# Forces exerted on legs of a 3-legged table

This is actually for a computer science homework problem, but I haven't studied any physics in five years so I'm having a bit of trouble. I'm given a table of three legs with their coordinates and I am told that a weight is placed at some coordinate $(x_0, y_0)$ on the table. (We are told to disregard the weight of the table itself in our calculations)

We have to first balance the forces; since this is a static system I would assume that, for $F_1, F_2, F_3, W$ the forces exerted by each leg and the weight, we have $F_1 + F_2 + F_3 + W = 0$. On the other hand, intuition is telling me that the forces exerted opposing the weight on the table would be related to the position of the weight, since it is not necessarily centered. I would think that they would be inversely proportional to the distance between the weight and each leg.

We need to solve a system of linear equations for the three forces after being given the value of W and its position. I have the two other linear equations which are calculating torque of the three forces over x and y axes. I am just unsure of how to represent the position of the weight in my system.

Torque equations as requested: $\frac{\cos(7\pi/6)}{\sqrt{3}}F_2 + \frac{\cos(11\pi/6)}{\sqrt{3}}F_3 = 0$ and $\frac{1}{\sqrt{3}}F_1 + \frac{\sin(7\pi/6)}{\sqrt{3}}F_2 + \frac{\sin(11\pi/6)}{\sqrt{3}}F_3 = 0$ for table-leg vertices at $(0,\frac{1}{\sqrt{3}}), (\frac{1}{2},\frac{-1}{2\sqrt{3}}), (\frac{-1}{2}, \frac{-1}{2\sqrt{3}})$.

The equations may be wrong, I'm pretty rusty with this stuff. Tips would be appreciated.

• can you add the torque balance equations to your post? Oct 21, 2014 at 17:05

If each leg $i=1\ldots 3$ is located at $\vec{r}_i = (x_i,y_i,0)$ and each force is $\vec{F}_i = (0,0,F_i)$ then the balance of forces and moments is

$$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = \vec{W}$$ $$\vec{r}_1 \times \vec{F}_1 + \vec{r}_2 \times \vec{F}_2 + \vec{r}_3 \times \vec{F}_3 = \vec{0}$$

where $\times$ is the vector cross product. From the two equations above, with 6 components total, only 3 are non-zero (planar case) which are solved for the three scalar leg forces $F_i$.

$$\begin{pmatrix} 0 \\ \frac{1}{\sqrt{3}} \\0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ F_1 \end{pmatrix} + \begin{pmatrix} \frac{1}{2} \\ -\frac{1}{2\sqrt{3}} \\0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\ F_2 \end{pmatrix} + \begin{pmatrix} -\frac{1}{2} \\ -\frac{1}{2\sqrt{3}}\\0 \end{pmatrix} \times \begin{pmatrix} 0 \\ 0 \\F_3 \end{pmatrix} = 0$$

The above torque balance together with the force balance yields the following 3 equations:

$$\frac{1}{\sqrt{3}} F_1 - \frac{1}{2\sqrt{3}} F_2 - \frac{1}{2\sqrt{3}} F_2 = 0$$ $$\frac{1}{2} F_3 - \frac{1}{2} F_2 = 0$$ $$F_1 + F_2 + F_3 = W$$

I get the solution $F_1=F_2=F_3=\frac{1}{3}W$ for this case. To generalize the problem you need to make this into a linear algebra problem.

$$\begin{bmatrix} 1 & 1 & 1 \\ y_1 & y_2 & y_3 \\ -x_1 & -x_2 & -x_3 \end{bmatrix} \begin{pmatrix} F_1 \\ F_2 \\ F_3 \end{pmatrix} = \begin{pmatrix} W \\ 0 \\0 \end{pmatrix}$$

Surprisingly, solution of this physical problem is ambiguous unless some restrictions are assumed for $\vec{F_1}, \vec{F_2}, \vec{F_3}$ (such as the floor has zero friction and hence all three forces are perpendicular to its plane).

Let our legs form vertices of an equilateral triangle and weight be placed precisely above its centre. Then $$\vec{F_1} = \vec{F_2} = \vec{F_3} = -\frac{1}{3} \vec{W}$$ is obviously a solution. Is it unique? No, because we can add to each of F vectors a force directed towards the centre, three horizontal vectors with equal magnitudes. Vectors of these additional forces not only sum to zero, but pass through one point (centre of the triangle) and hence have zero torque.

Is it physically possible? Yes, if magnitude of these additional forces will be less than $\frac{1}{3} W μ$ where μ is the coefficient of friction (assuming the floor is horizontal) and legs can withstand some bending moment.

• With three legs there is a unique solution that comes from the sum of the forces and two moment balance equations. Nothing ambiguous about it Nov 29, 2014 at 6:05