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A cylinderical watter tank with a diameter of 6 meters is placed so that the axis of the cylinder is horizontal. Find the fluid force on circular end of the tank if the tank is half full. The density of the watter is 1000kg/m^3, and the acceleration of the gravity is 9.8m/s^2.

A) Draw the picture?

B) Set up the integral to determine the force?

c) The total of the end of the cylinder is about 176400?

Explanation / Answer

The pressure is (3 - y)(9.8)(1000)

The length of the strip is 2* sqrt [9 - (3-y)^2]

So the integral is
int (0to3) of {9800*(3-y)*2*sqrt[9 - (3-y)^2] dy}
Let u = 9 - (3-y)^2.
Then du = 2(3 - y) dy

So, now your integrand is,
{9800 sqrt(u) du}
After integration we get ,
9800*(2/3)*u^(3/2)
= 9800*(2/3)*[9 - (3-y)^2]^(3/2)
= 9800*(2/3)*27 = 176,400 newtons      (kg*m/s^2 = newtons)-------------ans.

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