A 100-ton electric locomotive is driven by four electric engines operating with a potential
difference of 1000 volts and moves at a speed of 72 km/h. Assuming the friction coefficient between
the wheels of the locomotive and the rails equal to 0.5 and the free-fall acceleration
g=10 m/s
2, determine:
a) The electric current circulating on each engine;
b) If in each engine we have a current equal to 4000 amperes, how many railroad cars, 15 tons each,
will the locomotive be able to pull?
Problem data:
- Mass of the locomotive: M = 100 t;
- Mass of a railroad car: MR = 15 t;
- Speed of locomotive: v = 72 km/h;
- Coefficient of friction: μ = 0.5;
- Voltage line: U = 1000 V;
- Free-fall acceleration: g = 10 m/s2.
Problem diagram:
Solution
First, we convert the unit of mass given in tons (t) to kilograms (kg), and the speed is given in
kilometers per hour (km/h) to meters per second (m/s) used in the
International System of Units (
S.I.).
\[
\begin{gather}
M=100\;\cancel{\text{t}}\times \frac{1000\;\text{kg}}{1\;\cancel{\text{t}}}=100\times 1000\;\text{kg}=100000\;\text{kg}\\[10pt]
M_{R}=15\;\cancel{\text{t}}\times \frac{1000\;\text{kg}}{1\;\cancel{\text{t}}}=15\times 1000\;\text{kg}=15000\;\text{kg}\\[10pt]
v=\frac{72\;\cancel{\text{km}}}{1\;\cancel{\text{h}}}\times \frac{1000\;\text{m}}{1\;\cancel{\text{km}}}\times \frac{1\;\cancel{\text{h}}}{3600\;\text{s}}=\frac{720\;\text{m}}{36\;\text{s}}=20\;\text{m/s}
\end{gather}
\]
a) The power generated by one of the engines will be given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathscr{P}=Ui} \tag{I}
\end{gather}
\]
The total power generated by the four engines will be
\[
\begin{gather}
\mathscr{P}_{T}=4\mathscr{P} \tag{II}
\end{gather}
\]
substituting the expression (I) into (II), we have
\[
\begin{gather}
\mathscr{P}_{T}=4Ui \tag{III}
\end{gather}
\]
From
Classical Mechanics, we have the total power required to make the locomotive move itself
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\mathscr{P}_{T}=Fv} \tag{IV}
\end{gather}
\]
equating expressions (III) and (IV)
\[
\begin{gather}
4Ui=Fv \tag{V}
\end{gather}
\]
So the locomotive move, the engines should overcome the force of friction. In Figure 1, the
train wheels push the rails back with the force
\( \vec{F} \),
the rails react to the wheels with the force of friction making the train move itself
(
Newton's Third Law)
\[
\begin{gather}
F=F_{f}=\mu N \tag{VI}
\end{gather}
\]
substituting the expression (VI) into (V), we obtain
\[
\begin{gather}
4Ui=\mu Nv \tag{VII}
\end{gather}
\]
The normal
\( \vec{N} \)
and the weight
\( \vec{W} \)
of the locomotive cancel out (Figure 2)
\[
\begin{gather}
N=W \tag{VIII}
\end{gather}
\]
substituting the expression (VIII) into (VII)
\[
\begin{gather}
4Ui=\mu Wv \tag{IX}
\end{gather}
\]
the weight is given by
\[
\begin{gather}
\bbox[#99CCFF,10px]
{W=Mg} \tag{X}
\end{gather}
\]
substituting the expression (X) into (IX), we have
\[
\begin{gather}
4Ui=\mu Mgv\\
i=\frac{\mu Mgv}{4U} \tag{XI}
\end{gather}
\]
substituting the values given in the problem, we obtain
\[
\begin{gather}
i=\frac{0.5\times 100000\times 10\times 20}{4\times 1000}\\
i=\frac{10000000}{4000}
\end{gather}
\]
\[ \bbox[#FFCCCC,10px]
{i=2500\;\text{A}}
\]
b) Applying the expression (XI) to the current and using the data of this item, we have the total
mass
MT that can be driven by the engines
\[
\begin{gather}
M_{T}=\frac{4Ui}{\mu
gv}\\M_{T}=\frac{5\times 1000\times 4000}{0.5\times 10\times 20}\\
M_{T}=\frac{16000000}{100}\\
M_{T}=160000\;\text{kg}
\end{gather}
\]
Of this total mass, we have 100 000 kg represent the mass of the locomotive itself, then leftover
for the railroad cars
\( 1600000-100000=60000\;\text{kg} \),
\[
1600000-100000=60000\;\text{kg,}
\]
as each railroad car has a mass of 15 000 kg, the number of railroad cars will be
\[
n=\frac{60000\;\cancel{\text{kg}}}{15000\frac{\cancel{\text{kg}}}{\text{railroad car}}}
\]
\[ \bbox[#FFCCCC,10px]
{n=4\;\text{railroad cars}}
\]