Velocity and related frictional forces

Question

I will illustrate my question using a simple example.

Lets imagine the folowing situation: a big surface, a book on this surface and another smaller book located on the first one. There is friction between the books and between the bigger book. We apply a force which is big enough to accelarate the bigger book to a small velocity and then we decrease the force so the book continues to move with a constant velocity.

The smaller book moves with the bigger book having the same constant velocity. Why ? The forces that act on the bigger book are the force we applied, the kinetic friction from the floor, the static friction from the smaller book and the air resistance. The forces that act on the small book are the same static friction from the bigger book but just in opposite direction and the air resistance.

Why they have the same velocity? The forces that act on them are different and their masses are different so the books have different acceleration but in the end they both end up having the same velocity. What causes the upper book to reach the same velocity ?

Answer 1

First I want to point out that you can't have a constant velocity as you have mentioned that there is friction everywhere in your situation.

Second, we know that

Force of friction on surface = coefficient of friction * normal force of object.

Think this way. We take both books as our system and moved it by applying some force. Total mass of our system is the sum of the masses of both books and the force acting on our system is force of friction between the table and the big book, which just happen to be that case here, and the applied force.

Now you said both books moved, or say the system moved, that is force is applied so that it is greater than the force of friction on the table. Normal force is sum of the mass of both books * gravitational acceleration (g) and so you get the force of friction by multiplying it by coefficient of friction.

In short you have to apply the force such that

$$ g\times (m_b+m_s)\times \mu _{tb} +g\times (m_s)\times \mu _{bs}>f_{apply} > g\times (m_b+m_s)\times \mu _{tb}$$

So you see at this point you didn't applied force to move just the big book but rather to move both books.

However if you want to move only big book then you have to change your system to just that big book. And the forces are the forces of friction between the table and the big book, between small and big book and the applied force. And you have to appy the force such that

$$f_{apply} > g\times (m_b+m_s)\times \mu _{tb} +g \times(m_s)\times \mu _{bs} $$

Where

$m_b$= mass of big book

$m_s$ = mass of small book

$\mu _{tb}$ = coefficient of friction between table and big book

$\mu _{bs}$= coefficient of friction between small book and big book

So small book didn't just catch up, you just moved it.

BTW just try to make free body diagram. They help a lot.

Best regards.

Answer 2

The forces that act on them are different and their masses are different so the books have different acceleration

This is wrong logic. Remember Newton's 2nd law $\sum F=ma$. If both the total forces on and the masses of two objects are different, then that doesn't imply that the accelerations are different. One could easily have double the force but half the mass and thus have the exact same acceleration as the other.

And this is exactly what happens and what $f_s$ takes care of on the top book in your example. Had its mass been larger, then it would be tougher to induce the same acceleration in it as that of the bottom book. Then $f_s$ would increase the total force in order to accelerate it just as much. But in the constant-velocity situation with no acceleration anymore, the $f_s$ doesn't have to worry about that - it only takes care of the air resistance which is unrelated to the mass.

Answer 3

Both books will initially move together if the static friction force between the books is greater than the static friction force between the bottom book and the supporting surface. You will need the masses of the two books, the coefficient of static friction between the books, and both the coefficients of static and kinetic friction at the supporting surface to do the calculations and follow the analysis below.

Once the static friction at the supporting surface is overcome, there are two possibilities. One can maintain the force that overcame static friction and started the books in motion, in which case there will be a net horizontal force and the two books will accelerate. Or one can reduce the force to equal the sliding friction force at the supporting surface (which is generally less than the static friction force), in which case the sum of the horizontal forces will equal zero and the two books will continue to move at constant velocity (Newton's First Law).

All this ignoring air resistance.

Answer 4

The static friction force, acting on the top book, is automatically adjusted, as the acceleration changes, to keep the top book from moving relative to the bottom book or to keep it from accelerating faster or slower (if it can) than the bottom book.

The top book cannot accelerate faster than the bottom book because that would turn the static friction, pushing the top book forward, into dynamic friction, pulling it back. So, the top book would lose the only force (static friction) that could accelerate it and therefore would not accelerate at all, leave alone accelerate faster than the bottom book.

Can the top book slip and fall behind? Yes, it can, if the acceleration of the bottom book is too high and the maximum static friction force is not sufficient to support such acceleration, i.e., $f_{stat\_max}\lt m_{top}a_{bot}$.

In this case, the acceleration of the top book would be determined by the dynamic friction force (ignoring drag) and obviously it would be lower than the acceleration of the bottom book, since dynamic friction is less than the maximum static friction.

So, for the top book not to slip or not to accelerate slower than the bottom book, the following condition has to be met:

$a_{bot}\lt \frac {f_{stat\_max}} {m_{top}}$

Since, we've already shown that the top book cannot accelerate faster than the bottom book, we can state, that, with this condition met, the acceleration of the top book will be the same as the acceleration of the bottom book, i.e., the books move together.