Why Does Normal Force Increase with Theta in Banked Road Calculations?

[bthilton]

1. Hi, I've been reading through an introductory physics book and am having trouble understanding certain equations pertaining to centripetal force. The part of the book I am on is discussing banks and how they are used to create centripetal force on roads. It's been stumping me for months.

Specifically I am confused about the equation Fn= mg / cos theta since that reads to me "Normal force increases as theta increases(Up to 90 of course)". I thought that normal force was the force exerted on an object from the surface it rests on, which I also thought would naturally decrease as the incline steepens. It doesn't help my understanding when earlier in the book it states Fn = mg cos theta. Any help in understanding what this book is trying to tell me would be greatly appreciated.



2. Fn= mg / cos theta (Finds normal force for the equation for centripetal force, Fc=Fn sin theta = mv^2/r)



3. (Doesn't really apply.)

 

[rock.freak667]

I assume you are talking about a car banking around a corner at an angle θ.

Well for cosθ, as θ→π/2, cosθ→0. You can see this in the graph of y=cosx.
Cos(π/2)=0 and cos(0)=1. So logically from 0 to π/2, cosθ goes from 1 to zero (maximum to minimum)

 

[bthilton]

I understand that, that's why I'm confused. Yes I do mean a car banking around a corner at angle theta.

The thing I don't understand is how Fn= mg/ cos theta, which I assume means that normal force increases/decreases directly as theta increases/decreases. Which makes no sense to me.

 

[LBloom]

Edit: Normal force will not necessarily decrease in this case. All you have to know is the as theta increases, the normal force component along the x-axis will increase (since its F sin theta), and therefore that if the car is going faster and requires a greater centripetal force, the angle b/t the incline and the ground must be increased so more of the normal force is going along the x-axis and fulfilling its role as the normal force.
Yes, if you have object that doesn't move, and you increase the angle, the normal force will decrease, the friction will decrease, and more of the gravitational force will be pulling the object done, but that's not how you should think of it in this problem

btw, it says edit b/c i deleted my original post. It wasnt very helpful, or correct for that matter :)

Edit #2: this problem requires a lot of vectors, and is too hard to explain with words, so if you want, you can forget what i just said and just click on this link: http://www.batesville.k12.in.us/Physics/PhyNet/Mechanics/Circular Motion/banked_no_friction.htm

 

[kuruman]

Another thing you need to understand is that normal force is a "contact force" like tension in a string or static friction. Contact forces have the peculiar property that they adjust themselves to produce the necessary acceleration. For example, if a put a book on a horizontal table, the normal force exerted by the table on the book matches the book's weight. Now if I put two books on the table, the normal force matches the weight of two books. How does the table "know" to do that? It bends and flexes (imperceptibly so) a little more under the weight (like a spring) so that the force is what is needed to give zero acceleration in this case. Of course, the normal force like static friction and tension has an upper limit beyond which it can no longer "adjust itself." If I placed a battleship on the table, the table would collapse. Somewhere in between the weight of two books and a battleship is the upper limit beyond which the table can no longer support what is placed on it. With tension, strings break and with static friction things start sliding.

The bottom line is that because the normal force "adjusts itself to provide the observed acceleration", there is no specific equation that gives it. It depends on the problem. So the cosine multiplies mg for an object on an incline, but divides mg for an object sliding around a bank. Although the free body diagrams look the same, the normal force is different because the acceleration is different and the normal force has "adjusted" itself to "provide" that difference.

 

[bthilton]

Okay, thank you both. I'm pretty sure I get it now.

 

[harrinj4]

Okay, thank you both. I'm pretty sure I get it now.

They're the best aren't they?

 

[bthilton]

Not bad.