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In a front wheel drive car the friction acting on front wheel is in forward direction (say F1) and friction acting on rear wheel is backwards (say F2). So if the car accelerates forward F1> F2. Since static friction is acting in both cases if the mass of the car is equally distributed among front wheel and rear wheels the normal reaction would be same and so F1=F2. So in such a case how does the car accelerate?

Sorry if this is a stupid question. I am having trouble understanding the physics in front wheel drive cars and rear wheels drive cars. Hope my question is clear.

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  • $\begingroup$ Where did you get the impression that friction is opposite on the front and back wheels? $\endgroup$
    – S. McGrew
    Jan 29, 2021 at 16:07
  • $\begingroup$ @S.McGrew Since I am considering a front wheel drive car the engine rotates the front wheel in clock wise direction so there exists a relative motion between the road and the wheel. so to oppose that the road would exert a frictional force in forward direction.. then the front wheel moves forward so does the rear wheel .. then there is a relative motion between the rear wheel and road to prevent that the friction acts backwards $\endgroup$
    – user661645
    Jan 29, 2021 at 16:18
  • $\begingroup$ Do you know the difference between static and dynamic friction? $\endgroup$
    – CR Drost
    Jan 29, 2021 at 16:19
  • $\begingroup$ @CRDrost yes i know the difference $\endgroup$
    – user661645
    Jan 29, 2021 at 16:20
  • $\begingroup$ Then when I tell you that the problem is that the car’s wheels are engaged in static friction ($|F|\le \mu_s |F_N|$) and you are treating them like they are engaged in dynamic friction ($|F| = \mu_d |F_N|$) do you know enough to post here an answer to your own question? $\endgroup$
    – CR Drost
    Jan 29, 2021 at 16:24

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The coefficient of static friction together with the normal force give you a maximum value for the friction force. The magnitudes of $F_1$ and $F_2$ are free to vary within that range. When driving $F_1$ will be much larger, $F_2$ will be as small as the bearings on the wheels can make it.

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Actually, static friction is a very unusual force. Most forces have a force law that is an equality, like Hooke’s law for a spring: $F=-kx$, given $k$ and $x$ there is one and only one value for $F$. But static friction is given by an inequality: $F\le \mu N$, given $\mu$ and $N$ there are an infinite number of possible values for $F$.

The actual value that $F$ assumes is not given by $\mu N$. Instead, it is whatever value is needed for there to be no slipping at the surface. Cars accelerate by arranging it so that those values are much different on the two sets of wheels. The rear wheels are allowed to freely turn, so the required no-slipping force is very small. The front wheels are subject to a large torque, so the no-slipping force is large. Thus we get the required imbalance between the front and rear wheels.

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  • $\begingroup$ So that also means when decelerating the friction on front wheel becomes less than that on rear wheels .. right? $\endgroup$
    – user661645
    Jan 29, 2021 at 16:37
  • $\begingroup$ The brakes typically apply a large torque to all four wheels, so the no-slipping force is large on all four wheels. But in this case all four of the forces point in the same direction, backwards. $\endgroup$
    – Dale
    Jan 29, 2021 at 16:52

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