Question

Train $\text{A}$ runs from east to west and another train $\text{B}$ of the same mass runs from west to east at the same speed along the equator. $\text{A}$ presses the track with a force $F_1$ and $\text{B}$ presses the track with a force $F_2$ then

• A. $F_1>F_2$
• B. $F_1<F_2$
• C. $F_1=F_2$
• D. Data insufficient

Answer 1

The correct option is A $F_1>F_2$

We know that , earth rotates from west to east or counterclockwise as viewed from the North Pole star Polaris.


Top view for train $\text{A}$


We can see that linear speed of earth $\left(v\right)$ is opposite to the speed of train $\text{A}$ which is $u$. Thus, net speed of train decreases. Let $m$ be the mass of train.


From FBD,

$mg-N=\frac{m(u-v{)}^2}{R}$

$\Rightarrow F_1=N=mg-\frac{m(u-v{)}^2}{R}.....\left(1\right)$


Similarly, for train $B$



$F_2=N=mg-\frac{m(u+v{)}^2}{R}.......\left(2\right)$
From Eq.$\left(1\right)$ and Eq.$\left(2\right)$, we can see that,

Normal force applied by earth on train $\text{A}$ is greater than an train $\text{B}$. Thus, $F_1>F_2$.

Hence, option (a) is the correct answer.