Help solving Fluid Force problem

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Qbit42
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Homework Statement


A 2.20m wide tank is filled with water totaling 1.87m (H1). There is a hatch cosisting of a portion of a wall of the tank which is 1.15m(H2) in length (measured from the bottom of the tank). What is the Force on the hatch resulting from the water?

Homework Equations


F = INTEGRAL<UPPERBOUND = b, LOWERBOUND = a> ( wgp(b-y)dy ) = 0.5wgp[b^2 - a^2]

p = density of water = 1000 kg/m^3

The Attempt at a Solution


I've tried this problem many different times and I can't seem to get the correct answer. From my point of view the lower bound of integration would be 0.72 (1.87 -1.15) and the upper bound would be 1.87. This gives an answer of 32108 N, which is incorrect. Any help is greatly appreaciated!

Edit: Sorry about being unable to use the integral in the pallete correctly, I can't figure out how to but in bounds (if you can).
 

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Qbit42 said:
1.87m (H1). 1.51m(H2)

[From my point of view the lower bound of integration would be 0.72 (1.87 -1.15) and the upper bound would be 1.87.

Qbit42 said:
F = INTEGRAL<UPPERBOUND = b, LOWERBOUND = a> ( wgp(b-y)dy ) = 0.5wgp[b^2 - a^2]
I'm also not quite sure this integral is fine...
 
Fixed up that typo. The integral is one which i got from my physics textbook. Well actually the integral is just 0.5wpgb^2 since the problem in the text is asking about the pressure over the entire wall (ie from 0 to b) but all i did was make the lower bound non-zero.
 
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Well a is the point from which I'm measuring the liquid and b is the point at which I stop measuring. So a = 0.72 m and b = 1.87 m. As for how y is defined its the distance from the top (ie a) to a infindesimal strip (ie dA = wdy).