# Understanding force normal

1. Sep 1, 2011

### more_torque

This is my first post on PF. I've decided that I really want to learn physics, not for school, but because physics can be fun!

I'm trying to understand what FN is. I understand it is the force exerted by the ground working upwards contrary to the force of earth's gravitational pull.

I understand that FN is your weight when standing. I also know that it decreases while you bend down (during the duration of the bend), and stand or jump up (during the time you're pushing off the ground). However, I'm confused about situations that are a little more complicated.

Say you're standing, and drive one of your feet into the ground. It seems to me that all you're doing is shifting your weight from one foot the other. Is that what's really going on? Is it possible to increase your weight by pushing your foot into the ground below you?

What about stomping? In the previous case, we didn't lift the foot up off the floor, just pushed it downwards. What if we lift a foot off the ground and stomp? The ground has to exert a corresponding FN to counter the downward drive of the foot. But I'm not sure if a simple shift in weight is going on between one's two feet, or if one needs to add mg and the ma (where a is the upwards acceleration provided by the floor) to get FN. Simply put, is it possible to stomp greater than one's weight?

All of the above questions ultimately derive from an uncertainty about how forces act, not on a solid object, but on a person with joints and limbs.

2. Sep 1, 2011

### stallionx

The Normal Force is perpendicular to the surface the force is being exerted on.

Could be on a wedge , too, and points away from the surface in the direction of unit normal.

if you jump or stomp your feet you would get more than the regular normal opposing force you might get if you just stood on a wedge or planar surface.

By jumping you push the surface and the surface pushes you in response, You gain Kinetic energy as you jump. And as you land your mv=p ( momentum ) has changed and to nullify that momentum you need a greater force as a response or when you stomp your feet, you are using energy from your upper body to do the stomp. that energy translates into the momentum of your foot.

Think of it this way :

EQN :

p=mV
dp=v dm+m dv

dp = m dV for zero mass change

you can only zero that momentum in the way that you exert a counteracting force for t seconds.

m *Vf -m *Vi=integral ( F dt )

3. Sep 1, 2011

### more_torque

Thanks for the response. I'm still confused about the difference between 1) simply driving your foot into the ground without lifting it up before, and 2) stomping.

The way I see it, 1 and 2 are essentially the same, except that FN for 2 would probably be greater.

In the first instance, you're pushing down into the ground, so there must be a greater FN on that foot for the time period you're foot is going lower (say, on carpet . . . but if you were on a hardwood floor, all the pushing down you could do wouldn't make a difference, since their is no change in velocity, right?).

However, I also feel my weight shifting to that foot away from the stationary foot. How does one know that FN isn't constant, and that all that is occurring is a different division of FN? Instead of FN being apportioned equally to both feet, a more unequal division alone could result. (Or maybe that simply is what's occurring, and after the momentary period where one's foot is being driven down and the normal force is increasing, the normal force returns to one's weight and you're left with it's unequal apportionment among the feet.)

I'm just trying to see exactly when the normal force is increasing, when equal to one's weight, and how that corresponds the actual action of pressing one's foot down.

In the case of stomping, all of one's weight is on one foot, then the second foot is suddenly being brought down. At the time of impact, their is a downward acceleration of the foot, and, thus, two downward vectors (mg and ma), and a corresponding FN, equal to their sum, that lasts for the duration of the deceleration before being replaced by a normal force equal to mg. If the account given for 1 above is right, then one should be left with with an FN equal to one's weight, but simply divided unequally between one's two feet.

4. Sep 1, 2011

### stallionx

Well, when thinking try to keep these in mind

1) integral(F * dr ) = Change in Kinetic Energy ( here the dot product )

2)integral(F*dt)=Change in m*v ( here simply product )

These 2 eqns give us an intuition of how the Force changes.

All help I can do is this much, sorry :(

5. Sep 1, 2011

### stallionx

There is also the matter of elastic/inelastic collisions in one of your explanations ( ie : carpet, hard wood )

To be concise,

when 2 things collide, they deform each other which turns to heat energy ousted.

If they do not harm each other at all that is Perfectly Elastic

For the elastic and inelastic parts, momentum is conserved but only in the elastic part that is Kinetic Energy is conserved.

6. Sep 1, 2011

### meldraft

Keep in mind that in the case of stomping you are adding biochemical as well as potential energy into the system and then convert them to kinetic energy. This means that you do not only let your foot fall under the acceleration of gravity, you also expend biochemical energy to accelerate it with your muscles.

This manifests itself in the form of larger momentum since, the result of adding energy to your foot is increasing the speed with which it will collide with the ground.

So, to keep it simple, just keep in mind that there is not only mass at play here, but the explosive motion of the body as well

7. Sep 1, 2011

### more_torque

Thanks meldraft!

I understand that you're adding biochemical energy to the system, and that that's the reason for the increased velocity and KE of you're stomping foot at the time of impact. I think that's also why I was confused about this problem in the first place - not only are you a body with mass, but you're a massive body whose appendages can themselves exert forces independent of the earth's gravitational force.

I think that what's happening is that the normal force increases as my foot hits the ground and is decelerated from whatever vi it has to 0=vf over a small time interval. This is from the equation dF*dt=m*dv. It is only during this time interval that your FN is greater than your weight. Once your foot is stationary, FN
returns to your weight, and all you're left with is an unequal apportionment of FN
between your feet (since your weight has shifted due to the stomp).

Could you tell me if the above account is accurate?

In order to calculate it, what I'm thinking right now is that total force FN=m1g + m2a, where m1 is your body's mass (including the mass of the stomping foot) m2 is your foot's mass, and a is the acceleration of your foot as it hits the floor. The FN calculated would be the maximum FN the ground exerts on you (at initial impact). Is this true?