Question

A train $A$ runs from east to west and another train $B$ of the same mass runs from west to east at the same speed along the equator. A presses the track with a force $F_{1}$ and $B$ presses the track with a force $F_{2}$

• A. $F_{1}\mspace{6mu} > \mspace{6mu} F_{2}$
• B. $F_{1} ⥂ < \mspace{6mu} F_{2}$
• C. $F_{1} = F_{2}$
• D. The information is insufficient to find the relation between $F_{1}$ and $F_{2}$

Answer 1

Answer: (A)

$F_{1}\mspace{6mu} > \mspace{6mu} F_{2}$

Answer 2

We know that earth revolves about its own axis from west to east. Let its angular speed is $\omega_{e}$ and the angular speed of the train is $\omega_{t}$

For train A : Net angular speed = ($\omega_{e} - \omega_{t}$) because the sense of rotation of train is opposite to that of earth

So reaction of track $R_{1} = F_{1} = mg - m(\omega_{e} - \omega_{t})^{2}R$

For train B : Net angular speed = $(\omega_{e} + \omega_{t})$ because the sense of rotation of train is same as that of earth

So reaction of track $R_{2} = F_{2} = mg - m(\omega_{e} + \omega_{t})^{2}R$

So it is clear that $F_{1} > F_{2}$