# Homework Help: Giancoli ed 5 p 121

1. Sep 10, 2006

### jla2w

I am teaching myself physics using Giancoli's ed 5 text, and am confused on one of the examples. P 121 in the fifth edition, example 5-8 to be precise. In this a car is on an inclined plane (i.e a racetrack where the road banks to reduce the friction needed to keep the cars on the track). The solution explains that the normal force perpindicular to the track is greater than the force due to gravity directed downward. In previous examples, gravity was generally resolved into two components, but in this example it seems the normal force is resolved into two components, with the vertical component of the normal force set equal to gravity. This seems incorrect, what am I missing? Thanks

2. Sep 10, 2006

### Chi Meson

You can resolve any vector into componants along axes in any direction.

In this situation, due to the lack of friction, the only force that the track can exert on the car is perpendicular to the surface. THis is the definition of "normal" (direction perpendicular to the plane of the surface).

The normal force in this situation has to "accomplish" two things: first it must balance the weight (gravitational force), and simultaneously it must provide centripetal force.

These two "required componants" add up to the net force from the track, which is the normal force. Depending on the angle of pitch of the turn in the track, there is only one speed that will allow the car to make the turn without wiping out.

Last edited: Sep 10, 2006
3. Sep 10, 2006

### jla2w

thanks, that is a helpful explanation, but there is one aspect I'm still unclear on. If the car were at rest, then the normal force in a direction perpindicular to the track would be less than the force due to gravity directed downward. In this case the normal force is greater than the gravitational force. What is the principle or generalized explanation for why when the car is in motion the normal force exceeds gravity, and at rest it is less than gravity?

4. Sep 10, 2006

### mbrmbrg

Normal force remains the same regardless of whether the object is at rest or in motion (so long as the object remains in contact with the surface it's supposed to be on, at any rate. I don't know about you, but I like all four wheels of my car to be in contact with the road at all times, whether the car is parked or moving. )

PS I can't believe you're teaching yourself physics--more power to your elbow!

5. Sep 10, 2006

### Chi Meson

I'm sorry, but that is wrong.

Normal force will react to whatever the situation is. The normal force on a stationary car would be less than that of a car moving through the corner. In the situation of a car that starts at rest, the normal force would be balanced by (and therefore equal in magnitude to) a componant of the car's weight (the so-called "perpendicular componant" of the weight).

In the moving example, the normal force is the vector sum of the weight plus the "required" centripetal force.

6. Sep 10, 2006

### mbrmbrg

:blush:
My humblest apologies.

7. Sep 11, 2006

### jla2w

thanks, very helpful. So as the car moves around the banked road, the road exerts a force on the car (centripetal) directly proportional to the square of the velocity. In addition, since friction is not a factor, as it is in the resting situation, the vertical component of the normal force MUST equal precisely mg. Makes sense, but still pretty difficult to deduce without much practice.

8. Sep 11, 2006

### Chi Meson

The next step is to add friction (and thus make it a "real world" problem. Friction is the "surface force" that acts parallel to the surface (while normal is the perpendicular "surface force"). With friction, the speed of the car does not have to be exactly right for the given embankment.

Keep in mind, that if an object is moving in perfect circular motion, all forces must be balanced except for the force (or componant) that is "centripetal." In other words, in uniform circular motion, the centripetal force is the net force.

9. Sep 11, 2006

### jla2w

Thanks Chi, and by the way, wahoowa (SEAS '02)

10. Sep 11, 2006

### Chi Meson

Aha! Rugby Road to Vinegar Hill!