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Response Details:
For the system to be in equilibrium, the total torques need to be equal from both directions.
As for the top, this can be broken down into a square whose center of gravity is at the center of the beam, and a triangle whose centroid is located 2 m from the right side of the beam.
Fblock = 600N/m * 6m = 3600 N @ 3 m from right
Ftriangle = 1/2 (600)*6m = 1800 N @ 2m from right
as for the bottom supports, they are both squares whose center is at their center.
Fbc = Wbc * .8m = .8Wbc @ 5m from right
Fde = 1m* Wde = Wde @ 1m from right
Now, from point F, we set up all of the torques (clockwise is positive, counter is negative, sum must be 0)
0 = .8Wbc*5m - 3600N * 3 - 1800 N * 2 + Wde
So, we find Wde = 14400-Wbc
Now, setting Wbc's center as the pivot point, we find
3600*2m + 1800*3m = Wde*4
Wde = 3150
Wbc= 14400-3150 = 11250
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