91Ó°ÊÓ

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

Formula:

$F_{net}=ma$

At equilibriumwe can use the formulas as follows,

$∑F_x=0$

$∑F_y=0$

$∑\mathrm{τ}=0$

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,

$F_f=\mathrm{μ}mg$

From equation (2), we can write,

$F_{net}=\mathrm{μ}mg$

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

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

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

$\begin{array}{c}a=0.4×9.81\\ =3.92{\text{″¾/²õ}}^{\text{2}}\end{array}$

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

Using equilibrium conditions,

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

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

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,

$F_{Nr}+F_{Nf}=11000\text{ }N$

(3)

The kinetic frictional force acting on the car is,

$f_k=\mathrm{μ}mg$

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,

$3.92m×0.75+F_{Nr}×2.4−F_{Nf}×1.8=0$ (4)

The weight of the car is 11000 N.

So mass m can be calculated as follows,

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

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

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

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

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

And

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

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

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

Thus, the normal force on each rear wheel is $2000\text{ }N$.

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

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

Thus, the normal force on each front wheel is$3500\text{ }N$.

Braking force means friction force.

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

$F_{f1}=\mathrm{μ}×\text{normalforceoneachrearwheel}$

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

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

Thus, the braking force on each rear wheel is$7900\text{ }N$.

The braking force on each front wheel is as follows:

$F_{f2}=\mathrm{μ}×\text{²Ô´Ç°ù³¾²¹±ô f´Ç°ù³¦±ð o²Ô e²¹³¦³ó f°ù´Ç²Ô³Ù w³ó±ð±ð±ô}$

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

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

Braking force on each front wheel is$1400\text{ }N$.