Chapter 5, Problem 94P

(a)

To determine

To Find:The normal force.

Answer to Problem 94P

  $8.24\text{kN}$

Explanation of Solution

Given Information:

Radius of the curve is $150\text{m}$

Banking angle is ${10}^{\text{o}}$ .

Mass of car is $800\text{kg}$ .

Speed of the car is $85\text{ km/h}$ .

Formula Used:

The acceleration of the car along x- axis is the centripetal acceleration, that is,

  $a=\frac{v^2}{r}$

Here, $v$ is the speed of the car and $r$ is the radius of the curve.

Calculation:

The free-body diagram for the car on the curved road is as follows

  

In the free body diagram shown below, the normal force and the static friction force, both contribute to the centripetal force according to the situation described in this problem.

The speed of the car is

  $\begin{array}{c}v=85\text{km/h}\left(\frac{\frac{5}{18}\text{m/s}}{\text{km/h}}\right)\\ =23.61\text{ m/s}\end{array}$

The net force on the x -axis is

  $F_n\sin \theta +f_s\cos \theta =ma$

Substitute $\frac{v^2}{r}$ for a in above equation.

  $F_n\sin \theta +f_s\cos \theta =m\left(\frac{v^2}{r}\right)\mathrm{...}\left(1\right)$

Multiply the equation $\left(1\right)\text{ with sin}\theta$ .

  $f_s\sin \theta \cos \theta +F_n{\sin }^2\theta =m\left(\frac{v^2}{r}\right)\sin \theta .\mathrm{..}\left(2\right)$

The net force on the y-axis is,

  $F_n\cos \theta -f_s\sin \theta =mg\mathrm{...}\left(3\right)$

Multiply the above equation with $\cos \theta$ .

  $F_n{\cos }^2\theta -f_s\sin \theta \cos \theta =mg\cos \theta \mathrm{...}\left(4\right)$

Add the equations $\left(2\right)\text{ and }\left(4\right)$

  $F_n-mg\cos \theta =m\left(\frac{v^2}{r}\right)\sin \theta$

Rearrange the above equation for the normal force $F_n$ .

  $F_n=mg\cos \theta +m\left(\frac{v^2}{r}\right)\sin \theta$

Substitute $800\text{ kg for }m,9.81{\text{m/s}}^2\text{ for }g,{10}^{\text{o}}\text{ for }\theta ,23.61\text{ m/s for }v\text{ and 150 m for }r$ in above equation.

Conclusion:

The normal force exerts on the car is $8.2\text{4 kN}$ .

(b)

To determine

To Find:The frictional force exerted by the pavement on the tires.

Answer to Problem 94P

The frictional force exerted by the pavement on the tires is $1.56\text{ kN}$

Explanation of Solution

Given Information:

Radius of the curve is $150\text{m}$

Banking angle is ${10}^{\text{o}}$ .

Mass of car is $800\text{kg}$ .

Speed of the car is $85\text{ km/h}$ .

Formula Used:

The net force along the y-axis is

  $F_n\cos \theta -f_s\sin \theta =mg$

Calculation:

Rearrange the above equation:

  $f_s=\frac{F_n\cos \theta -mg}{\sin \theta }$

Substitute $8244.8\text{N for }{F}_n,{10}^{\text{o}}\text{for }\theta ,800\text{kg for }m\text{ and 9}\text{.81}{\text{m/s}}^2\text{ for }g$ in above equation.

  $\begin{array}{c}{f}_s=\frac{\left(\text{8244}\text{.8}\text{N}\right)\cos {10}^{\text{o}}-\left(\text{800}\text{kg}\right)\left(\text{9}\text{.81}{\text{m/s}}^{\text{2}}\right)}{\sin {10}^{\text{o}}}\\ =\frac{\left(\text{8244}\text{.8}\text{N}\right)\left(\text{0}\text{.9848}\right)\text{-}\left(\text{800 kg}\right)\left(\text{9}\text{.81}{\text{m/s}}^{\text{2}}\right)}{\left(0.1736\right)}\\ =1563.8\text{N}\\ =1\text{563}\text{.8 N}\left(\frac{\text{1}\text{kN}}{{\text{10}}^{\text{3}}\text{N}}\right)\\ =1.56\text{kN}\end{array}$

Conclusion:

The frictional force exerted by the pavement on the tires is $1.56\text{kN}$ .

(c)

To determine

To Find:The coefficient of the static frictional force.

Answer to Problem 94P

  $0.193$

Explanation of Solution

Given Information:

Radius of the curve is $150\text{m}$

Banking angle is ${10}^{\text{o}}$ .

Mass of car is $800\text{kg}$ .

Speed of the car is $85\text{ km/h}$ .

Formula Used:

The equation for the frictional force is

  $f_s={\mu }_{s,\min }{F}_n$

Here, $F_n$ is the normal force.

Calculation:

Rearrange the above equation for ${\mu }_{s,\min }$ .

  ${\mu }_{s,\min }=\frac{f_s}{F_n}$

Substitute $1.56\text{kN for }{f}_s,\text{ and 8}\text{.24}\text{kN for }{F}_n$ in the above equation.

  $\begin{array}{c}{\mu }_{s,\min }=\frac{\text{1}\text{.56}\text{kN}}{\text{8}\text{.24}\text{kN}}\\ =0.193\end{array}$

Conclusion:

The coefficient of the minimum static frictional force is $0.193$ .