Chapter 13.3, Problem 13.130P

The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of car A, causing it to slide on the track. The brakes are not applied on the wheels of cars B or C. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the time required to bring the train to a stop, (b) the force in each coupling.

To determine

Find the time required to bring the train (t) to a stop.

Answer to Problem 13.130P

The time required to bring the train (t) to a stop is $\underset{\_}{12.69\text{sec}}$.

Explanation of Solution

Given information:

The initial speed of the train $\left(v_1\right)$ is $30\text{mi}/\text{h}$.

The coefficient of kinetic friction $\left({\mu }_k\right)$ is 0.35.

The weight of the rail car A $\left(W_A\right)$ is $40\text{tons}$.

The weight of the rail car B $\left(W_B\right)$ is $50\text{tons}$.

The weight of the rail car C $\left(W_C\right)$ is $40\text{tons}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Show the impulse momentum diagram for the entire train as Figure (1).

Convert the initial speed of the train $\left(v_1\right)$ from $\text{mi}/\text{h}$ to $\text{ft}/\text{s}$:

$v_1={\left(v_1\right)}_{\text{mi}/\text{h}}\times \left(5280\frac{\text{ft}}{\text{mi}}\right)\times \left(\frac{1\text{hr}}{3600\text{sec}}\right)$

Here, ${\left(v_1\right)}_{\text{mi}/\text{h}}$ is the initial speed of the train in $\text{mi}/\text{h}$.

Substitute $30\text{mi}/\text{h}$ for ${\left(v_1\right)}_{\text{mi}/\text{h}}$.

$\begin{array}{c}{v}_1=30\times \left(5280\frac{\text{ft}}{\text{mi}}\right)\times \left(\frac{1\text{hr}}{3600\text{sec}}\right)\\ =44\text{ft}/\text{s}\end{array}$

Calculate the masses of the rail cars A $\left(m_A\right)$ using the relation:

$m_A=\frac{W_A}{g}$

Substitute $40\text{tons}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_A=\frac{40\text{tons}}{32.2\text{ft}/{\text{s}}^2}\\ =\frac{40}{32.2}\times \left(\frac{2000\text{lbs}}{1\text{ton}}\right)\\ =2484\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

Calculate the mass of the rail car B $\left(m_B\right)$ using the relation:

$m_B=\frac{W_B}{g}$

Substitute $50\text{tons}$ for $W_B$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_B=\frac{50\text{tons}}{32.2\text{ft}/{\text{s}}^2}\\ =\frac{50}{32.2}\times \left(\frac{2000\text{lbs}}{1\text{ton}}\right)\\ =3106\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

Calculate the mass of the rail car C $\left(m_C\right)$ using the relation:

$m_C=\frac{W_C}{g}$

Substitute $40\text{tons}$ for $W_C$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{m}_C=\frac{40\text{tons}}{32.2\text{ft}/{\text{s}}^2}\\ =\frac{40}{32.2}\times \left(\frac{2000\text{lbs}}{1\text{ton}}\right)\\ =2484\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

Calculate the frictional force acting on the car B after application of brakes $\left[{\left(F_f\right)}_B\right]$ using the relation:

${\left(F_f\right)}_B={\mu }_k{m}_{B}g$

Substitute $3106\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_B$, 0.35 for ${\mu }_k$, and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{\left(F_f\right)}_B=\left(0.35\right)\left(3106\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\\ =35000\text{lb}\end{array}$

Calculate the frictional force acting on the car C after application of brakes $\left[{\left(F_f\right)}_C\right]$ using the relation:

${\left(F_f\right)}_C={\mu }_k{m}_{C}g$

Substitute $2484\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_C$, 0.35 for ${\mu }_k$, and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{\left(F_f\right)}_C=\left(0.35\right)\left(2484\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\\ =28000\text{lb}\end{array}$

The brakes are not applied on the wheels of car A $\left[{\left(F_f\right)}_A\right]$. Hence, there is no frictional force acting on the car A that is zero.

Calculate the total mass of the train $\left(m\right)$ using the relation:

$m=m_A+m_B+m_C$

Substitute $2484\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_A$, $2484\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_C$ and $3106\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_B$.

$\begin{array}{c}m=\left(2484\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)+\left(3106\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)+\left(2484\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\\ =8074\text{lb}\text{.}{\text{s}}^2/\text{ft}\end{array}$

Calculate the frictional force acting on the car A after application of brakes $\left[{\left(F_f\right)}_A\right]$ using the relation:

${\left(F_f\right)}_A={\mu }_k{m}_{A}g$

Substitute $2484\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_A$, 0.35 for ${\mu }_k$ and $32.2\text{ft}/{\text{s}}^2$ for g.

$\begin{array}{c}{\left(F_f\right)}_A=\left(0.35\right)\left(2484\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(32.2\text{ft}/{\text{s}}^2\right)\\ =28000\text{lb}\end{array}$

The brakes are not applied on the wheels of car B and car C, so there is no frictional force acting on the car B and car C, that is ${\left(F_f\right)}_B=0\text{ and}{\left(F_f\right)}_C=0$.

Here, ${\left(F_f\right)}_B$ is the frictional force acting on the rail car B and ${\left(F_f\right)}_C$ is the frictional force acting on rail car C.

Calculate the total frictional force acting on the whole train $\left(F_f\right)$ using the relation:

$F_f=-\left[{\left(F_f\right)}_A+{\left(F_f\right)}_B+{\left(F_f\right)}_C\right]$

Substitute $28000\text{lb}$ for ${\left(F_f\right)}_A$, 0 for ${\left(F_f\right)}_B$, and 0 for ${\left(F_f\right)}_C$.

$\begin{array}{c}{F}_f=-\left[\left(28000\text{lb}\right)+0+0\right]\\ =-28000\text{lb}\end{array}$

The expression for the impulse acting on the train due to frictional force $\left({\text{Imp}}_T\right)$ as follows:

${\text{Imp}}_T=F_{f}t$

Here, t is the time taken by the train to come to rest.

Substitute $F_{f}t$ for ${\text{Imp}}_T$.

$\begin{array}{l}m{v}_1+{\text{Imp}}_T=m{v}_2\\ m{v}_1+F_{f}t=m{v}_2\\ F_{f}t=m\left(v_2-v_1\right)\\ t=\frac{m\left(v_2-v_1\right)}{F_f}\end{array}$

Substitute $8074\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for m, $44\text{ft}/\text{s}$ for $v_1$, 0 for $v_2$, and $-28000\text{lb}$ for $F_f$.

$\begin{array}{c}t=\frac{\left(8074\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(0-44\text{ft}/\text{s}\right)}{-28000\text{lb}}\\ =\frac{355256}{28000}\\ =12.69\text{sec}\end{array}$

Therefore, the time required to bring the train (t) to a stop is $\underset{\_}{12.69\text{sec}}$.

To determine

Find the force in each coupling.

Answer to Problem 13.130P

The force in AB $\left(F_{AB}\right)$ and BC $\left(F_{BC}\right)$ are $\underset{\_}{\text{19380lb}}$ and $\underset{\_}{8620\text{lb}}$ respectively.

Explanation of Solution

Given information:

The initial speed of the train $\left(v_1\right)$ is $30\text{mi}/\text{h}$.

The coefficient of kinetic friction $\left({\mu }_k\right)$ is 0.35.

The weight of the rail car A $\left(W_A\right)$ is $40\text{tons}$.

The weight of the rail car B $\left(W_B\right)$ is $50\text{tons}$.

The weight of the rail car C $\left(W_C\right)$ is $40\text{tons}$.

The acceleration due to gravity (g) is $32.2\text{ft}/{\text{s}}^{\text{2}}$.

Calculation:

Show the impulse-momentum diagram of rail car A as in Figure (2).

The expression for the impulse acting on the rail car A $\left({\displaystyle \sum {\text{Imp}}_A}\right)$ as follows:

${\displaystyle \sum {\text{Imp}}_A}=\left[{\left(F_f\right)}_A+F_{AB}\right]t$

Here, $F_{AB}$ is the coupling force acting between the rail car A and B.

The expression for the principle of impulse-momentum to rail car A alone as follows:

$m_A{v}_1+{\displaystyle \sum {\text{Imp}}_A}=m_A{v}_2$

Substitute $-\left[{\left(F_f\right)}_A+F_{AB}\right]t$ for ${\displaystyle \sum {\text{Imp}}_A}$.

$m_A{v}_1-\left[{\left(F_f\right)}_A+F_{AB}\right]t=m_A{v}_2$

Substitute $2484\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_A$, $44\text{ft}/\text{s}$ for $v_1$, 0 for ${\left(F_f\right)}_A$, 0 for $v_2$, and $5.64\text{sec}$ for t.

$\begin{array}{l}\left(2484\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(44\text{ft}/\text{s}\right)-\left[0+F_{AB}\right]\left(5.64\text{sec}\right)=0\\ \left(5.64\right)F_{AB}=109,296\\ F_{AB}=\frac{109,296}{5.64}\\ F_{AB}=19,379\text{lb}\\ F_{AB}\approx \text{19,380}\text{lb}\end{array}$

Show the impulse-momentum diagram of rail car C as in Figure (3).

The expression for the impulse acting on the rail car C $\left({\displaystyle \sum {\text{Imp}}_C}\right)$ as follows:

${\displaystyle \sum {\text{Imp}}_C}=\left[F_{BC}-{\left(F_f\right)}_C\right]t$

Here, $F_{BC}$ is the coupling force acting between the rail car B and C.

The expression for principle of impulse-momentum to car C alone as follows:

$m_C{v}_1+{\displaystyle \sum {\text{Imp}}_C}=m_C{v}_2$

Substitute $\left[F_{BC}-{\left(F_f\right)}_C\right]t$ for ${\displaystyle \sum {\text{Imp}}_C}$.

$m_C{v}_1+\left[F_{BC}-{\left(F_f\right)}_C\right]t=m_C{v}_2$

Substitute $2484\text{lb}\text{.}{\text{s}}^2/\text{ft}$ for $m_C$, $44\text{ft}/\text{s}$ for $v_1$, $28000\text{lb}$ for ${\left(F_f\right)}_C$, 0 for $v_2$, and $5.64\text{sec}$ for t.

$\begin{array}{l}\left(2,484\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(44\text{ft}/\text{s}\right)+\left[F_{BC}-28,000\text{lb}\right]\left(5.64\text{s}\right)=0\\ \left(5.64\right)F_{BC}-157,920=-109,296\\ \left(5.64\right)F_{BC}=48624\\ F_{BC}=\frac{48,624}{5.64}\\ F_{BC}=8,620\text{lb}\end{array}$

Therefore, the force in AB $\left(F_{AB}\right)$ and BC $\left(F_{BC}\right)$ are $\underset{\_}{\text{19,380}\text{lb}}$ and $\underset{\_}{8,620\text{lb}}$ respectively.