I’m studying for an upcoming physics exam and one of the sample questions essentially boils down to this: a book (not clamped on) is on a table that is moving to the left. Is there a force of static friction acting on the book? My reasoning is that since the book is not clamped on to the table, it “wants” to move to the right relative to the table, but static friction prevents it from doing so. I don’t really know, however.

  • $\begingroup$ Is this word for word what the question says? Saying the table is moving to the left isn't enough to give a unique scenario. $\endgroup$ May 8, 2020 at 18:41
  • $\begingroup$ The table is part of a ferris wheel rotating clockwise (basically it’s one of the Ferris wheel cars), and the problem asks to examine the forces on the book when the book/table is on the lowest position on the ferris wheel (closest to the ground). $\endgroup$ May 8, 2020 at 18:53
  • $\begingroup$ Yeah, that is a different scenario from just a table moving to the left. Your wording makes it seem like 1D motion $\endgroup$ May 8, 2020 at 18:54
  • 1
    $\begingroup$ I mean, I was only asking for the answer to my hypothetical question to understand the concept better $\endgroup$ May 8, 2020 at 19:04
  • 1
    $\begingroup$ I understand, and that is fine. But like I said earlier, "moving to the left" is not specific enough. That's why Gert's answer has to consider different cases. $\endgroup$ May 8, 2020 at 19:16

1 Answer 1


If the ensemble of table and book move to the left in a straight line and at constant speed ($v_{table}=\text{constant}$) no friction between book and table is needed ($F_f=0$) because no horizontal forces are acting on the table or book, acc. N2L.

But if the table is accelerating ($a=\frac{\text{d}v_{table}}{\text{d}t} \neq 0$), friction between the book and the table is needed ($F_f \neq 0$) to keep the book moving at the same speed as the table.

enter image description here

This friction force comes from the weight of the book: $W=mg$. The table (assuming it is strong enough to carry the book) now exerts an equal, upwardly reaction force, known as the Normal force: $F_N=mg$.

Now assuming there exists some friction (roughness) between the book and the table then the maximum static friction force is given by:

$$F_f=\mu_s F_N=\mu_s mg$$

where $\mu_s$ is the static coefficient of friction. $F_f$ points to the left because the book tends to move to the right with respect to the table and the friction force tries to prevent this motion.

As long as the maximum static friction force is equal (or potentially greater) than the inertial force $ma$ the book will move with the table.

But if the maximum static friction force is too low re. the inertial force, then the book will slide with respect to the table.

So the 'non-sliding condition' can be summarised as:

$$\mu_s mg \geq ma$$


$$\boxed{\mu_s g \geq a}$$


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