1
$\begingroup$

The classic introductory mechanics problem considers the motion of a box sliding down an inclined plane. As I'm reviewing the early chapters in Taylor's Classical Mechanics, I was struck by a question: in such a problem, we always say that the box has no motion in the y-direction (as is standard for such problems our coordinate system has axes parallel and perpendicular to the plane, the x- and y-axis, respectively) and therefore we can equate the normal force with the perpendicular component of gravity so that we can proceed with the analysis. For instance, in solving the problem, Taylor waves his hand, saying

Since the block does not jump off the incline, we know there is no motion in the y-direction[...]

What allows us, a priori, to say that the box will stay on the plane. Is it simply common sense, or is there some other way of showing it that is more rigorous?

$\endgroup$

1 Answer 1

1
$\begingroup$

The forces on the box in such questions are gravity, friction, and the reaction of the plane. The reaction of the plane is the key. Gravity has a component in the y direction, and would cause the object to fall through the plane if not for the reaction.

The plane is a rigid object. That means when a force tries to deform it, it pushes back with just enough force to prevent the deformation. An object sitting on it will not penetrate the surface. On the other hand, the plane will not push back strongly enough to lift the object off the surface.

This doesn't explain the reaction force. It is more or less a restatement of the fact that the plane is rigid. And that is as far as this kind of problem usually takes it.

To explain the normal force you need to explain atomic bonds. They have a more or less fixed length. They can stretch, but less than the separation between atoms. So you can model them as rigid links, or perhaps very stiff springs, and leave it at that.

Or you can look at the orbitals of molecules and calculate the configuration that results in the lowest energy. In spirit, this is like calculating orbitals of an atom. But instead of one nucleus, there are two for a diatomic molecule. Or a periodic array of them for a crystal.

There is a lowest energy at a particular distance. If the nuclei get too close, potential energy increases because they repel each other. If they are too far apart, energy increases because the electrons are not close enough to both nuclei.

$\endgroup$
3
  • $\begingroup$ I think I follow - essentially, we could theoretically show (using the solid-state physics techniques you're describing) that the normal force is such that it will prevent the box from penetrating the surface, but not push back strongly enough to lift the object off the surface. But for introductory mechanics, it suffices to say that the box stays on the plane as expected. $\endgroup$
    – EE18
    Commented Apr 24, 2020 at 13:06
  • $\begingroup$ That's right. But it also helps in introductory mechanics to know how the force behaves, even if you can't say why. $\endgroup$
    – mmesser314
    Commented Apr 24, 2020 at 13:17
  • $\begingroup$ Very fair, that's well-taken. That a box won't jump off a plane should be intuitively obvious. Thanks very much! $\endgroup$
    – EE18
    Commented Apr 24, 2020 at 13:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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