hydrostatic pressure

• August 19th 2009, 12:43 PM
dat1611
hydrostatic pressure
An aquarium 8 m long, 5 m wide, and 4 m deep is full of water. Find the following:
the hydrostatic pressure on the bottom of the aquarium 39200,
the hydrostatic force on the bottom of the aquarium 568000 ,
the hydrostatic force on one end of the aquarium?

i got the first 2 but i cant get the force on 1 end

the equation i got is integral from depth1 to d2of (density)(gravity)(length)x
what i tried was integral for 0 to 4 of (1000)(9.8)(4)x
but it did not work for me
• August 19th 2009, 01:18 PM
Mauritzvdworm
try
$\int^{5}_{0}\int^{4}_{0}\rho gydydx$
because $\rho gy$ is the pressure and to find the force you will need to consider one of the ends of the equarium and calculate the force and each differential area
• August 19th 2009, 11:31 PM
enjam
I don't think you calculated the hydrostatic force at the bottom of the aquarium correctly.

Water is 1000kg/m^3 = 9800N/m^3
So the weight of the water in the tank is 9800N/m^3 x (160m^3)
So F = 1568x10^3 N, this is the force that the bottom of the tank must oppose.
Or, you could say P at the bottom = 39,200Pa
Since P x A = F, we have 39,200 N / m^2 x 40 m^2 = 1568kN

By 'one end', do you mean the side of the tank, or one of the corners?
If you mean the side, it's simply F = rho x g x height to centroid
= 1000 kg/m^3 x 9.8 m/s^2 x 2 m x (1 N / kg.m/s^2)
= 19,600 N/m^2
Multiply this by the area of the side, and you have
19,600N/m^2 x either (4m x 5m) or (4m x 8m)
and you will get a value for force.