1. application

Theorem: The force on a submerged vertical plane area equals the product of the weight per unit volume, the submerged area, and the depth of the centroid of the area below the surface
F=wAy
1. What force must be withstood by a vertical dam 80ft long and 20ft deep? Ans. 500 tons
2. What force must be withstood by a trapezoidal dam 100ft long at the top, 80ft long at the bottom, and 20ft deep? Ans. 17,333w

2. What force must be withstood by a vertical dam 80ft long and 20ft deep? Ans. 500 tons

Is the dam for water? The dam is withstanding a body of water?

w for water is 62.4 lbs/cu.ft
1 metric ton = 2,204 lbs
1 short ton = 2,000 lbs
1 long ton = 2,240 lbs

Submerged 20 ft deep?
then centroid is at (20/2) = 10 ft below water surface.
So,
F = (62.4)(80*20)(10) = 998,400 lbs.

If your ton is metric ton,
F = 998,400 /2204 = 453 tons

If short ton,
F = 998,400 /2000 = 499 tons

If long ton,
F = 998,400 /2240 = 456 tons.

So your ton is short ton.
[It is easier to solve questions if they are defined very well, and are with complete information. ]

What force must be withstood by a trapezoidal dam 100ft long at the top, 80ft long at the bottom, and 20ft deep? Ans. 17,333w

Divide the trapezoid into
---a rectangle 80ft wide by 20ft high
---and an "inverted" triangle that is 20ft at at the top, zero at the bottom and 20ft high also.

Force on the rectangle is
F1 = w(80*20)(20/2) = 16,000w weight units.

Force on the "inverted" triangle is
[the centroid is (1/3)(20) from the top]
F2 = w*[(1/2)(20)(20)]*[(1/3)(20)] = 1,333w weight units

So, total force on the trapezoid = F1 +F2
= 16,000w +1,333w
= 17,333w weight units