So, time to post my calculations:
First, the load can either be perceived to be a point mass, OR have it's center of mass at the specified position. That's where the force will be concentrated in both cases.
Secondly, I'll give the load in mass, it's not what you'd officially do (just like I very much doubt you'd give a load in pressure, like BHS does) but since we're only talking about gravitational laod, and we'll assume a homogenous gravitational field, they'll be proportional anyways. Also, disputeone did it first so I just followed their lead.
So, the load of the floor on each bolt is obviously 25 kg.
Then we add the extra 100kg load. Let's ignore the load of the floor for now. Because the load is static and we'll assume the floor doesn't give or the bolts shear, the sum of the load on all bolts must be 100 kg.
But, the sum off all torques on all bolts must equal 0 kgm (kilogram meters, officially it is Nm, Newton meters), otherwise the floor or the bolts would twist. Torque is calculated around a pivot (can be static/rigid) and an axis by multiplying a force by it's distance to the force at a right angle from the axis you want to calculate the torque around.
As an example, let us take the south bolt, and let us work in one dimension at a time. We'll position ourselves so that we are aligned with the west and east bolt. From this position we can only deal with torque around the west-east axis, or basically torque that goes in a right angle from the west-east axis. The point load is 5+3 meters away from the south bolt, or 8 meters, since we are blind to distances that go west-east. That means that the torque on the south bolt around the west-east axis is 8m*100kg, or 800kgm. The west and east bolt are both 5 meters away, and from this position they will act like a single pivot. This single pivot will add a negative torque to the south bolt of 5*(-Loadwest-Loadeast). The north bolt is 10 meters away and will add a negative torque of 10*(-Loadnorth). The full torque around the west-east axis on the south bolt can be described as:
Torquesouth = 800kgm + 5*(-Loadwest-Loadeast) + 10*(-Loadnorth),
or 800kgm + 5*(-Loadwest-Loadeast) + 10*(-Loadnorth) = 0kgm,
since the torque on the south bolt must be 0 in all axis.
If we gather a few more relationships like these, we can compare them to each other in order to solve for all loads. However, there's a way I thought would be easier. Let's position ourselves so that the south and east bolt are aligned, and the north and west bolt are aligned. They are all evenly spaced along the circumference of a circle, so it is possible. From this northeast-southwest axis, the south and east bolts acts as one pivot and the north and west bolts act as another pivot. All cardinal directions are at a 45° angle to our axis, so we can calculate distances by adding their cardinal components (distance north + distance west - distance south - distance east, in this specific coordinate system) and multiplying with cos(45°).
Let's originate from the southeast pivot. The point mass is 3 meters north and 2 meters west away from from the center of the platform, or cos(45°)*5m. The southeast pivot is also cos(45°)*5m away from the center, but in the other direction. That means that the distance between the point mass and the pivot is 2*cos(45°)*5m, or cos(45°)*10m. That means that the torque is 100kg*cos(45°)*10m, or cos(45°)*1000kgm.
The north-west pivot is also cos(45°)*10m away from the south east pivot, and it's torque would be:
cos(45°)*10m*(-Loadnorth-Loadwest)
So the torque on the southeast pivot is:
Torquesoutheast pivot = cos(45°)*1000kgm + cos(45°)*10m*(-Loadnorth-Loadwest), and because the torque must equal 0:
cos(45°)*1000kgm + cos(45°)*10m*(-Loadnorth-Loadwest) = 0
Or:
cos(45°)*10m*(-Loadnorth-Loadwest) = -cos(45°)*1000kgm
10m*(-Loadnorth-Loadwest) = -1000kgm
(-Loadnorth-Loadwest) = -100kg.
Loadnorth+Loadwest = 100kg.
In order words, the north-west pivot takes all of the load of the 100kg load, because by dumb luck I managed to put it right on top of that pivot, or rather right in between the north and west bolt. That means that the south and east doesn't take any load from the 100kg weight, only the load of the floor.
Now it's easy, we just orient us at a right angle from the north-west axis, so that we have the north bolt, west bolt and the 100kg load in between. I'll calculate torque around the north bolt. Once again, I'll have to transform all directions with cos(45°) multiplied by (distance north - distance west - distance south + distance east).
The north bolt is cos(45°)*-5m away from the center (counting positive directions as towards the west bolt), the load is cos(45°)*(-3+2)m, or cos(45°)*-1m away from the center. The distance between the load and north pivot is therefore cos(45°)*5m-cos(45°)1m, or cos(45°)(5-1)m, or cos(45°)*4m.
The torque is cos(45°)*4m*100kg = cos(45°)*400kgm.
The west bolt is cos(45°)*5m away from the center, or cos(45°)*10m away from the north bolt. the negative torque is:
cos(45°)*10m*(-Loadwest)
The torque around north bolt is therefore:
Torquenorth bolt = cos(45°)*400kgm + cos(45°)*10m*(-Loadwest)
and because torque must be 0:
cos(45°)*400kgm + cos(45°)*10m*(-Loadwest) = 0
Which means:
cos(45°)*10m*(-Loadwest) = -cos(45°)*400kgm
10m*(-Loadwest) = -400kgm
(-Loadwest) = -40kg
Loadwest = 40kg
The west bolt takes 40 kg of the load of the 100kg mass. We know that the south and east doesn't take any of that load, and we know the sum of all loads must be 100kg. So the north bolt must take 60kg of load.
If we add in the 25 kg from the floor, the total loads become:
North: 85kg
East: 65kg
South: 25kg
East: 25kg
That means that even though we doubled the load on the floor, the load on the north bolt more than tripled and the load on the east bolt largely overshot double the load.
Had I placed the load even further northwest, the south and east bolt would experience less load as the north and east bolts acts as a pivot for a lever.
You can try inputting these values from any coordinate system, and it'll hold up. The sum of all torques will be 0, and the sum of all loads are obviously 100kg.
I don't know why BHS is trying to add variable that have nothing to do with this. Probably just as an excuse. As he says, this is pretty low-level for an engineer, but he still thought it's not solvable. Maybe it tells us something of his level?
Poll