Solving 9 variable Normal Reaction equations of Sudoku board

Question

We have a rigid Square board of negligible mass, which has been divided into a grid of 9 small squares(like a Sudoku Board), at centre of each square a point mass is attached. The gravity on the board is acting in downward direction and the board is supported by an array of 9 vertical pins, such that the tip of each pin lies exactly below the centre of respective squares/point masses.

Since the the masses of point particles are not same,the distribution of mass over the board is not uniform, hence the force acting on every pin need not to be same.

How can we calculate the force acting(Normal reaction from board) on every pin.

I think we can write the following equations:

$$\sum_{i=1}^9 N_i = \left(\sum_{i=1}^9 m_i \right)g$$ $$\overrightarrow\tau_{net}=0$$

From the above equations we can get total of four equation involving our unknown normal reactions: 1 from balancing weights with normal reactions and 3 in each space vector($\hat i,\hat j,\hat k)$ by making the net torque on the board equal to zero.

Using this how can we calculate a total of 9 unknown normal reactions

Its not homework question. I have nobody to ask right now except on stackexchange. It came to me as I was trying to develop a technique to solve sudoku.

Basically I want to know the pin which has the least normal reaction from board. Is there any way to do this without calculating all the normal reactions.

And the question is? And why should I do your homework for you? — Jon Custer, Jun 02 '16 at 13:29
@JonCuster Its not homework question, I can guarantee you that. I have nobody to ask right now except on stackexchange. It came to me as i was trying to develop a technique to solve sudoku. — Dost Arora, Jun 02 '16 at 14:09
Possible duplicate of Balancing Orthogonal Forces on 2D plane with arbitrary placement — John Alexiou, Jun 02 '16 at 14:41
Think of it this way. If you had 2 point masses on 2 pins, separated by a massless support, each normal force would be equal and opposite to the gravitational force on each mass. Then, if the massless support were extended to third pin, there would be no downward force until you added a third mass. The pin under that mass would push up at the gravitational force on the third mass, etc.. — Peter R, Jun 02 '16 at 17:37

sammy gerbil · Answer 1 · 2016-06-02T20:54:19.003

Your problem cannot be solved without further information. This is because the board and pins in your model are ideal, not real : they are perfectly flat, rigid and inflexible. As an ideal problem it is statically indeterminate : there are infinitely many solutions which are compatible with the given conditions. It is possible to increase some reactions and decrease others while maintaining the equilibrium conditions.

For the same reasons it is not even possible to identify which pin will bear the least load. If raised by an infinitessimal amount it could be made to bear the greatest load, of if lowered by an infinitessimal amount the load it carries could be reduced to zero.

In reality, the system will of course settle with definite values for the load supported by each pin. This is because of the flexibility of the wooden board and the pins. These provide additional conditions (constraints) relating internal stresses and deformations to the external loads and forces on the board and pins. If given additional information about the elastic properties of the board and pins you could take them into account to get a definite answer. You would also need to know about any deviations of the board from perfect flatness, and of the pins from perfect alignment in a plane, which could result in some pins bearing no load at all. Slight variations in elasticity between different pins, or between one part of the board and another, would also affect the solution.

https://en.wikipedia.org/wiki/Statically_indeterminate

@brucesmitherson : Please could you explain why you think so? The ideal conditions of the problem - board of negligible mass (while also presumably perfectly rigid), point mass loads, pins aligned exactly below each point mass - do not provide enough constraints. Other factors must be taken into account (ie the problem made more realistic) to provide more constraints and get a definite solution. — sammy gerbil, Jun 02 '16 at 14:37

Solving 9 variable Normal Reaction equations of Sudoku board

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