How Is Total Force Calculated on a Dam Wall?

[hannsparks]

Figure 2 below shows water at a dam. The depth of the water at the dam is
20 meters. The area of the wall of the dam which is in contact with the water
is 500 m 2. The atmospheric pressure is 1 atmosphere and the mass per unit
volume density of the water is 1000 kg/m^3 Find the total force exerted on the wall which is in contact with the water.


so far I've found p=2.97e5
p=1.013e5+1000*9.8*20

p=F/A
F=2.97e5*500
F=1.485e8

i found avg pressure : (1.013e5+2.97e5)/2
=1.991e5

Im not sure what to do next please help

 

[HallsofIvy]

I don't see any "figure 2 below". And, unfortunately, the shape of the dam may be important. The pressure upon parts of the dam 20 feet below the water level (where each the pressure will be 1000*20*9.8) is greater than the pressure upon parts of the dam only slightly below the water level (where there is very little water above it to exert pressure).
And where did you get that figure "1.013e5"? Is that due to atmospheric pressure? That should be very small.

 

[KKK]

You have asked the "Total" force exerted on the wall. It could be found by finding the average pressure acting on the wall and the area exposed to the pressure.
Depth of water is 20 metres. Pressure at top = 0 kgf/cm^2, pressure at bottom = 2 kgf/cm^2, average pressure = 1 kg/cm^2 = 10000 kg/m^2.
Area = 500 m^2. So "Total" force = 10000 x 500 = 5e6 kgf. = 5000 Tonnes