Engineering Statics: pressure on dam question

[thepatient]

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I just had a question regarding the determination of force P on the body of water CDE. For weight P, they used 1/2(18ft)(1ft)(18ft)(62.4lb/ft^3). I understand that this is because we are using volume*specific weight to find net weight P on body CDE. What I don't understand is why did they use 18ft as the base of the triangle? Is this because the pressure varies with the depth in the dam? Will this always be the case where there is a 1:1 ratio between height and width when considering the force per length on a distributed load due to pressure? Thanks. :]

 

[LawrenceC]

P, according to problem statement, is the force due to pressure. To go from pressure to force you must multiply by area. They calculate the average pressure by .5*18*62.4. That must be multiplied by an area to get force P. The area is 1.0*18.

 

[SteamKing]

The pressure at the bottom of the water is the depth * weight density of the water or
p = 18 * 62.4 = 1123.2 lbs / sq.ft.

Since the pressure of the water varies with depth, the average pressure is
0.5 p = 0.5 * 1123.2 = 561.6 lbs / sq. ft.

In order to calculate the total hydrostatic force acting on a 1-foot slice of the dam,
the average hydrostatic pressure must be multiplied by the area covered by the water.
The area is 18 feet deep by 1 foot wide,
so A = 18 x 1 = 18 sq. ft.

The hydrostatic force is equal to avg. pressure * area or
P = (p avg) * area = 561.6 * 18 = 10,108.8 lbs. or 10,109 lbs.