Distribution of normal force on a book resting on the edge of a table

Assume a uniform book is resting on a table. The book has length $1$. But not the entire book is resting on the table, a bit $\alpha$ is sticking out of the table.

Obviously, due to experience, we get that if $0\leq\alpha\leq 1/2$, then the book will just remain at rest. If $\alpha>1/2$, then the book will experience a nonzero torque and eventually fall.

But how do we obtain this same answer with physics? Here's a first attempt that apparently does not work out. Assume for simplicity that $\alpha=1/4$, then $3/4$ is resting on the table.

The book is experiencing a normal force and a weight force. To find the weight force, it is sufficient to take a weight force at the center of the book of magnitude $mg$. The contribution to the torque is then $mg(1/4)$ (since the center of mass is $(1/4)$ away from the rim of the table). But how to find the torque coming from the normal force? by Newton's law, the magnitude of the normal force equals the magnitude of the weight, namely $mg$. Now I try to split up the book into infinitesimal components and integrate. I obtain $$\int_0^{3/4} mgxdx = \frac{9mg}{16}$$ This does not at all equal the contribution of the torque, $mg/4$. So there will be a nonzero torque, which can't happen.

So what is going wrong? Is the normal force somehow not uniform over the surface of the table? If it is nonuniform, how is it distributed, and how do I solve this problem as above using integration?

By the way, just for information, I do know how to solve this problem the following way: if the book is somehow a lever resting on just the rim (and not the table), then we can solve this problem ver easily using the law of the lever and such. When we do this, we find eventually that if $0\leq\alpha\leq 1/2$, then the torque must be "negative". The constraint force of the table prohobits this and just sets the torque at $0$. I'm sorry if this doesn't make much sense, but it isn't my main question anyway. My question is why the naive integration answer does not work and if the normal force is somehow not uniform.

1 Answer

No, the normal force is not distributed uniformly across the contact area. As your calculation shows, a uniform distribution is not compatible with the conditions for equilibrium.

The weight of the book is distributed uniformly because its mass is distributed uniformly, but there is no reason why normal reaction should be distributed uniformly also. This is because the book is a rigid body, not a collection of identical disconnected masses. For disconnected masses, the normal reaction on each would equal the weight of each, which is the same. But there would be no force to support the part of book which overhangs the table. A rigid body redistributes forces internally, so that the part overhanging the table is supported by the part on the table.

Like static friction the normal force is a reaction force, a response to other forces or contraints. Wherever possible it adjusts to take whatever value is needed to keep the system in equilibrium, or a defined state of motion. This includes the equilibrium of torques as well as forces. It increases towards the edge of the table.

It is not possible to determine how the normal force is distributed if the book and table remain perfectly rigid and flat. The problem is statically indeterminate. See for example A simple (?) problem of static equilibrium.

If the book makes contact at only two points then we could solve the two conditions for equilibrium (zero net force, zero net torque) to determine the normal force at each. If the book makes contact at 3 or more points we cannot find the distribution of reaction forces between these points, because we do not have enough equations of constraint. If contact is made over a finite area, there are an infinite number of contact points, and therefore an infinite number of possible distributions of the normal reaction force which are compatible with the 2 conditions for equilibrium.

Real books and tables are elastic to some extent and deform on small scales - they are compressed or stretched or bent. The amount of deformation is related to the applied forces and torques and the material properties described by Young's Modulus $E$, Shear Modulus $G$ and bulk modulus $K$. When these extra equations describing the material properties are taken into account, the problem is no longer indeterminate.

One possible assumption is that the contact between the book and table can be modelled as a row of vertical springs obeying Hooke' Law between two perfect planes. The normal force must therefore increase linearly from one end of the contact surface to the other. See Estimate the reaction force on each leg of a 4-legged table for a similar example. If the surfaces are not smooth then lateral friction forces could be modelled with horizontal springs, possibly having a different spring constant.

However, this is only an assumption. It does not necessarily apply. The book or table could be warped so that the normal force varies unevenly.