Question

The minimum value of $l$ so that truck can avoid the dead end, without toppling the block kept on it is given by $\frac{h}{xb}\frac{v^2}{g}$. Find the value of $x$.

Answer 1

Let the acceleration of the truck be '$a$' .
Consider the truck to have undergone uniform retardation .
$0-v^2=-2al$
$a=\frac{v^2}{2l}$
From vertical force equation on block :
$N=mg$
The net horizontal force acting on the block is ma. This force acts at COM.
For finding minimum $l$, the normal would have shifted to the end of the block.
Therefore writing torque equation:
$\frac{hma}{2}=\frac{Nb}{2}$

$\frac{hm\frac{v^2}{2l}}{2}=\frac{mgb}{2}$

$l=\frac{h{v}^2}{2bg}$

$\therefore x=2$