How do you find to torque if force is applied over entire levl

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SUMMARY

The discussion focuses on calculating the torque exerted by water on a rectangular hatch in a box. The total force on the hatch is derived from the equation Force = Pressure x Area, with pressure varying with depth. The integration of pressure over the hatch surface is necessary to account for this variation, leading to the formula for total force: 1/2(density)(gravity)(width of the wall)(height of the wall)^2. Torque is then calculated using the integral of the product of distance from the reference point and pressure over the hatch surface.

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bjon-07
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Lets say you have a box that is filled with water. One of the side of the box contains retangular hatch. How much torques does the water exert on the hatch?


Do you just calculate the total force that the water is exerting on the square are of the rectangular hatch. Force=Pressure X Area ?
 
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bjon-07 said:
Lets say you have a box that is filled with water. One of the side of the box contains retangular hatch. How much torques does the water exert on the hatch?


Do you just calculate the total force that the water is exerting on the square are of the rectangular hatch. Force=Pressure X Area ?

Pressure depends on depth, so the force is not uniform. You well have to integrate over the surface, breaking it up into strips of constant force at constant depth. I assume you want the torque calculated about the top or the bottom of the hatch. The answer depends on the rotation axis you choose.
 
I found the total force that a fluid will expert on a wall using intergration,


since force is equal to pressure X times area and pressures = density x gravity x height = interal of PGH dh


which turned out to be 1/2(density)(gravity)(width of the wall)(height of the wall)^2

now that I have the total force exerted on the object, is there a way to find the total toruqe exerted on the object. Let's make the hing of the box be the folcrum
 
bjon-07 said:
I found the total force that a fluid will expert on a wall using intergration,


since force is equal to pressure X times area and pressures = density x gravity x height = interal of PGH dh


which turned out to be 1/2(density)(gravity)(width of the wall)(height of the wall)^2

now that I have the total force exerted on the object, is there a way to find the total toruqe exerted on the object. Let's make the hing of the box be the folcrum
HINT:
Since fluid pressure ⊥ Hatch Surface:

1: \ \ \ \ \textsf{Torque} \ \, = \, \ \int_{Hatch} r \, P \ dA

where "r" is the distance of Area Element "dA" from the Reference Point or Axis, and "P" is the fluid pressure on Area Element "dA". Both "r"and "P" will be functions of Area Element "dA" position on the Hatch Surface. You'll likely express "dA" in terms of thin rectangular area elements over the Hatch Surface.


~~
 
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