Centripetal Force: Why Static Friction is Used in Bank Curve Problems

AI Thread Summary
In bank curve problems, the coefficient of static friction is used instead of kinetic friction because static friction applies when the wheels are not sliding, ensuring stability and control. Once the wheels begin to slide, the dynamics become more complex and less predictable. The static case allows for a clearer understanding of forces acting on the vehicle. This distinction is crucial for accurately analyzing motion on banked curves. Understanding these principles is essential for effective problem-solving in physics.
oreo
Messages
81
Reaction score
2
Why do we use coefficient of static friction instead of coefficient of kinetic friction in bank curve problems
 
Physics news on Phys.org
When the wheels start sliding sideways things start getting complicated, and very exciting. When the part of the wheel in contact with the paved surface/track is not moving relative to the track (static case), things are under control.
 
Bystander said:
When the wheels start sliding sideways things start getting complicated, and very exciting. When the part of the wheel in contact with the paved surface/track is not moving relative to the track (static case), things are under control.[/QUOTE
 
Thanks for the clear answer.
 
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