Car Motion using Tire Friction

I am a bit confused regarding how car moves. Following is the explanation which I've come up with after reading but I would like if someone verifies and corrects it.

If a car is moving, the rotation of it's tyre will exert a backward force on the road. Due to newton's third law, the road also exerts a force on the tyre. The force will be in the forward direction. This force causes the tire and hence the car to move forward.

• I would say correct description. Note that in the ideal case (no air resistance, perfectly round wheel, perfectly rigid wheel and ground etc.) no force is needed to keep the car moving at constant speed - the described forces and reaction-forces are only needed for acceleration (speeding up, slowing down). Only ideally. For a non-ideal, real driving scenario, your description holds true all the time. – Steeven Aug 30 '18 at 8:21
• Yes, your analysics is correct in general. When we look in details it may be a little more complicated because the rotation of the tyres makes it a bit more complicated because the bit of tyre that is touching the ground and pushing on it is stationary. – tom Aug 30 '18 at 9:00

2 Answers

your explanation is correct, but I would add that the tyre that is rotating in your example is being forced to do so by the engine.

I think that this figure can answer your question

The tire force $f_R$ cause that the vehicle move $M\frac{dv}{dt}=f_R$.

The tire force is proportional to the slip between the tire and the road. No tire force without friction

Edit:

The equations of motion

\begin{align*} \text{Vehicle}\\ M\dot{v}&=-f_R&(1)\\ \text{Wheel}\\ \Theta\ddot{\varphi}&=f_R\,r+ \tau_E\,b+\tau_B &(2)\\ &\text{with:}\\ &f_R\quad \text{Constraint force}\\ &\tau_E\quad \text{Engine torque}\\ &\tau_B\quad \text{Brake torque}\\ &r\quad \text{Wheel radius}\\ &b\quad \text{Transmission between engine and wheel} \end{align*} I) Wheel rolling condition: \begin{align*} &\text{The equation for rolling wheel is:}\\ v&=\dot{\varphi}\,r\,,\quad \dot{v}=\ddot{\varphi}\,r&(3)\\ &\text{with equations (1) (2) and (3) we get}\\\\ &\left(\Theta+M\,r^2\right)\dot{v}=\left( \tau_{E}\,b+\tau_{B}\right)\,r \end{align*} II) Wheel slipping condition: \begin{align*} &\text{The wheel slipping equation is:}\\ &s_L=\frac{v-r\,\dot{\varphi}}{|v|}\\ &\text{for this case is the constraint force (tire force) $f_R$ is:}\\ &f_R=C_s\,s_L \end{align*}

• What is a tire force? And what "slip between the tire and the road" are we referring to? – Steeven Aug 30 '18 at 8:19
• The tire force is as you see is the force between the tire and the road $f_R=C_s\,s$ with the slip definition $s=\frac{v-a\dot{\varphi}}{|v|}$ – Eli Aug 30 '18 at 8:30
• I have never heard of a "slip" like this before. There is no actual slipping during the wheel's rolling when the car is driving, so what physically does this "slip" mean? – Steeven Aug 30 '18 at 11:00
• en.wikipedia.org/wiki/Slip_(vehicle_dynamics) – Eli Aug 30 '18 at 11:50