What is the Normal Force on Tires in a Banked Curve with Friction?

  • Thread starter Thread starter nns91
  • Start date Start date
  • Tags Tags
    Curve Friction
Click For Summary
SUMMARY

The normal force on tires in a banked curve with friction can be calculated using the equation ΣF = m * v²/r, where m is the mass of the car (800 kg), v is the speed (85 km/h), and r is the radius of the curve (150 m). The angle of the bank (10 degrees) affects the distribution of forces, including the gravitational force and the frictional force acting on the tires. The normal force cannot be simplified to Fn = mg/cos(10) due to the presence of horizontal forces from angular motion, which must be accounted for in the calculations.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with circular motion dynamics
  • Knowledge of frictional forces in physics
  • Basic trigonometry for resolving forces
NEXT STEPS
  • Study the derivation of the normal force in banked curves with friction
  • Learn about the role of centripetal force in circular motion
  • Explore the effects of varying angles on normal force calculations
  • Investigate real-world applications of banked curves in vehicle dynamics
USEFUL FOR

Physics students, automotive engineers, and anyone interested in understanding the dynamics of vehicles on banked curves with friction.

nns91
Messages
301
Reaction score
1

Homework Statement



Radius 150m, angle is 10 degree, mass of the car 800kg, speed 85 km/h.

Find the normal force on the tires exerted by the pavement. ( There is friction)



Homework Equations



[tex]\Sigma[/tex] F= m* v^2/r

The Attempt at a Solution



The friction makes it's so complicated. I know that Fn= mg/cos10 will not work.

Can anyone help me ?
 
Physics news on Phys.org
nns91 said:

Homework Statement



Radius 150m, angle is 10 degree, mass of the car 800kg, speed 85 km/h.

Find the normal force on the tires exerted by the pavement. ( There is friction)



Homework Equations



[tex]\Sigma[/tex] F= m* v^2/r

The Attempt at a Solution



The friction makes it's so complicated. I know that Fn= mg/cos10 will not work.

Can anyone help me ?

It's not that m*g*Cosθ doesn't work, it's just that there is a horizontal force from the angular motion that is acting as well.

If there is a horizontal force, what is its component that is also normal to the incline?
 
Thanks. I got it.
 

Similar threads

  • · Replies 9 ·
Replies
9
Views
2K
Replies
9
Views
4K
Replies
16
Views
3K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 6 ·
Replies
6
Views
5K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 9 ·
Replies
9
Views
7K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K