Center of mass and tipping force

[JoeMoss]

Alright, I have searched all over and I can't find the answer to what I thought was a simple problem.

I have designed a table in SW and have calculated the CoM, now I want to know what force is required to tip it over. I am disregarding friction at this point and assuming the table will not slide.

Any thoughts?

Along the same lines, is there a rule of thumb for the location of the CoM versus the width of the base? If the table is 72" tall and the base is 34x84, CoM is 48" from the ground (on center in two other directions). The table is symetrical across is depth and width. Is there an easy way to determine if your base is wide enough?

thanks!
Joe

 

[Theophilus]

Perhaps you could try this out in Cosmos or Simulation?

Otherwise, I'm not sure how to do the calculation, and you may want to ask it in the mechanical engineering forum.

Jeff Mowry

http://www.industrialdesignhaus.com

Reason trumps all. And awe trumps reason.

 

[smcadman]

I can't help you with your problem, but it would be interesting to see a family sitting on ladders eating dinner on a 72" tall table!

Flores
SW06 SP4.1

 

[JoeMoss]

Flores, that makes me laugh! Actually its more of an assembly easel, I try to simplify my examples but in the end its easier to just give the details!
Joe

 

[handleman]

You can consider the table weight as a point force acting straight down at the CG. The distance from the closest leg in the tipping direction is the moment arm. In your case, it looks like your moment arm is 17" in the least stable direction. So to tip the table, you have to create a moment about the same location that is equal to or greater than the moment caused by gravity.

Say your table weighs 100lbf. Your stabilizing moment for the least stable direction is 1700 in-lbf. To overcome that with a horizontal push at the top you need to push by 1700in-lbf/72in=23.6lbf.

 

[Heckler]

What is the weight of the table? Basic statics tells us that that weight can be moved to the Cg. Then draw a Free Body Diagram (FBD) with the reactions at the feet. Then solve the forces and moments.

Best Regards,

Heckler
Sr. Mechanical Engineer
SW2005 SP 5.0 & Pro/E 2001
Dell Precision 370
P4 3.6 GHz, 1GB RAM
XP Pro SP2.0
NVIDIA Quadro FX 1400
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_`\(,_
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Never argue with an idiot. They'll bring you down to their level and beat you with experience every time.

 

[JoeMoss]

Table weighs about 250lbs.
Joe

 

[nmaxwell]

If I’m not mistaken,

l = distance between fulcrum and center of mass
a = angle between horizontal and the line between CM and fulcrum
m = mass
g = grav. constant
t = angle that table makes with floor, has to go from 0 to a, in order for table to fall over.

Then evaluate the integral of m*g*cos(t)*l, (this should be the torque gravity applies to the CM at any given angle, t) with respect to t, from 0 to a

Think you get sin(a)*m*g*l

That ought to give you units in torque*angle, which you could use to compare two table dimensions, and get a good idea of how much more stable one is than the other, might not be too useful as an absolute scale. So if you have a real, tested table that is stable enough, work out that integral for it is, and use it as a minimum for your design. I think that’s valid at least.

 

[nmaxwell]

Sorry, meant to say,
a = angle table makes when CM is directly above fulcrum, so its 90 minus angle between the vector from fulcrum to CM, and horizontal of table)

 

[Heckler]

I don't have the time to work this out completely but a quick look at the geometry would tell me that this would tip easer along the short side of the base (34 inch). I would say that if you could make the base more symmetical say 60"x60". This would help with stability.

Best Regards,

Heckler
Sr. Mechanical Engineer
SW2005 SP 5.0 & Pro/E 2001
Dell Precision 370
P4 3.6 GHz, 1GB RAM
XP Pro SP2.0
NVIDIA Quadro FX 1400
o
_`\(,_
(_)/ (_)

Never argue with an idiot. They'll bring you down to their level and beat you with experience every time.

 

[handleman]

It doesn't matter how high the center of mass is. The only thing that matters is how far it is from one of the sides. There's no integral, cosine, nothing required. You say your table is symmetric across depth and width and weighs 250 lbs. To tip it towards the long side (easiest tip-over), you have to push horizontally at the top with a force of 250x(34/2)/72=59lbs. To tip it towards the short side (harder tip-over) you have to push horizontally at the top with a force of 250x(84/2)/72=146lbs. It really is that simple. The height of the CG only comes into play if you're trying to determine whether it will tip over or slide, but you've already assumed that it will tip rather than slide.

 

[CorBlimeyLimey]

The height of the CoG will affect the angle you have to tilt the table before it will fall. The higher the CoG, the less tilt angle required..


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[nmaxwell]

So you recon a table with CM 20 meters above the ground will be as stable as a table with CM 1 meter off the floor? Assuming same basses.

Confounded,
Nick

 

[handleman]

Stability and amount of force required to tip are two different questions. The original question was how much force is required, not how far it had to be tipped. Stability would have more to do with the amound of work required to tilt the table to a state of static, unstable equilibrium (center of mass directly over the point in contact with the ground). A 20 meter tall table will have a 20 meter lever arm to generate the required tipping torque. The force required to tip a 20 meter tall table would be 20x the force for a 1m tall table. Now, let's consider a hypothetical 1m cube with its center of mass in the bottom face - weight of 1N. It would take 0.5N-m to tip, but in order to reach unstable equilibrium you have to lift this mass to a height of 0.5m. Now move the CM to the cube center. Still takes 0.5N-m to tip, but now the CG only has to be lifted from its height of 0.5m to a height of 0.5 times sqrt 2, or a lift distance of 0.07m or so. The work required is much less, so a cube with CG in its center is much less stable than with its CG lower, even though both require the same tipping torque (and force) to start the tip.

 

[JoeMoss]

Man, I knew I asked the right group of guys! The fact that I may have mis-stated my question, is another issue. In reality, I am concerned with Stability, as handleman pointed out.

I will sort through this in the morning and see how it affects my design.

thanks!
joe

 

[nmaxwell]

Sorry, this is going to be lengthy, skip the first and second paragraphs if you haven’t the patients.

Yes, the amount of work required to bring your table to the point of tipping is, or at least I agree that it is, a good measure of how stable your table is. However considering how much force it will take to make the table move, round its fulcrum, is also valid, after all, you need to know what force you’re applying at any given moment in order to find work. For the time being, let’s simplify, and consider torque around the fulcrum rather than force applied to the side, as that’s another matter. This torque is just the torque applied by gravity at any given angle of your table. It can be expressed as cos(t+u)*m*g*l, using variables from before, and u being the angle between the leaver arm made from fulcrum to CM, and the bottom horizontal of the table (I forgot to add this in my equation before, thus, a, is 90-u).This is finding the component of gravity, acting perpendicular to what amounts to a leaver arm, from the fulcrum, to the center of mass, and multiplying it by the length of that leaver arm, l.

Now it’s not really meaningful to consider the torque alone, however it is somewhat meaning full to considered the amount of force required to get the table moving, this is simply the force you need to apply to overcome the torque applied by gravity. First you have to consider how you’re going to apply this force, F. I’m going to make the assumption that we’re considering the force to be purely horizontal, applied to some point on the side opposite to the fulcrum, some point that is, say, d, distance off the bottom of the table. Now because were applying a force to overcome a torque, we work out what torque that force becomes, so we have a second, as it were leaver arm, going from the fulcrum to the point at which were applying our force, so we have an angle that leaver makes with the horizontal, call it t2, and a length of that leaver, call it l2. t2 will be arctan(d/w), where w is the width of the table. L2 will be ((w)^2*(d)^2)^0.5. The force being applied perpendicular to the leaver, should be F*sin(t2). So the torque applied around the fulcrum by our horizontal force, F, acting on the side, w width form fulcrum, at distance, d, from the bottom, is F*sin(arctan(d/w))* ((w)^2*(d)^2)^0.5, which in order to move the table, has to be greater than the torque applied by gravity, cos(t+u)*m*g*l.

This is overly complicated; it’s not practical to evaluate stability based on this. However we can pick up on a few things, first, the force and torque required to overcome gravity does indeed rely on the height of the CM, more generally the position of CM relative to the fulcrum, as this is the only variable (assuming constant mass and g) which changes how much torque gravity applies, the torque we have to overcome. Now we can make the assumption that for any large variance in CM, the dimensions of your table, more importantly, w and d, change somewhat respectively, which is where I think handleman is coming from, that, practically speaking, for whatever height of your table, the CM will be roughly in about the same relative position. But we need to be careful with that kind of assumption, as it won’t always hold true, I’d say it basically never holds true, but that all depends on how much you want to reduce theory to rules of thumb. But the general point I’m trying to make here is that it’s impractical to consider this horizontal force, too much math, and too many assumptions. All we really have to consider is the torque gravity is applying on the CM, cos(t+u)*m*g*l. We can use this torque alone, and get an idea of how much more it takes to get one table going, than another, all relatively speaking. Or we can consider the about of work required to tip it over fully, which is that torque, cos(t+u)*m*g*l, integrated, with respect to t, from 0 to a, where, again, a is the angle t has to be in order for the CM to be above the fulcrum, the point at which the table will tip over. This integration gives us angle*torque, basically work, the amount of work we have to do throughout the range of 0 to a, to completely overcome gravity. The integral becomes sin(a+u)*m*g*l- sin(u)*m*g*l, again, I forgot to include u before. However, personally I would just go with evaluating cos(t+u)*m*g*l for some table of minimum stability, and using that as a minimum for your other tables, this is really easy, if you already know your CM, all you have to do is work out what angle it makes with the horizontal, or u, and how far it is from the fulcrum, l, and the mass, which I assume you know. Personally if I’m sitting at a table I’d rather the table not move at all, rather than just not tip far enough to fall completely over, it’s a bit more practical. So if you have some table that doesn’t move at all, using cos(t+u)*m*g*l for it, as a minimum, means that all tables following that, wont tip either.

Longwindedly yours,
Nick

 

[handleman]

Nick,

You're still doing a lot of unneccesary trigonometry. You're finding the angle and hypotenuse of the vector from fulcrum to CM and then using cosine to find the adjacent side length, which is the lever arm for the vertically acting force of gravity. Put the numbers given by Mr. Moss into your equations and you'll come up with the same numbers I did for a horizontal force applied to the top corner of the table. The tip force will always be maximum when the table level. Granted, your formula gives the tipping torque for any given tip angle, allowing you to integrate and get work required to actually tip the table over. However, there's no need for an integral to find the work required to tip. All you have to do is find the change in height of the CG from zero tip to unstable static equilibrium and multiply by mass and gravity to find the change in potential energy. Again using Mr. Moss's numbers, the distance from fulcrum to CG in the easiest tipping direction is sqrt(17^2 + 48^2), or approximately 50.9 inches. That gives a height difference of 48-50.9=2.9 inches, or 0.243ft. So we have about 60 ft-lbs potential energy difference between stable equilibrium (no tip) and unstable equilibrium, and we found it with no more than 8th grade math. I'm curious to see what numbers come from your integral, but I'm way too rusty on my calculus to attempt it.

 

[handleman]

Look at this formula, taken from the end of paragraph 2:

F*sin(arctan(d/w))* ((w)^2*(d)^2)^0.5

Guess what sin(arctan(d/w))* ((w)^2*(d)^2)^0.5 evaluates to.... d!

Think about what this is. It's a really complicated way to write F*d!

 

[nmaxwell]

Damn!

Yeah you're right, was thinking when I wrote that, that equation looked like it ought to reduce to something, that’s what I get for laziness. Actually if you want to get really technical, it evaluates to d*sign(w), just to be pedantic. And on the same note, l is just sqrt((1/2*w)^2+h^2), if h is the height of the CM, in that case, cos(t+u)*m*g*l reduces to 0.5*w*g*m, when t = 0. So the torque applied by gravity when the table is horizontal does indeed not depend on the height of the CM, interesting, its hard to wrap ones head around that.. Well at least now it’s been proven beyond doubt. Thanks for the enlightenment; it’s always great to see the math work out perfectly like that. I’ll mess around with that integral over the weekend, basic physics would tell us that the work will come out to the change in potential energy, I certainly don’t doubt it at least, but integrating torques is far slicker.

cheers,
nick