Car rounds a flat road-friction forces

In summary, a car with a mass of 2500kg is traveling at a speed of 15 m/s around a curve with a radius of 60m. The centripetal acceleration experienced by the car is 3.75 m/s^2 and the centripetal force is 9375N. To calculate the force of static friction between the tires and the road, you can use the formula Fs= μsN, where μs is the coefficient of static friction and N is the normal force. The normal force can be calculated by multiplying the mass of the car by the acceleration due to gravity. To find the coefficient of static friction, you can use a FBD and apply Newton's second law to find the normal
  • #1
Dannystu
16
0

Homework Statement



A car, mass m=2500kg, rounds a curve on a flat road at a speed v= 15 m/s. The radius of curvature of the curve is r= 60m. There is obviously (static) friction between the road and the car tires, or the car would not stay on the curve.


Homework Equations



a.) Compute the centripetal acceleration experienced by the car.

b.)Compute the centripetal force experienced by the car. What phenomenon is the cause of this centripetal force?

c.)Compute the force of (static) friction between the tires and the road.

d.) If the given speed of V=15 m/s is known to be the maximum speed for this curve for which a car will not skid, compute the coeffiecient of static friction between the tires and the road.



The Attempt at a Solution



Part a)

Centripetal acceleration= V^2/r

=3.75 m/s^2

b.) Centripetal force = M*(V^2/r)

=9375N

c.) I have no clue on how to do part C and D, I know the formula for Static friction is

Fs= Coefficient of static friction * normal force


I calculated the normal force to be 24,500N (mass times gravity) since there are no other forces acting on the vertical direction. Can someone give me a push in the right direction?
 
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  • #2
Draw a FBD and apply Newton's second law.
 
  • #3
Is the normal force equal and oppositely directed to the weight? I drew the FBD showing the a front view of the car.
 
  • #4
Write Newton's law in the vertical direction and find out.
 
  • #5
Yes because the car is not moving on the vertical direction, and the friction force in pointed in the opposite direction of motion.
 
Last edited:

1. What is the purpose of a car going around a flat road?

The purpose of a car going around a flat road is to demonstrate the effects of friction forces on the movement of the car. This can be used to understand the dynamics of a car's motion and how friction plays a role in controlling its speed and direction.

2. How does the force of friction affect a car's movement on a flat road?

The force of friction acts in the opposite direction to the car's motion, causing it to slow down. This is because the tires of the car are in contact with the road, creating friction that resists the motion of the car.

3. Can friction forces be beneficial in a car's movement on a flat road?

Yes, friction forces can be beneficial in a car's movement on a flat road. They provide the necessary traction for the car to grip the road and maintain control while turning or braking. Without enough friction, the car may slip or skid, resulting in loss of control.

4. How does the weight of the car affect the friction forces on a flat road?

The weight of the car affects the friction forces by increasing the normal force between the tires and the road. This results in a stronger friction force, which helps the car maintain its grip and control on the road.

5. Are there any other factors that can affect friction forces on a car going around a flat road?

Yes, there are other factors that can affect friction forces on a car, such as the type and condition of the tires, the speed of the car, and the surface of the road. These factors can all impact the amount of friction present and how it affects the car's movement.

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