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Finding Hydrostatic Force on the Plate

  • Thread starter McAfee
  • Start date
  • #1
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Before I begin I would like to say hello to everyone. This is my first post. I will follow and abide by all rules here.

Homework Statement


A rectangular plate measuring 2m by 5m is suspended vertically in a liquid with density 500 kg per cubic meter in such a way that one of its longer sides is on top and is 3 meters below the surface of the liquid. Find the hydrostatic force on the plate.



Homework Equations


The force(F) equals = Area*depth*density*acceleration

The Attempt at a Solution


F= A * depth * density * acceleration
The density and acceleration are 500 * 9.8 = 4900, so that goes outside the ∫
The bounds of the integral are 3 to 5 since the plate is 3 meters deep.
The Area I think is equal to 5 * x and the depth I think is (3-x)

So my final integral would be 4900 ∫5x (3-x) dx from 3 to 5

Not sure if I'm right or not
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
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Before I begin I would like to say hello to everyone. This is my first post. I will follow and abide by all rules here.

Homework Statement


A rectangular plate measuring 2m by 5m is suspended vertically in a liquid with density 500 kg per cubic meter in such a way that one of its longer sides is on top and is 3 meters below the surface of the liquid. Find the hydrostatic force on the plate.



Homework Equations


The force(F) equals = Area*depth*density*acceleration

The Attempt at a Solution


F= A * depth * density * acceleration
The density and acceleration are 500 * 9.8 = 4900, so that goes outside the ∫
The bounds of the integral are 3 to 5 since the plate is 3 meters deep.
The Area I think is equal to 5 * x and the depth I think is (3-x)

So my final integral would be 4900 ∫5x (3-x) dx from 3 to 5

Not sure if I'm right or not
Welcome to the forums! When you are integrating you are basically summing over small rectangular slices of the area. I think you mean x to be the depth of that slice. Picture a slice of your plate going from x to x+dx in depth. That makes the area of that slice dx * width of the plate, or 5*dx, right? And depth of the slice is just x. Try and set that integral up again.
 
  • #3
96
0
Welcome to the forums! When you are integrating you are basically summing over small rectangular slices of the area. I think you mean x to be the depth of that slice. Picture a slice of your plate going from x to x+dx in depth. That makes the area of that slice dx * width of the plate, or 5*dx, right? And depth of the slice is just x. Try and set that integral up again.
I decided to set up the integral differently so now I have:
4900∫5(3+x)dx from 0 to 2
=196000 Newtons

(3+x) is the depth
4900 is the density*acceleration due to gravity(both are givens)

or I could have done

4900∫5x dx from 3 to 5

Helped me realize I would get the same answer.
 
  • #4
Dick
Science Advisor
Homework Helper
26,258
619
I decided to set up the integral differently so now I have:
4900∫5(3+x)dx from 0 to 2
=196000 Newtons

(3+x) is the depth
4900 is the density*acceleration due to gravity(both are givens)

or I could have done

4900∫5x dx from 3 to 5

Helped me realize I would get the same answer.
Sure. Good job!
 

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