Consider book modeled by a rectangular prism with a square base of length $L$ and a height $e$. The book is placed on a table whose edge is parallel to an side of the rectangular base such that the center of the square base is a small distance $\delta$ from the edge of the table. In other words the overhang off the book is $L/2+\delta$. Call the

This configuration is unstable - the book will begin rotating about the pivot point. The forces acting on the book are gravity at its center of mass, the normal force at the pivot point which is perpendicular to the surface of the book, and the friction at the pivot point which acts in the radial direction. Note that the normal and friction forces change direction during the rotation.

The coefficient of static friction between the book and the table is $\mu_s$.

A priori depending on the parameters of the problem, the book can rotate about the pivot until the book leaves the surface of the table, or the the book can begin slipping before it leaves the table.

Call the angle which the book forms with the horizontal $\theta$ (so $\theta_0=0$) and define the corresponding polar coordinate unit vectors $\vec{u_r}$ and $\vec{u_\theta}$.

Suppose we have derived, based on the mechanics of the book's rotation (using moment of inertia etc) the expressions for the normal force $N\vec{u_\theta}$ and the friction force $f\vec{u_r}$ in function of $\theta$.

1. What is the condition for the book to fall off the table?

2. What is the condition for the book to begin slipping off the table?

I am not asking at what angle will the book fall off or being slipping etc, I am just asking for the condition so that I can do the calculations myself.

I guess the answers are:

  1. $N=0$

  2. $|f|=\mu_s|N|$

but I am not sure.

If this is indeed the condition, then I have found that it is impossible for the book to leave the table without first slipping, no matter the parameters of the problem.

  • $\begingroup$ The whole problem of whether the book will roll or slip of the table only makes sense when the edge of the table is rounded (finite radius of curvature) if it is sharp then the book will always fall off the table by a combination of toppling and slipping since the friction and gravity will always create a non zero moment $\endgroup$
    – alex
    May 21, 2017 at 16:11

1 Answer 1


I don't have any decent illustrating software so please bear with my hand made drawings. Apologies for the inconvenience

As can be seen in the sharp corner case, there is always a non-zero moment created by the weight and net reaction force(I have taken the height of the book to be $a$ so it's CG is at $a/2$):

sharp corner table

for this moment to be cancelled out, both the weight and reaction force should coincide at the CG. Also they should be equal and opposite for no net acceleration. This second case is possible if the table's edge has a finite radius. It is illustrated in this picture:round edge table

(note that I haven't drawn the full book here, only the relevant part). If you wish you can solve it yourself from here but I'll post the rest of the solution anyway for the benefit of others. We can decompose the Reaction force $R$ into the normal and tangential directions to the table. \begin{align*} \text{normal component:} &\; R \cos\theta \\ \text{tangential component:} &\; R \sin\theta \end{align*} Since the tangential component is nothing but the frictional force, it should be equal to $\mu_s$ times the normal force: \begin{align*} R \sin\theta &= \mu_s R \cos\theta \\ \mu_s &= \tan\theta \tag{1} \end{align*}

This is the same familiar condition as it would be on an inclined plane. But the line of forces also have to pass through the CG. For this observe that the length $\delta + \epsilon$ is equal to the arc length of the amount that it has rolled $r\theta$. $$\delta + \epsilon=r\theta$$ also observe from the geometry that $\epsilon= \frac a2 \tan\theta$. Therefore $$\delta + \frac a2 \tan\theta=r\theta \tag{2}$$ substituting eqn $(1)$ in $(2)$ we get:

$$\delta + \frac a2 \mu_s=r \tan^{-1}\mu_s\\ r=\frac{\delta + \frac a2 \mu_s}{\tan^{-1}\mu_s}$$

If $r$ is smaller than the above value then the book will fall off the table. In this case,the book will roll up till the angle of repose (given by $\tan^{-1}\mu_s$) and after that the book will simultaneously slip and rotate off the table.

  • 1
    $\begingroup$ I made several improvements to your math mark-up. Pointedly I set the trig functions as operators using \sin, \cos, \tan and so on which you will like because it gets the spacing you were working so hard for. I also used \tag for equation numbering, and the align environments for groups of lines. In any case, you might want to check my work to make sure I didn't mess anything up. $\endgroup$ May 21, 2017 at 18:50
  • $\begingroup$ @dmckee: Is it possible that the book falls off the table before it reaches the angle $pi/2$, even if we assume no slipping? $\endgroup$
    – math_lover
    May 21, 2017 at 20:13
  • $\begingroup$ @JoshuaBenabou It is the slipping that makes the book lose contact with the table. without slipping and just pure rolling it will always be in contact with the point about which it is rolling. So if we assume no slipping, it isn't going to fall off before $\pi /2$ $\endgroup$
    – alex
    May 21, 2017 at 20:23
  • $\begingroup$ @alex : that's weird, because the value for the normal force changes sign at some angle before $pi/2$. But we can't have the normal force be negative if the book is still in contact with the table, can we? $\endgroup$
    – math_lover
    May 21, 2017 at 20:33
  • $\begingroup$ @JoshuaBenabou The normal force here is given by $W \cos\theta$ that doesn't change sign. $\endgroup$
    – alex
    May 21, 2017 at 20:39

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