Uniform, identical bricks 20 cm long are stacked so that 4 cm of each brick extends beyond the brick beneath. How many bricks can be stacked in this way before the stack falls over?

6 bricks

Explanation:

For the bricks system to NOT fall over, then the center of mass of the system must lie within the touching area of the bottom brick.

Let the reference line be at the left end of the base brick, and the bricks are being stacking to the right direction.

- The 1st brick would have a center of mass at 20/2 = 10 cm or 10 + 0, this is < 20 cm so it stays

- 2 stacked bricks would have a center of mass at (10 + 14)/2 = 12 cm or 10 + 2, this is also < 20 cm so it stays

- 3 stacked bricks would have a center of mass at (10 + 14 + 18) / 3 = 14 cm or 10 + 2*2, this is also < 20 cm so it stays

- 4 stacked bricks would have a center of mass at (10 + 14 + 18 + 22)/4 = 16 cm or 10 + 2*3, this is also < 20 cm so it stays

- 5 stacked bricks would have a center of mass at (10 + 14 + 18 + 22 + 26)/5 = 18 cm or 10 + 2*4, this is also < 20 cm so it stays

- 6 stacked bricks would have a center of mass at (10 + 14 + 18 + 22 + 26 + 28)/5 = 20 cm or 10 + 2*5, this is also <= 20 cm so it stays before tipping over.

- The 7th brick would make everything fall down.

c) up

hope this

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