Why Is My Calculation of Vertical Hydrostatic Force Incorrect?

[The Real Nick]

MODERATOR'S NOTE: THIS THREAD HAS BEEN MOVED FROM ANOTHER FORUM, SO THERE IS NO TEMPLATE.

This is the problem here. I am having trouble making progress on this problem after getting the wrong answer for the vertical hydrostatic force.

The equations I am using are: fx= (density)(gravity)(Zcg)(Projected area)

For this I used
p= 1000 kg/m^3
g = 9.81 m/s^2
Zcg = 1.5m + (0.75/2)m
A = (0.75(1.2) m^2

this gave me 16,554.4 N, which is correct (for the horizontal component).

Now, trying to solve for the vertical hydrostatic force I used the same equation, but with Zcg being 1.5+0.75 meters as it is at the bottom of the quarter-circle. I am not sure what else to do, but this doesn't give me the right answer, it is short by about 3,000 N.

I have seen other solutions that solve for the vertical force by finding the "weight of the missing water" above the quarter-circle. I don't understand this method, and it was never explained in class or in my book, so I am weary to use it.

If anyone could help that would be great!

Thank you.

 

[The Real Nick]

Argh, I just realized I posted this in the wrong section. My apologies!

 

[SteamKing]

MODERATOR'S NOTE: THIS THREAD HAS BEEN MOVED FROM ANOTHER FORUM, SO THERE IS NO TEMPLATE.

This is the problem here. I am having trouble making progress on this problem after getting the wrong answer for the vertical hydrostatic force.

The equations I am using are: fx= (density)(gravity)(Zcg)(Projected area)

For this I used
p= 1000 kg/m^3
g = 9.81 m/s^2
Zcg = 1.5m + (0.75/2)m
A = (0.75(1.2) m^2

this gave me 16,554.4 N, which is correct (for the horizontal component).

Now, trying to solve for the vertical hydrostatic force I used the same equation, but with Zcg being 1.5+0.75 meters as it is at the bottom of the quarter-circle. I am not sure what else to do, but this doesn't give me the right answer, it is short by about 3,000 N.

I have seen other solutions that solve for the vertical force by finding the "weight of the missing water" above the quarter-circle. I don't understand this method, and it was never explained in class or in my book, so I am weary to use it.

If anyone could help that would be great!

Thank you.

For the hydrostatic force in the vertical direction, have you accounted for the variation in pressure due to depth, otherwise known as Pascals's Law:

https://en.wikipedia.org/wiki/Pascal's_law

Because pressure changes with depth, the c.g. and the center of pressure don't coincide in the vertical direction.

 

[The Real Nick]

For the hydrostatic force in the vertical direction, have you accounted for the variation in pressure due to depth, otherwise known as Pascals's Law:

https://en.wikipedia.org/wiki/Pascal's_law

Because pressure changes with depth, the c.g. and the center of pressure don't coincide in the vertical direction.

Well, I changed the Zcg to be the entire radius of the quarter-circle, as this would put it at the depth of the bottom of the circle, whereas for the horizontal component was the radius/2.

 

[SteamKing]

Well, I changed the Zcg to be the entire radius of the quarter-circle, as this would put it at the depth of the bottom of the circle, whereas for the horizontal component was the radius/2.

Yes, but as you found with your calculations, that didn't provide the correct answer.

 

[The Real Nick]

Yes, but as you found with your calculations, that didn't provide the correct answer.

This is true, but I'm not sure what else to do with your suggestion, as I feel I already tried that. The vertical component of the force would act on the horizontal plane of the bottom of the quarter-circle, so wouldn't the cp and cg coincide for that case?

 

[The Real Nick]

"The vertical component of pressure force on a curved surface equals in magnitude
and direction the weight of the entire column of fluid, both liquid and atmosphere,
above the curved surface."

This is from my textbook, and is adding to my confusion, because in this problem the surface technically has no water above it, but I know the pressure at the surface from the inside is affected by the column of water above it, which I thought my equation would handle, but it's not :/

 

[Chestermiller]

With regard to the vertical force, if the "missing water" region were actually filled with water, it's weight would just balance the upward force exerted on the arc from below. In other words, in this filled situation, if the arc weren't there, all the water would still be in equilibrium. It wouldn't know that the arc is there.

Chet

 

[The Real Nick]

With regard to the vertical force, if the "missing water" region were actually filled with water, it's weight would just balance the upward force exerted on the arc from below. In other words, in this filled situation, if the arc weren't there, all the water would still be in equilibrium. It wouldn't know that the arc is there.

Chet

I thought about that too, Chet. I tried to calculate the force by finding the volume of the water above the arc (as if there was water there), finding it's weight, and the force. This also gives me the wrong answer. I am starting to wonder if the answer in the back of the book is wrong. They get 22,600 N.

 

[Chestermiller]

I thought about that too, Chet. I tried to calculate the force by finding the volume of the water above the arc (as if there was water there), finding it's weight, and the force. This also gives me the wrong answer. I am starting to wonder if the answer in the back of the book is wrong. They get 22,600 N.

You can get the volume much more easily by just subtracting the area of the quarter circle.

 

[The Real Nick]

You can get the volume much more easily by just subtracting the area of the quarter circle.

That's what I did. Acted as if it were a rectangular cube, and then subtracted the area of the cylinder. This gives me 14,664.5N when converted to a force.

 

[The Real Nick]

That's what I did. Acted as if it were a rectangular cube, and then subtracted the area of the cylinder. This gives me 14,664.5N when converted to a force.

I am at my wits end with this problem. I will probably just go with this answer and hope the back of the book is wrong. If anyone has some suggestions otherwise, that would be great. Thanks again.

 

[Chestermiller]

My answer matches yours, plus I derived the equation by integrating the pressure from below over the surface, and got the same answer. So the book must be wrong.

Chet

 

[The Real Nick]

My answer matches yours, plus I derived the equation by integrating the pressure from below over the surface, and got the same answer. So the book must be wrong.

Chet

Thank you for being so thorough! I think I can finally move on to the next problem. FEWF! 1/5 complete, and only 3 hours later :D

 

[Chestermiller]

Thank you for being so thorough! I think I can finally move on to the next problem. FEWF! 1/5 complete, and only 3 hours later :D

Actually, I integrated it first because I couldn't figure out a simpler method of doing the problem. Then I realized that this was the same as the weight of the missing water.