
#1
Feb713, 09:41 AM

P: 9

I'm in the process of building a reef aquarium. It will have a movable sump which will slide out from under the tank. I would like to know if this will exceed the tipping force and cause the tank to fall over.
Components:
The sump when filled to the brim holds 20g @ 172lbs and sits 7" above the floor inside the cabinet. It extends two thirds of it's length from the cabinet. Below is a picture of the assembly. I'd like to know if the sump when filled will cause the tank to tip when extracted from the cabinet below. The sump will never be filled to capacity, but this would be the most extreme scenario and I should account for it. Any and all help is greatly appreciated. Note: the sump sits atop a sliding TV base rated for 225#. The tank will also contain sand and rock structure, but were not included as the bare tank with water would be the minimum load for the calculation. 



#2
Feb713, 12:34 PM

P: 9

Im wondering, did I post this to the wrong forum?




#3
Feb713, 12:35 PM

P: 2,861

I'm not sure there is quite enough information regarding the dimensions (eg which is the width vs length).
The way to approach the problem is to think of it as a teeter totter/seesaw with the pivot where the front wheels are (or front edge if no wheels). Then you work out the torques acting about that point. For example see this diagram which shows the set up from the side.. Mm = Mass of Main tank Ms = Mass of sump dm = distance from center of gravity of main tank to pivot ds = distance from center of gravity of sump to pivot. For most shape tanks the center of gravity will be in the middle of the tank (eg mid way front to back). Then if we define clockwise as positive the torque equation looks like this: Net torque = (Mm * g * dm)  (Ms * g * ds) If the result is +ve then the unit will try to rotate clockwise about the pivot (eg it's stable). If the result is negative the unit will try to rotate anticlockwise and will fall to the left (eg forwards). I've only shown the maths for the main tank and sump tanks but you can factor in the other items the same way. For example if the mass of the salt is Msalt and it is distance dsalt to the right of the pivot then the equation becomes.. Net torque = (Mm * g * dm) + (Msalt * g * dsalt)  (Ms * g * ds) Note that the height of items isn't a factor, or at least it isn't until the unit starts to tip over. If tipped over a few degrees that would move the center of gravity of the tanks and affect the calculation. I'm assuming that no amount of tipping is allowed! 



#4
Feb713, 12:41 PM

P: 2,861

Help with tipping force
oops sorry I miss read your post. I assumed there was also bag of salt that had to be stored in the unit.




#5
Feb713, 12:50 PM

P: 9

Am I correct in thinking the center of mass for the tank sitting on the stand is different than the center of mass for just the tank? Hope that makes sense. I'm also confused regarding the height of the sump in relation to the full assembly. Does the force required to tip the assembly change as the external mass moves upward on the assembly? In other words, if the sump were at the top, aligned with the top of the tank, would that change the equation? (assume the tank was affixed to the supporting base). I appreciate you help. Thanks. 



#6
Feb713, 12:55 PM

P: 9

Yes no movement or tipping is desired :) 



#7
Feb713, 01:14 PM

P: 2,861

The height of items only comes into play if the unit actually starts to tip. So raising the sump shouldn't make any difference. If the floor was soft carpet or something that might allow it to tip a little and then you might need to do a more detailed calculation involving the heights.
As I see it the numbers work out like this.. The center of gravity for stand, main tank and water is 12" behind the pivot point. The center gravity of the sump tank and it's water is 5" in front of the pivot point (when extended). Net Torque = {(100+385+45) * g * 12"}  {172 * g* 5"} = (6360  860) * g which is clearly +ve so it appears the sump can't tip it over. Looks like you would have to pull the sump out until the center of the sump is about three feet in front to the pivot before it might tip over (6360/172 = 37"). 



#8
Feb713, 01:20 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

This problem was originally solved in the sixteenth century by Sir Isaac the Newt.




#9
Feb713, 01:21 PM

P: 2,861

I just noticed that the weight of the empty sump tank hasn't been factored in. I don't think it will make a difference but do you know the empty weight?




#10
Feb713, 01:24 PM

P: 9

My reasoning behind thinking the height of the sump would matter was in terms of the masses vertical distance above the pivot point acting as a moment arm or lever. The closer to the floor the shorter the lever, the higher, the longer the lever. I guess that doesn't come into play here? Again, than you for your help. I didn't want to move ahead unless I had a better understanding of the physics. I basically want to avoid disaster :) 



#11
Feb713, 01:27 PM

P: 692

I have done some experiments with this sort of thing and found that in practice, theory and practice can differ. I would recommend using a safety factor, partly because the theory contains assumptions that may not be true, but also because the operative might stand on the open tv shelf without thinking.




#12
Feb713, 01:29 PM

P: 9





#13
Feb713, 01:32 PM

P: 9





#14
Feb713, 02:46 PM

P: 9

I see a similar question was posed before. Link That scenario took into account the height of all the components above the fulcrum (where the front legs meet the floor). However they were looking at lateral force exerted on the tank and structure. Does that change anything?




#15
Feb713, 05:28 PM

P: 2,861

Net Torque = {(100+385+45) * g * 12"}  {(172 + 32) * g* 5"} = (6360  1020) * g Still well on the safe side. That main tank is heavy! Lets see what happens if someone (190lbs) pulls out the sump and stands on the TV shelf say 10" from the front pivot.. Main tank  Sump  Person = {(100+385+45) * g * 12"}  {(172 + 32) * g* 5"}  {190 * g * 10") = (6360  1020  1900) * g Still seems ok. What happens when the main tank is empty and the sump tank is full... {(100+45) * g * 12"}  {(172 + 32) * g* 5"} = (1740  1020) * g Even that seems ok. 



#16
Feb713, 05:35 PM

P: 9

Again thank you for taking the time to respond and best of everything to you! 



#17
Feb713, 05:49 PM

P: 2,861

On the right you see the tank tipped over a bit. Now if you raise and lower the tank in the stand the vertical arrow acting through the center of gravity of the tank moves from one side of the pivot to the other. The diagram shows it on the left of the pivot contributing to tipping. If the tank was lowered the vertical arrow would move to the right. What this means is that the higher the tank the less you have to tilt it before it falls...but I think it will be fine. 


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