# The coin drop experiment

there is the experiment called coin drop. you have a glass and a coin on top of a surface such as a paper. so basically when you move the paper slowly the coin remains on it but when you move it really fast the coin drops into the glass. I asked my teacher why is that and he said that it is because of newton's first law, but the law talks about objects with no forces acting on them and in this case force do act on the coin. So I'm confused, why when I move the paper slowly the coin keeps with it but when I move it fast it falls into the glass. If it's because of newton's first law then should'nt the coin fall even if I move the paper slowly?

M2, is the paper and M1 is the coin. We are exerting a rightward force $F$ on the paper. It's obvious that the paper moves because of the force we are exerting on it, and it's acceleration is decreased by the friction force between the coin and the paper.(The paper is going rightwards, but the coin is exerting a friction force to the paper, leftwards). Based on Newton's third, because the coin is exerting a force on the paper leftwards, the paper is exerting a rightward force on the coin too. That's the reason the coin will stick to the paper when you're pulling the paper slowly. But the question is what happens when we increase this force. Firstly, the static friction force can not exceed a certain value. Which means if $F$ gets too high, the friction force can not keep up with $F$ and at some point the acceleration of the paper becomes more than the coin. So the coin can not keep up with the paper. So after a while, the coin just slips and falls into the glass.

I want to calculate the minimum force $F$ so that it causes the coin to slip. Imagine the maximum static friction force between the paper and the coin is $f_{k.max}$. So the maximum force that can be exerted on the coin is $f_{k.max}$. Which means the maximum acceleration of the coin is:

$a = F/M1 = \dfrac{f_{k.max}}{M1}$

So if you want to make the coin slip on the paper, you have to make the system move with an acceleration more that $a$. Which means:

$F > a * (M1 + M2) => F > \dfrac{(M1 + M2)f_{k.max}}{M1}$

If the force is below this value, then the coin doesn't slip, because the force exerted on it is not above $f_{k.max}$ but if the force exceeds this value, then $M1$ can not keep up with M2.

Think of the coin and the paper, as a whole system, only when the acceleration exceeds some value, the system breaks.

Forces on the paper:

$\sum{F_{paper}} = F - f_{k} = M_2 * a$

Forces on the coin:

$\sum{F_{coin}} = f_{k} = M_1 * a$

And we know that:

$f_k \leq f_{k.max} => M_1 * a \leq f_{k.max} => a \leq \dfrac{f_{k.max}}{M_1}$

$=> M_2 * a + f_{k} = F, => F \leq M_2 * \dfrac{f_{k.max}}{M_1} + f_{k.max} => F\leq \dfrac{(M_1 + M_2)f_{k.max}}{M_1}$

• but even if I move the paper slowly the force applying on the paper will still be greater than the force of friction applying on the coin, because I will always have to apply a greater force than the friction of the coin on the paper to even make it move? so theoreticaly even if I move it slowly for a long period of time, eventually it will fall down at some point? – Solomon May 2 '17 at 22:44
• I have updated the answer, please take a look at it. – Soroush khoubyarian May 3 '17 at 11:21
• If the force we apply on the paper be below the Fk max wouldn't the coin slip forward and the paper will try to get backwards? Because the paper will have a force backwards because of the friction force the coin applies on it backwards and the coin will go forward because of the friction force the paper apply on it forward(Newton's third law)? – Solomon May 3 '17 at 11:56
• No. The force that the coin exerts on the paper(which is backwards) is less than $F$. I have updated the question again... – Soroush khoubyarian May 3 '17 at 13:20
• The whole idea here is that I assume that the paper and coin are a whole system and they move together. But we know that the maximum force we can exert on the coin is $f_{k.max}$ and if the force exceeds that value, the system breaks(the coin slips) – Soroush khoubyarian May 3 '17 at 13:29

You have to consider friction on your problem.

When you "move" the paper, what you are really doing is accelerate it (at first is stopped, with no velocity, and then is moving, so you accelerate it). As a consequence, you are acting with a force on the system.

This force acts on the paper, but not directly on the coin: this is affected by the force through the frictional force (between the coin and the paper).

Now think on the action-reaction principle (Newton's third law). The coin will react to that frictional force with another force. Where does this reaction force come from? If you want to calculate the dynamics of the coin, you should move to the paper, because you are interested in the relative movement between the coin and the paper. But the paper is accelerating, so now you are not an inertial observer! The reaction force of the coin is a fictitious force (like the centrifugal force).

Frictional force does not depend on velocity, as it is $F_f=\mu N$. This means that there's a maximum frictional force which can act on the coin, and as a consequence of that, there is a maximum acceleration ($a_{max}=\frac{\mu N}{m}$) you can give to the paper to maintain the coin still with respect to it. If the acceleration of the paper is greater than $a_{max}$, then the coin will slide.

If the acceleration of the paper is not very big compared to the maximum acceleration $a_{max}$, the coin will slide but it won't continue still, it will move slower than the paper (from the lab reference frame), as the frictional force brakes it (if you are in paper reference frame) or moves it (if you are in lab reference frame). If the acceleration of the paper is very big compared to $a_{max}$, then the frictional force will be very small compared to the coin's fictitious force, so it will hardly move (if you are in lab reference frame).

• As reaction, frictional force (I guess is what you call $F_k$) direction is opposite to applied force ($F$). Also, frictional force is a reaction force, so it won't be bigger than $F$. As I say in the answer, there's a maximum frictional force, let's call it $F_{k,max}$. Now, if $F<F_{k,max}$, then both objects, coin and paper, move together, and the coin will remain still with respect to the paper. If $F>F_{k,max}$, then the coin will slide and it will move in the same direction of the movement of the paper. If $F>>F_{k,max}$, the coin will slide and it will hardly move. – falgenint May 3 '17 at 19:15
• The frictional force is not constant. It is a reaction force, so it grows when the applied force grows. However, it won't grow indefinitely, but until it reachs a maximum, which is $F_{k,Max}=\mu N$. You will never have a frictional force bigger than the applied force. – falgenint May 4 '17 at 13:01