# Determining forces on a table with four legs and an object placed on a random point

We have an object with weight $$D$$ at a given location on table wit with four legs ($$F_1$$ to $$F_4$$). What is the force applied on each leg? (for simplicity, I'm just using the same labels $$F$$ and $$D$$ for both the location and the force)

$$W$$, $$H$$, $$x$$, $$y$$ and $$D$$ are given. To find the forces on each leg, as far as I remember, I have to consider two general equations: $$\sum F=0$$ and $$\sum M=0$$. So I have:

$$F_1 + F_2 + F_3 + F_4 - D = 0$$

Also, considering the moments round the point $$F_1$$:

$$W(F_2+F_3) - xD = 0$$ $$H(F_3+F_4) - yD = 0$$

But this just give me 3 equations! I missing one more equation and cannot figure it out.

• Hi Shapul. Welcome to Phys.SE. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. Note in particular that the homework tag may apply even if it is not actual homework. Commented May 23, 2013 at 18:37
• Thanks Qmechanic. This is not a homework and I'm not a student. I searched the site and the web and could not find the answer to my question. I admit, this is a probably a very simple question but unfortunately after spending hours to figure it out, I'm still stuck. So that's why I figured out somebody here could help me. Commented May 23, 2013 at 18:40
• OK, I just added a homework tag as I'm sure this is such a home-work-like question. Commented May 23, 2013 at 18:45
• It probably is, but it's a good question because you're asking about the specific concept that confuses you, not asking for an answer to the question. :-) Commented May 23, 2013 at 18:52

As you have noticed yourself, your system is simply underdetermined. In order to find a unique solution you need to add some extra constraints in addition to Newton's equations. Imagine a table with more than four legs: the more legs you add, the more unknown forces you have. But the number of equations does not change. If we instead remove a leg we find a unique static solution.

• Oh, I see. I was so sure that I've made a mistake that didn't even think of that! Now, I've got an idea: I guess I could try to add some thickness to the table, solve it in 3D (we'll then have some shear forces applied to each leg) and then try to see what happens if the thickness becomes smaller and smaller... Any comments on that approach? Commented May 23, 2013 at 19:13
• I don't think adding thickness to the table helps. All the forces are parallel to the third axis, so the corresponding moment will be zero.
– Olof
Commented May 23, 2013 at 19:21
• Ah, I see. You are right. It wouldn't make a difference. dmckee suggested finite strain approach. I googled infinitesimal finite strain and what I found gave me headache! Commented May 23, 2013 at 19:27
• Thanks Olof. You have been most helpful. You gave the "correct" answer but Olin solved the very specific case that was stated in the question. I wish I could accept two answers. Commented May 23, 2013 at 19:32

The simple answer is that you can't fully solve this problem--because as you note it is under-constrained--under the assumptions that are made when you first start doing statics (that objects are completely rigid).

The introduction of finite strains bring in additional relationships.

• Thanks! How can I learn more about a solution without assuming a fully rigid body? Is there any textbook/site with a worked example for problem similar to this one? Commented May 23, 2013 at 19:14
• I never studied this class of problems in detail, so I am hesitant to offer a prescription. If you are taking a engineering statics class you'll see some finite-strain treatments later in the class. Commented May 23, 2013 at 19:58

This looks like a simple linear blending problem. It is two-dimensional, but each dimension can be considered independently.

The more to the right the weight is, the larger the fraction of it carried by F2 and F3. Basically, the fraction of the weight carried by F2 and F3 is X/W. Put more mathematically:

(F2 + F3) / (F1 + F2 + F3 + F4) = X / W

The same can be done for the Y direction.

This gets you 0-1 fractions for left/right and top/bottom, with left = 1-right, etc. Now multiply the fractions due to the X and Y ballance for each leg. For example, F3 = (X/W)(Y/H). You can write down the overall 0-1 fractions for each leg like this from inspection. Then to get the actual force instead of the fraction of the total, multiply each by the total weight, which it seems you are calling D. Therefore F3 = D(X/W)(Y/h), and the formula for the other legs follows similarly using the 1- rule.

• Interesting idea. Thanks. I'll give it a try. But is this accurate? I mean, where did the "linear blending" come from? The second order terms make be a bit nervous. Also, like what Olof said, what if we get 5 legs?! Still for this specific problem your solution might give a good, approximate result. Commented May 23, 2013 at 19:17
• @Shapul: Linear blending from inspection only works in this specific case of legs at the corner of a rectangle because that makes them nicely independent between axes. Actually you can use this concept on a parallelogram, since that's just a skewed rectangle. Basically you solve the rectagular case and transform. This method doesn't work with 5 legs. That is a underconstrained system that has to take deformation of the table into account. Commented May 23, 2013 at 19:22
• Very nice! So this is just a corner case with a very specific solution. Otherwise I need to look at the finite strain solutions. Commented May 23, 2013 at 19:29
• I'm going to accept your answer, although this just solves a very specific case of the problem. But at the end of the day, it is what was asked in the question! Now I have to read those tensor equations for the general solution... Commented May 23, 2013 at 19:31
• This answer will give good results. Car race teams use it all the time to estimate the weight each tire supports. Commented May 23, 2013 at 20:37