Euler [1] set a problem in St. Petersburg Academy of Sciences in 1773 in his dissertation De pressione ponderis in planum cui incumbit. The problem consisted of placing a weight on the surface of a rectangular table supported on four legs and determining the four reaction forces of the legs. Euler claimed that the solution to this problem “is not only much harder but also doubtful and misleading”. He said that the imperfections of the feet might force the body to rest on only three legs and the pressure on the fourth one may be null. In Section 3, we observe that the problem arises because the system of equations given by the mechanics of static equilibrium is a linear system of equations that consists of three equations and four unknowns, being these four unknowns precisely the reaction forces at the legs of the table. This linear system of equations is called an indeterminate system.

Euler then changed the problem: the support is no longer on a rigid plane but on a soil. Euler argued that to understand the problem on a rigid body, one must pose the problem on a soft body, and that once this problem has been understood there, the solution from one body to another can be carried out by a mathematical calculus argument (a limit argument), “in the case of a soft ground where four feet penetrate, one must assume that the pressures brought about by the feet are proportional to those movements”. This is the principle where Euler based his arguments. He stressed that even though here a soft ground is considered, this phenomena is in- dependent of the ground because the movements, that were hypothetically introduced, “happen with the support of our imagination”.

After Euler posed and discussed the problem, many people have studied it in depth, each contributing at some extent towards a richer interpretation of the problem. These people [2] include Jean Baptiste Lerond d’Alambert in 1780 who dedicated to the support problems part of a memoir entitled Sur quelques questions de Méchanique, Abbot Charles Bossut in 1788 who gives an interpretation of a triangular table in terms of two levers, Paolo Delanges of Brescia in 1790 who three times in 1790, 1798 and 1811 made a proposal for an experimental investigation, Pietro Paoli in 1792 who checked the validity of the computations done by others regarding the problem, Anton Maria Lorogna in 1794 who attempted to formulate a “reasonable” principle that can not be obtained from the principle of mechanics, Mariano Fontana [3] in 1798 who provides a formula to obtain the reactions on a parallelogramic table by imagining that the weight is transmitted to the sides of the parallelogram, later Fontana writes “I started to meditate more carefully about this problem and I have found that, the solution to the problem is not determined only by the principles of mechanics”. We will refer to Fontana later in this paper.

Other people that also worked on this problem (see [2] ) were Guiseppe Venturoli in 1806, Gabrio Piola in 1824, Barsotti in 1842 and Ottaviano Fabrizio Mossotti in 1858; Malfatti in 1805 published three memoirs, in one of them he stresses that if the object is placed at the center of a regular polygon, all the reactions at each of the vertices of the polygon will be equal and hence they can not be “assigned arbitrarily”, hence in this case there is no statics indetermination. It was finally Navier [4] in his Résumé des leçons données à l’Ecole royale des ponts et chaussées in 1825 who recognized that the indeterminate system will no longer be indeterminate if one takes into account the “elasticity of the reacting bodies”. Even though the solution to the problem of the rectangular table appears for the first time in Castigliano’s thesis in 1873, Augustin Cournot in 1828 is the one who describes an extrema point in the pressure involved in the problem, he talks about a minimum for the pressure exerted in the body. A derivation of the solution can be appreciated in Du Bois 1902 [5] , he considers that each foot of the table is hence a spring obeying Hooks’ Law and the work expressed in terms of the potential energy is computed. Using The Principle of Least Work, a fourth equation is obtained by obtaining the derivative of the total potential energy and setting it to zero. Therefore we now have a system of four in- dependent equations and four unknowns which can be solved in this case providing a unique solution.

There is a list of people who contributed to the solution of the problem during the 1830-1873 period, the list includes Moseley, Pagani and Menabrea, Massotti and Dorna among others. For a more detailed history of the problem you can read [2] .

The paper by Benvenuto [2] contains not only more details about the history of the problem itself but also gives a geometrical interpretation of the triangular table solution (as the one we have in this paper), it provides a short discussion and a geometrical interpretation of a solution given by Fontana to the general parallelogramic case (hence the rectangular case is included) which we will address in Section 5.

Recently, three NASA engineers Surya N. Patnaik, Dale A. Hopkins and Gary R. Halford published a paper in 2004 [6] where they discussed the problem and introduce a compatibility condition (a fourth equation) that they obtained based on the deformation displacements of the legs, these deformations are written in terms of the angles at which the table tilts. When this compatibility condition is expressed in terms of the forces, it is then concluded that the sum of the reactions of two opposite corners of the table matches the sum of the reactions of the other two opposite corners. After all, the solution to the new system of four equations with four unknowns that Patnaik, Hopkins and Halford obtained agrees with the one described in Du Bois [5] .

One of our main hypotheses (and here we differ from the results obtained by all the previous people) is that the legs of the table are completely rigid, i.e. they do not deflect or deform. The way we approach the problem is by considering the fact that the reactions on the legs are known and given in terms of two parameters. These parameters are obtained by first assuming that the four legs are replaced by a beam with end points at opposite edges of the table, then changing roles with the other two sides of the table. One important and natural assump- tion is a consistency condition on the forces in terms of these two parameters. We finally give a condition (and here is where our main contribution comes in) on the deflection of the table (not of the legs). This process allows us to minimize such deflections obtaining the values of the parameters we started with initially, providing therefore one more equation which, together with the three that one has by considering mechanical assumptions, one is able to find a solution for the table of rigid legs. In a forthcoming paper, we will generalize these results to the quadrilateral table and we are also considering increasing the number of legs on a future paper.

Regarding the structure of the paper, in Section 2, we discuss the Beam and the Triangular case, in each case we solve the system and we provide a geometrical interpretation that matches the ones given in Benvenuto [2] . In Section 3, we carefully analyze and discuss the indeterminate case of a rectangular table. Right after, in Section 4, we establish the main hypotheses under which we will solve the rectangular table case. It is precisely in Section 5 where we give conditions that will help us to solve the problem.

Finally, in Section 6 we provide some lab results that are consistent with the theoretical values obtained by the proposed theoretical model. A discussion on how these lab values are obtained is contained in this section too.

In this section we will discuss the beam (one dimensional table) and the triangular table case.

Consider a rectangular table of sides B and H and legs at each of its vertices and. A load of magnitude is placed at a point on the table’s surface and perpendicular to it. The point has distances to the sides x and y respectively as in Figure 4.

Here once more we have a rigid surface and we are assuming that the legs do not bend. For denote by the magnitude of the force. If we calculate the resultant torque on the body about an axis through and and we also calculate the resultant torque on the body about an axis through and we obtain the equilibrium equation system

This is all the mechanics says about the system. Systems of three equations and four unknowns might have no solutions or have an infinite number of solutions according to the consistency of the system and the existence of non-trivial solutions to the homogeneous system. It is the case that this system is consistent and the homo- geneous system has non-trivial solutions, hence it has an infinite amount of solutions. We call these type of systems indeterminate systems.

Patnaik and Hopkins [6] discussed the problem under different hypotheses. They assume that the legs deform and take in consideration the modulus of elasticity and the cross-sectional areal of the legs. By using the principle of least work, they arrive at a fourth equation which they refer to it as compatibility condition:

Adding this equation to the system 3.4 we obtain solutions where sometimes the reactions magnitudes are negative according to where the load is placed (a negative magnitude of a reaction means that the reaction force is pointing in the same direction as).

We should note that the problem we are approaching here is different. We will define the problem in the next section.

We are considering a rectangular table whose legs bent, shrink or deflect a sufficiently small amount that we are neglecting it, therefore in our case, the legs of this rectangular table deflect zero, i.e., they do not deflect at all. We are also assuming that the table’s surface is rigid, uniform and homogeneous enough to disregard any conditions about the material, hence we are also assuming that our table is an ideal rigid table. Let’s assume for the moment that the table weights 0 (at the end of the problem, to obtain the final reactions magnitudes at each leg, we could add 1/4 of the table weight to each reaction by symmetry).

The fact that in our case the legs do not deflect is the main difference with the work of Patnaik and Hopkins [6] . Whereas in the results of Patnaik and Hopkins, the reactions on some legs might point downwards because their flexible-legs table tilts, in our case, that does not happen, all the reactions point upwards.

Main Hypothesis 4.1. The rectangular table consists of a rigid flat surface and rigid legs, hence neither the legs nor the surface deform or deflect. An ideal object of weight is placed at a point on the surface of the table causing a force to point downwards and perpendicular to the table’s surface.

Consider a view from the top of the rectangular table described on Section 3. We are labeling the vertices with, starting with the bottom left corner and going around counter clockwise. These vertices are precisely where the reactions, perpendicular to the table’s surface, occur. Assume then that the reaction of magnitude occur at the vertex. For simplicity, we can assume that the vertices are in the cartesian plane

and they have the following coordinates, and Let be the segment from to for and let be the segment from to. Let be

the point where the load is placed as in Figure 5.

Since the legs are rigid, the table does not tilt, hence each magnitude of the reaction is non-negative (a negative magnitude would mean that the reaction points downwards).

We can imagine for a second that legs 3 and 4 disappear and that we want to equilibrate the table by placing an extra leg, leg 34 at a point on the segment in such a way that our new triangular table supported on leg 1, leg 2 and leg 34 is still in equilibrium and leg 34 has a reaction of magnitude equal to Where do we place such new leg 34? There is certainly a geometrical interval on the segment determined by leg 1, leg 2 and the point P where this new leg can be placed with those conditions. This geometrical interval is in fact found by drawing a straight line passing through and P and another one passing through and P. We extend these lines until they intersect also the prolonged segment. Such interval must be within the segment and between these two intersection points. The reasoning behind this is that the point P must be inside the triangle determined by and. This interval can be appreciated in Figure 5.

Let’s place such a leg at the point on the segment within the above interval, and let’s call t the

number between 0 and 1 such that In other words, is the vector plus t times the vector that goes from to. Let's call the reaction corresponding to the leg 34, and its magnitude, of course these are functions of the parameter t. Figure 6 exemplifies this situation.

From Figure 6, we can assume that the table we are dealing with is a triangular table, therefore the solved case applies (Figure 7).

If we calculate the resultant torque on the body about the x-axis as well as calculating the torque about the y-axis, we can also obtain an equilibrium equation system (equivalent to System 2.1) for this triangular table

We can easily solve for and Notice that all the magnitudes will be functions of the coordinates of namely:

If we now remove leg 34 and place back the leg 3 and leg 4, and can be computed using the case of the one dimensional beam as

To summarize, the solution to the rectangular case that depends on the parameter t, is given by

The solution to the rectangular case that we are looking for (and that does not depend on any parameter) satisfies the following conditions:

Condition 5.1.

The reactions and can be replaced by a reaction of magnitude at some point on the segment

The procedure that follows consists of finding such point by establishing some physical conditions on the net torque.

Similar to the description above, the reactions and can be replaced by a reaction of magnitude at some point on the segment within a geometrical interval. We might also say that the point

depends on a parameter r, where however, if we think that the reaction substi-

tutes the reactions and the reaction substitutes the reactions and, then the reactions are substituting and, and moreover, the table is still in equilibrium (we are assuming that the table weights zero), hence we do have a new condition.

Condition 5.2. The points and P are collinear! (as if the whole table were supported on a bar only).

Under this condition, the point does not depend on r anymore, the point determines the point, and hence depends on t, in fact, we have

When we do the same analysis as before, but this time using leg 3, leg 4 and leg 12 (at the point), we obtain the same values and as in the system 5.7.

Similarly, we can now think that the reactions and can be replaced by a reaction of magnitude at some point on the segment (within some interval) where as in Figure 8.

Performing the same analysis as before, we calculate the resultant torque on the body about the axis determined by and as well as calculating the torque about the axis determined by and We obtain the equilibrium system

Proceeding as before, we arrive at the solution to the system 5.8

Analogously to Condition 5.2, the point determines a point on the segment collinear with P and. In fact (Figure 9)

When doing the same analysis, such point also determines the same values and

These conditions of collinearity, do not provide any extra information, it only allows us reduce to two para- meters instead of four.

A natural assumption is that the solution given by considering the point will also match the solutions given by considering the point in other words.

Condition 5.3. The solutions to the system 5.9 agree with the solutions to system 5.7, in other words, the system is self-consistent.

This brings us to the equation

This does not solve the problem; however, whichever the solution is, there are parameters t and u such that the solution can be expressed as in the system 5.7 or 5.9, and those parameters should be related by Equation (5.10).

Since we are considering a rigid surface, we are expecting that the surface strains only a little bit, we really want the surface to strain the least to be in equilibrium, it is then a natural requirement to ask that the

table strains the least in two directions. Let’s consider only the line that goes from to Recall that and that

A way of measuring the deflection of the surface along this line is by considering the sum of the magnitudes of the torques at the points and with respect to the point P, this is

In our case, since and are collinear, we can use similar triangles to obtain that this previous quan- tity is

Since we are asking this quantity to be minimum, we need to minimize this as a function of t. The previous expression is minimum when the inside of the square root is minimum, hence to minimize the previous ex- pression, we set to zero the derivative of

with respect to t, which gives hence

therefore the point is located at Plugging these values in Solution 5.7, we obtain

If we now consider the other line that goes from to we can also measure the deflection of the surface along this line by considering the sum of the magnitudes of the torques at the points and with respect to the point P, this is

Once more, using collinearity and similar triangles, this quantity becomes

By a similar argument, minimizing the previous expression is equivalent to minimize the expression inside the square root as a function of u. Therefore, we need to set to zero the derivative of

which gives hence

Notice that this value of u agrees with the one obtained by substituting in Equation (5.10).

Therefore the point is located at

which also provides, according to System 5.7, the solutions

In a forthcoming paper, we will generalize these results.

We notice that the solutions achieved here are self consistent, the points are collinear as well as the points the net-torque quantities are minimum. We also have that the lines joining the collinear points determined four rectangles.

From Figure 10, it can be easily seen, by drawing the corresponding lines, that the points and lie within the geometrical interval discussed at the beginning of the previous section.

To each vertex of the table, among the four areas of those rectangles, one area is where the vertex is located, two areas are adjacent to the vertex and one area is opposite to the vertex. The formula obtained in Solution 5.11 or in 5.12 to compute the magnitudes of the reactions at any single leg in the rectangular table is given by

as proposed by Fontana [3] (when the parallelogram becomes a rectangle); however, Fontana imagined that this formula had to be true without any hypothesis and never justified such formula. We, in this paper, stated the conditions under which this formula is true.

Through this paper the well known beam case was provided, the triangular table case was also provided and the geometrical interpretations of the reaction forces at any given leg were given. The indeterminate case of the rectangular table, also well known, was discussed. The hypotheses to our problem in Section 4 clarified the situation of the problem we wanted to solve. By employing the method of introducing two parameters and supporting the table on ideal beams and minimizing a quantity that somehow measures the deflection of the table surface, we were able to solve the problem in Section 5. The main difference of our method and the one provided by Patnaik, Hopkins and Halford [6] is that we have different hypotheses. The results from Patnaik, Hopkins and Halford [6] and our result do not contradict each other, we just approach the same problem but with different conditions. The method that we provide here will be generalized in a forthcoming paper by the same authors. A physical experiment was performed and described in Section 6. In the same section, laboratory results were obtained and these results are consistent with the theoretical values obtained by our proposed theoretical model. Although the solution provided here has not been analyzed before, its novelty might con- tribute to some applications on the mechanics and engineering of solid structures.

We thank the Editor and the referee for their comments. The authors acknowledge the assistance of Phil West with the fabrication of several aluminum plates.