Years ago, I was told that the reason why you don’t sink into the ground when you walk is that the surface you’re walking on applies an equal but opposite force on your foot where you are standing. ie If my foot carries 80kg of force as I step forward, the ground under my foot applies an 80kg force back.
If that’s the case, how can the ground under my foot instantaneously apply that force, when the ground an inch either side of my foot doesn’t? And how does it instantaneously stop applying that force when I raise my foot?
There’s no such thing as an infinitely rigid material; even solid matter compresses very slightly when force is applied to it. So when you step on the ground you’re compressing it until the force stored in it is equal to the weight you’re applying. All happens in a millisecond or less.
Indeed, when one investigates the nature of matter on the quantum level, one finds that [thread=299054]there is no such thing as solid matter[/thread] either.
In response to the OP’s question, the notion that the ground “applies” a reaction force is a convenient simplification that allows us to draw an arrow to represent a resultant force vector and resolve forces in free body diagram notation. The force–actually applied as a pressure distribution–is merely a reaction to the force of the foot/ball-peen hammer/drunken frat boy being applied to it, and is, as Lumpy notes, effectively instantaneous for “rigid” materials and normal timescales, i.e. has an extremely large impulse in reaction. However, if you were to look at the sum pressure/force applied by a footstep using a very small timescale (on the order of, say, milliseconds) you’d see a sharp (but differentiable) increase in force, a plateau while the foot is dwelling in fully plantigrade presentation, and the a rapid decrease as the foot is raised from the ground. So even on an (effectively) rigid surface the application and reaction of force is not truly instantaneous. Just very, very quick. Good thing, too…Newton just hates it when forces aren’t balanced.
I heard even as a little kid that it was easier on the knees to run on grass or dirt than concrete. It stands to reason, as there’s got to be a significant increase in the time until the running step is counteracted.
An important distinction to be made here, which often confuses new physics students: The force of gravity and the force the ground exerts on your feet are not a pair of the equal and opposite forces referred to in Newton’s Third Law (“For every force, there is an equal and opposite reaction force”). What Newton’s Third Law means is that whenever A exerts a force on B, B exerts the same amount of force on A (I teach this to my students as the “revenge law”). Newton’s Third Law always holds, no matter what A and B are: It could be me jumping on a trampoline, or a bug splatting into the window of a 747, or a rock sitting on the ground (note that although the forces are the same, the effect that force has on the two objects may be very different).
What’s going on here is actually a special case of Newton’s Second Law, which states that F = ma. That is to say, the net force exerted on an object is proportional to the acceleration of that object. This, too, is always true. But now we look at a sub-class of physics problems where the acceleration is zero, such as a person just standing at rest on the floor. Since the person’s acceleration is zero, we know that the net force on the person must be zero. But we also know that there is a gravitational force acting on the person. So there must also be some other force which is counterbalancing the force of gravity, in this case the force of the floor on the person’s feet. Because it exactly counterbalances the force of gravity, this force must have the same strength, but be in the opposite direction.
But please note here: Newton’s Laws are always true, but this particular special case does not always apply. Suppose, for instance, I’m not just standing on the floor, but jumping. Now, my acceleration is not zero, so my net force is also not zero. That means that the floor force and the gravitational force do not exactly cancel each other out, and so must be different strengths (in this case, the force from the floor is greater than the gravitational force). There are also cases where the floor cannot produce a strong enough force to counter gravity, and so you accelerate the other way, presumably breaking through the floor in the process.
If I want to be perverse and analyze this in the most cumbersome fashion I can look at gravity as accelerating me down at g and the floor as accelerating me up at g with the net being zero.
Or, you can look at gravity as applying a force of mg downward, and the floor applying a force of mg upward, creating a net FORCE of zero, and then you end up with the exact same result. Chronos’ explanation is very good but one thing that is important to realize is that “F” is the NET force, meaning ALL the forces acting on an object. My physics teacher always puts a sigma in front of the F to remind us of this.
But… how can one piece of floor selectively exert a force, and an adjacent piece of floor not? And doesn’t the application of force by necessity involve an energy input. How can there be force without energy? The floor isn’t using any form of energy input to sustain its amazingly selective forceful activity!
You have to ignore my perverse analysis and stick to opposing forces. There can easily be large forces without any energy if there is no motion.
In this case though there is motion. When you first step on the floor it will sag slightly until the internal forces in the floor, stress, develop a force upward equal to your weight. During that short period of time while the floor is in the process of sagging, energy is being stored in the floor. After the sagging stops no more energy is being put into the system. When you step off the floor it returns to its original condition and the energy is recovered from the floor. It shows up as heat unless you have a linkage that converts the motion of the floor into some other forme of energy elsewhere.
I’m ignorant when it comes to this stuff but even before you walk on the floor, isn’t there already a force there? What about the Dead Load on the floor joists and the support beams before you step on the floor then you add a Live Load.?
Oh sure. A floor always has a load. Long boards like joists are never straight but always have a slight bow. Good framers sight along the joist and make sure they install them convex up so that the dead load tends to straighten them.
Here’s the problem; whether you realize it or not, you’re ascribing action and volition to the floor, i.e. “how can one piece of floor selectively exert a force…” In fact, it isn’t a little bit of the floor selectively exerting a force, but rather the entire floor, the joists upon which it sits, the foundation of the building that contains it, the soil under the foundation, et cetera, et cetera for as inclusive as you care to make the system which reacts. The floor “acts” upon the person only via his feet strictly because that is the point of contact, but for the purposes of analysis you could have the entire floor applying the force symmetrically upon him such that the sum of all “forces” is equal and opposite to the downward force applied to him via the gravity field. (This would require accepting the notion of action-at-a-distance between the bloke and the floor but if you’re prepared to accept that whole gravity thing–and as inconvenient and illogical as it is, it does keep you from being flung off into space–then the idea that the floor acts all around isn’t too far afield.)
We are, in this discussion, of course assuming that the floor is essentially rigid. Of course in reality, floors flex, joists shift, earthquakes…er…shake, and all that stuff, and so idealizing the floor and whatever else you care to clump into your “ground” system as being an immotile ignores internal forces, elastic and plastic deformation, hysteresis, blah blah blah. But in the end, once the system is in equilibrium, the result is the same; the floor–the whole thing–applies a reaction force exactly equal and opposite to that of the person standing on it. Any more or less and, as Ella sings, something’s gotta give.
It may help to visualize what is going on by standing on a surface that is NOT especially rigid. A trampoline for example. The place that is pushing up on your foot “knows” to push back because your foot has deflected it.
No. Forces only involve energy if the force moves something. For instance, if I lift a heavy object to a high shelf, then my arms have exerted an upward force on the object, and moved it up. In the process, I needed to spend energy, which turned into gravitational potential energy. Or if I’m pushing something some distance along a frictionless surface, then the force I’m exerting causes an acceleration, and the energy I spend turns into kinetic energy. Or if there is friction, then energy is converted to heat. But if I just hold something in place, I am not putting any energy into it, and therefore do not need to spend any energy.
It may help to understand that force and energy are not measured in the same units. Asking “How much energy is this force?” is like asking “How many ounces is it from here to Toledo?”. The question just doesn’t make sense, because they’re two different sorts of quantities.
It wouldn’t be like much of anything at all, since special relativity wouldn’t be in effect, taking quantum field theory and general relativity out with it, and pretty much nothing would work. Electrons would spiral into nuclei and everything would annihilate into energy, which would cool down right quick what with the universe expanding infinitely fast and taking radiation along with it.
