Car on an incline, kinetic/static friction when break applied confussion.

AI Thread Summary
When a car is on an incline and brakes are applied, static friction is relevant because the tires are rolling without slipping, even if the car is in motion. This means that the maximum braking force is determined by the static friction coefficient, which is typically higher than the kinetic friction coefficient. In scenarios where impending slipping occurs, static friction must be considered despite the car's movement. Conversely, if the brakes cause the tires to skid, kinetic friction would apply instead. Understanding the distinction between static and kinetic friction is crucial for solving problems involving braking on inclines.
alba_ei
Messages
38
Reaction score
1
I've been doing some problems relative to car on an incline (and also boxes on an incline).
Let's suppose we are given a problem with kinetic or static coefficient.
I'm confusing when should or should not apply kinetic friction to a car, for example in some problems if brakes are applied then friction is involved (as if it were a box), however if brakes are not applied friction is not taken into account(Am I right? I'm not sure, if it is true then why kinetic friction is not taken into account despite the car is moving).
Does this mean that the maximum brake force is (kinetic friction)*(normal)?

What really confused me was the other day a problem in which impending slipping was involved.

Homework Statement


Data is not important, the problem is concepts not the problem itself.
A car of weigh W is on an incline it has certain initial velocity then brakes are applied, knowing that slipping is impending find static coefficient.

The Attempt at a Solution


I'm not looking for an anwser to this problem, I just want to understand. why is it that when brakes are applied static friction coefficient is involved. Shouldn't kinetic friction be involved as the car is moving with respect to the ground? Why in other problems with horizontal surface surface kinetic friction was involvved and not static friction as in this problem? Does this happen only when impending slipping is involved?
 
Physics news on Phys.org
You have 2 basic kinds of friction represented by the static and kinetic coefficients.

When you are applying the brakes, presumably they aren't locked and you aren't skidding, for if you were you would be dealing with kinetic as opposed to static friction considerations.

When you step on the brakes of a car the tire - even though rolling is not slipping - the crucial phrase. So the maximum force the driver can apply will be based on static friction coefficient. Usually the static coefficient is larger than than the kinetic and offers the better braking opportunity.
 
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 '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