Question (body immersed in fluid)

[manal950]

Hi all

Can please check my answer in this Question ?
and Is my diagram correct ?

http://store2.up-00.com/Mar12/PD451302.jpg

thanks

 

[manal950]

where are you ?

 

[LawrenceC]

Do you have the plate upside down?

 

[manal950]

you mean that diagram is wrong ..

 

[LawrenceC]

Problem states base is 4 m long. It is not 1 m from surface...the way you have it drawn.

 

[manal950]

As Problem state the base of triangular is four m while the height is 6 m ... and from free srface is immersed at the one m

 

[manal950]

Now is the diagram correct

http://store2.up-00.com/Mar12/NjT38265.jpg

 

[LawrenceC]

Yes, the sketch is now correct.

 

[manal950]

ok what about other answers

 

[LawrenceC]

You should rework the problem. Let's see your work.

 

[manal950]

p = pgh
= 900 X 9.81 X 3
= 26.5

but the unit here is kpa or N/m^2

Now center of pressure
= (IG / AX ) + X

= bh^3/36 = (4X 6^3) / 36 = 24

so now (24/12X 1 ) +

= 5

 

[LawrenceC]

When the problem says "determine the total pressure", it is unclear whether they are seeking the total pressure force due to the pressure at a designated point.

The designated point is generally the center of mass of the triangle. You have not calculated that. The center of pressure is a function of the moment of inertia, the area, and the center of gravity of the plate.

So you need to compute those three items first. Your moment of inertia is incorrect.

 

[LawrenceC]

Is the plate not fully submerged? Did you quote the problem statement exactly?

 

[manal950]

yes plate fully submerged

The center of pressure is located at the centroid of the triangular shaped pressure field 2/3 from the top of the water line."

From this I may compute:

1 + 2/3 .6 = 5

 

[LawrenceC]

The location of the centroid is 1/3 the distance from the base of the triangle to the apex opposite the base along the height of the triangle. What is the height (altitude)? Ybar=h/3. First you must determine h, then add the 1.0 to it.

 

[manal950]

I have a little difficulty

seeee...

Now I want know Xbar is 3 or 5 ?

I will try ...

presure = pg(xbar)

= 2.6 kpa----------------
2)
Center of pressure

Center of pressure =( IG/Area(xbar) ) = xbar
IG = bh^3/36
IG = 4 X 6^ / 36 = 24

Area = 1/2 X 4 X 6
= 12 m^2

Now ..
Center of pressure = (24 / 12 X 3) + 3

= 3.6 m pleeeeeeeeeeeeese help me now what is the wrong ..

 

[LawrenceC]

You calculation for center of pressure looks good except for round off error. I get 3.66 which rounds to 3.7 m for the center of pressure. You use the distance that the centroid of the triangle is below the surface for the calculation. It is 3 m.

Your pressure computation is off by a factor of 10. It should be 26 KPa.