Question

The friction coefficient between a road and the type of a vehicle is 4/3. Find the maximum incline the road may have so that once had brakes are applied and the wheel starts skidding, the vehicle going down at a speed of 36 km/hr is stopped within 5 m.

Answer 1

Given,
initial velocity of the vehicle, u = 36 km/h = 10 m/s
final velocity of the vehicle, v = 0
s = 5 m, $\mathrm{\mu }=\frac{4}{3}$, g = 10 m/s²

Let the maximum angle of incline be $\theta$.

Using the equation of motion
$$\begin{gathered} a=\frac{v^2-u^2}{2s}=\frac{0-{10}^2}{2\times 5} \\ =-10\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

From the free body diagram

R − mg cos θ = 0
⇒ R = mg cos θ (1)
Again,
ma + mg sin θ − μ R = 0
⇒ ma + mg sin θ − μmg cos θ = 0
⇒ a + g sin θ − μg cos θ = 0
$\Rightarrow 10+10\sin \mathrm{\theta }-\left(\frac{4}{3}\right)\times 10\cos \mathrm{\theta }=0$
⇒ 30 + 30 sin θ − 40 cos θ = 0
⇒ 3 + 3 sin θ − 4 cos θ = 0
⇒ 4 cos θ − 3 sin θ = 3
$\Rightarrow 4\sqrt{\left(1-{\sin }^2\mathrm{\theta }\right)}=3+3\sin \mathrm{\theta }$
On squaring, we get
16 (1 − sin² θ) = 9 + 9 sin² θ + 18 sin θ
25 sin² θ + 18 sin θ − 7 = 0
$$\begin{gathered} \Rightarrow \sin \mathrm{\theta }=\frac{18+\sqrt{{18}^2-4\left(25\right)\left(-7\right)}}{2\times 25} \\ =\frac{-18+32}{50}=\frac{14}{50}=0.28\left(\mathrm{Taking}\mathrm{positve}\mathrm{sign}\mathrm{only}\right) \\ \Rightarrow \mathrm{\theta }={\sin }^{-1}\left(0.28\right)=16° \end{gathered}$$
Therefore, the maximum incline of the road, θ = 16°.