In Fig.12-48, the driver of a car on a horizontal road makes an emergency stop by applying the brakes so that all four wheels lock and skid along the road. The coefficient of kinetic friction between tires and road is $\mathbf{0}\mathbf{.40}$.The separation between the front and rear axles is $\mathit{L}\mathbf{=}\mathbf{}\mathbf{4}\mathbf{.2}\mathbf{}\mathit{m}$, and the center of mass of the car is located at distance $\mathit{d}\mathbf{=}\mathbf{1}\mathbf{.8}\mathbf{}\mathit{m}$ behind the front axle and distance$\mathit{h}\mathbf{=}\mathbf{}\mathbf{0}\mathbf{.75}\mathbf{}\mathit{m}$ above the road. Theweight of car is $\mathbf{11}\mathbf{}\mathit{k}\mathit{N}$. Find the magnitude of (a) the braking acceleration of the car, (b) the normal force on each rear wheel, (c) the normal force on each front wheel, (d) the braking force on each rear wheel, and (e) the braking force on each front wheel. (Hint:Although the car is not in translational equilibrium, it isin rotational equilibrium.)

• The coefficient of kinetic friction between tires and road is, $\mu =0.4$.
• The separation between the front and rear axle is,$L=4.20m$.
• The center of mass is at height,$h=0.75m$and distance,$d=1.8m$.
• The weight of the car is,$W=11kN=11000\text{\hspace{0.17em}N}$ .

Formula:

$F_{net}=ma$

At equilibriumwe can use the formulas as follows,

$\sum F_x=0$

$\sum F_y=0$

$\sum \tau =0$

The free Body Diagram can the drawn as,

Braking acceleration of the car:

According to Newton’s second law, the net force can be written as,

$F_{net}=ma$

(1)

But, from the diagram,

$F_{net}=F_f$ (2)

We can write the force of friction as,

$F_f=\mu mg$

From equation (2), we can write,

$F_{net}=\mu mg$

Substitute the value from equation 1 in the above expression, and we get,

$\begin{array}{c}ma=\mu mg\\ a=\mu g\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}a=0.4\times 9.81\\ =3.92{\text{\hspace{0.17em}m/s}}^{\text{2}}\end{array}$

Thus, the braking acceleration of the car is$3.92\frac{m}{s^2}$.

Using equilibrium conditions,

As acceleration is only along a horizontal direction, for the vertical direction,we can write,

$\begin{array}{c}\sum F_y=0\\ F_{Nr}+F_{Nf}-mg=0\end{array}$

Here $F_{Nf}$,$F_{Nr}$are the normal force on the front wheel and rare wheel.

We know the weight of the car, and then the above expression can be written as,

$F_{Nr}+F_{Nf}=11000N$

(3)

The kinetic frictional force acting on the car is,

$f_k=\mu mg$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{f}_k=0.4\left(9.8\right)m\\ =3.92m\end{array}$

Now torque at the center of mass is zero; thus, we can write the equation as,

$f_k\left(h\right)+F_{Nr}(L-d)-F_{Nf}\left(d\right)=0$

Substitute the values in the above expression, and we get,

$3.92m\times 0.75+F_{Nr}\times 2.4-F_{Nf}\times 1.8=0$ (4)

The weight of the car is 11000 N.

So mass m can be calculated as follows,

$\begin{array}{c}m=11000/9.8\\ =1122.45kg\end{array}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}3.92\times 1122.45\times 0.75+F_{Nr}\times 2.4-F_{Nf}\times 1.8=0\\ 330000+F_{Nr}\times 2.4-F_{Nf}\times 1.8=0\end{array}$

So after solving the vertical equilibrium condition equation (3) and torque equation (4), we get:

$\begin{array}{c}{F}_{Nr}=3929N\\ ~4000N\end{array}$

And

$\begin{array}{c}{F}_{Nf}=7071\text{\hspace{0.17em}N}\\ ~7000N\end{array}$

The normal force on each rare wheel will be half of the normal force on the rare wheels, which can be calculated as,

$\begin{array}{l}=\frac{4000}{2}\\ =2000N\end{array}$

Thus, the normal force on each rear wheel is $2000N$.

The normal force on each front wheel will be half of the normal force on the front wheels, which can be calculated as,

$\begin{array}{l}=\frac{7000}{2}\\ =3500N\end{array}$

Thus, the normal force on each front wheel is$3500N$.

Braking force means friction force.

So,the braking force on each rear wheel is as follows:

$F_{f1}=\mu \times \text{normalforceoneachrearwheel}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{f1}=0.4\times 3929\\ =7900\\ =7.9\times {10}^{3}N\end{array}$

Thus, the braking force on each rear wheel is$7900N$.

The braking force on each front wheel is as follows:

$F_{f2}=\mu \times \text{normal\hspace{0.17em}force\hspace{0.17em}on\hspace{0.17em}each\hspace{0.17em}front\hspace{0.17em}wheel}$

Substitute the values in the above expression, and we get,

$\begin{array}{c}{F}_{f2}=0.4\times 3535.5\\ =1400\\ =1.4\times {10}^{3}N\end{array}$

Braking force on each front wheel is$1400N$.