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. $F_1>F_2$
• B. $F_1<F_2$
• C. $F_1=F_2$
• D. The information is insufficient to find the relation between $F_1$ and $F_2$

Answer 1

Answer: (A)

Earth rotate about its axis from west to east

For Train A : $V_A=V-\omega r$

For Train B : $V_B=V+\omega r$

For Train A

$mg-N_A=\frac{m{\left(v-\omega r\right)}^2}{r}$



$N_A=mg-\frac{m{\left(v-\omega r\right)}^2}{r}$

Similarly for Train B: $N_B=mg-\frac{m{\left(v+\omega r\right)}^2}{r}$

Clearly $N_B<N_A$ and

$F_1>F_2$