# Help me understand how deformations work

1. Mar 16, 2014

Ok, umm these may sound like dumb questions, and im sorry but here goes. (I think i have a lot of misconceptions)

1. If I push myself against a wall and exert a constant force, the wall will exert a reaction force against me and therefore net force = 0 and no motion. But if I throw myself against the wall, at the instant i hit the wall I'll be pushed back and likely fall over.

This is where im kind of confused. Regardless of whether I'm applying a constant force or an instantaneous force, won't the reactionary force be the same and hence the net force = 0? Why is it for the latter I will tend to experience a force/acceleration away from the wall (the reactionary force), when it should have been cancelled out by the initial force I exerted (i.e net force should be zero?)

2. I was hoping that someone could help me understand how deformations work. This is again related to my understanding of Newton's 3rd Law. Say I place a mass on a wooden plank, the plank will deform until it is bent to support the weight of the mass. Is it right to say that when I first place the mass on the plank, the plank is 'unable' to supply an equal reactionary force to the mass' weight, and hence has to bend until a point where the reaction force is equal to that of the weight?

How does the deformation of the plank in this case help to support the weight of the mass (I assume something to do with tension?), also (if i'm right) why is it that it is initially unable to supply an equal, opposite reaction force? Does this imply that Newton's 3rd Law only applies for 'hard' objects, or those that don't deform?

I'm also thinking that, if on the molecular level, at the point of contact between the mass and the plank, if all the forces on the molecules (I use this for lack of a better word) have equal and opposite reactionary forces, won't deformation not take place at all?

2. Mar 16, 2014

### Jilang

In your first question when you are pushing against the wall you are applying a force that counteracts the opposite force from the wall (your muscles are straining against it). When you throw yourself against it this is not the case. The length of time involved is really not important. Imagine you had so something coming up behind you that would counteract the force. You wouldn't bounce off then (but you might get squished!).
In your second question the restoring force increases with the amount of deformation of the plank. Forces are only equal and opposite for static situations, like a weight sitting on a plank.

3. Mar 16, 2014

### Staff: Mentor

Pretty much, yes.

All objects deform under pressure and, over a very wide range, the amount of force needed to produce the deformation increases with the amount of deformation (google for "modulus of elasticity" for a more complete/accurate description). Thus, "hard" objects are those that require a large force to deform even slightly; push a block of rubber and you can see and feel it compress, but it takes very sensitive instruments to detect the compression of a block of steel under the same force.

In all three cases (my block of rubber and steel, your plank) basically the same thing is happening: We apply a force to the object and it deforms; the more the deformation the more the opposing force; when the opposing force is equal to the applied force we're at equilibrium, no further deformation happens, nothing moves, and Newton's 3d law is saved.

4. Mar 16, 2014

### Staff: Mentor

This analysis is incorrect. You can't cancel out action and reaction forces like this. The reason the net force is zero is that the floor is pushing you forward and the wall is pushing you back. These are the forces which cancel out and sum to zero. Try pushing on the wall if you are standing on ice.
The plank is like a spring (a leaf spring). If you hang a weight from an ordinary spring, the spring will stretch until is develops enough tension to support the weight. The plank does the same thing by bending.
The way the beam works is that the top half of the beam stretches to develop tension, and the bottom half of the beam compresses to develop compression force. This creates a so called bending moment. You need to learn about beam bending to see how this is able to support forces perpendicular to the beam. The stiffer the beam is, the less it has to deform to provide the required bending moment. In the limit of very high stiffness, it doesn't deform at all. None of this has anything to do with the law of action-reaction, since, as I said before, an action force on one body and a corresponding reaction force on another body do not cancel each other (for either body). You need to go back and study the free body diagrams on each of the objects involved in action-reaction to get a correct understanding of how Newton's 3rd law works.