# Finding Hydrostatic Force on the Plate

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

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Dick
Homework Helper
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.

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.

Dick
Homework Helper
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!