin\beta_i, \quad &i = 1,\ldots,n; \end{aligned} $$

We have $2+n$ variables ($a_x$, $a_y$, $N_1,\ldots,N_n$) and $2+n$ equations. This should be enough.

Using the first two equations one can remove $a_x$ and $a_y$ and get a linear system of $n$ equations with $n$ variables: $$ \sum_{j=1}^n \left( \frac{1}{M} \cos\beta_i \cos\beta_j + \frac{1}{M} \sin\beta_i \sin\beta_j + \frac{1}{m_i} \delta_{ij} \right) N_j = \frac{F}{M} \cos\beta_i, \qquad (3) $$ $$ i = 1,\ldots,n. $$

Whew. Now the most interesting part.

  1. When does the system have a solution?

  2. If one of the small circles is fixed we can just put its mass to infinity and remove the term $\frac{1}{m_i}\delta_{ij}$ from the corresponding equation.

  3. If there are external forces (remove assumption 4.) we can add them to eq. (1) or/and (2). They will contribute to the right part of (3) if they do not depend on $\vec{N}_i$ or to the left part of (3) if they depend on $\vec{N}_i$.

  4. If $|\beta_i| > \pi/2$ the small circle can be glued (negative $N_i$). I say "can" because $\vec{a}$ is not obliged to be directed along $\vec{F}$. If one just refuse the circles with negative $N_i$ and solve the equations again the direction of $\vec{a}$ can change and some of the refused circles will get positive $N_i$.
    What to do then?

  5. EDIT: Assumption 3. is the main point of the solution because it adds $n$ equations (2) and makes the system complete. This can not be done in the case of collision when the interaction should be considered as instant.

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Thanks. I solved the problem for N-balls and M-touching point. You may see my result link (recommended browser Crome). In page balls can move by mouse and for all balls you can change force by mouse. Sample - link. --- And now i have another problem. - How to complete system when balls are collided with other. Conservation of momentum and Conservation of energy - is two equations. and it solved for two balls. How to be with three and more ball? –  tvolodimir Jan 2 '12 at 14:39
    
@tvolodimir, wow! Looks nice! As for collision... Interaction of 3 and more bodies is a very complicated problem that is not solved so far. Usually collision is treated as an instant process. You can always assume that only two bodies collide at any moment of time. The possibility of two instant process to take place simultaneously is zero. –  Maksim Zholudev Jan 2 '12 at 16:05
    
> The possibility of two instant process to take place simultaneously is zero. Let the case. But in case when a lot of balls lying on the ground and another flying at them from above. Then the momentum of the ball must immediately redistribute all other and your proposal is not suitable, because let falling tangent to each other. then that is another tangent to at least two, and so on ... Tips in this case what to do? sample link –  tvolodimir Jan 2 '12 at 16:28
    
@tvolodimir Let's assume that the time of collision is enough for distribution of the forces among the bodies. In this case you can use your program to calculate the forces: 1) put the incoming ball to the collision point; 2) apply some external force to it in the direction of the velocity; 3) calculate the forces; 4) remove the external force. The change of momentum for each body will be proportional to the force applied to it. The absolute value can be calculated with the energy conservation law. –  Maksim Zholudev Jan 2 '12 at 20:45
    
1. > The change of momentum for each body will be proportional to the force applied to it. --- can get some laws or other substantiation for this? 2. > Interaction of 3 and more bodies is a very complicated problem that is not solved so far. --- what kind of problem? cant complete (or solve) of mathematical model (or physical model)? –  tvolodimir Jan 2 '12 at 21:52
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