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Q.
On a railway curve the outside rail is laid higher than the inside one so that resultant force exerted on the wheels of the rail car by the tops of the rails will
Laws of Motion
Solution:
If the outside rail is h units higher than inside of rail track as shown in figure then
$N \cos\,\theta = mg ......$(i)
$ N \sin \theta=\frac{m v^2}{r} ....$(ii)
& $\tan \theta=\frac{ v ^2}{r g} ....$(iii)
where $\theta$ is angle of banking of rail track, $N$ is normal reaction exerted by rail track on rail.
It is clear from the equation (i) & (ii) that $N \cos \theta$ balance the weight of the train & $N \sin \theta$ provide the necessary centripetal force to turn.
If width of track is $\ell( OB ) \& h ( AB )$ be height of outside of track from the inside, then $\tan \theta=\frac{ h }{\ell}=\frac{ v ^2}{ rg }$ or $h =\frac{ v ^2 \ell}{ rg }.....$ (iv)
So it is clear from the above analysis that if we increase the height of track from inside by $h$ metre then resultant force on rail is provided by railway track & whose direction is inwards.
