0
$\begingroup$

When a paper over a glass is pulled suddenly, the coin over it falls to the water. But on pulling it slowly, the coin gets pulled along with the paper. Why does the coin in the former case not feel the force applied to the paper, but in the latter case does? Does force take time to transfer between the not-rigidly connected bodies? Otherwise why in the second case, inertia of both bodies is overcome, while in the first,only that of the paper?

$\endgroup$
2
$\begingroup$

The forces are the same (assuming the the coin is moving relative to the paper in both cases), is the friction force between the coin and the paper, and is in a first approximation at least, independent of the relative speed between paper and coin. The difference between the two cases is that is that the force acts during a shorter time in the fast motion case, just because it took less time to remove the paper.

Because $F=ma$, then you have the same acceleration of the coin but acting during a shorter time $t$, thus the change in position of the coin $\Delta x=\frac{1}{2}at^2$ the coin will move a smaller distance in the fast case.

$\endgroup$
  • $\begingroup$ so because pseudo force is more in faster pulling case, the frictional forces (same in both cases) can't balance it, that's why the coin falls into the water. I think i got it. $\endgroup$ – Nabin Jun 12 '16 at 20:50
  • $\begingroup$ I am not sure what pseudo force you refer. In my explanation there is only one horizontal force acting on the coin, and is the friction force. It is the force that makes the coin move horizontally. $\endgroup$ – Wolphram jonny Jun 12 '16 at 20:54
  • $\begingroup$ As the paper is pulled, the coin is in the accelerated frame of reference , so experiences a fictitious force $\endgroup$ – Nabin Jun 12 '16 at 20:56
  • $\begingroup$ I think it is simpler to see in the inertial system of reference were the coin moves, where there are no fictitious forces. But it is correct that in accelerated frame of reference a pseudoforce appears $\endgroup$ – Wolphram jonny Jun 12 '16 at 20:58
1
$\begingroup$

There is a limit on how much frictional force the penny can apply to the paper in order to remain attached to it. It is limited by its mass based on $F_f = \mu _s F_N = \mu _s * m_{penny}*g$. If you accelerate the paper, you are applying a force $F = ma $ and the penny's frictional force will try to oppose to keep the penny on the paper (inertia). For a certain amount of acceleration, the force will be greater than the force attaching the coin to the paper, leaving it behind.

Basically, if $m_{paper}* a_{paper} > \mu _s* m_{penny}*g $ then the penny will stay.

$\endgroup$
1
$\begingroup$

This happends because of the friction generated between the paper and the coin.

Remember, friction, at low speed, is not affected by the speed. So when you pull the paper, at any speed, the friction will have always the same value. The distance covered by the coin is always $x=\frac{1}{2}\frac{F}{m}t^2$, where, $F$ is the force of the friction (constant), $m$ is the mass of the object (constant) and $t$ is the time that the coin moves.

So you can see that the distance covered by the coin is relative to the time. So small time equals small distance for the coin.

If the distance covered by the coin is small and the distance covered by the paper is big (because you pulled the paper really really hard), the coin will not be pulled away with the paper.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.