Chapter 13.1, Problem 13.17P

(a)

To determine

Find the distance (x) required to bring the train to stop.

Answer to Problem 13.17P

The distance (x) required to bring the train to stop is $\underset{\_}{124.07\text{ft}}$.

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 train A $\left(W_A\right)$ is $40\text{tons}$ or $\text{80kips}$.

The weight of train B $\left(W_B\right)$ is $50\text{tons}$ or $\text{100kips}$.

The weight of train C $\left(W_C\right)$ is $40\text{tons}$ or $\text{80kips}$.

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

Calculation:

Show the free body diagram of the train A, B and C with the forces as in Figure (1).

Convert the unit of initial velocity of trailer truck from $\text{mi}/\text{h}$ to $\text{ft}/\text{s}$:

$\begin{array}{c}{v}_1=30\times 1.4667\\ =44\text{ft}/\text{s}\end{array}$

Calculate the force at A $\left(F_A\right)$ using the relation:

$F_A={\mu }_k{N}_A$

Substitute 0.35 for ${\mu }_k$ and $\text{80kips}$ for $N_A$.

$\begin{array}{c}{F}_A=0.35\left(80\right)\\ =28\text{kips}\end{array}$

Calculate the force at B $\left(F_B\right)$ using the relation:

$F_B={\mu }_k{N}_B$

Substitute 0.35 for ${\mu }_k$ and $\text{100kips}$ for $N_B$.

$\begin{array}{c}{F}_B=0.35\left(100\right)\\ =35\text{kips}\end{array}$

Calculate the force at C $\left(F_C\right)$ using the relation:

$F_C={\mu }_k{N}_C$

Substitute 0.35 for ${\mu }_k$ and $\text{80kips}$ for $N_C$.

$\begin{array}{c}{F}_C=0.35\left(80\right)\\ =28\text{kips}\end{array}$

Calculate the total weight (W) using the relation:

$W=W_A+W_B+W_C$

Substitute $\text{80kips}$ for $W_A$, $\text{100kips}$ for $W_B$, and $\text{80kips}$ for $W_C$.

$\begin{array}{c}W=80+100+80\\ =260\text{kips}\end{array}$

Calculate the mass of the truck (m) using the formula:

$\begin{array}{l}W=mg\\ m=\frac{W}{g}\end{array}$

Substitute $260\text{kips}$ for W and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{260}{32.2}\\ =8.075\text{k}\cdot \text{ft}/{\text{s}}^{\text{2}}\end{array}$

Calculate the initial kinetic energy $\left(T_1\right)$ using the formula:

$T_1=\frac{1}{2}m{v}_1^2$

Substitute $8.075\text{k}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for m and $44\text{ft}/\text{s}$ for $v_1$.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 8.075\times {44}^2\\ =7816.6\text{kips}\end{array}$

The final kinetic energy $\left(T_2\right)$ is zero.

Calculate work done $\left(U_{1-2}\right)$ for the cars B and C using the formula:

$U_{1-2}=-\left(F_B+F_C\right)x$

Substitute $35\text{kips}$ for $F_B$ and $\text{28kips}$ for $F_C$.

$\begin{array}{c}{U}_{1-2}=-\left(35+28\right)x\\ =-63x\end{array}$

Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

Find the distance (x) required to bring the train to stop:

$T_1+U_{1-2}=T_2$

Substitute $7816.6\text{kips}$ for $T_1$, 0 for $T_2$ and $-63x$ for $U_{1-2}$.

$\begin{array}{l}7816.6\text{kips}-63x=0\\ 63x=7816.6\\ x=124.07\text{ft}\end{array}$

Therefore, the distance (x) required to bring the train to stop is $\underset{\_}{124.07\text{ft}}$.

(b)

To determine

Find the force in each coupling.

Answer to Problem 13.17P

The force in coupling AB is $\underset{\_}{19.38\text{kips}\left(\text{Tension}\right)}$.

The force in coupling BC is $\underset{\_}{\text{8}\text{.62kips}\left(\text{Tension}\right)}$.

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 train A $\left(W_A\right)$ is $40\text{tons}$ or $\text{80kips}$.

The weight of train B $\left(W_B\right)$ is $50\text{tons}$ or $\text{100kips}$.

The weight of train C $\left(W_C\right)$ is $40\text{tons}$ or $\text{80kips}$.

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

Calculation:

Consider car A:

Show the free body diagram of the train A with the forces as in Figure (2).

Assume $\left(F_{AB}\right)$ to be in tension.

Calculate the mass of the truck (m) using the formula:

$\begin{array}{l}{W}_A=mg\\ m=\frac{W_A}{g}\end{array}$

Substitute $80\text{kips}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{80}{32.2}\\ =2.4845\text{k}\cdot \text{ft}/{\text{s}}^{\text{2}}\end{array}$

Calculate the initial kinetic energy $\left(T_1\right)$ using the formula:

$T_1=\frac{1}{2}m{v}_1^2$

Substitute $2.4845\text{k}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for m and $44\text{ft}/\text{s}$ for $v_1$.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 2.4845\times {44}^2\\ =2404.996\text{kips}\end{array}$

The final kinetic energy $\left(T_2\right)$ is zero.

Calculate work done $\left(U_{1-2}\right)$ for the cars B and C using the formula:

$U_{1-2}=-F_{AB}x$

Substitute $35\text{kips}$ for $F_B$ and $\text{124}\text{.07ft}$ for x.

$U_{1-2}=-124.07F_{AB}$

Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

The expression for the principle of work and energy is as follows;

$T_1+U_{1-2}=T_2$

Substitute $2404.996\text{kips}$ for $T_1$, 0 for $T_2$ and $-124.07F_{AB}$ for $U_{1-2}$.

$\begin{array}{l}2404.996\text{kips}-124.07F_{AB}=0\\ 124.07F_{AB}=2404.996\\ F_{AB}=19.38\text{kips}\left(\text{tension}\right)\end{array}$

Consider car C:

Show the free body diagram of the train C with the forces as in Figure (3).

Calculate the mass of the truck (m) using the formula:

$\begin{array}{l}{W}_A=mg\\ m=\frac{W_A}{g}\end{array}$

Substitute $80\text{kips}$ for $W_A$ and $32.2\text{ft}/{\text{s}}^{\text{2}}$ for g.

$\begin{array}{c}m=\frac{80}{32.2}\\ =2.4845\text{k}\cdot \text{ft}/{\text{s}}^{\text{2}}\end{array}$

Calculate the initial kinetic energy $\left(T_1\right)$ using the formula:

$T_1=\frac{1}{2}m{v}_1^2$

Substitute $2.4845\text{k}\cdot \text{ft}/{\text{s}}^{\text{2}}$ for m and $44\text{ft}/\text{s}$ for $v_1$.

$\begin{array}{c}{T}_1=\frac{1}{2}\times 2.4845\times {44}^2\\ =2,404.996\text{kips}\end{array}$

The final kinetic energy $\left(T_2\right)$ is zero.

Calculate work done $\left(U_{1-2}\right)$ for the cars B and C using the formula:

$U_{1-2}=\left(F_{BC}-F_C\right)x$

Substitute $28\text{kips}$ for $F_C$ and $\text{124}\text{.07ft}$ for x.

$\begin{array}{c}{U}_{1-2}=\left(F_{BC}-28\right)124.07\\ =124.07F_{BC}-3473.96\end{array}$

Use work and energy principle which states that kinetic energy of the particle at a displaced point can be obtained by adding the initial kinetic energy and the work done on the particle during its displacement.

Find the force in coupling BC:

$T_1+U_{1-2}=T_2$

Substitute $2404.996\text{kips}$ for $T_1$, 0 for $T_2$, and $124.07F_{BC}-3473.96$ for $U_{1-2}$.

$\begin{array}{l}2,404.996\text{kips}+124.07F_{BC}-3,473.96=0\\ 124.07F_{BC}=1,068.964\\ F_{BC}=8.62\text{kips}\left(\text{Tension}\right)\end{array}$

Therefore, the forces in coupling AB and BC are $\underset{\_}{19.38\text{kips}\left(\text{Tension}\right)}$ and $\underset{\_}{\text{8}\text{.62kips}\left(\text{Tension}\right)}$ respectively.