# Why is the magnitude of friction acting on the rear wheels greater than the magnitude of friction acting on the front wheels?

When a car is accelerating on a horizontal road , friction acts in the forward direction on the rear wheel and in the backward direction on the front wheel . But I am not able to understand why the friction acting on the rear wheel is greater in magnitude than the friction acting on the front wheel ? Is this because the car is accelerating forward and the net force should act in the forward direction ?

• Have you considered the moment of the centre of mass about the bottom of the rear tire, during acceleration? – Declan Apr 17 '19 at 22:07

I think you have misunderstood how friction slows cars down. The friction between the tires and road allow the car to move (think about it). The friction between the axles and wheel are what slow the car down.

• But then why is the friction due to the road on the rear tyres greater than the friction on the front tyres , when a car is accelerating ? – Pratham Yadav Apr 17 '19 at 22:11
• @PrathamYadav: But then why is the friction due to the road on the rear tyres greater than the friction on the front tyres , when a car is accelerating ? What makes you think that is the case? – Gert Apr 17 '19 at 22:39
• @Gert I read on a website that the friction acting on the rear tyres is greater than , equal to , or less than the friction acting on the front tyres accordingly if the car is accelerating , moving in uniform motion , or decelerating. – Pratham Yadav Apr 18 '19 at 8:02
• @PrathamYadav: yes, that website is correct. I've added the dynamic case which shows it. – Gert Apr 18 '19 at 14:58
• @Gert Why didn't you take into account the friction acting on the front wheel? – Pratham Yadav Apr 18 '19 at 15:16

Let's look a the case of rear wheel drive and designate all the forces acting during acceleration (air drag, ball bearings friction and other non-conservative forces are ignored)

In the $$y$$-direction where there's no acceleration we get:

$$N_F+N_B-mg=0\tag{1}$$

In the $$x$$-direction:

$$ma=F_B-F_F$$ This the Equation of Motion of the car.

Readers may wonder why $$F_F$$ points in the $$-x$$ direction. When the car accelerates, friction is needed on the front wheels to sustain the increase in angular velocity $$\omega$$.

For rolling without slipping in uniform motion (constant velocity), with $$R$$ the wheel's radius:

$$v=\omega R$$

In acceleration, rolling without slipping:

$$a=\frac{\mathbf{d}\omega}{\mathbf{d}t}R=\dot{\omega}R$$

To get this angular acceleration clockwise torque $$\tau$$ needs to act on the wheel:

$$\tau=I\dot{\omega}$$

This torque is provided by the friction force $$F_F$$, so that:

$$\tau=F_F \times R$$

The balance of torques about the CoG must be zero, to prevent the car from starting to rotate, so:

$$N_BL_1+F_F h=N_F L_2+F_Bh$$

And with: $$F_F=\mu N_F$$ and $$F_B=\mu N_B$$, then:

$$N_BL_1+\mu N_F h=N_F L_2+\mu N_Bh$$

$$N_B(L_1-\mu h)=N_F(L_2-\mu h)$$

Combined with $$(1)$$ and some reworking we get:

$$N_B=\frac{L_2-\mu h}{L_1+L_2-2 \mu h}\tag{2}mg$$ $$N_F=\frac{L_1-\mu h}{L_1+L_2-2 \mu h}\tag{3}mg$$

Now compare this to the static case, where $$a=0$$ and $$v=\text{constant}$$. It can beshown easily that in that case:

$$N_B=\frac{L_2}{L_1+L_2}mg\tag{A}$$ $$N_F=\frac{L_1}{L_1+L_2}mg\tag{B}$$

Comparing $$(2)$$ to $$(B)$$ it is clear that in the case of the accelerating car, the normal force on the rear wheels and thus also the friction, is higher than in the static ($$a=0$$) case.

• But sir , in the case V= constant that you stated above , Nb is still greater than Na ( as l2 > l1 ) , but in a non accelerating car , the friction on both the tyres is the same. – Pratham Yadav Apr 19 '19 at 16:46
• @Pratham: No, that was where I said entirely correctly that in the static case $N_B>N_F$ IF and only IF the CoG is more towards the back than the front. The current derivation compares static v. dynamic, all other things being equal. – Gert Apr 19 '19 at 17:27