To solve the problem of finding the elevation of the outer rail above the inner rail for a train moving on a curved railway line, we can follow these steps:
### Step 1: Convert the speed from km/h to m/s
The speed of the train is given as \( V = 36 \, \text{km/h} \). To convert this to meters per second (m/s), we use the conversion factor:
\[
V = 36 \times \frac{1000 \, \text{m}}{3600 \, \text{s}} = 10 \, \text{m/s}
\]
### Step 2: Identify the radius of curvature and the distance between the rails
The radius of curvature \( r \) is given as \( 1000 \, \text{m} \) and the distance between the two rails \( d \) is \( 1.5 \, \text{m} \).
### Step 3: Set up the equations for forces
When the train is moving along the curved path, the normal force \( N \) acting on the train can be resolved into two components:
1. \( N \sin \theta \) provides the centripetal force required for circular motion.
2. \( N \cos \theta \) balances the gravitational force \( mg \).
From the centripetal force equation, we have:
\[
N \sin \theta = \frac{mv^2}{r}
\]
From the balance of forces in the vertical direction, we have:
\[
N \cos \theta = mg
\]
### Step 4: Divide the equations to eliminate \( N \)
Dividing the first equation by the second gives:
\[
\tan \theta = \frac{v^2}{rg}
\]
### Step 5: Substitute known values
Substituting \( v = 10 \, \text{m/s} \), \( r = 1000 \, \text{m} \), and \( g = 9.8 \, \text{m/s}^2 \):
\[
\tan \theta = \frac{10^2}{1000 \times 9.8} = \frac{100}{9800} = \frac{1}{98}
\]
### Step 6: Relate \( \tan \theta \) to the elevation \( h \)
The elevation \( h \) of the outer rail above the inner rail can be expressed in terms of \( \tan \theta \):
\[
\tan \theta = \frac{h}{d}
\]
where \( d = 1.5 \, \text{m} \). Therefore:
\[
h = d \cdot \tan \theta = 1.5 \cdot \frac{1}{98} = \frac{1.5}{98}
\]
### Step 7: Calculate the elevation \( h \)
Calculating \( h \):
\[
h = \frac{1.5}{98} \approx 0.0153 \, \text{m} \text{ or } 15.3 \, \text{mm}
\]
### Final Answer
The elevation of the outer rail above the inner rail is approximately \( 0.0153 \, \text{m} \) or \( 15.3 \, \text{mm} \).
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