Using Normal Force on inclined plane vs banked curve

In summary, when considering forces on an inclined plane, the equations "\SigmaFy = N - mg cos ø = ma = 0" and "\SigmaFx = mg sin ø = ma" (frictionless) can be used to find the normal force (N) in the y direction. However, on a banked curve with circular motion, the equations "\SigmaFy = N cos ø - mg = ma= 0" and "\SigmaFx = N sin ø = ma = mv²/r" (frictionless) must be used, taking into account the additional centrifugal force. This explains why "N = mg cos ø" does not work to find the tangent of the angle (ø) in the equation
  • #1
Rick4
2
0
On an inclined plane, "[tex]\Sigma[/tex]Fy = N - mg cos ø = ma = 0".
And "[tex]\Sigma[/tex]Fx = mg sin ø = ma" (frictionless).
So, in the y direction, "N = mg cos ø".


But on a banked curve, "[tex]\Sigma[/tex]Fy = N cos ø - mg = ma= 0".
And "[tex]\Sigma[/tex]Fx = N sin ø = ma = mv²/r" (frictionless).
So, in the y direction, "N cos ø = mg".
I know how to get "tan ø = v²/rg".
The two free body diagrams look identical to me so why doesn't "N = mg cos ø" (from inclined plane) work to get "tan ø = v²/rg"?
 
Physics news on Phys.org
  • #2
On a banked curve, circular motion is normally involved and so there is an extra centrifugal force. I don't see the problem except you've neglected to take that into account on the free body diagram.
 
  • #3


The equations for normal force on an inclined plane and a banked curve may look similar, but they are actually describing two different scenarios. On an inclined plane, the normal force is equal to the force of gravity acting on the object, which is why N = mg cos ø. This is because the object is only moving in one direction, along the incline.

On the other hand, on a banked curve, the normal force is not equal to the force of gravity. This is because the object is moving in two directions - along the curve and perpendicular to it. In this case, the normal force is equal to the force of gravity minus the centripetal force needed to keep the object moving in a circular path, which is why N cos ø = mg.

To understand why N = mg cos ø does not work to get tan ø = v²/rg on a banked curve, we need to consider the forces acting on the object in the x-direction. On an inclined plane, the only force acting in the x-direction is the force of gravity (mg sin ø). But on a banked curve, there are two forces acting in the x-direction - the normal force (N sin ø) and the centripetal force (mv²/r). This is why the equation for the x-direction is different for the two scenarios.

In summary, while the two equations may look similar, they are describing different situations and cannot be used interchangeably. It is important to carefully consider the forces acting on an object in different scenarios to accurately determine the normal force and other forces involved.
 

1. What is the difference between using normal force on an inclined plane and a banked curve?

The main difference is that on an inclined plane, the normal force is equal to the component of the weight of an object perpendicular to the plane, while on a banked curve, the normal force is equal to the centripetal force needed to keep an object moving in a circular path. In other words, the normal force on an inclined plane is used to counteract the weight of an object, while the normal force on a banked curve is used to keep an object from sliding off the curve.

2. How does the angle of inclination or banking affect the normal force?

The angle of inclination or banking affects the magnitude of the normal force. On an inclined plane, as the angle of inclination increases, the normal force also increases. On a banked curve, as the angle of banking increases, the normal force decreases. This is because the normal force is dependent on the component of the weight or centripetal force that is perpendicular to the surface.

3. Can the normal force be greater than the weight of an object on an inclined plane or banked curve?

Yes, the normal force can be greater than the weight of an object on both an inclined plane and a banked curve. This can occur if there is an additional force acting on the object, such as a pushing or pulling force, or if the angle of inclination or banking is steep enough to require a greater normal force to keep the object in place.

4. Is the normal force always equal to the weight of an object on a banked curve?

No, the normal force is not always equal to the weight of an object on a banked curve. As mentioned earlier, the normal force on a banked curve is equal to the centripetal force needed to keep an object moving in a circular path. This force can be greater or less than the weight of the object, depending on the angle of banking and the speed of the object.

5. How does friction play a role in the normal force on an inclined plane or banked curve?

Friction can affect the normal force on both an inclined plane and a banked curve. On an inclined plane, friction can decrease the normal force by acting in the opposite direction of the weight of an object. On a banked curve, friction can increase the normal force by acting in the same direction as the centripetal force, helping to keep the object in its circular path.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
692
  • Introductory Physics Homework Help
Replies
12
Views
2K
  • Introductory Physics Homework Help
Replies
27
Views
3K
Replies
24
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
667
  • Introductory Physics Homework Help
2
Replies
41
Views
2K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
3K
  • Introductory Physics Homework Help
Replies
7
Views
1K
Back
Top