Calculating Force Needed to Topple Over 900 lb Aquarium

[dan2x38]

Sorry folks been centuries since I was in physics class but hoping you can help me. Also hope I have the right forum?

I am an aquarist (aquarium hobbyist). Someone I know is placing a large 48"l x 12.5"d x 18"h aquarium on a piece of furniture. It will be filled to within 1.5" of the top. It will also have approx. 2.5" of gravel & some decorations like wood, etc. I know the weight of the water/gal. and the weight of tank 140 lb. and can calculate the gravel weight approx. The person is placing this aquarium on a non-standard aquarium stand that is much taller by almost a foot than a standard aquarium stand. I told them it is not safe with a toddler cruising around clutching at things to walk. So my question how do I calculate the force it would take to topple over the aquarium? Please keep in mind I a layman in all physics terms now. :-(

Thank you for any help I want to show this person it is a risky placement of so much weight approx. 900 lbs I figure. It is for the safety of the child.

cheers,
Dan

 

[JaredJames]

Honestly, I'd say the quickest and easiest way is to load it up with an equivalent weight and then 'rock' it gently to find which direction it will go in and with what force.

A foot really isn't that much and I think you may be over panicking here. Providing it is nice and sturdy (good footing on the ground), the weight alone should prevent minor movement from things such as a child or animal bump.

It's impossible to calculate the force from what you have given, there are simply too many factors unknown. For example, the primary factor is the stability of the stand itself. Followed by things such as where the force is applied.

 

[dan2x38]

With just the basics such as:

stand height = 38"
depth of stand = 12.5"
weight of aquarium & contents = 900lbs

Is it not possible to calculate the force needed to tip it? Not saying an exact number just a rough calculation - I know physics is an exact science! But say it is just a weight sitting on top the stand that is 18"h x 12"d x 48"w and weighs 900lbs with a centre of gravity = 6.25" at a height of 38" from the bottom of the object. Can't it be calculated how much force it would take to push over? If not what other facts are needed?

thx,
Dan

 

[cjl]

Is the stand footprint 12.5 inches deep? If so, the force can indeed be calculated. The center of mass of everything is 44.25 inches up, based on the numbers in your post above, and I'll assume that it's centered over the stand (so 6.25 inches from the edge). Based on this, for it to tip over, the center of mass would need to be lifted along a path inclined 8.04 degrees to the horizontal in order to tip it. Based on the given weight of 900 pounds, the aquarium would resist this tipping with a force of 126 pounds. If the aquarium were pushed at the very top, which by your numbers would put it 56 inches up, it would take just under 99 pounds of force to begin tipping the aquarium. If it were pushed from lower down, such as where a toddler would likely grab, the force should be much larger.

Of course, this is assuming a perfectly rigid stand and a solid footing on the ground.

 

[dan2x38]

Is the stand footprint 12.5 inches deep? If so, the force can indeed be calculated. The center of mass of everything is 44.25 inches up, based on the numbers in your post above, and I'll assume that it's centered over the stand (so 6.25 inches from the edge). Based on this, for it to tip over, the center of mass would need to be lifted along a path inclined 8.04 degrees to the horizontal in order to tip it. Based on the given weight of 900 pounds, the aquarium would resist this tipping with a force of 126 pounds. If the aquarium were pushed at the very top, which by your numbers would put it 56 inches up, it would take just under 99 pounds of force to begin tipping the aquarium. If it were pushed from lower down, such as where a toddler would likely grab, the force should be much larger.

Of course, this is assuming a perfectly rigid stand and a solid footing on the ground.

Thanks that was the kind of numbers I was curious of. I had an aquarium of approx. 500lbs on a stand 12" deep, 28"h and the tank was 18" tall and when the kids or folks walked by it it wobbled that always worried me. When they played on the floor near it the water really splashed around and I was big time concerned. I eventually drained the tank and moved it.

This one above is much lighter and lower body of mass so when she talked of her plans I was concerned.

Would you be able to show how you calculated the 99 lbs & 8* to tip? Thx again...

 

[pergradus]

If you assume the stand is perfectly rigid, and approximate the aquarium as a point-mass centered on the center of mass, then you can calculate the minimum torque needed to move it past the equilibrium point, using the ground as the axis of rotation.

Grab a physics 1 textbook and work it out.

additional thought: because the water is fluid, the center of mass will shift toward the axis of rotation as you begin to tilt the system, which actually means it would require significantly less force than the above calculation would show. You'd probably have to work out a differential equation to get a good estimation.

 

[dan2x38]

If you assume the stand is perfectly rigid, and approximate the aquarium as a point-mass centered on the center of mass, then you can calculate the minimum torque needed to move it past the equilibrium point, using the ground as the axis of rotation.

Grab a physics 1 textbook and work it out.

additional thought: because the water is fluid, the center of mass will shift toward the axis of rotation as you begin to tilt the system, which actually means it would require significantly less force than the above calculation would show. You'd probably have to work out a differential equation to get a good estimation.

Thank you but I was hoping to not have to study a textbook. I did a bit of surfing to find the answer and it was either unavailable or so convuted I would have to go back to classes to understand the answer. :-( Was hoping someone with the knowledge already could just set me on the right path. Hey if anyone needs suggestions or help with their aquariums freshwater or marine corals or plants I would gladly offer up that info. and no books... ;-)

thx,
Dan

 

[JaredJames]

stand height = 38"
depth of stand = 12.5"
weight of aquarium & contents = 900lbs

Is it not possible to calculate the force needed to tip it?

It is, but only with the addition of that info, no need to repeat the above though so shall let it go as is.

 

[DickL]

I hope I don't set off alarms, but if the stand isn't setting on a solid foundation/floor it can get to rocking. Once rocking sloshing of the water can make the situation worse, more likely to tip. Still from the numbers above, it doesn't look too bad.

What you may want to do are: 1) make certain it is on a firm footing, best to not be on carpet and padding, 2) tie it back to the wall (that is code for earthquake areas anyway), and 3) add weight at the bottom of the stand, such as books (a lot of them) on a low shelf.

 

[sophiecentaur]

It strikes me that the practical solution is to hold the stand against the wall with a bracket. It needn't be large or obtrusive but it would solve any worries you might have about stability.
The fish might feel better about it too!

 

[Borek]

Thanks that was the kind of numbers I was curious of. I had an aquarium of approx. 500lbs on a stand 12" deep, 28"h and the tank was 18" tall and when the kids or folks walked by it it wobbled that always worried me. When they played on the floor near it the water really splashed around and I was big time concerned. I eventually drained the tank and moved it.

That's problem with the floor, not with the tank itself.

 

[dan2x38]

It strikes me that the practical solution is to hold the stand against the wall with a bracket. It needn't be large or obtrusive but it would solve any worries you might have about stability.
The fish might feel better about it too!

Thing is this isn't my setup. I would like to have as much data as possible to convince her to secure it. Using a scrap to secure it was my solution but i fear without further proof that there is an inherent risk of it tipping she will do nothing. By doing nothing her young one might be at risk.

So a if you guys could show me the simple solution & formula I am sure that would be enoug. To me it is simple why risk it when a child is involved.

I found a formula online and below is my effort to implement it. Is it correct?

water: 31 gal. * 8.5lbs = 263.5lbs
tank weight = 140 lbs
gravel @ 2.5" (approx) = 97lbs online calculator
decorations = 25lbs
cabient weight (guessing) = 150lbs
Total weight = 527lbs
tank = 48" x 12.5" x 12"

stand is 36" high & 12.5" deep & 48" wide

T=Fl, where T = torque, F = force, l is lever arm
F(mass) * l(mass) = F(applied) * height
F(mass) = 527
l (mass) = 6.25"
height = 36" + 12" = 48"

Gives,
F(applied) = F(mass)*l(mass) / height
F(applied) = 527lbs * 6.25"/48" = 68.6lbs

force to tip the tank 69 lbs on a level surface with gravel level and over centre of gravity

but the water moves as it starts to tip this changes the center to mass calculations?

 

[I like Serena]

force to tip the tank 136.7 lbs

but the water moves as it starts to tip this changes the center to mass calculations?

I can confirm the method and the result of your calculations based on a total weight of 1049.5 lbs.

And yes, once the aquarium starts tipping the center of gravity will shift, and the force required will become less and less, but only after it starts tipping.

There are a couple of other things to consider.

You are assuming that the center of gravity is in the middle.
This is only true if the gravel, ornaments and other stuff are more or less centered.
Otherwise it becomes easier to tip it.

The same holds if the aquarium is rocked before tipped.
Rocking the aquarium will make the center of gravity slide back and forth, making it easier to tip.

You're also assuming the aquarium won't slide.
Is the bottom anchored somehow?
Or are you just considering a possible worst case scenario?EDIT: I just tried to check your numbers for the volume and the weight, and they seem to be off.
I have to admit that I'm not really used to gallons and pounds (being from Europe), so I'm converting as I go.
It seems to me that a volume of 48" x 12.5" x 12" corresponds to 31.1 liquid US gallons.
And if I fill it to 18", I get 46.7 gallons.
How did you get 75 gallons?

EDIT2: That brings me to another inconsistency. At first you said the aquarium was 18" high and filled to 1.5" from the top.
In your calculation you made it 12" high.
What is it?

 

[dan2x38]

EDIT: I just tried to check your numbers for the volume and the weight, and they seem to be off.
I have to admit that I'm not really used to gallons and pounds (being from Europe), so I'm converting as I go.
It seems to me that a volume of 48" x 12.5" x 12" corresponds to 31.1 liquid US gallons.
And if I fill it to 18", I get 46.7 gallons.
How did you get 75 gallons?

EDIT2: That brings me to another inconsistency. At first you said the aquarium was 18" high and filled to 1.5" from the top.
In your calculation you made it 12" high.
What is it?

Sorry about that I was thinking of my tank it is 75 gal. with dia. 48 x 18 x 18. I think I corrected the formula and results. Does it look correct now? I was at work when I calculated and hurrying.

 

[I like Serena]

Sorry about that I was thinking of my tank it is 75 gal. with dia. 48 x 18 x 18. I think I corrected the formula and results. Does it look correct now? I was at work when I calculated and hurrying.

Assuming your measurements and guesses are correct, the total weight still seems to be off.
I get 677 lbs (instead of your 527 lbs), resulting in a 88 lbs tipping weight.
Did you leave out the cabinet weight?

 

[DaveC426913]

The tank is 48x18x18. It's much easier to work out in cm/litres.
30cm per foot.

So the tank is 120x45x45
= 243,000 cm^3
= 243 L
= 243 kg
162x2.2 = 535lbs.
Add 75 lbs for tank and decorations and you're around 620lbs.

Most furniture is not designed to hold 620lbs. Don't know what the legs are like but if they're spindly, if they don't actually punch through the floor they may well well dent it enough to tip the tank. You will not need 88lbs of force to tip the tank; it may collapse under much less force.

Forget trying to convince her with numbers. This is a bad idea. You just don't mess with 620lbs of water.

 

[dan2x38]

Assuming your measurements and guesses are correct, the total weight still seems to be off.
I get 677 lbs (instead of your 527 lbs), resulting in a 88 lbs tipping weight.
Did you leave out the cabinet weight?

That's why they pay you the big bucks your right I was wrong once again.

What sort of force could a 2 year old boy say 30 lbs exert pulling themselves up on the stand? How would you estimate that force? Lost there even more than above... LOL

 

[DaveC426913]

That's why they pay you the big bucks your right I was wrong once again.

What sort of force could a 2 year old boy say 30 lbs exert pulling themselves up on the stand? How would you estimate that force? Lost there even more than above... LOL

I still say this is the wrong way to go about it. If she's got a 2-year-old and is thinking about putting a 600lb. aquarium on top of a dresser, and she sees no problem with that, well, no force calculations are going to convince her otherwise.

Maybe a call to CAS would smack some sense into her...

 

[I like Serena]

I still say this is the wrong way to go about it. If she's got a 2-year-old and is thinking about putting a 600lb. aquarium on top of a dresser, and she sees no problem with that, well, no force calculations are going to convince her otherwise.

Maybe a call to CAS would smack some sense into her...

I can't judge the sanity or insanity of the OP and I won't.
I can see that dan2x38 asked an honest question about physics.
That is something I can help with.
What the OP does with it is up to the OP.

 

[dan2x38]

I still say this is the wrong way to go about it. If she's got a 2-year-old and is thinking about putting a 600lb. aquarium on top of a dresser, and she sees no problem with that, well, no force calculations are going to convince her otherwise.

Maybe a call to CAS would smack some sense into her...

Thanks I posted my thoughts and calculations with all your corrections - I should say your calculations. ;-) Again thank you and I hope in some way my ramblings will wake her up.

Cheers and thanks again seriously!

Any idea how to calculate the child's potential exerted force pulling themselves up if they weighted say 30lbs. I have no idea how to consider that one?

 

[I like Serena]

Any idea how to calculate the child's potential exerted force pulling themselves up if they weighted say 30lbs. I have no idea how to consider that one?

I find it difficult to say what a 30 lbs child might be able to achieve.

It does seem unlikely to me that the child could for instance exert double her weight.

Another possibility is the child hanging full weight on the side of the aquarium, generating a moment of 30 lbs times the maximum distance its center of gravity would be to the center of gravity of the aquarium.
A quick calculation shows that it would not be enough to tip the aquarium (not without tools).

What happens if the child climbs on the stand and starts rocking everything remains conjecture.

 

[DickL]

620 lbs on 6 sq ft, over 100 lbs/sq ft. That is getting close to (if not over) the standard floor loading for houses. And that is for a distributed load. This will more likely be 4 nearly point loads. That floor is likely to cause problems, either adding to the instability or as noted by Dave resulting in localized failures.

 

[dan2x38]

620 lbs on 6 sq ft, over 100 lbs/sq ft. That is getting close to (if not over) the standard floor loading for houses. And that is for a distributed load. This will more likely be 4 nearly point loads. That floor is likely to cause problems, either adding to the instability or as noted by Dave resulting in localized failures.

LOL I've had as many as 16 aquariums at one time. At this time I only have one a marine tank. Now this is off topic but hey I am the OP so jacking my own thread... LOL Any ways my marine tank is 75g. full to within just over an in. from the top. It has approx. 90 of live rock plus 50 or 60 lbs of substrate. There is the stand weight too. Inside the stand is a sump which is 2/3 full of water and on end has 4" of very fine sand approx. 20lbs I guess. There is also a lot of plumbing it must weigh another 15lbs or so. Near the tank approx. 16" is a Rubbermaid bin roughly 8 gallons that holds water - it is always between full and 1/2. Right beside the tank is a tower (protein skimmer) approx. 24" tall 3/4 full of water and it is 3" diameter and a heavy water pump attached to it.

Now calculate that for floor load and stress. ;-)

 

[DaveC426913]

It has approx. 90 of live rock plus 50 or 60 lbs of substrate.

People (not you!) often forget that weight of interior stuff is not a straight addition to the total weight. Interior stuff displaces water.

So, if a rock weighs 10kg. but displaces 2kg of water, then you only add 8kg. Likewise gravel.

Worse, if a 10kg chunk of wood has only a small negative bouyancy, it displaces almost its own weight of water, so adds a negligible amount to the total weight...



Never had the guts to start a marine aquarium.

 

[dan2x38]

People (not you!) often forget that weight of interior stuff is not a straight addition to the total weight. Interior stuff displaces water.

So, if a rock weighs 10kg. but displaces 2kg of water, then you only add 8kg. Likewise gravel.

Worse, if a 10kg chunk of wood has only a small negative bouyancy, it displaces almost its own weight of water, so adds a negligible amount to the total weight...



Never had the guts to start a marine aquarium.

Yeah I realize that the rock displaces water so I have less water. It is actually something we must watch out for... this is why a sump is really handy adds water volume to the over all setup. Either way it is well over 1/2 a ton and my floor supports it... or so far... :-S

 

[AlephZero]

620 lbs on 6 sq ft, over 100 lbs/sq ft. That is getting close to (if not over) the standard floor loading for houses. And that is for a distributed load. This will more likely be 4 nearly point loads. That floor is likely to cause problems, either adding to the instability or as noted by Dave resulting in localized failures.

There seems to be something wrong with those numbers. A 200-lb man would put more than 100lbs/sq foot of loading on the floor when walking, when all of his weight was on one leg.

If he was sitting on a 4-legged chair with legs say 1 inch square, the local load from each chair leg would be more like 7000 lbs/sq ft.

I agree with DaveC that there may be issues with a "non-standard" stand collapsing sideways if it is not properly designed to carry the weight. Ask the common-sense question, would it be safe for say 5 adults to stand on top of it. If the answer is no, it isn't safe to put 900 lbs (or even 560 lbs) of aquarium on top of it.

That said, a 30 lb toddler will probably not be able to apply a sustained force of more than 30lb, unless he/she can brace himself against a wall or something while pushing, because the coefficient of friction between the toddler and the floor is unlikely to be more than about 1.0 and his/her feet will slip. And when calculating the lever arms, remember the relevant "height" is the distance from the floor where the toddler is pushing, not the height of the tank.

 

[dan2x38]

There seems to be something wrong with those numbers. A 200-lb man would put more than 100lbs/sq foot of loading on the floor when walking, when all of his weight was on one leg.

If he was sitting on a 4-legged chair with legs say 1 inch square, the local load from each chair leg would be more like 7000 lbs/sq ft.

Yeah we've had this discussion before many times. I say if 5 men stood looking at a picture during a party they would exert way more force on the floor then an aquarium on a stand with a flat bottom setup level ever would.

 

[DaveC426913]

Yeah we've had this discussion before many times. I say if 5 men stood looking at a picture during a party they would exert way more force on the floor then an aquarium on a stand with a flat bottom setup level ever would.

I would not be worried about the pressure per square foot; I'd be worried about the pressure per square inch.

i.e. I don't think the floor would collapse under the weight, but the legs could dent the floor material. Considering how overbalanced the tank setup is, it would not have to dent it a lot to start a runaway effect.