# Very confused about normal Force

1. Aug 21, 2015

### mahrap

I'm having a very hard time understanding the concept of a "Normal Force." The only framework with which I've come to terms with is imagining the Normal Force as the force exerted when object A comes into contact with object B which keeps object B from losing its macroscopic structural integrity. In another words, if I'm standing on the ground, the ground exerts a normal force on me to keep its structural integrity. This force has to be equal and opposite to the force I'm exerting on the ground, my weight. So the normal force the ground exerts on me is equal to my weight.

Now the part that becomes very confusing to me is a scale. Say I'm on a scale and I see my weight. Then I jump up and then land on the scale. Well in order to jump up there has to be a force exerted on me that overcomes the force of gravity. So I have to use my muscles and convert chemical energy into mechanical energy to push down on the ground. Now the ground has to exert a greater normal force than my weight to keep its structural integrity and since this force will be greater than the force of gravity pulling me down I'll fly off the scale. At the moment which this happens of course the scale will say I'm fatter. Which makes sense since the scale measures the normal force exerted by the scale on me. Now when I'm landing the scale will also say I'm fatter right at the moment I land. This is confusing to me. On my way down to the scale from the top of my projectory, the only force that's acting on me is gravity. Therefore, when I land the force I'll impart onto the scale should be the force of gravity on me. But that's not the case. So F = ma = mg does not hold. Instead what we have is F = mg + ma. Where's that other ma coming from?

2. Aug 21, 2015

### brainpushups

The situation is not static - you accelerate as you land! I'd prefer to write Newton's second law this way for the landing on the scale:
$$∑F=ma$$
$$N(t)-mg=ma(t)$$
where N(t) is the normal force as a function of time.

3. Aug 21, 2015

### SteamKing

Staff Emeritus
I would forget about this "losing structural integrity" spiel, whatever that means, because that's not why there is a normal force acting on your feet when you stand on the ground. If that explanation were true, we could walk on water or across a bed of quicksand: both of these things have "structural integrity".

Just like earth's gravity attracts your body to it, the earth, your body is also attracting the earth to you. The forces created by this mutual attraction are equal and opposite: we call the force of attraction the earth has on your body weight, and the attraction your body has on the earth is called normal force. Because these two forces are equal and opposite, neither your body nor the earth goes flying off into space.

'Projectory' is not a word. What you are looking for is call 'trajectory'.

4. Aug 21, 2015

### SammyS

Staff Emeritus
You are coming down, approaching the scale with some velocity. In order to come to rest, you change velocity to zero. This implies some acceleration. That's the " a ".

5. Aug 21, 2015

### William White

huh? liquid water and quicksand are fluids and cannot resist shearing;

the defintion of structural integrity is a structure's ability to withstand loads (and therefore withstand shearing)

liquid water is easily defomed, you just pour it into another container or stir it about!

not a good example!

I think we know what he meant: structures and surfaces have a certain integrity that resists deformation under load.

well, if you are going to be picky about the OPs language, then be precise. By definition, something cannot be equal AND opposite.

the forces are equal in magnitude and opposite in direction

Last edited: Aug 21, 2015
6. Aug 21, 2015

### SteamKing

Staff Emeritus
The magnitude of the load is what makes the difference, not the nature of the load. There are creatures which can walk across water, just not humans.

"Structural integrity" is too vague a concept here to be meaningful.

I thought my extended verbal description of the forces implied this.

7. Aug 21, 2015

### William White

no, the definition of a fluid is its lack of resistance to shear; therefore it has no structural integrity. Creatures that walk on water have nothing to do with this; that is to do with surface tension, not the (lack of) shear strength of the liquid

Structural integrity is well defined branch of engineering. It is not a vague concept at all.

don't want to start a fight, but you were being unreasonably picky with the OP; whilst yourself using imprecise terms and hoping "implication" works. (sorry I missed the verbal description, I can't hear this forum).

When somebody (such as the OP) is confused about weight, normal forces, reaction forces, or whatever, its very important not to rely on implication when teaching them what these things are. "Equal and opposite" is wrong and it is confusing. end of.

8. Aug 21, 2015

### SteamKing

Staff Emeritus
I don't think the OP is a structural engineer, but that's just a guess on my part.

Notice, I said "here", as an explanation of normal force, which is a nice weasel word to catch the unwary.

Would you like the five-minute argument, or do you want the half-hour version?

'Verbal' means 'with words', of which, I believe, my description was composed. You are conflating 'verbal' to mean 'oral', I suppose.

I took my shot. If you have a better description, don't argue with me; tell the OP what he should know about 'normal force'.

9. Aug 21, 2015

### FactChecker

This is essentially correct, although the term "structural integrity" is covering a lot of things. Let's just say that there is enough "structural integrity" so that the ground is a constraint. It will apply enough force to stop you, no matter how much you weigh, within reason.

Gravity has had time to act on you and you have accelerated downward. So when you hit the scale, you have a velocity that must be slowed to 0. As @SammyS has said, the scale has to supply enough force not only to oppose gravity but also to cause a deceleration from your downward velocity to 0 within the distance of the scale spring compression.

10. Aug 21, 2015

### Jeff Rosenbury

The normal force (and all calculated forces to some extent) is a bookkeeping fiction. The reality is that all the forces add together, but that doesn't help us tell which engineering items contribute what effects.

The normal force is the force of one object resting on another. It is called the normal force because it is always 90º to the surface on which the object is resting, which is different for slopes than flat surfaces. It is the force that keeps the object from moving through the surface.

On a slope, the normal force plus the forces keeping an object from sliding (friction, or maybe a bolt or something) typically add up to total the gravitational force.

Let's say you are driving down a hill and you step on the brakes. Your car has forces acting on it. By convention, the force of the tires gripping the road is called friction and is parallel to the road surface. Gravity is pushing straight down. It pays no heed to the surface angle. To make the forces balance (with the deceleration/acceleration) another force is needed that vector adds to equal the gravity plus the (force of) acceleration. This force is exerted by the road on the car. By convention it is normal to the road and called the normal force.

One might choose to combine the friction and the normal force as they tend to be in nature, but that doesn't separate out important information like how well a particular brand of tires grips the road. So we don't typically do that. Instead we divide the forces in ways that make sense for designing things.

11. Aug 22, 2015

### CWatters

As others have said.. the scale also has to decelerate you. That's where the extra ma comes from.

Consider also..

1) A roller coaster. You feel heavier than normal at the bottom of a loop because the track has to accelerate you towards the centre (upwards). In this case the ma is equal to the centripetal acceleration. eg..

ma = mV2/r

At the bottom your total weight becomes...
W = mg + mV2/r

which is greater than mg.

2) A rigid object falling on hard rather than soft earth. It's more difficult to compress hard earth so the effective stopping distance will be shorter and the deceleration larger on hard earth than soft. The rate of deceleration on concrete can be very high - hundreds or even thousands of times that of gravity.