Calculating Force for Breaking a Window with a Lego Brick: A Realistic Approach

[davee123]

I recently caught a question that I realized was far more difficult than expected. Someone asked how hard they would have to throw a Lego brick in order to break a window. The mass of a normal run-of-the-mill 2x4 Lego brick is around 2.3 grams. And looking at wikipedia, they guessed the tensile strength of compressed glass is 50 MPa (which is probably more than normal glass, comparing the value of concrete which was 3 MPa). My instinct was of course that this was sufficient data to calculate the speed necessary with which to hurl the brick at the window in order to break it.

Alas, F=ma, so we've actually got to know the acceleration of the Lego brick as it slows down-- the mass and speed are insufficient data (not even counting the fact that if it hits at some angle other than a perpendicular, it'll throw things all out of whack).

So what else would I need? I imagine there's some fancy calculations I might be able to perform if I knew the elastic constants of the materials involved, plus the thickness of the glass (that at least I could approximate), and possibly constants of elasticity for the specific shape of the Lego brick. But who's got time for that? How would you go about solving this problem in reality? Is there no other solution than by trial? Perhaps I could even get a high-speed camera to determine the approximate acceleration values, or some variety of fancy force meter for the glass, but that still requires some degree of trial (as well as equipment).

Are there any "reasonable" numbers you'd use to approximate? I realized quickly that I had no idea whether the time it took to decelerate would be in hundredths of a second or millionths of a second or somewhere in between-- which is quite a few orders of magnitude in difference!

DaveE

 

[Tomtom]

I'm just thinking a bit here. Firstly, as MPa = 10^6*N/m^2, you need the area of the glass, hit by the lego brick. Let's call this 0.01 m * 0.02 m = 0.0002 m^2.

Multiplying 10^6N/m^2 and 0.0002 m^2 gives 20N. That is the force, which if exerted by the brick on the glass, that will break the glass.

You also need the time the brick impacts on the glass. (I'll be back)

 

[Tomtom]

From P [momentum]= m*v, and F = change in P / change in time, you can insert the second equation for the force needed, and the measured time. Solving for v, you get your answer.

 

[t3chn0n3rd]

Where did you get the tensile strenght of glass?

 

[Tomtom]

From Wikipedia, he says.

 

[Topher925]

From P [momentum]= m*v, and F = change in P / change in time, you can insert the second equation for the force needed, and the measured time. Solving for v, you get your answer.

there's no way your going to solve an inelastic collision problem that way. At least not with any kind of accuracy.

This is not a simple problem and to solve it with any type of accuracy your probably going to need an FEA analysis to determine the stress placed on it by an impact. This involves the geometry of the glass, how its fixed or restrained, and the geometry of the object hitting it. You don't want to approach this with typical F=ma or conservation of momentum principles. I would go with energy conservation and apply Castigliano's theorem to the glass plate.

Collision problems are hard, very hard. Your not going to be able to solve this with simple hand calculations. You also need to be more specific on the geometry of the glass. For example are you breaking a large storm glass window or a small window in a boxed frame?

 

[davee123]

From P [momentum]= m*v, and F = change in P / change in time, you can insert the second equation for the force needed, and the measured time. Solving for v, you get your answer.

So, basically, I'd still need the change in time in order to solve it?

Collision problems are hard, very hard. Your not going to be able to solve this with simple hand calculations.

Ahh, that's what I was afraid of. Effectively, it sounds like I'd be better off solving it by trial than by theory.

DaveE

 

[Topher925]

t sounds like I'd be better off solving it by trial than by theory

You got it. Just make sure everyone is wearing close toed shoes and safety glasses.

 

[Tomtom]

You could try dropping a lego brick on a force-meter behind a glass plate, and using the equation in reverse to calculate the change in time. If you drop it from rest from a certain height, you can roughly calculate the velocity (and momentum) of the brick.

 

[russ_watters]

Hold on a sec. Though it was correctly pointed out that it is very difficult to solve an inelastic collision problem via force, what wasn't said is that it is very easy to solve such a problem via energy. It's just a matter of finding the fracture energy of a pane of window glass, then using the kinetic energy equation for the lego.

Of course, I'm having trouble finding the fracture energy via google. There are a bunch of articles that look promising, but most require you to pay.

This is the type of thing you need (this is for glasses, though): http://www.asse.org/foundation/research/docs/vinger.pdf

 

[Topher925]

Though it was correctly pointed out that it is very difficult to solve an inelastic collision problem via force, what wasn't said is that it is very easy to solve such a problem via energy

Oh no, minus 10 smart points for not reading my post well enough. j/k Anyways,...

I would go with energy conservation and apply Castigliano's theorem to the glass plate.

While you are right, it is easier to use the energy conservation (what I suggested) it still doesn't make it "very easy". It is extremely difficult to model the energy transferred into and throughout the glass from the lego brick. there's just to much going on there for a single (or 10) equation that can be solved by hand.