How Do You Solve for Normal Force in Different Dynamics?

Click For Summary
SUMMARY

The discussion focuses on solving for the normal force in dynamic scenarios, specifically when a car rolls over a hill. The key conclusion is that the normal force is not always equal to the perpendicular component of gravitational force, as demonstrated in cases like vertical loops and banked curves. Instead, one must first determine the motion of the object and then set the normal force to match the required forces for that motion. In the example provided, the normal force is less than the gravitational force when the car rolls down the hill, confirming option "b".

PREREQUISITES
  • Newton's Second Law of Motion
  • Understanding of gravitational force and its components
  • Basic principles of dynamics in circular motion
  • Concept of normal force in contact mechanics
NEXT STEPS
  • Study the effects of centripetal force on normal force in circular motion
  • Learn about the dynamics of objects on inclined planes
  • Explore the concept of net force and acceleration in dynamic systems
  • Investigate normal force variations in different contact scenarios, such as banking and loops
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of forces acting on objects in motion, particularly in scenarios involving curves and inclines.

Impulse
Messages
19
Reaction score
0

Homework Statement



The problem motivating this post is: "An out-of-gas car is rolling over the top of a hill at speed v. At this instant,
a. n > FG
b. n < FG
c. n = FG
d. We can't tell about n without knowing v." (Associated figure attached.)

Homework Equations


[/B]
Newton's second law

The Attempt at a Solution



What I'd like to understand is in what framework to think about the normal force.

I currently understand the normal force to be equal and opposite to the perpendicular force exerted by an object on the surface with which it is in contact. I still believe this to be correct.

I'm wrestling with how to solve for the normal force. Originally, I assumed it was always equal and opposite to the perpendicular component of the force of gravity, such as is the case in a standard incline plane problem. To solve for the normal force, I would first solve for the perpendicular component of the force of gravity, then flip the sign. However, the motivating problem I posted above, as well as examining a car doing a vertical loop-the-loop and driving around a curve on a banked track, has shown that this framework for solving for the normal force can be faulty.

It now appears that, to solve for the normal force acting on an object, I should first determine the resulting motion of the object known to be true in a given scenario and then set the normal force equal to whatever force is required to make that motion result. So, in the motivating problem above, because I know the car will roll down the other side of the hill, the downward force of gravity must exceed the upward normal force, and the answer must be "b".

Is this line of thinking how you understand to solve for the normal force?

The paradigm of "first determine how you know the system to act, then set a force equal to whatever is required to make the system act in that way" feels foreign to me. I'm used to "first solve for all the forces, then determine how the system will act as a result of those forces."
 

Attachments

  • Car Rolling Over Hill.jpg
    Car Rolling Over Hill.jpg
    9.1 KB · Views: 897
Physics news on Phys.org
Is the direction of the car's velocity vector constant as the car goes over the hill, or is it changing? What does this tell you about the car's acceleration? What does that tell you about the net force acting on the car?

Chet
 
What normal force did you have in mind?

There is the normal force between the wheel and the road;
the normal force between the air and the car;
the normal force between the axle and the car body,
etc.
 
@Old: come on, a clear picture is provided !

@IMP: Chet's chat makes good sense to me !
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 13 ·
Replies
13
Views
2K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 3 ·
Replies
3
Views
9K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K