Answer:
Explanation:
Let's find the center of mass of the load assuming it is of uniform construction
The rectangular portion has an area of
(4.32)(2.16) = 9.3312 m²
The triangular portion has an area of
½(2.16)(2.16) = 2.3328 m²
so the weight of the rectangular portion is
W(9.3312 / (9.3312 + 2.3328)) = 0.8 W
and the weight of the triangular portion is
W(2.3328 / (9.3312 + 2.3328)) = 0.2 W
Balance moments about the front of the load to find the distance to the load center of gravity.
W(d) = 0.8W(4.32 / 2) + 0.2W(4.32 + 2.16/3)
d = 0.8(4.32 / 2) + 0.2(4.32 + 2.16/3)
d = 2.736 m
which is equivalent to 3.6 - 2.736 = 0.864 m in front of the rear tires
Knowing that the loaded weight of the rear wheels is twice that of the front wheels, we can deduce that the CG of the loaded truck is a third of the distance from the rear wheels to the front wheels.
CG loaded = (3.6 + 0.6 + 1.8) / 3 = 2.0 m
Now we can balance moments about the loaded CG to find the load weight
Measurements are to a rear wheel ground contact reference point
W[2.0 - 0.864] = 30000[3.6 - 2.0 + 0.6]
W = 58,098.6 N
58.1 kN is probably close enough
