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 F1 and B presses the track with a force F2

To understand the forces with which the two trains press against the tracks, we need to analyze their motion in a rotating frame of reference (the Earth) or from an inertial frame of reference. Let us analyze this from an inertial (non-rotating) frame of reference looking down at the Earth from above the North Pole.

The Earth rotates from west to east with an angular velocity $\omega$. The radius of the Earth at the equator is $R$. Thus, any stationary object on the equator moves from west to east with a speed of $v=\omega R$ relative to the center of the Earth.

Let the speed of both trains relative to the Earth's surface be $u$.
Let $m$ be the mass of each train.
The gravitational force pulling each train toward the center of the Earth is $F_g=mg$, where $g$ is the acceleration due to gravity if the Earth were not rotating.
The track exerts an upward normal force $N$ on each train, which is equal to the force with which the train presses the track (by Newton's third law).

Now, let's find the net velocity of each train with respect to the inertial frame:
1. Train A runs from east to west (opposite to the direction of Earth's rotation).
Its velocity relative to the center of the Earth is:
$v_A=\omega R-u$
2. Train B runs from west to east (in the direction of Earth's rotation).
Its velocity relative to the center of the Earth is:
$v_B=\omega R+u$

For any object moving in a circle of radius $R$ with speed $v$, the net radial force toward the center of the circle provides the necessary centripetal acceleration:
$F_g-N=\frac{m{v}^2}{R}$
Rearranging this gives the normal force (the force with which the train presses the track):
$N=F_g-\frac{m{v}^2}{R}$

Using this relation for both trains:
For train A (pressing the track with force $F_1$):
$F_1=mg-\frac{m{(\omega R-u)}^2}{R}$
For train B (pressing the track with force $F_2$):
$F_2=mg-\frac{m{(\omega R+u)}^2}{R}$

Comparing the two expressions, since $\omega R+u>|\omega R-u|$ (assuming standard operational train speeds where $u>0$), we have:
${(\omega R+u)}^2>{(\omega R-u)}^2$

Since we are subtracting a larger term in the case of $F_2$, it follows that:
$F_1>F_2$

Therefore, the train running from east to west presses the track with a greater force than the train running from west to east.