Question

Four identical planks each of length a are arranged one above the other over a table as shown. Each projects a distance beyond the edge of the one that is below it. The maximum value for L so that no block topple is

• A. $\frac{a}{2}$
• B. $\frac{3a}{4}$
• C. $\frac{25a}{24}$
• D. $\frac{11a}{12}$

Answer 1

The correct option is D $\frac{11a}{12}$

Let the weight of each brick be $W$ and length $a$ . As bricks are homogeneous, the centre of gravity of each brick must be at the midpoint. Therefore, the topmost brick will be in equilibrium if its centre of gravity lies at the edge of brick below it, i.e., II brick. Thus the topmost brick can have maximum equilibrium extension of $a/2$ .

$C_1$ is the centre of mass of the top two bricks which lies on the edge of the third brick.

$C_2$ is the centre of mass of the top three bricks which lies on the edge of the fourth brick.

Thus, the maximum overhanging length to top from the edge of bottom brick is

$\frac{l}{2}+\frac{l}{4}+\frac{l}{6}=\frac{11}{12}a$