I'm going to assume both walls of that skyscraper are at 45 deg angle to the screen as it clearly looks that way
Letter 14 px - 9.14 m
Length of the building: - 131 px - 85.52 m
True length: 85.51/cos45 deg = 120.92 m
Height 1 - 107 px - 69.86 m
Width 1, 46 px - 30.03 m
True width 1 - 30.03/cos45 deg = 42.47 m
Heigh 2 - 173 px - 112.94 m
Width 2 - 60 px - 39.17 m
True width 2 - 39.17/cos45 deg = 55.39 m
Height 3 - 134 px - 87.48 m
Width 3 - 65 px - 42.44 m
True width 3 - 42.44/cos45 deg = 60.02 m
Innitial support: 11 px - 7.18 m
Volume 1:
120.92*69.86*42.47 = 358764.10 m^3
Volume 2:
120.92*112.94*55.39 = 756444.88 m^3
Volume 3:
120.92*87.48*60.02 = 634896.45 m^3
Total volume: 634896.45+756444.88+358764.10 = 1750105.43 m^3
Using density of 316 kg/m^3 we will get...
1750105.43*316 = 553033316 kg
Cross sectional area of the building:
69.86*42.47+112.94*55.39+87.48*60.02 = 14473.25 m^2
Rotational volume 1 (as 3 cilinders, axis across the building's corner):
1) 69.86*pi*42.47^2 = 395861.24 m^3
2) 112.9*pi*55.39^2 = 1088194.55 m^3
3) 87.48*pi*60.02^2 = 990035.19 m^3
Total rotational volume: 990035.18+1088194.55+395861.24 = 2474090.97 m^3
Now the can use Pappu's centroid theorem to find the center of mass:
2474090.97/(2*pi*14473.25) = 27.21 meters from the front of the building
For the second rotational volume (as 3 cilinders, axis as the base of the building) we need to subtract width of each section of the building to get the hight of each of the cilinders
60.02-55.39 = 4.63 m
55.39-42.47 = 12.92 m
Rotational volume 2
1) 42.47*pi*(69.86+112.94+87.48)^2 = 9746752.86 m^3
2) 12.92*pi*(69.86+112.94)^2 = 1356328.13 m^3
3) 4.63*pi*69.86^2 = 70988.50 m^3
70988.50 + 1356328.13 + 9746752.86 = 11174069.50 m^3
So according to the Pappu's centroid theorem the center of mass is also:
11174069.50/(2*pi*14473.25) = 122.88 meters above the ground.
Knowing the location of the center of mass we can find the angle of it in relation to the opposite side of the building and the distance from it
60.02-12.92 = 47.10 meter away from the back of the building
Angle: atan(47.1/122.88) = 0.366 rad
Distance from the opposite corner (hypotenuse): sqrt(47.1^2+122.88^2) = 131.60 m
I don't have any problem this the second scan from my old calc so I gonna use his values:
Total Falling Distance: 7.18 - (1.78+1.381) = 4.02 m
Tipping angle: asin(4.02/60.02) = 0.0670 rad
Finally, we can find the displacement of the center of mass and total potential energy
0.0670+0.366 = 0.433 rad
131.60*(1-cos0.366) = 8.72 m
131.60*(1-cos0.433) = 12.15 m
12.15-8.72 = 3.43 meters
PE = 553033316*9.81*3.43 = 1.86086309e10 Joules, 4.45 tons (Large Building level)
Now lifting strength:
131.60*sin0.433 = 55.22 m
553033316*(55.22/60.02) = 508805393 kgf (Class M)