Hydrostatic force on submerged rectangle

[Urmi Roy]

Homework Statement

(Referring to attached diagram)

Find the hydrostatic force F2 on the rectangular part and the point of action of this force on rectangular area.
The triangular part is submerged in oil of specific gravity 0.8, while the rectangular part is submerged in water. The whole structure is upright with no inclination with fluid surface.

Homework Equations

Hydrostatic force= ρghA, where h=perpendicular distance of centroid of object from fluid surface.
ρ= water density
A= area of surface submerged

The Attempt at a Solution

According to me:
F2= ρxgx(height of triangle+further distance to get centroid of rectangle)x(area of rectangle)
= 1000x9.81x(3+1)x(2x4)

However the solution in the same book gives
F2= 9.81x1000x(3x0.8 + 1)x(2x4)

I don't understand why 3 is multiplied by 0.8.

For point of action of force F2,
According to me,
y_p (vertical distance from fluid surface of point of action)
= ∫(integral between 0 to 5m) ρgz(0.8xarea of triangle+4x(z-3))

where z= total vertical distance downward from fluid surface

In the solutions on my book,
it gives the right answer as:
y_p (vertical distance from fluid surface of point of action)
= ∫(integral between 3 to 5m) ρgz(3x0.8+(z-3))(4dz)z

Again, there's that 0.8x3...Also, it's as if they're ignoring the triangle since the integral is between 3 to 5 m...but then why would the 3x0.8 be in there?

 

[SteamKing]

Do you have the rectangle submerged in just water, or is there something else on top of the water? Is that something else massless?

 

[Urmi Roy]

Hi SteamKing,

As you can see from the figure I attached, above the rectangle is a triangle submerged in oil of sp. gravity 0.8. So there is basically a layer of oil above the water layer.

Thanks!

 

[Chestermiller]

What is your equation for the hydrostatic pressure at the base of the triangle?

Chet

 

[Urmi Roy]

Hi Chet!
So for the triangle, I just used the equation F=ρghA

h= (2/3)x3..because for a triangle the centroid is at 2/3rds of the height

A=1/2 x base x height

I didn't find the hydrostatic force at the base of the triangle in particular. However if I consider a narrow horizontal strip at the base, it will be
F=ρgx h x A
F= ρg (3) (basexdz)
(z is vertical downward coordinate)
(the narrow strip will have its centroid at z=3

 

[Chestermiller]

Actually, I was asking for the pressure at the depth of the base of the triangle, not the force on the triangle.

Chet

 

[Urmi Roy]

Wouldn't that just be ρgh? The same as I said above, without multiplying in the area...

 

[SteamKing]

Wouldn't that just be ρgh? The same as I said above, without multiplying in the area...

But then, isn't it reasonable to assume that the presence of the oil on top of the water influences the pressure found below the waterline?

 

[Urmi Roy]

The pressure on the triangle at the base is ρgh, where h is the total vertical length of the triangle.

That pressure doesn't work on the rectangle, so I don't understand why the force on the rectangle would include just that ρghxA_rectangle

I would understand if one said that since the triangle and rectangle are joined, the force acting on the triangle would act on the rectangle too...but then the force due to the oil on the rectangle would just be ρghxA_triangle...(not area of rectangle)..and we would add that to the force on the rectangle die to water...as I described in the opening post, that's not what they do.

 

[Chestermiller]

Yes. I understand your concern, but please bear with me.

When you say that the pressure at the base of the triangle is (pho)gh, what values are you using for rho and h?

From your result for the pressure at the base of the triangle (top of the rectangle), what is the pressure at the base of the rectangle? What is the average pressure over the face of the rectangle?

Chet

 

[Urmi Roy]

Yes. I understand your concern, but please bear with me.

When you say that the pressure at the base of the triangle is (pho)gh, what values are you using for rho and h?

From your result for the pressure at the base of the triangle (top of the rectangle), what is the pressure at the base of the rectangle? What is the average pressure over the face of the rectangle?

Chet

Hmmm...when you say base of triangle, I get confused. I know that if its the whole of the triangle, the h is the vertical coordinate of the centroid of the triangle.

As for your other questions, I don't see how the pressure at the base of the rectangle is affected by pressure at base of the triangle.
As for the average pressure over the rectangle, the h in ρgh is the coordinate of the centroid of the rectangle i.e. 4m...and rho (ρ) is density of water.

 

[Chestermiller]

Hmmm...when you say base of triangle, I get confused. I know that if its the whole of the triangle, the h is the vertical coordinate of the centroid of the triangle.

As for your other questions, I don't see how the pressure at the base of the rectangle is affected by pressure at base of the triangle.
As for the average pressure over the rectangle, the h in ρgh is the coordinate of the centroid of the rectangle i.e. 4m...and rho (ρ) is density of water.

OK. Please don't mention the word centrioid again. What I'm asking about is not about centroids and areas.

When I say the base of the triangle, what I am referring to is the liquid pressure at the depth of the base of the triangle. The pressure at this location is the same, irrespective of whether the triangle is present or not. It just depends only on the depth of the oil (which, in this problem, happens to be the same as the depth of the base of the triangle).

Chet

Chet

 

[Urmi Roy]

Okay, so at the base of the rectangle it would be the (rho x g x h) in oil layer, h is total depth of oil layer + (rho x g x h) in water layer till base of the rectangle.

Average pressure over the rectangle's face is the above expression divided by area of rectangle face?

 

[Chestermiller]

Okay, so at the base of the rectangle it would be the (rho x g x h) in oil layer, h is total depth of oil layer + (rho x g x h) in water layer till base of the rectangle.

Average pressure over the rectangle's face is the above expression divided by area of rectangle face?

Nope. You used the forbidden word "area."

The pressure at the depth of the oil layer (top of rectangle) is
$$p=ρ_{oil}gh_{oil}$$
and the pressure at the depth of the base of the rectangle is
$$p=ρ_{oil}gh_{oil}+ρ_{water}gh_{water}$$
So, the average pressure acting on the rectangle is:
$$p_{ave}=\frac{ρ_{oil}gh_{oil}+(ρ_{oil}gh_{oil}+ρ_{water}gh_{water})}{2}=ρ_{oil}gh_{oil}+\frac{ρ_{water}gh_{water}}{2}$$

So, now, what is the pressure force acting on the rectangle (finally, this time, you are allowed to use the forbidden word "area")?

Chet

 

[Urmi Roy]

I get it so far.

So the net force on the rectangle is the P_avg x area of rectangle, I think.

 

[Chestermiller]

Yes.

As far as the point of application is concerned, the pressure force on the rectangle is a distributed load, so there is no specific point of application. But maybe the problem is asking you to find the depth at which the moment above that depth is equal to the moment below that depth.

Chet

 

[Urmi Roy]

Yes, that is what they're asking. Thanks for your help! I'm much clearer about this now.

 

[Chestermiller]

So please show us how you determined the point of application of the force, and what your result was.

Chet