Friction on a banked road and max velocity

Click For Summary
SUMMARY

The discussion focuses on calculating the maximum speed of an automobile on a wet banked roadway with a banking angle of 25˚, a static friction coefficient of 0.300, and a kinetic friction coefficient of 0.250. The radius of the curve is 50.0 m. Participants utilize Newton's second law, ΣF=ma, to set up equations based on the forces acting on the vehicle, including static friction and normal force. The key challenge is solving the system of equations derived from the forces in both the x and y directions to find the maximum velocity before sliding occurs.

PREREQUISITES
  • Understanding of Newton's second law (ΣF=ma)
  • Knowledge of static and kinetic friction coefficients
  • Familiarity with free body diagrams
  • Basic trigonometry, particularly sine and cosine functions
NEXT STEPS
  • Study the derivation of equations for circular motion on banked curves
  • Learn how to solve systems of equations in physics problems
  • Explore the effects of varying friction coefficients on vehicle dynamics
  • Investigate the impact of banking angles on maximum velocity calculations
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and dynamics, as well as educators looking for practical examples of applying Newton's laws in real-world scenarios.

spark706
Messages
2
Reaction score
0

Homework Statement


Consider a wet banked roadway, where there is a coefficient of static friction of 0.300 and a coefficient of kinetic friction of 0.250 between the tires and the roadway. The radius of the curve is 50.0 m. If the banking angle is 25˚, what is the maximum speed an automobile can have before sliding up the banking?

Homework Equations


Free body diagram-http://s3.amazonaws.com/answer-board-image/e89af301a0a3642873d7794ad22fba83.jpg

f=static friction
θ=25˚ angle
μ=friction coefficient=.300 static
r=50.0m
m=mass
g=gravity

Using ΣF=ma Newton's 2nd law:
x-> ΣF=μmgcos(25˚) + N*sin(25˚)=mv^2/r
y-> ΣF=Ncos(25˚)=μmgsin(25˚) +mg

The Attempt at a Solution


masses cancel I know.

x-> .300*9.8cos(25˚) + 9.8*sin(25˚)=v^2/50.0
y-> 9.8*cos(25˚)=.300*9.8sin(25˚) + 9.8

and that's as far as I get. Can anybody please explain what I need to do next?

Thanks
 
Physics news on Phys.org
I guess what I'm asking is do I need to treat this as a system of equations? Do I need to divide, and if so is it x/y or y/x?
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
16
Views
3K
  • · Replies 4 ·
Replies
4
Views
17K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 10 ·
Replies
10
Views
1K
  • · Replies 4 ·
Replies
4
Views
8K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 19 ·
Replies
19
Views
4K
Replies
6
Views
2K