Calculating Force Exerted by Water on a Tilted Plate in a Tank

[NoobDoingMath]

Hi I have a midterm study guide question. This one has stumped me for a while and probably the only one undone.

Homework Statement

Suppose there is a semi-circular plate of radius 5 ft that rests on its
diameter and is tilted at 45 degree angle to the bottom of a tank lled with water to depth
6 feet. Find the force exerted by the water against one side of the plate. (The
weight-density of water is 62.4 lb=ft^3)

Homework Equations

So I'm reading the book and I know that to solve the problem Force is weight-density of water (62.4) times the depth (6-y) and the area.

Now the problem is I'm not quite sure how to approach the area. I just can't seem to grasp the image of the tank. Not to mention the 45 degree angle really confused my approach. I was under the assumption that its a 6ft tall cylinder with length 10ft and a plate on the bottom tilted at 45 degrees

The Attempt at a Solution

Problem seems simple, but I can't seem to figure out the 45 degree plate to find the area.

What i have is:

Integral from 0 to 6 of (62.4)(6-y)(area)

Now this is assuming that I approached this correctly.

 

[tiny-tim]

Welcome to PF!

Hi NoobDoingMath! Welcome to PF!

What i have is:

Integral from 0 to 6 of (62.4)(6-y)(area)

Yes, that's basically correct.

You seem to be confused about the area …

I suggest in future you always use the slicing method.

In this case, slice the plate into horizontal slices of vertical distance dy …

then find the area of that slice (it'll be dy√2 times the width, won't it?)

 

[NoobDoingMath]

Still a little bit confused, and I want to see if I'm understanding correctly. My "math English" isn't too good. :shy:

So √2 is a result of the 45-45-90 triangle right?
Therefore the slice is √2dy*width

The width is 10ft since its radius is 5 and the plate rest on its diameter?

Resulting my solution to be:

∫(62.4)(6-y)(10√2)dy a=0, b=6

My answer would become:

11232√2

 

[tiny-tim]

Hi NoobDoingMath!

(just got up :zzz:)

The width is 10ft since its radius is 5 and the plate rest on its diameter?

Nooo, you're not thinking straight.

Or, rather, you are thinking straight, and you should be thinking circular! ​

The width has to be the width of the slice …

that's the whole point of slicing …

you add the area of each slice, and that depends on y, doesn't it?

Try again! ​