In a circus show are used large numbers of light boards, each of which can rotate around a fixed fulcrum. Fulcrum of reach board divides the length of the board in ration `2:1`. At the one end of the left most board is placed a small block of mass `30 kg` and a team of clowns stand keeping their feel at the ends of adjacent boards as shown in figure. the mass of each clown is `80 kg`. What maximum number of clowns can keep balance in this way?

Explanation

Method 1 Consider the last board.
the clown on the last board will keep his total weight on the extreme end.

Balancing torque about fulcrum of last board.
`W_(1)xx(2l)/3=800xxl/3`
`W_(1)=400N`
Hence `W_(2)=800-400=400N`
Now balancing the board next to last (second last board)

Again balancing torque about fulcrum of second last board
`W_(3)xx2/3=l=W_(2)xxl/3`
`W_(3)=(W_(2))/2=200N`
Hence `W_(4)=800-200=600N`
Now cosidering next to second last board. Balancing torque about fulcurm

`W_(5)=(2l)/3=W_(4)xxl/3`
`W_(5)=(W_(4))/2=300N`
But on 1st board on extreme end `300 N` is applied.
Hence total number of boards are three.
Method 2: On extreme left board at extreme end `300 N` is placed and on extreme right board on extreme right end the clown will apply his total weight (`800 N`).
For balancing any board the ratio of force applied on lef and right side on the board should be `1:2`.
If `W=300 N, W_(2)=600N`
Here number of boards should be three.