forcesnewtonian-mechanics
Suppose that I have a situation like the figure below, with several circular objects in contact. I know:
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?
They don't point toward the center.
You are thinking of the special case for circular motion with constant speed. In that case, the velocity and acceleration vectors are perpendicular, implying the acceleration points toward the center.
However, an object moving freely in a vertical loop will slow down on the way up and speed up on the way down. This is not constant speed, so one shouldn't expect the acceleration to be perpendicular to the velocity. In fact, it isn't.
You can always break up (decompose) the acceleration vector into a component parallel to the velocity ($a_\perp$, which changes speed) and perpendicular to the velocity ($a_{||}$, which changes direction). When you're slowing down, the component parallel to the velocity will be opposite to the velocity; when you're speeding up, the parallel component will be in the same direction.
This can be applied to the vertical hoop situation. Slowing on the way up, speeding on the way down. Only when the speed is not changing (e.g., at the very top of the loop, or at the very bottom) will there be no parallel component, and hence the acceleration vector at those special points would only have a perpendicular component.
I find that an easy, never-fail way to analyze the forces acting on an object in circular motion is:
(forces toward the center) - (forces away from the center) = mv²/r.
mv²/r is the amount of center-directed force necessary to make the object, which constantly wants to move in a straight line, move in a circle instead. It's often called the centripetal force, but that's just a label we put on the combination of identifiable forces like gravity, normal force, tension, etc., that gives us mv²/r.
Here is a force diagram (a.k.a. free body diagram) for any car-on-a-(circular)-hill scenario:
Notice that there are no forces labeled as centripetial force, or mv²/r. Our equation becomes $$mg(sin\theta) - F_N = mv²/r,$$ where $\theta$ is the angle that the position vector of the car makes with the horizontal, and FN is the normal force exerted on the car by the hill, or road.
At the top of the hill, where $\theta$ is 90°, the equation reduces to $$mg - F_N = mv²/r$$
So yes, in the case of a car going over a circular hill, we see that the centripetal force is the difference between the car's own gravitational force and the normal force exerted on it by the hill, as you suggested.
If we rearrange the equation: $$F_N = mg(sin\theta) -mv²/r$$
We see that only when v = 0 is the normal force exerted by the road on the car equal to the gravitational force on the car. In that situation they are equal and they oppose each other, so they look like a Newton's Third Law pair. But they are not. Note that they act on the same object.
There are two Newton's Third Law pairs in the situation. One pair consists of the force mg(sin$\theta$), exerted on the car by the mass of the Earth, and an equal force exerted on the mass of the Earth by the car. The other pair consists of FN exerted on the car by the road, and an equal normal force exerted on the road by the car. That pair gets smaller as the velocity increases, as shown by the latest equation above. When FN gets to zero, the car leaves the road tangentially.
Best Answer
Some assumptions:
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.