Confused about N*sin(theta) = (m*v^2)/r in banked curve

In summary, if there is a car resting on a banked curve with angle theta, velocity v = 0, but N (normal)*sin(theta) > 0. So N*sin(theta) =/= (m*v^2)/r with v = 0. But my physics textbook just defined N*sin(theta) = (m*v^2)/r in banked curve. What is going on here?
  • #1
annamal
381
33
If there is a car resting on a banked curve with angle theta, velocity v = 0, but N (normal)*sin(theta) > 0. So N*sin(theta) =/= (m*v^2)/r with v = 0. But my physics textbook just defined N*sin(theta) = (m*v^2)/r in banked curve. What is going on here?
 
Physics news on Phys.org
  • #2
##\vec N\sin\theta## is the horizontal component of the normal force. The Normal force, N, is the component of ##\vec F_g = -m\vec g## that is perpendicular to the surface of the road.

1648141858274.png

This component of the normal force provides contributes to the centripetal force required to keep the car turning with the road. So the bank reduces the need for friction between the road and tires to supply all the force needed for the car to make the turn successfully.

AM
 
  • #3
Andrew Mason said:
##\vec N\sin\theta## is the horizontal component of the normal force. The Normal force, N, is the component of ##\vec F_g = -m\vec g## that is perpendicular to the surface of the road.

View attachment 298870
This component of the normal force provides contributes to the centripetal force required to keep the car turning with the road. So the bank reduces the need for friction between the road and tires to supply all the force needed for the car to make the turn successfully.

AM
Yes, I understand that part, but my question is if the car were at rest. v = 0, but N*sin(theta) > 0
 
  • #4
There is a horizontal component trying to slide the car down the slope, but that is from the weight of the stopped car, not from the centrifugal effect.
 
  • #5
##\frac{mv^2}{r}## is the centripetal force that the designers want the horizontal component of normal force to provide for the expected or recommended car speed rounding that curve. The designers first determine what the appropriate range of speed should be for cars rounding that curve. The coefficient of static friction factors into this appropriate range because they don't want cars to slide up or down when rounding the curve because they are traveling in the upper or lower parts of that range. Then they determine what ##\theta## should be such that for v in the middle of that appropriate range ##F_N\sin\theta=mv^2/r##.

AM
 
  • #6
If the car is at rest on the slope there must be some friction. Otherwise the car will slide down the slope. You should include the friction components in the fee body diagram and the Newton's laws. And of course that there is no $$ mv^2/r $$ term as there is no centripetal acceleration in this case. You should realize that the normal force depends on the situation, there is no universal formula that gives the normal force for any system and any state of motion.
 
  • Like
Likes Lnewqban

Related to Confused about N*sin(theta) = (m*v^2)/r in banked curve

1. What is the equation N*sin(theta) = (m*v^2)/r used for in banked curves?

The equation N*sin(theta) = (m*v^2)/r is used to calculate the normal force on an object moving in a banked curve. The normal force is the force perpendicular to the surface that prevents the object from falling through the curve.

2. How does the angle of the banked curve affect the normal force?

The angle of the banked curve, represented by theta in the equation, directly affects the normal force. As the angle increases, the normal force decreases. This is because a steeper angle allows for more of the force of gravity to be directed towards the direction of motion, reducing the need for the normal force to counteract it.

3. What does the mass and velocity of the object have to do with the normal force in a banked curve?

The mass and velocity of the object are both factors in determining the normal force in a banked curve. The greater the mass and velocity of the object, the greater the centrifugal force acting on it, and therefore, the greater the normal force needed to keep it on the curve.

4. Can the normal force ever be greater than the weight of the object in a banked curve?

Yes, the normal force can be greater than the weight of the object in a banked curve. This occurs when the angle of the banked curve is steep enough to create a centripetal force greater than the force of gravity acting on the object.

5. How does the radius of the curve affect the normal force in a banked curve?

The radius of the curve, represented by r in the equation, also affects the normal force. As the radius decreases, the normal force increases. This is because a tighter curve requires a greater centripetal force, and therefore, a greater normal force to keep the object on the curve.

Similar threads

Replies
24
Views
4K
  • Introductory Physics Homework Help
Replies
12
Views
2K
Replies
9
Views
1K
Replies
19
Views
5K
Replies
3
Views
125
  • Introductory Physics Homework Help
Replies
1
Views
276
Replies
2
Views
4K
Replies
5
Views
811
Replies
8
Views
2K
Replies
8
Views
529
Back
Top