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

AI Thread Summary
In a banked curve scenario, the normal force's horizontal component, N*sin(theta), contributes to the centripetal force required for a car to turn. When a car is at rest (v = 0), N*sin(theta) can still be greater than zero due to gravitational forces, but there is no centripetal force acting since mv^2/r equals zero. The design of banked curves considers the expected speed range for vehicles, factoring in static friction to prevent sliding. If the car is stationary, friction must be present to prevent it from sliding down the slope. Understanding these dynamics clarifies the relationship between normal force, friction, and the design of banked curves.
annamal
Messages
393
Reaction score
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
##\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
 
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
 
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.
 
##\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
 
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.
 
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Thread 'Beam on an inclined plane'
Hello! I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below. Here is how I wrote the condition of equilibrium forces: $$ \begin{cases} F_{g\parallel}=F_{t1}+F_{t2}, \\ F_{g\perp}=F_{r1}+F_{r2} \end{cases}. $$ On the other hand...
Back
Top