Q100P - University Physics with Modern Physics

Question

Consider a wet roadway banked as in Example 5.22 (Section 5.4), where there is a coefficient of static friction of 0.30 and a coefficient of kinetic friction of 0.25 between the tires and the roadway. The radius of the curve is R = 50 m . (a) If the bank angle is $\mathit{\beta }\mathbf{=}\mathbf{25}\mathbf{°}$ , what is the maximum speed the automobile can have before sliding up the banking? 

(b) What is the minimum speed the automobile can have before sliding down the banking?

Step-by-Step Solution

Answer

(a) The maximum velocity is 20.90 m/s .

(b) The minimum velocity is 8.46 m/s .

Step 1: Identify the given data

• The coefficient of static friction is ${\mu }_s=0.35$ .
• The coefficient of kinetic friction is ${\mu }_k=0.25$ .
• The radius of the curve is $R=50m$ .
• The bank angle is $\beta =25°$ .

Step 2: Concept/Significance of Static friction force

The static force occurs when the object is at rest. The expression of the static friction force is given by,

 $\mathit{f}\mathbf{=}{\mathit{\mu }}_{\mathbf{s}}\mathit{N}$

Here, ${\mathit{\mu }}_{\mathbf{s}}$ is the coefficient of static friction, and N is normal force.

Step 3: Find the maximum speed of the automobile (a)

Draw the free-body diagram for sliding up.

The net force in the vertical direction is given by,

 $$\begin{gathered} \underset{ }{ \sum }{F}_y=0 \\ N \cos \beta =mg+{\mu }_s\sin \beta \\ N \cos \beta =mg+{\mu }_{s}N\sin \beta \\ N\left( \cos \beta -{\mu }_s\sin \beta \right)=mg ........\left(1\right) \end{gathered}$$

 

The net force in the horizontal direction is given by,

 

$$\begin{gathered} \sum _{ }{F}_x=m{a}_c \\ f\cos \beta +N\sin \beta =m\frac{v_{max}^2}{R} \\ {\mu }_{s}N\cos \beta +N\sin \beta =m\frac{v_{max}^2}{R} \\ N\left( {\mu }_s\cos \beta +\sin \beta \right)=m\frac{v_{max}^2}{R} .........\left(2\right) \end{gathered}$$ 

 

Here, $v_{max}$ is the maximum velocity, m is the mass of the object, g is the acceleration due to gravity, $\beta$ is given angle, R and is the radius of the curve.

 

Divide equations (2) and (1).

 

$\frac{{\mu }_s\cos \beta +\sin \beta }{\cos \beta -{\mu }_s\sin \beta }=\frac{v_{max}^2}{Rg} .......\left(3\right)$ 

 

Substitute 0.3 for ${\mu }_s$ , $25°$ for $\beta$ , 50 m for R , and $9.8 m/s^2$ for g in equation (3).

$$\begin{gathered} \frac{\left(0.3\right)\cos \left(25°\right)+\sin \left(25°\right)}{\cos \left(25°\right)-\left(0.3\right)\sin \left(25°\right)}=\frac{V_{max}^2}{\left(50m\right)\left(9.8m/s^2\right)} \\ V_{max}^2=436.562m^2/s^2 \\ V_{max}^{ }=\sqrt{436.562m^2/s^2} \\ =20.90 m/s \end{gathered}$$

 

Therefore, the maximum velocity is 20 .90m/s .

Step 4: Find the minimum speed of the automobile (b)

Draw the free-body diagram for sliding down.

The net force in the vertical direction is given by,

 $$\begin{gathered} \sum _{ }{F}_y=0 \\ N \cos \beta =mg+{\mu }_s\sin \beta \\ N \cos \beta =mg+{\mu }_{s}N\sin \beta \\ N\left( \cos \beta -{\mu }_s\sin \beta \right)=mg ........\left(4\right) \end{gathered}$$

 

The net force in the horizontal direction is given by,

$$\begin{gathered} \sum _{ }{F}_x=m{a}_c \\ -f \cos \beta +N\sin \beta =m\frac{v_{min}^2}{R} \\ -{\mu }_{s}N\cos \beta +N\sin \beta =m\frac{v_{min}^2}{R} \\ N\left(-{\mu }_s\cos \beta +\sin \beta \right)=m\frac{v_{min}^2}{R} .......\left(5\right) \end{gathered}$$ 

 

Here, $v_{min}$ is the maximum velocity, and is the radius of curvature.

 

Divide equation (5) by (4).

 

$\frac{-{\mu }_s\cos \beta +\sin \beta }{\cos \beta +{\mu }_s\sin \beta }=\frac{V_{min}^2}{Rg} .........\left(6\right)$ 

 

Substitute 0.3 for ${\mu }_s$ , $25°$ for $\beta$ , 50 m for R , $9.8 m/s^2$ and  g for in equation (6).

 

$$\begin{gathered} \frac{-\left(0.3\right)\cos \left(25°\right)+\sin \left(25°\right)}{\cos \left(25°\right)+\left(0.3\right)\sin \left(25°\right)}=\frac{v_{min}^2}{\left(50 m\right)\left(9.8m/s^2\right)} \\ v_{min}^2=71.563m^2/s^2 \\ v_{min}=\sqrt{71.563m^2/s^2} \\ =8.46 m/s \end{gathered}$$

 

Therefore, the minimum velocity is 8.46 m/s .