I derived that from moments.

Take this, where X and Y are left sensor and Y is the right sensor.

W is the weight and A is the force applied.

From moments, we know that W is equally shared between X and Y, hence on X, you will have W/2 as force that W applies and on Y, you will have W/2 again. But knowing that there are four corners and that the weight acts through the centre of gravity at the geometric centre of the glass sheet, each corner will have first a force of W/4 due to the weight.

Now, when you consider the force A (in red), you have to work it out. The net moment is 0 (the sheet is not turning clockwise nor anticlockwise in the picture I uploaded. I will work with A only, and now, you can ingnore the W, since I already explained that to you and in all the calculations, you will have to include W/4 (which you shouldn't forget after the following steps)

So, clockwise moment about X gives you:

A*43.5 - Y*46.5 = 0

Clockwise moment due to A (A makes the sheet move clockwise) added with the moment due to Y (Y tends to 'counter' A, or make the sheet go counterclockwise, which is a reaction force. Remember Newton's 3rd law of motion, saying that every action has an equal and opposite reaction) equals the total moment, which is zero. Y is what you want. This gives: Y = 43.5A/46.5 = (43.5/46.5)A

Then you do the same about Y:

A*3 - X*46.5 = 0

Anticlockwise moment due to A added with clockwise moment due to X equals zero. X is what you want this time, and you get X = 3A/46.5 = (3/46.5)A

Note that this is in one dimension, you have to account for the other dimension because the (43.5/46.5)A represent all the force due to A only applied to the left side in your drawing, and (3/46.5)A represent all the force due to A only applied to the right side in your drawing.

You will further have to distribute that force between the upper and the lower corners, once with each part. Let's call (43.5/46.5)A = F1

Working through the same procedure, you should get the force at the top left corner as (32/35)F1. Substitute F1 and you get (32/35)(43.5/46.5)A.

And doing the same with the other part gives you (32/35)(3/46.5)A

If you want to get the other forces at the last two corners, you should get (3/35)(43.5/46.5)A for the lower left corner and (3/35)(3/46.5)A for the lower right corner.

Well, when you have done a lot of practice, you can grasp some shortcuts to all those calculations. For example, A is nearer to Y in my sketch. The only two values that I have are 43.5/46.5 and 3/46.5 from the lengths that I have. So, the force at Y is obviously the larger of the two, giving the force at Y (43.5/46.5)A. I repeat with the other.

And now, the TOTAL force at upper left corner is Sr = W/4 + (A)(32/35)(43.5/46.5) while at Sl, it's W/4 + (A)(32/35)(43.5/46.5)

I hope I didn't tell you too much at a time

If you think you need explanation on something, let me know