Question

A boy of mass $m$ with his mass centered at height $H$ is standing in a train moving with constant acceleration $a$. If his legs are wide spread with a distance $2d$ and he is not taking the help of any support then normal reactions at his feet are given by

• A. $\frac{m}{2}(g+\frac{Ha}{d});\frac{m}{2}(g-\frac{Ha}{d})$
• B. $m(a+g);mg$
• C. $\frac{Hma}{d};\frac{m}{2}(g-\frac{Ha}{d})$
• D. $mg;mg$

Answer 1

Answer: (A)

$\frac{m}{2}(g+\frac{Ha}{d});\frac{m}{2}(g-\frac{Ha}{d})$

Let $N_1$ and $N_2$ are the normal reaction at one foot and the other foot.
Thus, $N_1+N_2=mg....\left(1\right)$
For equilibrium, $maH=N_{1}d-N_{2}d$
or, $N_1-N_2=\frac{maH}{d}.....\left(2\right)$
solving (1) and (2), $N_1-mg+N_1=\frac{maH}{d}$
or, $N_1=\frac{m}{2}(g+\frac{aH}{d})$
Substituting this in equation 1 we get
$\frac{m}{2}(g+\frac{aH}{d})+N_2=mg$
or, $N_2=\frac{m}{2}(g-\frac{aH}{d})$