Question

A train $A$ runs from east to west and another train $B$ of the same mass and speed runs from west to east along the equator. $A$ presses the track with a force $F_1$ and $B$ with $F_2$. Then:

• A. $F_1=F_2$
• B. $F_1<F_2$
• C. $F_1>F_2$
• D. $F_1\le F_2$

Answer 1

The correct option is C $F_1>F_2$

Figure shows the train moving along earth's equator, comparing it with motion on a circular path with radius of curvature of circular path equal to earth's radius $\left(R\right)$.


At every point on the equator, applying equation of dynamics in radial direction,
$\Rightarrow mg-N=\frac{m{v}_n^2}{R}$
$[{\overrightarrow{v}}_n={\overrightarrow{v}}_t+{\overrightarrow{v}}_{tp}]$
Here,${\overrightarrow{v}}_t=$velocity of train
${\overrightarrow{v}}_{tp}=$ velocity of train's position, due to rotation of earth (Earth rotates in anticlockwise direction i.e from West to East)
$\Rightarrow N=mg-\frac{m{v}_n^2}{R}...\left(i\right)$
[$m$ and $R$ are $\text{constant}$]

Case $1$ :
Train running from east to west


$\Rightarrow v_n=v_t-v_{tp}...\left(ii\right)$

Case $2$ :
When train runs from west to east


$\Rightarrow v_n=v_t+v_{tp}...\left(iii\right)$
From Eq.$\left(i\right)$, we can see that higher value of $v_n$ will yield a lower normal reaction $\left(N\right)$ from earth on train.
From Eq.$\left(ii\right)$ and $\left(iii\right)$,
$v_n\text{(case-2)}>v_n\text{(case-1)}$
$\Rightarrow N\text{(case-1)}>N\text{(case-2)}$
From Newton's $3^{\text{rd}}$ law, train $A$ presses the track with larger force than that train $B$.
$\therefore F_1>F_2$