Estimate the reaction force on each leg of a 4-legged table
Assume you have a table with four legs with a weight placed somewhere inside the boundary created by the legs. The task is to determine the reaction force at each leg. Here are the problem constraints...
1. The sum of the four reaction forces must equal the total weight set on the table.
2. The table has no weight of its own.
3. You cannot directly solve for the reaction of each leg using sum of forces and sum of moments because the structure is statically indeterminate.
4. We are not allowed to introduce deflection in order to develop a 4th independent relationship between the four reactions
There are methods to estimate the reaction force on each leg. 151
One such method uses ratios to determine the four leg reactions.
\begin{eqnarray} P_1 & = & \left(\frac{y_w-y_b}{y_f-y_b}\right)\left(\frac{x_w-x_l}{x_r-x_l}\right) \ P_2 & = & \left(\frac{y_w-y_b}{y_f-y_b}\right)\left(\frac{x_w-x_r}{x_l-x_r}\right) \ P_3 & = & \left(\frac{y_w-y_f}{y_b-y_f}\right)\left(\frac{x_w-x_r}{x_l-x_r}\right) \ P_4 & = & \left(\frac{y_w-y_f}{y_b-y_f}\right)\left(\frac{x_w-x_l}{x_r-x_l}\right) \end{eqnarray} Here is my question. The above solution is only good for a square leg configuration.
Can this method be modified to work for a trapezoidal leg configuration, like this? Better yet, can this method be modified to work for any leg configuration?
Edit 1
<PERSON>, This is in response to your answer. Please forgive me. It's been 15 years since I've worked with summations and linear algebra. I'm having trouble solving the two torque equations for $\psi$ and $\varphi$. Please look at what I have below and let me know if I'm on the right track. First I found $F_i$ using your second equation.
\begin{equation} F_i=-\frac{W+k(\psi(x_i-x_w)+\varphi(y_i-y_w)}{3k} \end{equation} Now I substitute this into your two torque equations and attempt to solve. 151