Fluid force--calculus question?
Find the fluid force on the vertical side of the tank, where the dimensions are given in feet. Assume that the tank is full of water. (then there's a diagram of a rectangle with a height of 3 ft and a length of 6 ft.)
Find the fluid force on the vertical side of the tank, where the dimensions are given in feet. Assume that the tank is full of water. (then there's a diagram of a rectangle with a height of 3 ft and a length of 6 ft.)
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If the tank is rectangular in horizontal section, with perpendicular sides, the tank can be divided up into horizontal segments the full length (horizontal measure) of a side of the tank and width, height (vertical measure) of each segment. The force on each segment is the pressure at the depth of the segment, multiplied by the area of the segment. The sum of the forces of all segments is equal to the total force on the side of the tank. Pressure on each segment,
p = rho x g x d. where rho is the density of water, g is accel. of gravity and d is the depth of the segment from surface of the water. The force on the segment is the pressure times length (L) times height (h) of the segment:
The total pressure of the side = L x rho x g x integral[h (dh)] with limits between top and bottom of the tank. The solution is:
F = L x rho x g x h^2 / 2, calculated from 0 to depth of tank.
p = rho x g x d. where rho is the density of water, g is accel. of gravity and d is the depth of the segment from surface of the water. The force on the segment is the pressure times length (L) times height (h) of the segment:
The total pressure of the side = L x rho x g x integral[h (dh)] with limits between top and bottom of the tank. The solution is:
F = L x rho x g x h^2 / 2, calculated from 0 to depth of tank.
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The water pressure starts at zero at the top of the tank and goes up by a standard formula you have to look up. This pressure then presses on the wall of the tank. You just find the average pressure and multiply it by the area of the wall.