Question

1. A square tile of side 2x mm and thickness y mm rests on a horizontal platform. A second tile of the same type is placed over the first so that only one side of the second protrudes over a parallel side of the first by r mm. A third tile is placed over the second so that only one side of the third protrudes over a parallel side of the second by r mm. A fourth is placed over the third in a similar way and so on. (Cross-section of the tiles is shown below.)
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Assume that the centre of gravity of each tile is at the centre of the tile. If

Answer 1

A is the lowest corner of the first tile, denote the horizontal distance of the centre of gravity of the kth tile by xk, so that x1 = x, x2 = x+r, x3 = x + 2r, . . . The centre of gravity of n slabs placed in the above manner is at a horizontal distance x from A given by x = (x1 + x2 + . . . xn)/n. The nth slab will topple over if x > AB = 2x.
(a) Write down an explicit expression for x in terms of x, r and n.
(b) Find the maximum number of slabs that can be so placed without toppling over in the following cases.
(i) x = 50, r = 5.
(ii) x = 50, r = 8.
(iii) General case of x, n.