Would a fall from 53 centimeters break the glass from a window?

[AkioHiromi]

TL;DR

I was studying the physics involving glass and got curious while looking to my window.
Details:
The window weights 3kg.
The “ground” which the window would fall is made from wood, the same wood that holds all four sides of the glass like a frame. (common, not tempered glass)
The glass has an area of 0,27 square meters, and has a width of 3mm.
The fall would be straight down, same movement as closing the window, but instead of a supported lowering, it’d be a free fall from it’s stand.

 

[phinds]

I can imagine it breaking or not breaking, depending on exactly how it hit. How the force gets transferred to the glass will make all the difference. I wouldn't test it out though.

 

[Baluncore]

What happens will depend largely on how long it takes to stop at the end of the fall. The forces and deceleration are proportional to the fall time divided by the stop time. A soft weather strip at the bottom, to seal the window when it is closed, could also lengthen the stopping time and so reduce the chance of breakage.

The energy of the fall will go into making noise and vibration. The difference between the speed of sound in glass and in wood will scatter the impulse and so dissipate the impulse of energy in time and in space. The state of the putty or sealant that holds the glass in the wooden frame will also make a difference because energy will be coupled across that interface several times at different angles and in different directions.

Travelling energy would be concentrated at a gap in the sealant or at a poor joint in the timber frame. That could start a fracture through the glass. If the glass is old, or is stressed in the frame, it may fracture more easily.

 

[256bits]

What happens will depend...

Frame might not even be square with the sill.
One corner of the frame could hit first rotating the frame with extra jar and adding stress to the glass.
The weather strip is a good solution to alleviate possible breakage if the frame drops on its own.

 

[fresh_42]

If e.g. a tiny stone is in the window's frame, and the image suggests that the window wasn't built under laboratory conditions, such that all forces accumulate to a single point of contact with the glass, then it can easily break falling from any height.

 

[Baluncore]

For each catastrophization there is a counter.

Frame might not even be square with the sill.
One corner of the frame could hit first rotating the frame with extra jar and adding stress to the glass.

That would provide two impulses of half the amplitude. It would also cause a rotational mode to better scatter the energy. A skew frame is more likely to hit the sash on the way down, which would increase the fall time and so reduce the impact energy.

If e.g. a tiny stone is in the window's frame, and the image suggests that the window wasn't built under laboratory conditions, such that all forces accumulate to a single point of contact with the glass, then it can easily break falling from any height.

A tiny stone will be punched into the wooden sill or frame. That will lengthen the time the window takes to stop, so will reduce the deceleration and the forces.

 

[fresh_42]

A tiny stone will be punched into the wooden sill or frame.

You bet?

 

[Baluncore]

You bet?

No bet, but it is an alternative with a Baysian probability of 50%.
How small, or how large, is a tiny stone ?
How hard is the low density wood normaly selected for sash window frames ?
And how did the stone get to rest on the window sill while the window was open ?

 

[sophiecentaur]

We really don't know enough about the situation here. I know that you can drop a large sheet of (window) glass by just a few cm onto a concrete floor, on its corner and a great chunk will come off or there could be a crack all the way across. Otoh, we have all got away with some pretty savage treatment of similar pieces and they have been unscathed.
If the OP is about what could happen to the glass in the picture then just how slippery the track is will make all the difference.

An experiment would be out of the question so I can only suggest reasonable care and the use of window wedges if the existing felt (?) is very worn. If the OP is 'investigating' some damage by tenants or a young family member then what happened is anyone's guess. A repair wouldn't;t be too expensive.

 

[pbuk]

The fall would be straight down, same movement as closing the window, but instead of a supported lowering, it’d be a free fall from it’s stand.

That's why we have sash cords: if one breaks the other one stops the window falling.

 

[sophiecentaur]

That's why we have sash cords: if one breaks the other one stops the window falling.

Correct for corded windows but the image seems to be of a smaller, lightweight type with a friction brake. There doesn't seem to be enough room fora weight channel.

 

[diogenesNY]

As per phinds, I can see it breaking or not breaking depending on any number of unspecified (and possibly hidden) variables.

In my one personal experience with this scenario that I can clearly recollect, I observed such a fall. Rotten sash cords broke (I do not recall the _exact_ circumstances the led to this). The window dropped to the bottom of the sill with quite a loud clatter, but no breakage. It is a reasonable assumption that the friction between the window frame and the window's frame somewhat arrested the acceleration of the fall, but it still looked like it fell without a great deal of resistance and landed with a loud crash (but no actual glass breakage)... so that is my one anecdotal, sort of empirical data point.

--diogenesNY

 

[Arthead_66]

Having broken massive amounts sheet glass, there is an easy way to do it. It's cleavage, glass breaks from the edge, all it needs is a tap to crack it. Conversely I have thrown hammers at large glass plates they bounce right off it.

 

[Baluncore]

As per phinds, I can see it breaking or not breaking depending on any number of unspecified (and possibly hidden) variables.

I would consider the aspect ratio of the window to be important in it's ability to fall freely. I expect tall windows will fall more freely than wide windows that will tend to jam.

 

[sophiecentaur]

I would consider the aspect ratio of the window to be important in it's ability to fall freely. I expect tall windows will fall more freely than wide windows that will tend to jam.

When I lived in a rented place (never any maintenance by the landlord or me!) the sash windows were nearly all down to just one cord. Lowering them involved lifting the non-corded side and letting gravity do the work. On disaster involved a very large, wide window with the thickest glass I have ever come across. It went down with a bang when the remaining cord snapped. A scary experience, aamof. The (legal) replacement glass made the window a lot easier to lift and lower and I never replaced a cord. The place was demolished shortly after so no harm done.

 

[JackCatDaily]

Summary:: I was studying the physics involving glass and got curious while looking to my window.
Details:
The window weights 3kg.
The “ground” which the window would fall is made from wood, the same wood that holds all four sides of the glass like a frame. (common, not tempered glass)
The glass has an area of 0,27 square meters, and has a width of 3mm.
The fall would be straight down, same movement as closing the window, but instead of a supported lowering, it’d be a free fall from it’s stand.

View attachment 295121

You could test 25 times(or any number) and the results will vary each time. A pattern may surface, but not much more. There are many many factors/parameters that would apply.

 

[N1206]

Like everyone said, in practicality there are NUMEROUS factors that play into what will actually happens.
As a pure physics question, you are dropping a sound piece of ordinary glass , 0.27 sq metres, 3 mm thickness, 53 cm and want to know if it would break.

I think that is a Young's modulus question.
Young's modulus for glass seems to be ~68 GPa
The force you develop from the drop is mg --> 3kg * 9.81 m / s*s --> 29.43 N
You are applying that force on one edge of the glass
That's be the square root of 0.27 sq metres time 3 mm --> 0.52 m * 0.003 m --> 0.00156 sq metres
The pressure developed is F/A --> 29.43 N / 0.00156 sq metres --> 18880 Pa --> 18.9 KPa --> 0.189 MPa --> 0.000189 GPa
And that's WAAAAAY below 68 GPa, so if the glass hit the bottom squarely, it would survive.

Hit a little rock 1mm x 1mm wide at the bottom?
The pressure developed is F/A --> 29.43 N / (.001 *.001) --> 29430000 Pa --> 29430 KPa --> 29.43 MPa --> 0.2943 GPa and that too is still probably survivable. You might chip the spot of impact. Glass is tougher stuff than we give it credit for!

If it were a pointy little rock 0.1 mm x 0.1mm? Well then you'd be at 29.43 GPa, so then it might break if there were flaws, etc.

Glass is tough stuff. That whole Hollywood smash-a-beer-bottle-on-the-counter-as-an-impromptu-weapon thing? That is actually MUCH MUCH harder to do that it looks!

 

[Baluncore]

I think that is a Young's modulus question.
Young's modulus for glass seems to be ~68 GPa

But it is the frame that hits, not the glass.
So you must model the low density softwood frame, with a filler seal to the glass.

 

[pbuk]

So you must model the low density softwood frame, with a filler seal to the glass.

I think that is the least of @N1206 's problems: their attempt at a solution implies that a fall from 1 cm is as catastrophic as one from 1m.

 

[N1206]

But it is the frame that hits, not the glass.
So you must model the low density softwood frame, with a filler seal to the glass.

There are a boatload of caveats about real-life behavior. I took the OPs post as a straight physics question. Would glass falling from a 53 cm height, landing on a 3mm wide by 52 cm long edge survive? You can calculate THAT without any further (nor not much after googling up typical glass tensile strength) information than what was provided.

And the glass is in the frame. It is subject to acceleration, and when the frame hits, it will be subject to deceleration. If it is a Young's modulus question--and I think it is--then the mass of the glass, g = 9.81, and the area that it impacts on are all that come into play. And as I noted, since the F=mg is given, the real rub is how much surface area is going to be affected at the sudden stop at the end.

 

[N1206]

I think that is the least of @N1206 's problems: their attempt at a solution implies that a fall from 1 cm is as catastrophic as one from 1m.

Tensile strength is rated in units of pressure--not energy.
A fall from a greater height will certainly generate more energy that will have to be dissipated--but whether the energy gets dissipated or causes failure is NOT dependent upon the quantity of energy, but rather upon the force applied to an area. The glass was subjected to an acceleration. Whether it survives depends on how much surface area the force of deceleration is spread over.

Where things get tricky is in deciding what the force of deceleration is. The mass is constant, but you are going from some velocity to zero. If that happens instantaneously, the force involved is infinite. How that plays out when you are dealing with Young's modulus questions is a good question, that someone may be able to answer.

But I reiterate: Tensile strength is rated in units of pressure, and breakage requires the tensile strength to be exceeded. So, propose a numerical answer from what the OP gave, if you know or believe I have erred.

 

[pbuk]

If it is a Young's modulus question--and I think it is

You are wrong.

then the mass of the glass, g = 9.81, and the area that it impacts on are all that come into play.

So the height from which the glass falls, whether it is aluminium oxide glass or regular silica, whether it lands on a stone lintel or a rubber draft excluder... none of these things make any difference?

 

[pbuk]

So, propose a numerical answer from what the OP gave, if you know or believe I have erred.

As @Baluncore, @diogenesNY, @fresh_42, @JackCatDaily, @256bits... (sorry if I have missed anyone out) and I have all said, there too many unspecified parameters to make an attempt at a numerical answer meaningful.

 

[pbuk]

The glass was subjected to an acceleration. Whether it survives depends on how much surface area the force of deceleration is spread over.

And do you think that the force of deceleration is constant regardless of the impact velocity?

Spoiler

The answer 'yes, it is clearly always infinite' is not correct.

 

[sophiecentaur]

Tensile strength is rated in units of pressure--not energy.

True but Energy is needed to do damage. Work is actually what counts so distance moved during a collision has to be taken into account. This make things much harder (more difficult).

 

[fresh_42]

As @Baluncore, @diogenesNY, @fresh_42, @JackCatDaily, @256bits... (sorry if I have missed anyone out) and I have all said, there too many unspecified parameters to make an attempt at a numerical answer meaningful.

Indeed. Beginning with the production process and the amount of pre-stress put into the glass. It will probably never break under perfect conditions, but the reality is far from having those. So the question behind the question is: What are possible disturbances and to which degree are they relevant such that a certain experiment has a predicted outcome? And the answer to that question is a long list of parameters together with a list of margins of which many depend on each other.

 

[N1206]

You are wrong.

So the height from which the glass falls, whether it is aluminium oxide glass or regular silica, whether it lands on a stone lintel or a rubber draft excluder... none of these things make any difference?

Excellent!
If it is NOT a Young's modulus question, then what is it?
You didn't say.
If you had all the parameters you wanted specified, how would you calculate it?

And as I have noted, the force exerted depends very much on the the deceleration involved.
a = (Vf-Vi)/t
If t = 0 then a is infinite.
And things that cushion the blow (i.e. extend the period of deceleration) matter of course, and different materials have differing Young's moduli

You could assume the wood involved is oak, and work out how much the oak would elastically deform, and then use the distance you got from that to work back to a value for deceleration. And from there you could work back to the force involved -- and then you'd apply that force over an area you calculated and see if it exceeded the Young's modulus for the glass.

As another poster with practical experience has noted, busting glass is much, much more sensitive to WHERE you hit the glass than to how hard you hit it. It is also RIDICULOUSLY sensitive to existing stress risers. So, no matter how you spit-ball the math, don't go dropping any windows!

But you could <try> to run with the OPs numbers and come up with some math that makes sense.

 

[Frabjous]

Young‘s modulus is used for theoretical strength of materials. Actual strength is orders of magnitude smaller. I’d use tensile strength.

I would use impedence matching to determine the change in particle velocity and stress=density×sound speed×change in particle velocity.