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

a)	The top brick overhanging the one below by
b)	The second brick from top overhanging the one below by
c)	The third brick from top overhanging the bottom one by
d)	The total overhanging length on the edge of the bottom brick is

(a,b,c,d)

Let the weight of each brick be  and length  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

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

 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