Impact Force on stack of concrete slabs!
I was watching a martial artist breaking a stack of 9 out of 10 concrete bricks. I noticed that the last didn't break. It made me wonder if this would allow calculating the force dealt to be simplified.
The line of thought being it was enough kinetic energy for 9 but not 10, giving a limit. Rather than if it just went all the way through, you'd only know the minimum dealt force, but not maximum. Upon traveling down this rabbit hole I realized just how many variables there are.
So my question is, if I wanted to calculate the force dealt, what would the formula look like?
I understand there are different ways to attack this, but there exists some things I'm not going to know and that could be used to hopefully refine the angle of attack. 1018
For example I can weigh a brick, but I have no idea how I would measure how much mass he put behind the blow, so we can rule out the F=M*A approach.
edit: First, thank you for the responses. I appreciate any and all input to work towards better understanding on this. This is my first time on this site and based on the answers it seems my expectation of there being more of a conversation working towards an answer was incorrect. I also think I was expecting/hoping for something along the lines of "oh you just need to know the Joules needed for one brick and multiply it out because of x,y,z.." With all that said Ill gladly supply as much info as needed for an answer, or an idea of an answer. -the bricks were spaced about half an inch by small wooden pegs at either end -L:15" W:7.5" H:1.5" Weight: 50lbs ~mass: ~22.68kg -I don't know how this would help but it seemed the strike starting from rest traveled ~4.5ft in ~.5s
Two blocks each of thickness h acting independently [note: Ze is Z sub e, Zc is Z sub c, My is M sub y, Ze is Z sub e, b is width]
Elastic Modulus of a single block(beam)= Ze=I/y=(bh^2)/6
Combined elastic Modulus of two block acting separately= Zc=2Ze=(bh^2)/3
Bending Stress f=My/I=M/Zc=(3M)/(bh^2)
f=My/I=M/Zc=(3M)/(bh^2)
Now if the there was a single block of thickness 2h,then
Elastic Modulus of a block having thickness 2h: Ze=I/y=(b(2h)^2)/6=(2bh^2)/3
Bending Stress f=My/I=M/Zc=(3M)/(2bh^2)
Thus double bending stress will be generated in separated blocks as compared to non separated, thus Half force will be required for separated configuration in comparison to not.
This can be easily extrapolated to n number of blocks.
Edit again: really the biggest question here is: Does the fact that the last brick doesn't break reveal some key to the force or energy dealt? BUT a formula for the scenario with explanation of how you derived would be very enlightening and.. well fun 441