Three Legged Table

The weight supported by leg B is zero .

If there were any upward force by the floor on leg B, it would create a torque around axis AC. Thus the table would topple. It can only remain in (unstable) equilibrium if there is no such force.

Here I have only considered a symmetric table top supported by the legs. If you consider the table including (massive) legs, then leg B supports precisely its own weight. The presence of the legs also makes the table more stable: it shifts the center of mass downward and toward leg B.

Let $a\sqrt 2, M$ be the side and mass of the table top; let $\ell, m$ be the height and mass of each leg, in the corner of the table. Call the center of the table top $(0,0,0)$ and let the legs run in the $+z$ -direction. Leg B is attached at the tabletop at point $(a,0,0)$ ; legs A and C, at points $(0,\pm a,0)$ . The center of mass has coordinates $x = \frac{ma}{M + 3m},\ \ \ y = 0,\ \ \ z = \frac{m\ell}{M + 3m}.$ Because of symmetry, we only need to consider torques about the $y$ -axis. The torque of the gravitational force about the origin is $\tau_g = +(M + 3m)gx = +mga;$ this is no surprise, as the lines of the gravitational forces on the tabeltop, leg A, and leg C all run through the $y$ -axis.

Let $F_B$ and $F_{AC}$ be the vertical forces by the ground on the legs, then $\tau_B = -F_Ba;\ \ \ \ \tau_A = \tau_C = 0.$ Equilibrium requires that $\tau_g + \tau_A + \tau_B + \tau_C = 0$ , so that $mga -F_Ba = 0;\ \ \ F_B = \frac{mga}{a} = mg.$ This proves that the upward force by the ground on leg B only carries the weight of leg B. Each of the other two legs also carries its own weight, but also half of the weight of the tabletop. This follows from the fact that $F_A + F_B + F_C = (M + 3m)g$ , so that $F_A = F_C = \frac{(M + 3m)g - mg}2 = \left(\frac M 2 + m\right)g.$

Actually, I think

The weight on leg B = the weight of leg B itself.

Why? Well, imagine if only the table itself had mass, and all its legs were weightless... Then the table's center of mass would be exactly in the middle of the square, which means the table would 'pivot' about that point, and all its mass would be equally distributed onto legs A and C.

This means that as long as the table is parallel to the ground (or tipping toward the missing leg), none of the table's mass will be resting on leg B.

Bonus: Now imagine chopping half of the length off of leg B. If you are extremely skilled, you could in theory balance the table solely on legs A and C if you were able to keep the table exactly parallel to the floor. But as soon as it starts tipping to one side, it will fall. If it fell to the side with the halved leg B, it would land on B and now finally some of the table's weight would be resting on leg B.

There are three unknown upward (z-direction) forces (A,B,C). We can write three balance equations (the torque vector inherently has no z-component):

1) Force in z-direction
2) x-component of torque vector (about the table top center)
3) y-component of torque vector (about the table top center)

These equations turn out to be:

$A + B + C = W \\ A + B - C = 0 \\ -A + B + C = 0$

Solving these results in:

$A = \frac{W}{2} \\ B = 0 \\ C = \frac{W}{2}$

This result holds no matter how long the table legs are (assuming that they are massless).

Common sense should give you the right answer: Imagine trying to lift up on the B corner. Even if the table is very heavy, it would lift easily and tip over with almost no effort. Conversely, it would obviously take much more effort to lift up on leg A or C. Other solutions posted here give the correct analytical reason why this is true.

Let the weight of the table is 40 N. When there were all the 4 legs, the whole weight was supported by all the four legs equally means 10N per leg and it was in angular equilibrium as well. As soon as the 4th leg is removed, the whole weight would lie upon the two diagonally situated legs i.e. A and C because there is no counter support to the leg B which would balance the torque created by the leg B. So now, each of the leg A and C would support 20 N weight. So better remove the leg B as well if you are removing the leg D. Would save you the extra material. :-p

Think of the B leg as "hanging off". the *only * force used is that for the leg itself, anything more would be virtually non-existent.

If foot B is removed, the table balances the rest of the legs.

The mass of table is being supported only between the legs a and c. Leg b only supports a small part of table

Well as it's mentioned that the table is well built so,

First remove the leg B also, theoretically the table shouldn't fall.

Now add leg B so now there is small torque due to that and also the contact force from the ground acting on the leg B so overall the table shouldn't fall {as normal force is self adjusting force}---(1)

Let's see some math

Mg∆=N∆ {∆=length of diagonal/2} (as (1) is true)

So Mg=N

Also the mass of leg is much less than the table top so we get our answer

If the table was split in half diagonally such that it was triangular with a leg at each corner, the weight distribution would be even (or very close). However, by adding this other half, completing the square, a moment is created relieving force off leg B thus meaning that the weight on B would be less than A

Reuben - age 14

When the leg diagonally opposite B is removed the torque due to the weight of that edge of the table would cancel the torque acting at B (due to that edge of the table) - all the torques are about the AC axis- and so the leg B wouldn't be supporting any weight.