Chapter 13.3, Problem 13.129P

The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track. The brakes are not applied on the wheels of car A. 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 13.129P

The time required to bring the train (t) to a stop is $\underset{\_}{5.64\text{sec}}$.

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 $\left[{\left(F_f\right)}_A\right]$ so 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 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 0 for ${\left(F_f\right)}_A$, $28000\text{lb}$ for ${\left(F_f\right)}_C$ and $35000\text{lb}$ for ${\left(F_f\right)}_C$.

$\begin{array}{c}{F}_f=-\left[0+\left(35000\text{lb}\right)+\left(28000\text{lb}\right)\right]\\ =-63000\text{lb}\end{array}$

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.

Use the principle of impulse-momentum to the entire train to find the time taken by the train to stop by application of brakes.

The expression or the principle of impulse-momentum as follows:

$m{v}_1+{\text{Imp}}_T=m{v}_2$

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

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

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 $-63000\text{lb}$ for $F_f$.

$\begin{array}{c}t=\frac{\left(8074\text{lb}\text{.}{\text{s}}^2/\text{ft}\right)\left(0-44\text{ft}/\text{s}\right)}{-63000\text{lb}}\\ =\frac{355256}{63000}\\ =5.64\text{sec}\end{array}$

Therefore, the time required to bring the train (t) to a stop is $\underset{\_}{5.64\text{sec}}$.

To determine

Find the force in each coupling.

Answer to Problem 13.129P

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.

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

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.