How to determine minimum weight to tip over a single pedestal dining table?
We have a client who wants a rectangular table, 72" L x 42" W, and the MDF top will weigh approximately 100 lbs and the hollow cast bronze base measures 18.5" diameter at the base with a slight taper moving up to 14.5" where it meets the top. The base will weigh approximately 150 lbs but we can add additional weight at the bottom to stabilize it if necessary. The question is how much weight would need to be exerted on one of the 72" ends to make it tip over? (If someone sits on the table will it fall over?)
Best Answer
This is the diagram you need:
The table will tip if the net torque is clockwise. At the moment of tipping, $F_2 + mg + F_1 = 0$, and the net torque is zero:
$$F_2 d_1 + F_1 d_2 = 0$$
Putting $F_2 = -(mg + F_1)$ and substituting, we get
$$-(mg + F_1) d_1 + F_1 d_2 = 0\\ F_1 = \frac{mg d_1}{d_2-d_1}$$
This shows that when $d_1 = d_2$, you can support any load; as the overhang becomes bigger, you need a heavier and heavier support.
In your case, you have a 18.5" diameter pedestal and a 72" length table; the overhang is $\frac12(72 - 18.5) = 26.75"$, so the "force magnifier" $\frac{d_1}{d_2-d_1} = 2.9$. Your table (top plus base) needs to be 3x heavier than the weight you want to be able to support.
Note - the height doesn't really come into it if the forces are properly vertical; also note that from a safety perspective you will need a considerable safety factor (in other words, while you may be designing "for a 175 pound person", when the 250 pound person stumbles into the table while piggy-backing someone else, the table will still tip over).
The above is purely "the physics of tipping". The "good engineering design" question is something that is off topic for this forum (but I would start by thinking hard about that overhang...)