HYDROSTATIC FORCE ON PLANE SURFACE
TITLE: HYDROSTATIC FORCE ON PLANE SURFACE
INTRODUCTION:
- The determination of force which are exerted by liquid which are at rest on surface immersed in liquids.
- From the study by hydrostatic, the following principles have been established :
ü There are no shear stresses present when the fluid is not in motion.
ü The pressure exerted by a fluid under hydrostatics conditions. This pressure acts perpendicular to an immersed surface.
ü Hydrostatic pressure various linearly, increasing with an increase in depth.
OBJECTIVE:
- The objective of this experiment are:
ü To determine experimentally the magnitude of the force of pressure (hydrostatic force) and its point of action (center of pressure) a plane surface.
ü To compare the experimental result with the theoretical values
HYDROSTATIC FORCE ON PLANE SURFACE
- A plane surface of total area, A which is immersed in a liquid at depth, h is subjected to hydrostatic pressure P.
- If the plane surface is horizontal as shown, then anywhere on the plane surface, the pressure is given;
P= ρgh
- Another explanation is shown in the THEORETICAL content.
THEORY
TOTAL PRESSURE
For a static fluid, the shear stress is zero and the only stress is true normal stress. Pressure which when in contact with solid surface exerts a normal force towards the surface. It is also call hydrostatic force.
F=ρghA
Where;
F=total pressure (hydrostatic force)
P=mass density of the liquid
g=gravitation acceleration
A=submerge area of the plane surface
H=vertical distance from the liquid surface to the centroid of the submerge plane surface
CENTRE OF PRESSURE
Centre of pressure is general below centroid since pressure increases with depth. Centre of pressure is determine by equating the moments of the result and distributed force about any arbitrary axis. The application point of the total pressure (hydrostatic force) on the surface.
h= Ia sin O + h
hA
Where;
h=vertical distance from the liquid surface to the centre of pressure
Ig=moment of inertial of the plane surface about its centroid G
PROCEDURE
Part A: vertical plane surface (O=0)
Counter balancing the water vessel.
- The water vessel is set to angle of O = O by using the detent
- The unit was counterbalance with using the rotating cylinders the stop pin is precisely in the true middle of the hole
Measurement
- The rider then mounted and the level arm is set at L=22 mm. The level arm (the distance from the rider to the centre of rotating of the water vessel ) was recorded
- The appended weight was placed and the value is recorded
- Water is added into the water vessel until the unit is balanced (until the stop pin is at the centre of the hole). The level of water in the vessel was recorded
- The water shape was determine for a different profile of pressure distribute (triangular profile or trapezoidal profile)
- Steps 4 and 5 were repeated for at least 5 value of appended weight.
Part B: Inclined plane surface (O=O)
· Water rider and appended weight was removed during counter balancing process.
· Steps 1 and 2 of the counterbalancing process were repeated for an angle O=O.
· Steps 3 to 6 of part A are repeated for measurement.
Results from the Experiment
Part A α=0°
| lever arm | Appended weight | Water Level | | Hydrostatic Force | |||
| L(mm) | Fc(N) | S1(mm) | S2(mm) | S(mm) | ℓ(mm) | h*(mm) | F(N) |
1 | 22 | 1 | 0 | 58 | 58 | 254.167 | 112.167 | 0.0866 |
2 | 22 | 2 | 0 | 82 | 82 | 176.042 | 58.042 | 0.25 |
3 | 22 | 3 | 0 | 104 | 104 | 165.432 | 69.432 | 0.399 |
4 | 22 | 4 | 0 | 124 | 124 | 161.261 | 85.261 | 0.5456 |
5 | 22 | 5 | 0 | 144 | 144 | 158.865 | 102.865 | 0.6924 |
Part B α=10°
| lever arm | Appended weight | Water Level | | Hydrostatic Force | |||
| L(mm) | Fc(N) | S1(mm) | S2(mm) | S (mm) | ℓ(mm) | h*(mm) | F(N) |
1 | 22 | 1 | 2 | 60 | 58 | 243.688 | 101.024 | 0.09 |
2 | 22 | 2 | 2 | 86 | 84 | 173.609 | 58.01 | 0.2534 |
3 | 22 | 3 | 2 | 106 | 104 | 164.986 | 69.518 | 0.4 |
4 | 22 | 4 | 2 | 126 | 124 | 160.977 | 85.57 | 0.5467 |
5 | 22 | 5 | 2 | 146 | 144 | 158.661 | 103.289 | 0.6933 |
CALCULATION
Part A
A= 7.5x10 m
h= 0.028 m
F= ghA
= (1000)(9.81)(0.028)(7.5x10 )
= 2.0233 N
Ig= bd
12
= 75(100)
12
= 6.25x10 mm
h=Ig sin O + h
hA
= (6.25x10) sin O + 28
(28)(7500)
= 28 mm
= 200 – (55-0)
3cos 0
= 181.667 mm
Part B
Case: triangular profile of pressure distribution
F=L Fe
= 200 (1)
180.490
= 1.11 N
h= 2s
3
= 2 (65 – 10)
3
= 36.67
Case: trapezoidal profile of pressure distribute
h= ( - 200) cos + s
= (165.657 – 200 ) cos 20 + (1070- 10)
= 64.73 mm
= 150 + (100)
12(s - 50 )
cos X
= 150 + (100)
12 (s - 50 )
cos X
= 165.657 mm
DISCUSSION
- From the experiment that were done;
ü The data is applied to the formula from theoretical value
ü From the data, it has a different between vertical plane surface and inclined plane surface.
CONCLUSION
- Conclusion from the experiment we can get that the fluid static it is the study of fluids which are not in motion.
- The are no relative motion between the fluid particles. The only stress will be normal stress which is equal to the pressure.
RECOMMENDATION
- There was an error when doing this experiment. The error is;
ü When the reading is taken, parallax error has occurred at the left and right water level
ü When pouring the water it may be over the limit that we want.
- So, we recommended that;
ü Change to the new equipment and apparatus.
ü Built the platform when taking the parallax error.
1 Response to HYDROSTATIC FORCE ON PLANE SURFACE
What is the difference between the total force and resultant force?
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