Chapter 5, Problem 95P

(a)

To determine

To Find:The normal force.

Answer to Problem 95P

  $F_n=7.83\text{kN}$

Explanation of Solution

Given Information:

Mass of the car $m=800kg$

Angle of incline, $\theta ={10}^{\text{o}}$

Radius of curvature, $r=150m$

Speed of car is $\text{38 km/hr}$

Formula Used:

Newton’s second law of motion:

  $F_{net}=ma$

Calculation:

The free body diagram shows the forces acting the car.

  

Speed of the car $v=38\text{km/h}$

  $\begin{array}{l}\text{=}\left(\text{38}\text{km/h}\right)\left(\frac{\text{1000}\text{m}}{\text{1}\text{km}}\right)\left(\frac{\text{1}\text{h}}{\text{3600}\text{s}}\right)\\ \text{=10}\text{.56}\text{m/s}\end{array}$

Apply ${\displaystyle \sum F_x=m{a}_x}$ to the car.

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

Apply ${\displaystyle \sum F_y}=m{a}_y$ to the car

  ${\displaystyle \sum F_y}=F_n\cos \theta -f_s\sin \theta -mg=0\mathrm{....}\left(2\right)$

Multiply the equation $\left(1\right)$ by $\sin \theta$ and the equation $\left(2\right)$ by $\cos \theta$

  $\begin{array}{l}{F}_n{\sin }^2\theta +f_s\sin \theta \cos \theta =m\frac{v^2}{r}\sin \theta \mathrm{....}\left(3\right)\\ F_n{\cos }^2\theta -f_s\sin \theta \cos \theta -mg\cos \theta =0.\mathrm{...}\left(4\right)\end{array}$

Add the equation $\left(3\right)\text{ and }\left(4\right)$ to eliminate $f_s$

  $\begin{array}{l}{F}_n{\sin }^2\theta +f_s\sin \theta \cos \theta +F_n{\cos }^2\theta -f_s\sin \theta \cos \theta -mg\cos \theta =m\frac{v^2}{r}\sin \theta +0\\ F_n\left({\sin }^2\theta +{\cos }^2\theta \right)-mg\cos \theta =m\frac{v^2}{r}\sin \theta \\ F_n-mg\cos \theta =m\frac{v^2}{r}\sin \theta \end{array}$

Solving the above equation for $F_n$ yields.

  $\begin{array}{l}{F}_n=mg\cos \theta +m\frac{v^2}{r}\sin \theta \\ F_n=m\left(g\cos \theta +\frac{v^2}{r}\sin \theta \right)\mathrm{....}\left(5\right)\end{array}$

Substitute all the known values in equation $\left(5\right)$ , the normal force acting on the tires by the pavement is given by:

  $\begin{array}{c}{F}_n=m\left(g\cos \theta +\frac{v^2}{r}\sin \theta \right)\\ =\left(\text{800}\text{kg}\right)\left(\left(\text{9}\text{.81}{\text{m/s}}^{\text{2}}\right){\text{cos10}}^{\text{o}}\text{+}\frac{{\left(\text{10}\text{.56 m/s}\right)}^{\text{2}}}{\text{150 m}}{\text{sin10}}^{\text{o}}\right)\\ \text{=7}{\text{.83×10}}^{\text{3}}\text{N}\\ \text{=7}\text{.83}\text{kN}\end{array}$

Conclusion:

Therefore, the normal force is $F_n=7.83\text{ kN}$

(b)

To determine

To Find:The frictional force exerted.

Answer to Problem 95P

  $f_s=-0.78\text{kN}$

Explanation of Solution

Given Information:

The speed of car on the curve road is $38\text{km/hr}$

Formula Used:

Newton’s second law of motion:

  $F_{net}=ma$

Calculation:

The free body diagram shows the forces acting the car.

  

Apply ${\displaystyle \sum F_x=m{a}_x}$ to the car.

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

Apply ${\displaystyle \sum F_y}=m{a}_y$ to the car

  ${\displaystyle \sum F_y}=F_n\cos \theta -f_s\sin \theta -mg=0\mathrm{....}\left(2\right)$

Solving the equation $\left(2\right)$ for $f_s$ yields.

  $\begin{array}{l}{F}_n\cos \theta -f_s\sin \theta -mg=0\\ f_s\sin \theta =F_n\cos \theta -mg\\ f_s=\frac{F_n\cos \theta -mg}{\sin \theta }\\ =\frac{\left(\text{7}{\text{.83×10}}^{\text{3}}\text{N}\right){\text{cos10}}^{\text{o}}-\left(\text{800}\text{kg}\right)\left(\text{9}\text{.81}{\text{m/s}}^{\text{2}}\right)}{{\text{sin10}}^{\text{o}}}\\ =-\text{788}\text{.69N}\\ \text{=}-\text{0}\text{.78}\text{kN}\end{array}$

Calculation:

Therefore, the frictional force is $f_s=-0.78\text{kN}$

Here the negative sign indicates that force $f_s$ points upwards along the incline plane rather than as shown in the diagram.