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

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 ?

Last edited:

## Answers and Replies

OlderDan
Science Advisor
Homework Helper
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. Lets make the hing of the box be the folcrum

xanthym
Science Advisor
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. Lets 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.

~~

Last edited: