Why does vertical component of normal force

In summary: The summary is that in order to make a car turn on a banked curve on the road, the vertical and horizontal components of the normal force must be equal but opposite in direction to the force of gravity. The only possible speed at which this force will be equal to the weight is v^2=r*g*tan(alpha).
  • #1
boris16
46
0
hi
For car to be able to drive trough banked curve on the road with no friction, the vertical component of a normal force must be equal in size but opposite in direction to the force of gravity.

Horizontal component F[h]

F[h] = m * g * tan(alpha) = m * v^2 / r

So there is only one speed at which F[h] will equal centripetal force at specific angle of the curve and and specific radius.
But why does vertical component of normal force equal the force of gravity when F(h) = m * v^2 / r ?

thanx
 
Physics news on Phys.org
  • #2
The vertical componant of the normal force has to balance the weight, otherwise the car will move up or down the banked curve, thus changing the radius of the turn. There is only one speed at which F[h] will provide centripetal force while simultaneously F[v] will balance weight. That's why this problem is so unrealistic
 
  • #3
What do you mean by unrealistic?

Forget that you ever heard of car making a turn on a banked curve.
You only know that for a car to make that turn:

A- vertical force must cancel out force of gravity

B- only possible centripetal force at certain angle of a curve and certain radius must equal m * v^2 / r and thus speed must be v^2=r*g*tan(alpha).

But your problem is that you don't know if A and B can happen at the same time ( obviously they do since cars do make the turns but suppose you never knew that )
Based on what would you figure out if those two things can happen at the same time?
 
Last edited:
  • #4
First of all the problem is unrealistic because you always have to stay at the same radius and cause there's no friction.
>How do you tell if the two things can happen? - make two equations and compare them.
 
  • #5
Perfect uniform circular motion is very difficult. For constant speed in a perfect circle to happen, the following things must be in perfect harmony:

the relationship between force, mass, speed, and radius of curvature MUST be exactly thus: F = (mv^2)/r . If anyone of these variables is a tiny bit higher or lower, then there is no circular motion. It's a nearly impossible balance to achieve for anything other than a very controlled experimental setup.

Also, the force must contiunally change direction so as to always be exacly perpendicular to the direction of motion at one instant. This is also nearly impossible to attain.

For the car to turn on a *Frictionless* banked turn at a constant radius, it must start the turn at the perfect speed, and remain at this perfect speed, that is predetermined by the given situation. The given situation (through balancing two necessary formulas, one for the vertical compnant, one for the horizontal componant of normal force) dictates that there will be only one speed for a given radius on a given incline where the exact centripetal force will be supplied AND the exact balancing force to gravity will be supplied by the normal force.

It's possible, but there is no real situation that uses this, hence: "unrealistic." A real banked turn will use static friction to allow for other speeds.
 
  • #6
I forgot that normal force (when equal to gravity force) is already included in

F[h]=m*g*tan(alpha) with N*sin(alpha)= ( m*g/cos(alpha) ) * sin(alpha)

thanx
 

1. Why does the vertical component of normal force exist?

The vertical component of normal force exists due to the fundamental laws of motion. When an object is on a flat surface, the force of gravity pulls it downward. This creates a normal force that acts perpendicular to the surface to balance out the force of gravity and prevent the object from falling through the surface.

2. How is the vertical component of normal force calculated?

The vertical component of normal force is equal to the weight of the object, which is calculated by multiplying the mass of the object by the acceleration due to gravity (9.8 m/s²). This is because the normal force must be equal and opposite to the force of gravity for the object to remain at rest.

3. Can the vertical component of normal force change?

Yes, the vertical component of normal force can change. It depends on the weight and position of the object. If the weight of the object changes, the vertical component of normal force will also change. Additionally, the angle of the surface can also affect the vertical component of normal force.

4. What happens to the vertical component of normal force if the object is on an inclined plane?

On an inclined plane, the vertical component of normal force is still present, but it is smaller than the weight of the object. This is because some of the force of gravity is acting parallel to the surface, causing the object to slide down the incline. The normal force now has an additional horizontal component to counteract this motion.

5. How does the vertical component of normal force affect the motion of an object?

The vertical component of normal force plays a crucial role in keeping an object at rest or in motion on a flat surface. It helps to counteract the force of gravity and maintain equilibrium. On an inclined plane, the vertical component of normal force also helps to support the object and prevent it from sliding down the incline. Without this force, an object would not be able to maintain its position on a surface.

Similar threads

  • Introductory Physics Homework Help
Replies
7
Views
264
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
984
  • Introductory Physics Homework Help
Replies
6
Views
2K
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
996
  • Introductory Physics Homework Help
Replies
8
Views
976
  • Introductory Physics Homework Help
3
Replies
97
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
1K
Back
Top