How Do You Calculate Fluid Force on a Rotated Square Tank?

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To calculate the fluid force on a rotated square tank filled with oil, one must set up a definite integral based on the tank's dimensions and the oil's weight-density. The tank has a height and width of 12 feet, with oil at a depth of 4 feet. The fluid force can be expressed using the formula f = pA, where pressure (p) and area (A) are functions of depth. The area function must account for the tank's 45-degree rotation, affecting how pressure varies with depth. Properly defining these functions is crucial for accurately calculating the fluid force.
tnutty
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A metal oil tank has cross-section that is a square rotated 45 degrees as shown in the figure above. Its height is 12 ft and width is 12 feet. The oil in the tank has a weight-density of 57 lb/ft3.
Write a definite integral that expresses the fluid force on the end of the tank when the oil is 4 feet deep?

Here is a link to the photo :
http://img17.imageshack.us/img17/2294/76984288.jpg

hopefully you can make it out.

I am not sure how to set up the definite integral.
 
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Well, I think you can figure out out that
f = pA
Right?

Area is a function that you have. Pressure is a function as well. What are these functions?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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