# Homework Help: Why does the centripetal force subtract from total force toward center?

1. Feb 11, 2012

### Jacobim

Text of the problem:

Because the Earth rotates about its axis, a point on the equator experiences a centripetal acceleration of 0.033 7 m/s2, whereas a point at the poles experiences no centripetal acceleration. Assume the Earth is a uniform sphere and take g = 9.800 m/s2.

(a) If a person at the equator has a mass of 68.0 kg, calculate the gravitational force (true weight) on the person. (Give your answer to four significant figures.)

(b) Calculate the normal force (apparent weight) on the person. (Give your answer to four significant figures.)

I have already used the tutor application on this webassign and solved the problem, it just doesnt make sense to me. Here is the tutor text:

At the equator, the centripetal acceleration, ac is downward. Taking as positive the direction away from the center of the Earth, we have
n − mg = −mac.

Solving for the normal force from this equation, gives the person's apparent weight as
n = mg − mac
= m(g − ac)

I do not understand why it should not be n = mg + mac

It seems to me the magnitude of the normal force should equal the sum of the forces caused by gravitational acceleration and centripetal acceleration.

usually n=mg so the net Fy is zero if on a flat surface. so adding in force from acceleration it should simply be
n=mg+mac, and then the person would feel heavier not lighter.

Thanks for any help!

Last edited: Feb 11, 2012
2. Feb 11, 2012

### PeterO

The centripetal Force is the net force - if you like the answer to the situation - so the other forces add up to give it.

If you stand on the equator - and don't fly off, nor sink into the earths surface, then you must be accelerating at of 0.033 7 m/s2 down, so have a net force of 68*0.0337 = 2.3N.

The way you have defined direction, that means -2.3

The two forces that actually are acting (mg and N) must add up to that answer.

so N + mg = mac (this is a vector equation - I just can't put the vector signs on the N, g & a)

To remove that vector issue, they have introduced negative signs to the formula, so that you can use just the magnitudes of those, so you start with:

N - mg = -mac

So N = mg - mac

Note: I think your idea N = mg + mac comes from a belief that the upward force (N) has to match the downward forces ( mg & mac ).

If that was true, the body would have a net force of zero, and thus (Newtons 1st law) travel in a straight line - which is a pity, since the surface of the Earth, at the Equator, is travelling in a circle.)

3. Feb 11, 2012

### Jacobim

Thanks!

So it sounds like the 'centripetal force' is generally the net force.

Is the centripetal force a seperate system than the system at the surface where n=mg?

Im having a hard time visualizing why a person would feel lighter because both gravity and the centripetal acceleration are directed towards the center of the earth.

4. Feb 11, 2012

### PeterO

When you consider that g = 9.81, and ac = 0.0337, you should see that the effects of rotation make a 0.3% difference from equator to pole.
Most of us don't live on the equator [unless you are Brazillian] the effect is even smaller.
We generally quote answers to only 2 or 3 figures, so our uncertainty is often greater than the effect of this circular motion - so to ignore it is common.
Note that we often also ignore friction because it is too complicated to allow for it accurately.

EDIT: Note that in secondary school we often take g = 10 since we have no desire to get bogged down with decimals in introductory problems. The other alternative is 9.8 - but you have to wonder about use of 9.81, since the error in not using the extra 1 is less than the error brought about by ignoring centripetal acceleration.

5. Feb 12, 2012

### PeterO

The Earth is not a perfect sphere - the poles are close to the earth's centre than is the equator.
It would be interesting to see of the radius variation actually gives rise to a greater variation that this centripetal effect.

6. Feb 12, 2012

### Sefrez

If you are on a ferris wheel, at what point to you feel lightest? If you have been on one, you should remember how you feel lightest at the top. At the top the centripetal acceleration is directed down. In my mind, having centripetal acceleration in the direction of g essentially makes for less difference in acceleration frames. For example, if you were in free fall (accelerating at nearly 9.8m/s^2) you would feel weightless.

7. Feb 12, 2012

### PeterO

Red Herring Sefrez. It was exactly the lack of feel on the Earth's surface, compared to when on a Ferris Wheel that worried OP. It is the relative size of the Acceleration due to gravity, and the acceleration needed for circular motion at the equator that is the key.

8. Feb 12, 2012

### vela

Staff Emeritus
You might notice the problem statement never refers to a centripetal force but to a centripetal acceleration. This is done to try avoid the very misconception you had that the centripetal force is another force on an object.

You can think of it this way. Part of the force of gravity gets used up causing the centripetal acceleration, and the rest of it goes into balancing the normal force. Consequently, the normal force, which is your apparent weight, is always going to be less than or equal to your true weight.

9. Feb 12, 2012

### Sefrez

I guess so. I was pretty much trying to imply what vela said in a different way though. You have gravity giving your weight and some is simply going into this "centripetal acceleration." It ends up as I said though, the difference between the two. m(g - ac)

10. Feb 12, 2012

### Jacobim

I just about came to that reasoning my self, that the centripetal acceleration is a component of the gravity, so less of the gravity is in opposition of the normal force.

I should completely grasp this concept in about 3 months :(

Very tricky, why is physics so tricky? Im 29 and this is killing me. I wish I had cared about education when I was a teenager. Imagine if everyone were able to comprehend the logical discoveries of science, and what effect that would have on our societies creativity. The road block for me is that it is just so difficult learning the math and language to understand the scientific models. The pace of college is really too much as well. Dont know if I'll make it, I may switch to one of the engineering tech degrees, or go to a trade school.

Can a person get a job in engineering with a C average?

Ha ok enough innappropriate first coffee of the morning rambling, time for more physics immersion.

11. Feb 12, 2012

### Jacobim

I think I kind of just visualized this...

The tangential velocity of the person is causing the fictitious centrifugal force, which feels as though you are being pushed upward... and this would be in the same direction of the normal force also pushing upward.

So this tangential velocity is opposing gravity along with the normal force, so the normal force has to do less work.

Am I on the right track?

Thanks again !

12. Feb 12, 2012

### vela

Staff Emeritus
I wouldn't say "the tangential velocity is opposing gravity"; the fictitious centrifugal force is.

This is a valid way of looking at the situation, but you should be able to understand it both ways.

One other comment: Physics isn't really tricky. The problem is that people come in with preconceptions about how objects move, developed over the years by experience. When these preconceptions are in fact misconceptions, it takes a bit of effort to recognize and unlearn the wrong things and learn the concepts correctly. Sometimes your intuition can help you, but it can also hinder you. You just need to be aware that your intuition will be wrong sometimes.

Last edited: Feb 12, 2012
13. Feb 12, 2012

### Jacobim

Yes that is a HUGE problem. Maybe the main problem. If I had a blank slate it would be easy.

I declare we should teach all the kiddos proper physics logic in the first place!