Is There Any Difference between Reaction Force and Normal Force?Date: 01/26/2008 at 12:22:26 From: Mike Subject: Is there any difference between reaction force and normal fo Do reaction force and normal force mean the same kind of force... followed from Newton's third law? Must the normal force always be perpendicular to the surface acted upon? What about reaction force? If a force of, say, 10 N acts on a surface at a 45 degree angle, what are the degrees of normal force and reaction force? Should the normal force always be 90 degrees, regardless of the angle of the original force? Date: 01/26/2008 at 17:32:23 From: Doctor Rick Subject: Re: Is there any difference between reaction force and normal fo Hi, Mike. Normal force and reaction force are different. In order to talk coherently about reaction force, we need to consider *two* free body diagrams, for two different bodies: +---+ +---+ R <-------| A | | B |-------> F +---+ +---+ Body A exerts some force F on body B. (The way I have drawn it, A is pushing on B.) Newton's third law tells us that body B therefore exerts a force R (the reaction force) on body A; R is of equal magnitude and opposite in direction to force F. When we talk about normal force, generally we aren't interested at all in a free body diagram for body B; it is essentially the earth. Its mass can be well approximated as infinite relative to A, and therefore its acceleration is well approximated as zero. We don't care, or need, to add up all the forces acting on the earth! You refer to "a force of, say, 10 N act[ing] on a surface ...". This is not a clear way to talk about it: generally we have a body A, sitting on some surface, and the force F acts on body A, *not* on the surface. N ^ | | +---+ F <------| A | f +---+ --------------------\------------- \ B \ _| F To reiterate, F is *not* the force with which A acts on "body B", the object whose surface is shown. We do not draw that force, since we are showing the free body diagram for body A only. We *do* show the reaction force that body B exerts on body A, but we resolve that reaction force into two components: the normal force N (perpendicular to the surface), and the friction force F_f (parallel to the surface). If the surface is assumed frictionless, then F_f = 0 and the "reaction force" is equal to N, strictly normal (perpendicular) to the surface. In that case, it could be said that the reaction force and the normal force are the same force. However, this can lead to confusion; it's best not to think of a "reaction force" in this problem at all. That's because if you call it a reaction force, you'll be tempted to think that Newton's third law is being invoked somewhere here--but it isn't, because we are only considering one free body diagram. If you aren't careful, you might fall into the trap of thinking that N (or N + F_f) is equal and opposite to F. KEEP THIS STRAIGHT: F is an external force acting on A, NOT a force with which A acts on B. - Doctor Rick, The Math Forum http://mathforum.org/dr.math/ |
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