To Calculate Force at a Ponit

Just use the same analysis you would for a concrete slab floor with a live load on it.

yah the reaction on 4 supports will be different.Thinkin case of a simply supported bean on which aload is applied but not at the middle.If you will provide the data i can solve it for u.

Hello

Thanks for the reply. The dimensions are as below:
Square shaped plates dimension = 152mm X 152 mm X 10mm (L X B X H)
The Support is places at a distance of 25 mm from each corner ( from
both sides)
Small Apparatus dimension = 75 mm X 65 mm X 24 mm

this small apparatus is placed at exactly in one corner.

I agree with you about the first but not about the second part of your figures.

If you consider center of gravity as an equilibrium point between to forces, where you will have 550 g in let's call F1 (one leg of the table) and 900 g in the other F2 (exact opposite corner) you just will have the others two corners in equilibrium.

Whit this concept in mind and considering:

Z= diagonal between the uneven forces corners

A= size of the square

Z= A*V ¯ 2¯ (square root of 2) Pythagoras = A*1.4142

Without apparatus center of gravity is Z/2, but with apparatus it will be at X mm from the heavy corner and Y from the other, it means Z=X+Y= A*1.4142 è X= A*1.4142-Y

Therefore,

550*Y = 900*X è Y= (900*(A*1.4142-Y))/550

And so, you will get with simple mathematic,

Y= A*2.31/2.64

And with that getting the new center of gravity

This is assuming that the added weight is directly and evenly spread over the support in that corner.

There was no mention of where the added weight was exactly, just "in the corner" just as it does not exactly state where the supports are in accurate terms.

Especially seeing this was home work in my opinion, you better make sure the original poster is made to think about all aspects rather than spoonfed the correct answer.

Hello case491

Here the small apparatus is placed "evenly spread " at the corner. so i hope Marzuk64 2 calculation is correct.

many thanks

No it is not and your returned details point that out rather aptly.

You say 75x65x24. Which size is what and which of the footprint is on what side of the plate. As the footprint will always have unequal sides, given your details are 3 different numbers, your point of gravity for the added weight will not be on the cross lines between the support points assuming the support points are on the diagonals and 25 mm inward from the corner.

You see how easy it is to convey a message you think is complete but which means many different things to those who are left to assume too many variables. This is how the mars space lander completely missed something the size of a planet.

Thank you for helping him with his homework.

That is much the same question that i used on my students when teaching them force calculations.

If you have a square with a side dimension A, its CG shall be located at intersection of its diagonals at point m, and if we add a weight of 350 gramme at one end of the square like point a, the resultant CG of the whole system shall be located at the diagonal at point c in between points m and a.

Square side length = A

Diagonal length = √2 A

Half of Diagonal, am = A/√2

By taking moments about point c,

350(ac) = 2200(cm)= 2200( am-ac)= 2200(A/√2-ac)

(350+2200)(ac)=2200 A/√2

ac = 2200 A /(√2*2550) = 0.610053 A

i.e the CG of the whole system is located at a distance of ac = 0.610053 A

Here is the easy way, use AutoCAD and build the shapes in 3-Dimensions. Create blocks for each part with accurate dimensions and placement of the added weight. Join the two pieces together and do a query on the CG for the single object. It can do it for you, just define all your parameters. If you are looking for the reactive forces from the supports you will have to use your parallel-axis theorem which can be found in every statics/dynamics text book.

Use your math skills from Calc and have fun with the problem, you can do it.

I made no assumptions about the distribution of the apparatus or the footprint of the apparatus. I assumed that the centroid of the apparatus was directly over one corner of the plate and that the plate was supported by four corner supports located precisely at each corner. If that is not correct, the center of gravity of applied loads is unknown and the problem has no solution.

I assumed that the settlement of the support under the apparatus was negligible. Otherwise, the second part of the problem, namely the magnitude of the four reactions is not possible.

If the centroid of the apparatus is somewhere other than the corner of the plate, then the rigidity of the plate and the rigidity of the supports must be considered. It is still possible to determine the center of gravity of the applied load, but it is not possible to determine corner reactions without more information.

If one support is removed, then the problem becomes statically determinant and the three reactions can be calculated by simple statics.

As presented, the structure is indeterminate to the first degree unless simplifying assumptions are made about support stiffness.

Or, if you want to do this with good precision, and are willing to spend some time in looking at construction and calibration details, you should look at Force Platforms commonly used in biometrics. They do exactly the measurement you describe.