# How much force to tip over an umbrella stand?

Summary:
Well, not really an umbrella stand, but same principle!
I'm working on a project where I have an upright post secured to a flat square base (similar to an umbrella stand). I want to know how much force would be required to start it to tip toward its left edge. Here's an illustration where the base is 10" square, 1" thick and weighs 50 lbs. There is an upright pole of insignificant weight anchored to the base assuming the right edge is centered exactly 5" from the edge of the base.

I want to know the formula to calculate how much force would be needed to move (lift) the base if it's applied from various elevations along the upright pole (for example 30" above the base).

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berkeman
Mentor
Welcome to PF.

Are you familiar with the concept of moments (torques)? What is your math background so far? We are happy to help, but knowing a little more about your background will help us tailor our responses in the discussion. Thanks.

sysprog
Sorry for the lack of profile info! I'm only a basic, algebra, trig guy. No theoretical physics training. I was hoping for something like:
(basewidth * baseheight) / baseweight) x sin( pole height squared) = TADA! lbs of force.

I know I'm in this forum way above my paygrade!

sophiecentaur
Can you show us what you think are the acting torque forces around the tipping point?

berkeman
Mentor
No worries at all; like I said, we are happy to help. You said that the vertical pole part has negligible mass compared to the base, right?

Also, will the force applied to the pole always be straight horizontal, or will it angle over as the vertical pole angles over so that the force is always at a right angle to the pole?

And I'm assuming that the friction between the edge of the base and the ground is high enough to keep the base from slipping along the ground at any point?

If the pole mass is negligible and the force on the pole is always at a right angle to the pole, then the maximum force to lift the base to start tilting it is the main force that you need to find. That force will depend on the distance up the pole (from the center of mass of the base) that the force is applied.

Precisely! Your assumptions are all correct. Friction coefficient is very high, force applied to pole is horizontal. Here's the project: I have 50lb slabs of granite that I attached a poles to in order to make a railing alongside a walkway. The railing will be mostly for visual control and not to support a lot of force. I can make it any height I want but I want to get an idea if it could be used for crowd control at a rock concert or if it will blow over in the wind, which could influence how high I can make it.

What precision are you looking for?

I just need a rough number. Like less than a pound, 5lbs, 10lbs, etc. If a person grabs the top of the rail are they going to think "this is going to fall over now", or "this at least gives me a bit of guidance so I won't fall over", or "I can count on this to keep me from falling". I realize there are tons of considerations to make this a precise number, just looking for a general guideline on how to measure that initial force based upon the length and weight of the base, and the length of the lever.

berkeman
Mentor
Your assumptions are all correct. Friction coefficient is very high, force applied to pole is horizontal.
No, the simplest assumption is that the force stays at a right angle to the pole as the base lifts and tips over.

Here's the project: I have 50lb slabs of granite that I attached a poles to in order to make a railing alongside a walkway. The railing will be mostly for visual control and not to support a lot of force. I can make it any height I want but I want to get an idea if it could be used for crowd control at a rock concert
So I think what you are really asking is how much force will it take to *start* tipping over the railing (because once started tipping, it will keep tipping if the force is perpendicular to the post, which is what it will be if somebody is leaning into the fence and it starts tipping over).

Can you show the dimensions of the slab? That will determine the Center of Gravity of the slab and the moment arm of tipping it with respect to the ground. It's a simple equation after that.

Yes, I should have made it more clear that I was looking for the INITIAL force to lift the slab. The dimensions of the slab are 10"x10"x1" as shown on the image. But I don't understand why the 3rd dimension of the slab (depth) is needed. As long as we know it's 10" from the pivot point and 1" thick with a given weight, why would it matter if it were 10" deep or a mile deep? (I also understand that the 1" height would need to be considered but only has effect after initial motion has taken place, no?)

Baluncore
But I don't understand why the 3rd dimension of the slab (depth) is needed.
Don't say you think something is not required, answer the question. If you understood the problem you would not have asked the original question. You need to find the puzzle piece you are missing.

You must specify the length of the post, the force you will apply, or understand torque.

The post is a lever arm. The overturning is hard at first but then becomes easier as the edge begins to lift.

The slab thickness is only important because it adds to the post height.

If the post mass really was negligible, the analysis would not need to know the point of attachment of the post to the base, only that it remains perpendicular. You need the total weight of the post and slab.

sysprog
berkeman
Mentor
The dimensions of the slab are 10"x10"x1" as shown on the image. But I don't understand why the 3rd dimension of the slab (depth) is needed.
Oops, sorry. I have a short memory.

The depth makes a difference because it raises the COM (center of mass) above the ground by some amount. That changes the angle of the COM with respect to that lower left contact point with the ground, and changes the moment arm length of the force on the pole.

But let's just do an approximate initial calculation to get you going. Let's ignore the 1" thickness and use the 5" horizontal distance between the pole and the fulcrum point at the lower left of the base, and let's just use the vertical height of the force point on the pole with respect to the ground.

Torques are the product of the force multiplied by the lever arm length (assuming that the two are orthogonal (at a right angle), just like when you use a torque wrench. So the torque required to start lifting the 10" slab that weights 50 pounds is approximately (50 pounds of force with a 5" lever arm):
$$\tau = 50 * \frac{5}{12} = 20.8 ft-lbs$$
Then the torque that you apply with your horizontal force some distance up the pole is that force multiplied by the vertical distance from the base. To see what force it takes 3' up the pole to generate that same torque, you would do:
$$\tau = 20.8 ft-lbs = F * 3$$
$$F = \frac{20.8}{3} = 6.9 lbs$$
Which is pretty small. That setup won't be very good for crowd control, IMO. Take a look at the weight and foot-span of standard crowd control barriers to get an idea of what you will need...

Baluncore
For safety reasons you also need to consider how the posts are joined to each other, and possibly the distance between posts. If a rope is used, then pressing down on the rope can pull posts together, pinching and trapping a body between the top of two falling posts. The posts need safe upper ends.

sysprog and berkeman
Hmm. Very interesting and informative. 6.9lbs seems like a small amount of pressure to start it tipping. But I did the same calculation on the commercial grade unit you show above. They're touted as 34lbs and from the man in the picture I would estimate 3.5' high and 1' leg on each side. With these numbers your formula comes up with 4.85lbs of force to start them tipping! Did I do something wrong?

Maybe in addition to heeding the advice of @berkeman and the caveats of @Baluncore, you might do well to recognize that you are approaching the regions of parametric equations ##-## to assist you with the specific calculations, perhaps this might be useful to you: https://www.geogebra.org/3d?lang=en

berkeman
Mentor
Hmm. Very interesting and informative. 6.9lbs seems like a small amount of pressure to start it tipping. But I did the same calculation on the commercial grade unit you show above. They're touted as 34lbs and from the man in the picture I would estimate 3.5' high and 1' leg on each side. With these numbers your formula comes up with 4.85lbs of force to start them tipping! Did I do something wrong?
The example that I quickly picked with a Google Images search does appear to be a lighter one:
These economy barricades have the same dimensions as our Heavy Duty model but at 34lbs are light enough for one person to carry easily
So it would not take too much force to start it tipping. The regular ones that I'm used to would take more, but I agree that stand-alone even they may not take too much force. The big claim to fame for these crowd control systems is the way that the fence modules interlock on each end. You can see the male/female linking system at each end that is used to tie these fence pieces together. That lets them work as a system to provide much higher tilt-over resistance.

Can you think of any way to incorporate that type of tie-together system into what you are thinking about?

sysprog
sophiecentaur
Gold Member
2020 Award
The big claim to fame for these crowd control systems is the way that the fence modules interlock on each end.
And the extra strength you get by zigzagging or curving a wall of them.

I would imagine you could get far better performance with a wider base. 10 inches diameter would (obvs) only have 5 inch radius. As long as the base doesn't trip people up then the diameter should be as great as possible.

There's also an issue with the line / chain between the posts. Only a small lateral force, half way between two posts can produce a higher tension in the chain to tip the two posts, on either side.

I reckon some homework on-line to find what's commercially available would be time well spent.

sysprog
berkeman
Mentor
As long as the base doesn't trip people up then the diameter should be as great as possible.
Yeah, the trip hazard is a real issue. I was working as a Medic at the finish line of an Ironman Triathlon a couple years ago (the finish line was at a high school), and got a radio call for a female spectator who was down. I found her fairly quickly, and it turned out that she had tripped on a leg of the barriers that defined the finish line chute (the barriers looked like the one I posted a picture of in post #13). She stumbled and fell, but was mostly unhurt from that fall. But she was a bit disoriented and distracted and embarrassed by the fall, and when she got up she turned and walked quickly away, right into the school's flagpole. That knocked her silly, and she laid down and waited for help to arrive.

In the end she was mostly okay, and recovered enough that I could send her home with her friends instead of needing a transport to the local hospital. So yeah, you have to consider all aspects of crowd control barriers when designing and deploying them.

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Tom.G, sophiecentaur and sysprog
sophiecentaur
Gold Member
2020 Award
Summary:: Well, not really an umbrella stand, but same principle!

I'm working on a project where I have an upright post secured to a flat square base (similar to an umbrella stand).
We've all piled into this thread (as we do, when it's interesting) as if it's a project that will actually be brought to fruition. We're imagining we are actually producing a number go these things. That's fine if that's the actual context of your post and you will be snowed under with helpful practical comments.
However, it has struck me that it may be a 'thought' project, in which case you may just need the numerical solution to one particular design like the one in the OP. If that's what your question is really about then you just need the basic maths associated with levers and the DIY / crowd control aspect can be ignored.

is your question as straightforward as it looks from the OP?

berkeman
I'm learning so much in this forum! It's rewarding to find so many people still with science on the brain instead of politics.
Anyway, sophiecentaur is right: this is pretty straightforward issue. Here's exactly what I'm looking to do. I want to build a small handrail for my pool steps without having to dig into the decking. So I'm picturing a portable solution with 2 granite bases and a white 1.5" PVC pipe bent to shape and slid over short upright steel tubes which are attached to the slabs (as in these pictures). So the height of this structure is about 30" and it's just for a bit of stability (mostly mental), and not subject to extreme forces. I just want to get an idea of how stable this will be. I should have been more clear in my original post but I do appreciate the tangents of creativity that resulted.

Yes, but I was kind of hoping to make the bases a bit less obtrusive and more aesthetic.

sophiecentaur
berkeman
Mentor
Maybe see if you can find them in white. I saw a couple of different colors in my Google Images search. I agree that black is not an attractive color for a pool accessory.

sophiecentaur
The base I found in post #21 is 15" square and 55lbs. So how much force applied horizontally from a height of 30" would be needed to START this base tipping?