How do I find the force an object exerts on adjacent objects when I push on it? 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 http://i.minus.com/jbggsUzWrXFbzS.png
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?
 A: Some assumptions:


*

*The masses of the circles are known

*There is no friction between the circles (no tangential forces)

*If the circles move the small ones do not loose contact with the big one

*There is no external forces but $\vec{F}$

*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.


*

*When does the system have a solution?

*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.

*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 $\vec{N}_i$.

*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?

*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.
