Question

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

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

Answer 1

Answer: (A)

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

Answer 2

If N₁ and N₂ are the normal reactions at one foot and the other foot, then

N₁ + N₂ = mg ……….. (i)

For equilibrium,

(ma) H = dN₁ – dN₂ = d(N₁ – N₂)

(taking moments about centre of mass)

or N₁ – N₂ = (ma) $\frac{H}{d}$……….. (ii)

From (i) and (ii)

2N₁ = mg + ma $\frac{H}{d}$= $m\left( g + \frac{aH}{d} \right)$

or N₁ = $\frac{m}{2}\left( g + \frac{aH}{d} \right)$

and N₂ = $\frac{m}{2}\left( g - \frac{aH}{d} \right)$