An object sits on an inclined plane. The weight of the object will have a normal and parallel component. I always thought that the reaction of the plane was simply the negative of the normal component of the weight.

Similarly, an object swings on a pendulum. The weight of the object can be decomposed into a radial and tangential component. I assumed that the reaction (tension) of the string must be the negative of the radial component.

But several examples have made me doubt my assumptions. One exercice in my textbook involves determining the reaction on a mass as it slides down a parabolic surface, as a function of $\theta$, the angle the vertical makes with the surface at any given point. Under my assumptions about reaction force, the answer is so trivial as to make the exerice pointless (since they give you $\theta$), but instead the exerice launches into a lot of complicated reasoning involving the Frenet-Serret base and comes out with a very different answer to just $-mg\ sin(\theta)$.

Also, consider the second example I gave. If the object is swinging in a circle (to make it even simpler, say it has uniform circular motion), then it must have radial acceleration (centripetal, specifically). But that's impossible if the reaction is exactly the negative of the radial weight.

But then... how can I determine a reaction force? In the absence of the simple rule I gave in the first paragraph, what else is there? Can it be done without making assumptions about the motion (eg. if you suppose the motion of an object is uniform circular, you have a formula all ready for the centripetal force and may thus be able to deduce the reaction force).

  • $\begingroup$ There may be an issue of terminology here. As far as I know, the standard terminology is that "reaction force" means "Newton's-third-law partner." However, ja72's answer seems to take it as meaning "force of constraint." Is this perhaps a different usage in physics than in engineering or something? $\endgroup$ – Ben Crowell May 2 '13 at 13:33

Newton's third law (1) relates forces that are of the same type, and (2) always involves exactly two objects.

Let's call the object resting on the inclined plane A. You proposed linking the following two forces through Newton's third law:

  • earth's gravitational force on A
  • inclined plane's normal force on A

These are of different types (gravitational and normal), and there are three objects involved (earth, A, and plane). The correct pair is:

  • earth's gravitational force on A
  • A's gravitational force on the earth

Another correct pair is:

  • plane's normal force on A
  • A's normal force on plane

A similar analysis applies to the pendulum.

There is no notion of a "reaction force" as distinguished from any other type of force. Forces occur in pairs, and they have types such as gravitational and normal. In such a pair, neither force is a reaction to the other. Neither occurs first in time; they are equal at all moments in time. Neither is the cause of the other.

The real issue here is not how to compute the strength of a third-law partner of a force $F$ -- it's simply equal to $|F|$ in magnitude. The real issue is how to compute unknown quantities such as the tension in the string or the normal force between the plane and the object resting on it. The answer is that you write down Newton's second law and produce multiple equations in multiple unknowns. Then if luck is with you, there are enough equations to determine all the unknowns.


Here are two simple rules about reaction forces. 1) they do no work and 2) their job is to enforce a motion constraint.

In the simplest term when an object of mass $m$ is sliding on a horizontal plane with velocity vector $\vec{v}$, there is a motion constraint of the form $y=0$. In Pfafian form this constraint is $\dot{y}=0$ and $\ddot{y}=0$.

The reaction force vector $\vec{N}$ has to do no work so $\vec{N}\cdot \vec{v}=0$ or $N_x\,\dot{x} + N_y\,\dot{y} = 0$ and since $\dot{y}=0$ and $\dot{x}\neq 0$ you must have $N_x=0$ and $N_y \neq 0$. That is how you find the direction of any reaction force. The direction is always orthogonal to the motion.

To find the value of the reaction magnitude you do the force balance perpendicular to the motion

$$ N_y - m\,g = m \ddot{y} = 0 $$

With solution $N_y=m \, g$. Now I know this is simplistic example, but it illustrates the process exactly on how to deal with reaction forces for every case.

  • $\begingroup$ What you're describing is what I would call a force of constraint, not a reaction force. $\endgroup$ – Ben Crowell May 2 '13 at 13:31
  • $\begingroup$ Same difference. $\endgroup$ – ja72 May 2 '13 at 13:56
  • $\begingroup$ As far as I know, the standard meaning of "reaction force" is "Newton's-third-law partner." Is there some field in which it's common to use "reaction force" to mean "force of constraint?" $\endgroup$ – Ben Crowell May 2 '13 at 15:53
  • $\begingroup$ @BenCrowell, in robotics and machine design. $\endgroup$ – ja72 May 2 '13 at 20:23
  • $\begingroup$ Interesting. Good to know that there is such a different use of the term in that field. $\endgroup$ – Ben Crowell May 2 '13 at 22:55

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