Help solving Fluid Force problem

In summary, the conversation discusses a problem involving a tank filled with water and a hatch located 1.15m from the bottom of the tank. The goal is to determine the force on the hatch resulting from the water. The conversation also mentions the use of the integral F = 0.5wgp[b^2 - a^2] where w is the width of the tank, g is the gravitational constant, and p is the density of water. After several attempts, the calculated force of 32108 N was found to be incorrect and further clarification is needed.
  • #1
Qbit42
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0

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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  • #2
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...
 
  • #3
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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  • #4
How are you defining y? And what are you using now for a and b?

The key here, which you seem to be aware of, is that the pressure is p*g*depth.
 
  • #5
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).
 

1. What is fluid force?

Fluid force, also known as hydrostatic force, is the force exerted by a fluid on an object placed in it. It is caused by the pressure of the fluid acting on the surface of the object.

2. How do you calculate fluid force?

The formula for calculating fluid force is F = ρghA, where F is the fluid force, ρ is the density of the fluid, g is the acceleration due to gravity, h is the depth of the fluid, and A is the surface area of the object.

3. What is the difference between fluid force and buoyant force?

Fluid force is the overall force exerted by a fluid on an object, while buoyant force is a specific type of fluid force that is equal to the weight of the fluid displaced by the object.

4. Can fluid force be negative?

Yes, fluid force can be negative if the object is pushing against the fluid rather than being pushed by it. This can happen when an object is moving upwards in a fluid.

5. How is fluid force used in real life?

Fluid force is used in various applications, such as designing buildings and bridges to withstand water pressure, calculating the force on ships and submarines, and understanding the movement of ocean currents. It is also important in fluid mechanics and hydraulic engineering.

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