Consider a collection of identical boards each with mass m l

Consider a collection of identical boards, each with mass m, length L, and uniform mass density. Clearly lacking other forms of entertainment, we decide place a board on a ledge, then a second board on top of the first board, then a third board on top of the second board, and so on. Let x_n represent the greatest possible horizontal distance between far edge of the uppermost board and the ledge, such that the assembly of boards may remain stationary. Prove the recursive relation for x_n, x_n= X_n-1 + L/2n Prove the explicit licit form of x_n Imagine Romeo is attempting to sneak into Juliet\'s nd story room. Lord Capulet has coated the ding wall with a material so slippery that is has no

Solution

a) Length of the identical Blocks is L and Mass of of the Blocks is M

We have to calulate the center of mass of the structure . The distance between the centre of mass of the lower blockand right hand edge of the bottom blockwe\'ll define as L/2

From this we can write the equation

X₂(2M)=(X₁)M

Here X₁=L/2

X₂(2M)=(L/2)M

X₂=L/4

at this point over hang of the block is X₁+X₂=L/2+L/4=3L/4

For third block

X₃(3M)=(L/2)M

X₃=L/6

at this point over hang of the block is

X₁+X₂₊X₃=L/2+L/4+L/6=11L/12

This value is close to L.

For fourth Block

X₄(4M)=(L/2)M

X₄=L/8

At this point over hang of the Block X₁+X₂₊X₃+X₄=L/2+L/4+L/6+L/8=25L/24. This is greater than one

Xn(nM)=(L/2)M

The displacement for each block is L/2n

Hence we can write The recursive relation

X_n=X_n-1+L/2n

b) Over hang =L/2 (Summation of (k =1 to n))