How do I find the force an object exerts on adjacent objects when I push on it?

Question

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

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?

Answer 1

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.