Why Is Normal Force Less Than mg at the Top of a Hill?

[corey2014]

So in class today we were talking about Normal Force, and I was really confused on one diagram.

Here is a link about the question, but I don't understand why n<mg because if it is less than mg wouldn't that make it go through the earth? not sure but help understanding would be great thanks!

http://www.colorado.edu/physics/phys1110/phys1110_sm11/ConceptTests/pdf/Lecture6_Using%20Newton_FreeBodyDiagrams_Normal_tension_gravity.pdf

And if the link doesn't work the question is, a car traveling at a constant speed goes over a hill, at the top of the hill what is the normal force?

Thanks for the help!

 

[Doc Al]

Here is a link about the question, but I don't understand why n<mg because if it is less than mg wouldn't that make it go through the earth? not sure but help understanding would be great thanks!

The normal force would equal mg (at least on a horizontal surface) if there were no acceleration normal to the surface the surface. But if there's acceleration in the direction of the surface, the normal force can be less than or greater than mg. Think of an elevator. When the elevator accelerates upward, the normal force from the floor of the elevator on you is greater than your weight.

And if the link doesn't work the question is, a car traveling at a constant speed goes over a hill, at the top of the hill what is the normal force?

When the car travels over a hill, it's actually accelerating. (Think of it as moving in a circle.) So ƩF = ma, thus weight - normal force = ma. So normal force = weight - ma. (You can plug in the value of the centripetal acceleration for a.)

 

[Lsos]

Here is a link about the question, but I don't understand why n<mg because if it is less than mg wouldn't that make it go through the earth?

Why would it go through the earth? The car going over the hill actually tends to go AWAY from the earth. You know how in movies if the car goes fast enough, it actually jumps off the end of the hill? Well in that situation, the normal force would be 0 (or even upwards).

Think of the normal force as what the reading would be if you placed a scale underneath the car. Since the car would be in the air, it wouldn't even be touching the scale.

 

[sophiecentaur]

When the car is going over the hill it is NOT in equilibrium, is it? So you can't insist on using an equilibrium equation to describe what's going on.