Checking answer to stacking problem

[timetraveller123]

Homework Statement

A uniform brick of length L is laid on a smooth horizontal surface. Other equal bricks are now piled on as shown, so that the sides form a continuous plane, but the ends are offset at each block from the previous brick by a distance 0.15L. How many bricks can be stacked in this manner before the pile topples over?

Homework Equations

centre of mass = (m₁x₁ + m₂x₂) / (m₁ + m₂)

The Attempt at a Solution

i did the solution and got total number of blocks when toppling occurs is 7 is it correct i did it stating that toppling occurs when center of mass is at least l/2 away from the center is it correct?
one more the question states smooth floor is it important?

 

[haruspex]

total number of blocks when toppling occurs is 7

Yes.

when center of mass is at least l/2 away from the center

The centre of mass of which set of bricks, and L/2 from the mass centre of what?

 

[timetraveller123]

i started calculating the centre of mass from on top of the stack so if at any point the centre of mass a block and the blocks above it lies outside of it it topples is the logic correct

 

[kuruman]

... i did it stating that toppling occurs when center of mass is at least l/2 away from the center is it correct?

The idea is correct, but I don't see how you got 7 bricks. Can you show me?

 

[haruspex]

i started calculating the centre of mass from on top of the stack so if at any point the centre of mass a block and the blocks above it lies outside of it it topples is the logic correct

That's still a bit garbled. I assume you mean that if the centre of mass of the bricks above a given brick lies beyond that given brick then they will topple.
So when it topples, where is the tipping point?

 

[timetraveller123]

so i assumed n blocks were stacked
so the centre of mass of top two blocks measured from the centre of bottom block is
(0 * m + 0.15L *m)/2m
then using this the centre of mass top two blocks and the third block from top once again measured from centre of third block is

( 0.15L *m)/2m + 0.15L)*2m/3m here 3m is the mass of all three blocks and the third block itself is not contributing to the numerator

so for n blocks the centre of mass from the centre of nth block is (the equation is expanded out and cancelling m and l)

(0.15 + 0.15(2) ...0.15*n)/(n+1) ≥ 0.5 here 0.5 is representive of half the length of the block as centre of mass was measured from centre of block half the length means that it is tipping over

solving for n gives 6.67 or 7 blocks when it topples
is it logical or have i made i mistake

 

[kuruman]

So when it topples, where is the tipping point?

You did not explicitly answer the question that @haruspex asked, but the answer is implicit in the inequality that you set up. Anyway, it seems you understand the basics of the problem although your solution is a bit unorthodox.

solving for n gives 6.67 or 7 blocks when it topples

The problem is asking, "How many bricks can be stacked in this manner before the pile topples over?"

 

[timetraveller123]

oh 6 then?

 

[haruspex]

oh 6 then?

Right.

 

[timetraveller123]

thanks