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 ?
[Physics] Why is the magnitude of friction acting on the rear wheels greater than the magnitude of friction acting on the front wheels
rotational-kinematics
Best Answer
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.