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A weight is placed on the top of a rectangular rigid ideal table with four legs, each leg is placed at each vertex of the rectangular table. It is assumed that the legs do not bend when the weight is added. The reactions are computed by assuming the table is supported on a beam, introducing two new beam parameters and minimizing a deflection function of the new parameters. A physical experiment is performed in the lab and the reactions on each leg are provided. The experimental results match the theoretical ones obtained by the proposed model. Geometrical interpretations of the results are given.

Euler [

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 [

Other people that also worked on this problem (see [

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 [

The paper by Benvenuto [

Recently, three NASA engineers Surya N. Patnaik, Dale A. Hopkins and Gary R. Halford published a paper in 2004 [

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 [

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.

A load

Here we are interested in finding the magnitudes

has the following solution

Given a leg of the beam, the point where the load is placed divides the beam into two segments, one adjacent to the given leg and one opposite to it, in each case, the solution takes the form

Consider a triangular table of sides

Here we have a rigid surface and we are assuming that the legs do not bend. For

When we solve the system we find that

By adding the three small areas of the triangles

Hence

From Equation (2.3) and Equation (2.2) we can also conclude that

On the table’s surface, let’s draw three lines from the point where the load is located to the three corners of the table. We obtain three smaller triangles (shaded differently on the figure) inside the big triangle as in

Among these three triangles, the triangle

We observe that

From the solution to the triangular case given by Equation (2.2) and the previous observation, we can establish a formula to obtain the magnitude of the reaction forces at any single leg in the triangular table according to

Consider a rectangular table of sides B and H and legs at each of its vertices

Here once more we have a rigid surface and we are assuming that the legs do not bend. For

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 [

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 [

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

Consider a view from the top of the rectangular table described on Section 3. We are labeling the vertices with

and they have the following coordinates,

the point where the load

Since the legs are rigid, the table does not tilt, hence each magnitude

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

Let’s place such a leg at the point

number between 0 and 1 such that

From

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

If we now remove leg 34 and place back the leg 3 and leg 4,

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

The solution

Condition 5.1.

The reactions

The procedure that follows consists of finding such point

Similar to the description above, the reactions

tutes the reactions

Condition 5.2. The points

Under this condition, the point

When we do the same analysis as before, but this time using leg 3, leg 4 and leg 12 (at the point

Similarly, we can now think that the reactions

Performing the same analysis as before, we calculate the resultant torque on the body about the axis determined by

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

Analogously to Condition 5.2, the point

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

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

A natural assumption is that the solution given by considering the point

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

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

In our case, since

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

therefore the point

If we now consider the other line that goes from

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

Notice that this value of u agrees with the one obtained by substituting

Therefore the point

which also provides, according to System 5.7, the solutions

In a forthcoming paper, we will generalize these results.

Geometric InterpretationWe notice that the solutions achieved here are self consistent, the points

From

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 [

To test the predictions of the Equation (5.11) above, load weights of mass 0.200 kg up to 3.50 kg were sus- pended from a rectangular aluminum plate. Reaction forces were measured by digital scales affixed to the cor- ners of the plate. A detailed procedure is described below. The resulting experimental reaction forces agree well with the theory, as seen in

In

mental fraction reaction force is the slope of a linear fit such as those in

Each reaction force in

In

On

On

On

The slopes and y-intercept values in the data

Force | y-int (oz) | slope (oz) | f(exp) | f(theory) | unc |
---|---|---|---|---|---|

Units: | oz | kg/oz | - | - | - |

0.30 | 6.62 | 18.9 | 17.9 | 0.51 | |

-0.29 | 9.15 | 26.2 | 27.1 | 0.29 | |

0.28 | 11.98 | 34.3 | 33.2 | 0.20 | |

-0.29 | 7.213 | 20.6 | 21.9 | 0.23 |

Loc | Support | f(exp) | f(theory) | unc(exp) | unc(theory) |
---|---|---|---|---|---|

2 | 18.9 | 17.9 | 0.51 | 0.87 | |

2 | 26.2 | 27.1 | 0.29 | 1.2 | |

2 | 34.3 | 33.2 | 0.20 | 1.2 | |

2 | 20.6 | 21.9 | 0.23 | 0.94 | |

6 | 6.8 | 9.1 | 0.69 | 0.60 | |

6 | 23.2 | 21.1 | 0.82 | 0.97 | |

6 | 46.8 | 48.7 | 0.80 | 1.5 | |

6 | 23.3 | 21.0 | 0.68 | 1.3 | |

3 | 10.9 | 10.2 | 0.48 | 1.2 | |

3 | 56.3 | 56.7 | 0.53 | 1.5 | |

3 | 28.5 | 28.0 | 0.62 | 1.1 | |

3 | 4.2 | 5.1 | 0.53 | 0.60 |

Label | x (cm) | y (cm) |
---|---|---|

Loc 2 | 20.7 | 12.6 |

Loc 6 | 10.4 | 6.9 |

Loc 3 | 23.0 | 3.5 |

0 | 0 | |

34.35 | 0 | |

34.3 | 22.9 | |

0 | 22.9 |

uncertainties of 10 grams due to the limitations of the accuracy of the scales (this corresponds, roughly, to the size of the symbols in

In calculating the theoretical rigid plate reaction forces, from Equation (5.11), above, the dominant source of uncertainty is the location of the load weights. These are constrained by the size of the nuts, which are 8.0 ± 0.2 mm across (from corner to corner, the longest distance across the hexagonal nut) thus the uncertainty of the load weight position is taken to be ±4 mm, which produces the horizontal error bars in

The goal of the procedure is to measure the distribution of weight from a single load weight onto supports places at the four corners of the table. To do so, it is important that the table remain level during measurements. Several methods were investigated to reach optimal conditions the most successful of which are detailed below (

Four PASCO SE-9372 laboratory jacks are places close to, but outside, where the four corners of the rec- tangular plate will lie. Height of each jack is set to near the middle of the range of travel, allowing future up or down adjustment. At approximately equal height, a right angle clamp is affixed to each ring stand. Steel cross- bars are connected to pairs of ring stands. A carpenter’s level is used to check for levelness. With meter stick, ensure the crossbars are roughly parallel and separated by the width of the plate.

The longer side of the rectangle is here termed the length; the shorter side the width. The longer sides should be directly under the crossbars. PE scales are clamped to crossbars. Scales are connected to the plate via steel machine hooks with threads which screw into the plate. To avoid stripping the threads, washers are placed below the plate and tightened. The threaded screws give a second method (after the lab jacks) to level the plate. At this point, level the plate adjusting only the screws. Ensure the hooks below the PE scales are vertical. To detect any deviation from levelness, one or more carpenter’s levels can be left on the plate. The four PE scales are tarred at this point.

Now the load weight can be suspended from the plate and the reaction forces can be measured by the four scales. However, if the load weight is not placed near the center of the plate, large load weights cause the plate to rotate out of level. For these larger load weights, re-level the plate using the lab jacks, then remove load weight. Plate will rotate somewhat out of level. Tare scales. Add load weight (which will rotate the plate back very close to level) and record reaction force readings.

Plates are fabricated from 6061 aluminum with a thickness of 3.18 mm with threaded holes for the load weights and measuring reaction forces of inner diameter 2.69 mm, outer diameter of 3.51 mm. Steel washers connecting

screws to plate have a width from corner to corner of 8.0 mm and a width from flat side to parallel flat side of 7.0 mm and a height of about 2.5 mm.

Weights and scales are affixed to the plate via steel machine screws, clamped to plate via washers. Scales are West-Boao Science & Technology Co., Ltd. brand ETWT001 portable electronic scales, with maximum load of 50 kg and accuracy to 10 grams.

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 [

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.