# Hydrostatic force problem

1. Mar 7, 2012

### matthew_hanco

Using the picture added Q4, work out the hydrostatic force applied to AB when the water level is at 9cm and 12.8cm.

50g of weight was added to balance the setup.
And 260g and 460g were used respectively.

width of container was 75mm

Also need the moments

Heres what ive done

F when level of water is 9cm
= 0.09^2 x 0.075 x 9.81 x 1000
= 5.95N

Acting at 2/3 of height so 30mm or 172 from the centre.

total moment = (260 x 9.81) x202 = 515Nmm

I would use the same principal for the 12.8cm but i think that is totally wrong due to the curve in the apparatus.

Can anyone help?

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2. Mar 8, 2012

### matthew_hanco

could really use some help?

3. Mar 10, 2012

### LawrenceC

Pressure/force exerted by a stationary fluid is always normal to the wetted surface. Therefore any forces on the radii, outer or inner, pass through O. What does that say about any moment about O resulting from the curved surfaces?

4. Mar 10, 2012

### matthew_hanco

Are you suggesting that the pressure applied to the curved outersurface is equal to the pressure applied to the line AB. Obviously times by the leaver arm which is 202mm.

Also how would i calculate where the Resultant Force F would act, as for a straight line it would be 2/3 of the height, but im not sure for the curved section.

Thanks for the help its greatly appreciated.

5. Mar 10, 2012

### LawrenceC

No, I'm not saying that. All I am saying is that the pressure on the curved inner surfaces do not contribute to a moment about point O because their line of force passes through point O. The surface AB contributes to the moment, however. If you sum moments about A, the curved surfaces contribute complicating the problem.

You should be able to solve this problem by either of two approaches (assuming I perceive the problem correctly). You could compute the weight of the water in the semi-annular volume. Then find its center of gravity and determine the moment in that manner. This moment is balanced by the addition of weight.

Or you could solve it by determining the resultant force on AB, where it is applied, and the resulting moment that is balanced by weight addition.

6. Mar 10, 2012

### matthew_hanco

So as im just taking moments about O, then the force applied by the mass should be equal to the force applied by the force on ab.

Would i then simply take 12.8cm and use it in the equation
0.128/2 * .128*.075 * 9.81 * 1000 = 6.02 N and would act at 159mm from 0.

then the moment equation would be 6.02 * 159 - 4.5 * 203 = 0
But when i calculate it it is 40Nmm out. the 6.02 is out.

Have i still done something wrong?

7. Mar 10, 2012

### LawrenceC

I've worked out the case for a depth of 9 cm balanced by 260 g mass. You came up with 5.95N force. I don't compute that force.

Where is your center of pressure (Ycp) where the pressure force may be considered applied? The formula is:

Ycp = Icg/(Ycg*A) + Ycg

where

Icg is moment of inertia
A is area
Ycg is center of gravity

When I run the numbers for your first case I am off by less than 1% on the moment balance.

Last edited: Mar 10, 2012
8. Mar 10, 2012

### LawrenceC

OK I see your center of pressure for a depth of 9 cm so ignore my comment about center of pressure. Your force is incorrect.

9. Mar 10, 2012

### matthew_hanco

ive recalculated the force for 9cm

i get 9cm, using the formula, h/2 * h * w * g * p = F
where h =height, w = width g= 9.81 and p = 1000

so i get 0.09/2 * 0.09 * 0.075 * 9810 = 2.98N
Acting at 2/3 from top of water so 30mm
So the level arm turns out to be 172mm

Putting into the moment equation
FabD - FwZ = 0
Where Fab is the force acting on the line ab and Fw is the force of the mass
where D and Z are the respective lever arms

2.98 x 172 = 512.5 (FabD)
0.260 x 9.81 = 2.55 (Fw) 0.260kg is the mass added to balance the system, as the total weight is 0.310kg but 0.05Kg was needed to balance the system before the water is added.

There fore 2.55 x 203 = 517.65

i think this is right as 517.6 is roughly equal to 512.5.

What do you think?

10. Mar 10, 2012

### LawrenceC

That's it. I also computed 2.98N.

11. Mar 10, 2012

### matthew_hanco

ok so now we have the 9cm correct what about the 12.8cm

From what you have said, i should just take the Force acting on AB and when calculated im out by 40nmm which is a little too much, i mean it could be right but i feel i may have made a mistake.

So here's what i did using the same formula

0.128/ 2 * 0.128 *0.075* 9810 = 6.027N
acting at 2/3 from the top of the water
0.128/3 =0.043
0.202-0.043 = 0.159

Putting into the same equation FabD - FwZ = 0

FabD = 6.027* 159 = 958.30

Fw= 0.460*9.81 = 4.51
FwZ = 4.51 * 203 = 916.06

difference of 42.24 which is a little less than 5%

Im wondering if ive made an error.

12. Mar 11, 2012

### LawrenceC

I worked out the second one where the liquid is 12.8 cm and was off on the moment by about 44 N-mm so I agree with your figures. Incidentally, I worked it this time by determining the weight of the water and its center of mass. I used xbar as the moment arm and the weight as the force. Gauss quadrature was implemented for the integrations.

So two methods give the same answer. I suspect an error in measurement.

13. Mar 11, 2012

### matthew_hanco

i guess that must be the case thankyou for all your help its being most appreciated. Could i trouble you to look at my other problem. It is the orifice question, although i just need confirming that i have done the right thing.

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