# Can you calculate the force of chair legs on floor?

I have kinda formalistic question. Eg. we have normal symmetric chair with 4 legs, and mass $$M$$. Can we calculate forces of chair legs on floor surface $$F_1,F_2,F_3,F_4$$ with only this data and why not. How does the assumption, that chair is completely rigid or that it is elastic affect the problem. Is such calculation possible with chair with 3 legs?

A little background to this question. I barely remember statement from lectures from my 1st year classical physics, that such solution is possible for chair with 3 legs, but not for chair with 4 legs unless we don't know how the chair deforms. I am looking for formal reason, why do we need to take deformations into account with 4 legs, but not with 3 (although we could use it even for 3 legs).

• Three points describe a plane, so 3 legs are always in plane, 4 legs may not all be coplanar if the chair is not perfect. So the chair would deform with weight until all 4 touched the floor. Even if the weight were centered the legs could have different weight on the floor Oct 12, 2020 at 12:37
• What happens, if you put a point of mass M at one corner of the chair? What happens if you put it in the center instead? Oct 12, 2020 at 12:40

If the chair has three legs then there are three equations (net sum of forces equals zero, moments about two horizontal axes) in three unknowns, which determine the forces on each leg unambiguously. If we introduce a fourth leg then we have to take account of the stiffness and deflection of the chair to get a solution.

This is the two dimensional analogue of the following one-dimensional situation. If a horizontal beam with a known load is supported at two points then we can determine the force on each support without worrying about the deflection of the beam. But if we add a third support then we have to take the stiffness and deflection of the beam into account.

There are two aspects why you can't calculate the forces from the given data.

## Geometry

The geometry is not given to sufficient degree:

• Are the legs forming a rectangle? Maybe that's what "normal symmetric" wants to express, but real-world symmetric chairs often have their legs arranged in a trapezoid shape instead.
• Where is the center of gravity of M located with respect to the four legs?

Typically, all relevant dimensions should be given, or at least the relative placement of M's center of gravity with respect to the legs' positions.

## Solving the force equations

There are four unknown values, but only three independent equations available:

• The forces must sum up to the weight of M.
• The total moment around some front-back axis must be zero.
• The total moment around some left-right axis must be zero.

To compute a moment, besides the forces you also need the geometry, meaning the distances from the axis (see above).

Even if the geometry is known, there are still four unknowns out of three equations, meaning that infinitely many solutions can be found.

You mention the concept of deformation. This means that you can compute the deformation that the chair will experience when applying a given set of F1 ... F4 forces. We assume the ground to be perfectly flat and the legs to be perfectly aligned. Then you have an additional equation from the fact that all four legs are supposed to reach the ground, meaning zero deformation, so only force combinations solving that no-deformation constraint can be solutions.

The exact physics and math behind deformation can become quite complex, so you typically can't write down this fourth constraint as a simple concise equation.

• When you say "momentum" I think you might mean "moment". Oct 12, 2020 at 15:09
• Sorry, I'll correct that (English is not my native language). Oct 13, 2020 at 10:03

This general problem is essentially finding the barycentric coordinates (the baryweights) of a point inside general polygon (or rectangle specifically). This problem is sometimes called the force on legs of table problem.

## Solution

Consider the 5 sided table below

Each leg $$i=1\ldots 5$$ has known cartesian coordinates $$\boldsymbol{r}_i$$ and the center of mass C also has know cartesian coordinates $$\boldsymbol{r}_C$$.

The problem now is what percent of the weight $$w_i$$ of the table does each leg bear.

$$\sum_{i=1}^n w_i = 1$$

But also by the definition of center of mass

$$\sum_{i=1}^n w_i \boldsymbol{r}_i = \boldsymbol{r}_C$$

The solution to this problem proceeds by solving the following system of equations. This is least squares solution that minimizes the "energy" of the system.

1. Build a (3×n) matrix with each point as a column

$$\mathbf{G} = \left[ \begin{array}{c|c|c|c} \boldsymbol{r}_1 & \boldsymbol{r}_2 & \boldsymbol{r}_3 & \boldsymbol{r}_4 \end{array} \right]$$

2. Form the coefficient matrix $$\mathbf{M}$$ by adding a matrix filled with ones to the product of transpose of $$\mathbf{G}$$ with itself

$$\mathbf{M} = \begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \\ 1 & 1 & 1 & 1 \end{bmatrix} + \mathbf{G}^\top \mathbf{G}$$

$$M_{ij} = \{ 1 + \sum_k G_{k i} G_{k j} \}$$

3. Form the constant vector $$\boldsymbol{p}$$ by adding a vector filled with ones to the product of the transpose $$\mathbf{G}$$ with the center of mass point

$$\boldsymbol{p} = \pmatrix{1 \\ 1 \\ 1 \\ 1} + \mathbf{G}^\top \boldsymbol{r}_C$$ $$P_{i} = \{ 1 + \sum_k G_{ki} c_k \}$$

4. Find the vector $$\boldsymbol{w}$$ of corner weights by solving the system

$$\mathbf{M}\, \boldsymbol{w} = \boldsymbol{p}$$

5. Check that the solution is indeed what is expected by evaluating the following

$$\sum \boldsymbol{w} \equiv 1$$ $$\mathbf{G} \boldsymbol{w} \equiv \boldsymbol{r}_C$$

or

$$w_1 + w_2 + w_3 + w_4 + \ldots \equiv 1$$ $$w_1 \boldsymbol{r}_1 + w_2 \boldsymbol{r}_2 + w_3 \boldsymbol{r}_3 + w_4 \boldsymbol{r}_4 + \ldots \equiv \boldsymbol{r}_C$$

## Example

Consider the following 4 random points

\begin{aligned} \boldsymbol{r}_1 & = \pmatrix{ 7.45578 \\ 1.97323 } & \boldsymbol{r}_2 & = \pmatrix{ 9.29723 \\ 0.994307} \\ \boldsymbol{r}_3 & = \pmatrix{ 8.63861 \\ 5.30554} & \boldsymbol{r}_4 & = \pmatrix{ 3.79046 \\ 8.75215 } \end{aligned}

and the specific center of mass $$\boldsymbol{r}_C = \pmatrix{ 8.0 \\ 4.0 }$$

1. $$\mathbf{G} = \begin{bmatrix} 7.45578 & 9.29723 & 8.63861 & 3.79046 \\ 1.97323 & 0.994307 & 5.30554 & 8.75215 \end{bmatrix}$$
2. $$\mathbf{M} = \begin{bmatrix} 60.4823 & 72.2801 & 75.8766 & 46.5308 \\ 72.2801 & 88.4271 & 86.5905 & 44.9431 \\ 75.8766 & 86.5905 & 103.774 & 80.1792 \\ 46.5308 & 44.9431 & 80.1792 & 91.9678 \end{bmatrix}$$
3. $$\boldsymbol{p} = \pmatrix{ 68.5391527922694 \\ 79.35505307751927 \\ 91.3310423486012 \\ 66.33226226099814 }$$
4. $$\boldsymbol{w} = \mathbf{M}^{-1} \boldsymbol{p} = \pmatrix{ -0.030933867991783423 \\ 0.4914450326148808 \\ 0.33345705320796537 \\ 0.20603178216894402}$$
$$-0.031 + 0.491 + 0.333 + 0.206 = 1.00$$ $$\mathbf{G} \boldsymbol{w} = w_1 \boldsymbol{r}_1 + w_2 \boldsymbol{r}_2 + w_3 \boldsymbol{r}_3 + w_4 \boldsymbol{r}_4 = \pmatrix{ 8.00 \\ 4.00 } \; \checkmark$$
Summary, leg (1) carries $$-3.1\%$$ of the weight, leg (2) carries $$49.1\%$$ of the weight, leg (3) carries $$33.3\%$$ of the weight and leg(4) carries $$20.6\%$$ of the weight.