Question

Four identical planks each of lengths L are arranged one above the other over a table as shown. Each projects a distance a beyond the edge of the one that is below it. What is the maximum possible value of a for the system to be in equilibrium without tripping forward?

• A. L/5
• B. L/ 4
• C. L/3
• D. L

Answer 1

Answer: (B)

L/ 4

In the given system, the center of mass of a plank must lie beyond the consequent plank's edge to be in equilibrium.

The length of a plank is $L$

Therefore, for the $n^{th}$ planck, the position of the center of mass has to be such that the system remains in equilibrium.

So, we have-

$\frac{3L}{2}=na+\frac{L}{2}$

$\Rightarrow a=\frac{L}{n}$

For, $n=4$, we have $a=\frac{L}{4}$