Chapter 13.2, Problem 13.71P

To determine

The normal reaction at point $B$.

Answer to Problem 13.71P

The minimum normal reaction is $731\text{ N}$ at point $B$ and maximum normal reaction is $5518.25\text{N}$ at point $D$

Explanation of Solution

Given:

Radius of arc $AB$ is $27\text{ m}$.

Radius of arc $CD$ is $72\text{ m}$.

Mass of car and occupants is $250\text{}\text{kg}$.

Velocity at point $A$ is $0\text{ m/s}$.

Angle of Arc from $A$ to $B$ is $40°$.

Concept used:

Write the expression for the displacement of car from $A$ to $B$.

$h=R\left(1-\cos \theta \right)$   ...... (1)

Here, $h$ is the displacement of car from $A$ to $B$, $R$ is the radius of $AB$ and $\theta$ is the angular displacement of car from $A$ to $B$.

Write the expression for kinetic energy at point $A$.

$T_A=\frac{1}{2}m{v}_A{}^2$

Here, $T_A$ is the kinetic energy at point $A$, $m$ is the mass of car and occupants and $v_A$ is the velocity of car at point $A$.

Write the expression for kinetic energy at point $B$.

$T_B=\frac{1}{2}m{v}_B{}^2$

Here, $T_B$ is the kinetic energy at point $B$ and $v_B$ is the velocity of car at point $B$.

Write the expression for the potential energy at point $A$.

$V_A=mg{h}_A$

Here, $V_A$ is the potential energy at point $A$, $g$ is the acceleration due to gravity and $h_A$ is the elevation of point $A$ from ground.

Write the expression for the potential energy at point $B$.

$V_B=mg{h}_B$

Here, $V_B$ is the potential energy at point $B$ and $h_B$ is the elevation of point $B$ from ground.

Write the expression for conservation of energy for point $A$ to point $B$.

$T_A+V_A=T_B+V_B$

Substitute $\frac{1}{2}m{v}_A{}^2$ for $T_A$, $\frac{1}{2}m{v}_B{}^2$ for $T_B$, $mg{h}_A$ for $V_A$ and $mg{h}_B$ for $V_B$ in above equation.

$\frac{1}{2}m{v}_A{}^2+mg{h}_A=\frac{1}{2}m{v}_B{}^2+mg{h}_B$

Simplify the above expression.

$\frac{1}{2}m\left(v_B{}^2-v_A{}^2\right)=mg\left(h_A-h_B\right)$

Substitute $0$ for $v_A$ and $h$ for $\left(h_A-h_B\right)$ in above expression.

$\frac{1}{2}\left(v_B{}^2\right)=gh$

Solve the above expression for $v_B$.

$v_B=\sqrt{2gh}$   ...... (2)

Write the expression for the normal acceleration of car.

$a_{Bn}=\frac{v_B{}^2}{R}$   ...... (3)

Here, $a_{Bn}$ is the normal acceleration at point $B$.

Apply Newton’s Law of motion for carat position $B$.

$mg\cos \theta -N=m{a}_{Bn}$   ...... (4)

Here, $N$ is the normal reaction at point $B$.

Write the expression for kinetic energy at point $D$.

$T_D=\frac{1}{2}m{v}_D{}^2$

Here, $T_D$ is the kinetic energy at point $D$ and $v_D$ is the velocity of car at point $D$.

Write the expression for the work done at point $A$ and $D$.

$U_{AD}=mg\left(d_{AD}\right)$

Here, $U_{AD}$ is the work done from $A$ to $D$ and $d_{AD}$ is the distance from $A$ to $D$.

Write the expression for the work energy principle for point $A$ to $D$.

$T_A+U_{AD}=T_D$

Substitute $\frac{1}{2}m{v}_A{}^2$ for $T_A$, $\frac{1}{2}m{v}_D{}^2$ for $T_D$ and $mg\left(d_{AD}\right)$ for $U_{AD}$ in above equation.

$\frac{1}{2}m{v}_A{}^2+mg{d}_{AD}=\frac{1}{2}m{v}_D{}^2$

Substitute $0$ for $v_A$ in above equation.

$g{d}_{AD}=\frac{1}{2}{v}_D{}^2$

Rearrange the above expression for $v_D$

$v_D=\sqrt{2g{d}_{AD}}$   ...... (5)

Write the expression for the normal acceleration at point $D$.

$a_{Dn}=\frac{v_D{}^2}{r}$   ...... (6)

Here, $a_{Dn}$ is the acceleration at point $D$ and $r$ is the radius of section $CD$.

Apply Newton’s Law of motion for carat position $D$.

$N_D-mg=m{a}_{Dn}$   ...... (7)

Here, $N_D$ is the normal reaction at $D$.

Calculation:

Substitute $27\text{ m}$ for $R$ and $40°$ for $\theta$ in equation (1).

$\begin{array}{c}h=27\left(1-\cos 40°\right)\\ =6.317\text{ m}\end{array}$

Substitute $6.317\text{ m}$ for $h$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (2).

$\begin{array}{c}{v}_B=\sqrt{2\left(9.81\right)\left(6.317\right)}\\ =11.133\text{ m/s}\end{array}$

Substitute $11.133\text{ m/s}$ for $v_B$ and $27\text{ m}$ fro $R$ in equation (3).

$\begin{array}{c}{a}_{Bn}=\frac{{\left(11.133\right)}^2}{27}\\ =4.59{\text{ m/s}}^2\end{array}$

Substitute $250\text{ kg}$ for $m$, $4,59{\text{ m/s}}^2$ for $a_{Bn}$, $40°$ for $\theta$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (4).

$\begin{array}{c}\left(250\right)\left(9.81\right)\cos 40°-N=\left(250\right)\left(4.59\right)\\ 1878.724-N=1147.5\end{array}$

Simplify above expression for $N$.

$N=731\text{N}$

Substitute $45\text{ m}$ for $d_{AD}$ and $9.81{\text{ m/s}}^2$ for $g$ in equation (5).

$\begin{array}{c}{v}_D=\sqrt{2\left(45\right)\left(9.81\right)}\\ =29.71\text{ m/s}\end{array}$

Substitute $29.71\text{ m/s}$ for $v_D$ and $72\text{ m}$ for $r$ in equation (6).

$\begin{array}{c}{a}_{Dn}=\frac{{\left(29.71\right)}^2}{72}\\ =12.263\text{}{\text{m/s}}^2\end{array}$

Substitute $250\text{ kg}$ for $m$, $12.263{\text{ m/s}}^2$ for $a_{Dn}$, and $9.81{\text{ m/s}}^2$ for $g$ in equation (7).

$N_D-\left(250\right)\left(9.81\right)=\left(250\right)\left(12.263\right)$

Simplify above for $N_D$.

$N_D=5518.25\text{N}$

The minimum normal reaction is $731\text{ N}$ at point $B$ and maximum normal reaction is $5518.25\text{N}$ at point $D$

Conclusion:

Thus, the minimum normal reaction is $731\text{ N}$ at point $B$ and maximum normal reaction is $5518.25\text{N}$ at point $D$