Car rounds a flat road-friction forces

  • Thread starter Thread starter Dannystu
  • Start date Start date
  • Tags Tags
    Car Flat Forces
AI Thread Summary
The discussion centers on a physics problem involving a car rounding a curve on a flat road. The car has a mass of 2500 kg, a speed of 15 m/s, and a curve radius of 60 m. The centripetal acceleration is calculated to be 3.75 m/s², and the centripetal force is determined to be 9375 N. The participant struggles with calculating the static friction force and the coefficient of static friction, seeking guidance on applying Newton's second law and understanding the relationship between normal force and weight. The conversation emphasizes the importance of free body diagrams (FBD) in solving the problem.
Dannystu
Messages
16
Reaction score
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?
 
Physics news on Phys.org
Draw a FBD and apply Newton's second law.
 
Is the normal force equal and oppositely directed to the weight? I drew the FBD showing the a front view of the car.
 
Write Newton's law in the vertical direction and find out.
 
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:
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top