Question

Four bricks, each of length l, are put on the top of one another in such a way that part of each extends beyond the one beneath. For the largest equilibrium extensions,

Answer 1

Correct option is D. The total overhanging length on the edge of the bottom brick is $$(11/12)l$$
Let the weight of each brick be W and length l. 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 $$l/2$$.
$$C_1$$ 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 thee bricks which lies on the edge of the fourth brick.
Thus, the maximum overhanging length of top from the edge of bottom brick is
$$\dfrac{l}{2}+\dfrac{l}{4}+\dfrac{l}{6}=\dfrac{11}{12}l$$.