CENTER OF PRESSURE
INSTRUCTED BY:
Mr. A. SIRIWARDENA
NAME: T.D.D. SAMARANAYAKE
REGISTRATION NO: EN/97651
INDEX NO: 20/ENG/123
GROUP: B4
DATE OF PRACTICAL: 31/08/2021
DATE OF SUBMISSION: 07/09/2021
AIM
The practical was conducted with the aim to broaden the understanding concerning hydrostatic forces and
pressure acting on fully or partially submerged plane surfaces. Furthermore the practical was also performed
with the intention of comprehending the distinctive natures of the center of gravity and the center of pressure
and to determine the center of pressure of a submerged body using an appropriate apparatus.
APPARATUS
(8)
(11)
(10)
(9)
(1)
(7)
(2)
(6)
(3)
(5)
(4)
Figure 01: P6237 – Centre of Pressure Apparatus (Claydon, 2015)
(1) Level indicator
(2) Weight hanger
(3) Spirit level
(4) Drain valve
(5) Adjustable feet
(6) Quadrant
(7) Scale
(8) Adjustable counterbalance
(9) Knife edge pivot
(10) Balancing arm
(11) Clamping screw
01
THEORY
When a plane surface is submerged fully or partially in a fluid of which’s relative motion between its layers
is zero (that is the fluid is at rest or in equilibrium) hydrostatic pressure caused due to the gravitational force
acting on the fluid particles acts normal to the submerged surface.
To determine the total pressure and the center of pressure of a submerged plane surface the following
situation is considered as shown in Figure 02.
G – Center of gravity
P – Center of pressure
Figure 02: Hydrostatic pressure force on an inclined arbitrary plane
(Hamill, 2011)
Pressure acting on the submerged surface
Considering the top part of the submerged surface the force acting on its top surface is a result of the weight
of the water particles above it. However as the plane surface is inclined there is difference in magnitude of the
forces acting on the shallow parts when compared to parts that are deeper.
For an elemental strip of an area of δA at a distance of L from O at a depth h, submerged in a fluid of a
density of ρ the pressure acting on it can be identified as
dP = ρgh = ρgLsinα
Therefore the force acting on the strip is
dF = dPdA = (ρgLsinα)dA
Total force acting on the whole surface can be found by integrating
F = ∫ (ρgLsinα)dA
However as ∫ LdA = 1st moment of area about O = LGA
F = ρgsinα LGA
P = F/A = = ρghG
[As hG = LG sinα]
02
Center of Pressure of the Submerged Surface
To determine the point at which the total center of pressure acts moments about point O is taken
Moment = ∫ (ρgh)dA × L
Moment = ∫ (ρgLsinα)dA × L
Moment = (ρgsinα) ∫ L2 dA
Moment = (ρgsinα) I0
Equation 1
[ ∫ L2 dA = Second moment of area = I0 ]
Furthermore the total moment due to the hydrostatic pressure
Moment = FLP
Moment = ρgsinα LGALP
Equation 2
Equating the equations 1 and 2
ρgsinα LGALG = (ρgsinα) I0
LGALP = I0
LP = I0 / LGA
Using parallel axis theorem
I0 = IG + A(LG)2
L P = (IG + A(LG)2) / (LGA)
L P = (IG / (LGA) + LG
The distance between the center of pressure and the center of the area can be derived as follows
L P - LG = (IG / (LGA)
(hP - hg) /sinα = (IG / (LGA)
hP - hg = (IG / (LGA) × sinα
Application of the theory to the experiment
The experiment uses a three dimensional quadrant of which two levels of being submerged is witnessed
namely partial (case 1) and fully (case 2) submerged. As seen in Figure 03.
L
Figure 03: Center of pressure apparatus its two levels of evaluation
(Texas Education, 2021)
03
When submerged the pressure due to hydrostatic forces acts normally on the quadrant’s surface
However
The forces acting on the upper and lower curved surfaces of the 3 dimensional quadrant will not be
considered as their lines of action go through the pivot point around which the moment is considered
The forces acting on the sides if the quadrant cancels each other out as they are equal in magnitude and
opposite in direction
Therefore only the forces acting on the vertical rectangular cross sectional surface would be considered
when taking the moment
As the hydrostatic force exerted on the vertical rectangular surface is counteracted with the counterbalance
weights the force on the vertical rectangular surface can be calculated by the balance weights and the depth
of water
04
PROCEDURE

Equipment preparation
1. The apparatus was positioned on the work surface of the hydraulic bench and its base was leveled by
adequate adjustments made to the leveling screws
2. A hose of a considerable length was attached to the drain cock at one end and the other end was
directed to the overflow pipe of the volumetric measuring tank.

Experimental procedure
1. The dimensions a, b and d of the quadrant as seen in figures 04 and 05 were measured
2. The distance between the pivot and the weight hanger (L) was measured
3. A wetting agent was applied to the scale to reduce the effects of surface tension on the practical
4. The toroid was inserted to the tank locating the balance arm on the knife edges
5. The counter balance weight was adjusted until the balance arm was horizontal as it was indicated on
the datum level indicator
6. All the weights that were supplied were added to the carrier
7. The tank was filled with water until the balance beam tipped lifting the weights a small quantity of
water was drained out and the balance arm was brought back to horizontal (This was done without
adjustment to the counter balance weight as that would result in a changed datum setting)
8. The water level shown in the scale was recorded
9. One or more weights were removed from the weight carrier and the balance arm was leveled by
draining out a certain amount of water
10. The depth of immersion was recorded from the scale on the quadrant when the arm was leveled
11. The readings were repeated for reducing weights of the weight carrier
05
OBSERVATIONS
L = 29.9 cm a = 10 cm b = 7.9 cm d = 9.7 cm
M (grams)
120
Table 01: Observations from the experiment
y (mm)
Surface Partially or fully
submerged
loading
unloading
70
70
P
140
76
76
P
160
83
84
P
180
89
89
P
200
95
95
P
220
101
100
P
240
104
104
F
260
109
108
F
280
114
113
F
300
118
117
F
320
124
123
F
340
129
129
F
360
134
134
F
380
138
138
F
400
144
144
F
Name:
T.D.D. Samaranayake
Index No:
20/ENG/123
Date:
31/08/2021
Instructor’s Name: Mr. A. Siriwardena
Signature:
06
CALCULATIONS AND RESULTS
 Partially immersed condition
Figure 04: Quadrant and vertical surface in the partially submerged condition
Theoretical Pressure Force (FT) = ρghGA
= (ρgby2) / 2
Practical Pressure Force (FP)
Taking moment at knife edge point
Mg × L = Fp × (a + d – y + hp)
FP = (Mg × L) / (a + d – y + 2y/3)
FP = (Mg × L) / (a + d – y/3)
Center of pressure
LP = I0 / LGA
hp = I0 / hGA [As the plane is vertical ]
= IG + A(hG)2 / hGA
= [ by3/12 + by(y/2)2] / [y/2×(by)]
= 2y/3
Sample calculation for partial immersed condition:For
M = 120 × 103- kg , yavg = (70 +70)/ 2 = 70 mm
Theoretical Pressure Force (FT) = (ρgby2) / 2
= [(1000 Kgm3-) × (9.81 ms2-) × (7.9 × 102- m) × (70 × 103- mm)2 ] / 2
= 1.9 N
07
Center of pressure
hP = 2y/3
= (2 × 0.7)/3
= 46.67 mm
Practical Pressure Force (FP) = (Mg × L) / (a + d – y + hp)
= [120 × 103- Kg × 9.81 ms2-× 29.9 × 102- m] / [0.1 + 0.097 – 0.07 + 0.0467]
= 2.03 N
Table 02: Calculations for the partially submerged condition
M (grams)
Loading
y1 (mm)
Loading
y2 (mm)
Average
yavg= ( y1 + y2)/2
(mm)
Center of
pressure
(mm)
Practical
Force (N)
Theoretical
Force (N)
120
70
70
70
46.67
2.023
1.90
140
76
76
76
50.67
2.39
2.24
160
83
84
83.5
55.67
2.77
2.70
180
89
89
89
59.33
3.16
3.07
200
95
95
95
63.33
3.53
3.50
220
101
100
100.5
67.00
3.95
3.91
 Fully immersed condition
hG
G
P
Figure 05: Quadrant and vertical surface in the fully submerged condition
Theoretical Pressure Force (FT) = ρghGA
= ρgbd (y – d/2) [As hG = (y – d/2)]
Practical Pressure Force (FP)
Taking moment at knife edge point
Mg × L = Fp × (a + d – y + hp)
FP = (Mg × L) / (a + d – y + hp)
08
hp
Center of pressure
LP = I0 / LGA
hp = I0 / hGA
[As the plane is vertical ]
= [IG + A(hG)2] / hGA
= [ bd3/12 + bd(y –d/2)2] / [bd × hG]
[hG = ( y – d/2)]
= (y - d/2) + d2/ 12hG
Sample calculation for partial immersed condition:For
M = 240 × 103- kg , yavg = (104 +104)/ 2 = 104 mm
Theoretical Pressure Force (FT) = ρgbd (y – d/2)
= [(1000 Kgm3-) × (9.81 ms2-) × (7.9 × 102- m) × (9.7 × 102- m) × ((104 – 97/2) × 103- mm) ]
= 4.17 N
Center of pressure
hP = (y - d/2) + d2/ 12hG
= (104 – 97/2) × 103- m + (9.7 × 102- m) 2 / 12(104 – 97/2× 103 m)
= 69.62 mm
Practical Pressure Force (FP) = (Mg × L) / (a + d – y + hp)
= [120 × 103- Kg × 9.81 ms2-× 29.9 × 102- m] / [0.1 + 0.097 – 0.104+ 0.0696]
= 4.33 N
Table 03: Calculations for the fully submerged condition
M (grams)
Loading y1
(mm)
Loading y2
(mm)
Center of
pressure
(mm)
69.63
Practical
Force (N)
Theoretical
Force (N)
104
Average
yavg= ( y1 + y2)/2
(mm)
104
240
104
4.33
4.17
260
109
108
108.5
73.07
4.72
4.51
280
114
113
113.5
77.06
5.12
4.89
300
118
117
117.5
80.36
5.50
5.19
320
124
123
123.5
85.45
5.90
5.64
340
129
129
129
90.24
6.30
6.05
360
134
134
134
94.67
6.70
6.43
380
138
138
138
98.26
7.09
6.73
400
144
144
144
103.71
7.49
7.18
09
Table 04: Average depth and Practical force for partially submerged conditions
Average
yavg= ( y1 + y2)/2
(mm)
Practical Force (N)
70
2.023
76
2.39
83.5
2.77
89
3.16
95
3.53
100.5
3.95
Table 05: Average depth and Practical force for fully submerged conditions
Average
yavg= ( y1 + y2)/2
(mm)
Practical Force (N)
104
4.33
108.5
4.72
113.5
5.12
117.5
5.50
123.5
5.90
129
6.30
134
6.70
138
7.09
144
7.49
10
Table 06: Practical force and theoretical for partially submerged conditions
Practical Force (N)
Theoretical Force (N)
2.023
1.90
2.39
2.24
2.77
2.70
3.16
3.07
3.53
3.50
3.95
3.91
Table 07: Practical force and theoretical force for fully submerged conditions
Practical Force (N)
Theoretical Force (N)
4.33
4.17
4.72
4.51
5.12
4.89
5.50
5.19
5.90
5.64
6.30
6.05
6.70
6.43
7.09
6.73
7.49
7.18
11
Graph 01: Hydrostatic Forces (Practical Values) vs Average Depth Partially Submerged Conditions
4
3.9
3.8
3.7
3.6
3.5
Practical Values Hydrostaic Force Partially Submerged (N)
3.4
3.3
3.2
3.1
3
2.9
2.8
2.7
2.6
2.5
2.4
2.3
2.2
2.1
2
65
70
75
80
85
Average Depth (mm)
12
90
95
100
105
Graph 02: Hydrostatic Forces (Practical Values) vs Average Depth Fully Submerged Conditions
8
7.5
Practical Values Hydrostaic Force Partially Submerged (N)
7
6.5
6
5.5
5
4.5
4
100
105
110
115
120
125
130
Average Depth (mm)
13
135
140
145
150
Graph 03: Theoretical Vs Practical Hydrostatic Forces for Partially Submerged Conditions
4.5
Theoretical hydrostatic force for partially submerged conditions (N)
4
3.5
3
2.5
2
1.5
1.5
2
2.5
3
3.5
4
Practical hydrostatic force for partially submerged conditions (N)
14
4.5
Graph 04: Theoretical Vs Practical Hydrostatic Forces for fully Submerged Conditions
Theoretical hydrostatic force for partially submerged conditions (N)
7
6.5
6
5.5
5
4.5
4
4
4.5
5
5.5
6
6.5
Practical hydrostatic force for partially submerged conditions (N)
15
7
7.5
DISCUSSION
 The experimental and theoretical hydrostatic force values
 Comparison of theoretical and practical hydrostatic force values

Table 08: Difference between theoretical and practical hydrostatic forces
Partially or fully
submerged
Practical Force FP
(N)
Theoretical Force FT
(N)
Difference
(FP – FT) N
P
2.023
1.90
0.123
P
2.39
2.24
0.150
P
2.77
2.70
0.070
P
3.16
3.07
0.090
P
3.53
3.50
0.030
P
3.95
3.91
0.040
F
4.33
4.17
0.016
F
4.72
4.51
0.210
F
5.12
4.89
0.230
F
5.50
5.19
0.310
F
5.90
5.64
0.260
F
6.30
6.05
0.250
F
6.70
6.43
0.270
F
7.09
6.73
0.360
F
7.49
7.18
0.310
All the deviations between the theoretical and practical values are positive. The deviations between
the theoretical and practical values under partially submerged conditions are lower than the
deviations occurred under fully submerged conditions. The highest deviation recorded as 0.360 N
which was for a weight of 380 g under fully submerged conditions and the lowest of deviations was
recorded as 0.030 N which was for a weight of 200g under partially submerged conditions.

Gradient values of graphs 03 and 04
o Gradient of graph 03 – Graph between theoretical vs practical hydrostatics forces for partially
submerged conditions
Gradient of graph 03 = (3.8 – 2.1) / (3.87– 2.24)
= 1.04
o Gradient of graph 04 – Graph between theoretical vs practical hydrostatics forces for fully submerged
conditions
Gradient of graph 03 = (6.7 – 4.2) / (7.04 – 4.39)
= 0.94
16
Considering the two gradients it can be noticed that under both conditions the gradient between the obtained
theoretical and practical values are closer to one that is the hydrostatic force values are more or less equal to
the theoretically obtained values of them the values that are more closer to the theoretical values are the ones
obtained under partially submerged conditions.

Possible reasons for the slight disparities between the practical and theoretical values
o The theoretical calculation depends on the density of the fluid in which the quadrant was immersed
However there is a possibility that the theoretical density used in the theoretical calculation might not
be the same as the actual density of the fluid used due to any impurities that the fluid contain
o Furthermore density also varies with temperature therefore a variation of temperature while
conducting the practical may also result in a erroneous calculation. As the temperature increases the
density decrease and vice versa.
Graph 05: Density of water at an ambient pressure of 1,013 bar (Greber, 2021)
o Another possibility is due to errors occurred while measuring the dimensions and when reading
the water scale. The water depth is read from the scale placed on the quadrant however and the
sensitivity of the scale would prove to be insufficient at times therefore the degree of precision that can
be expected from the readings would be insufficient to obtain the desired result.
o Even though a wetting agent was applied to the quadrant to minimize the effect of surface tension on
the experiment the effect was not nullified therefore when taking moments the effect of surface tension
would also be included in the value obtained for the magnitude of the pressure force.
 Discuss the difference between fully submerged and partially submerged surfaces.

Fully submerged
This means that a body or a surface lies completely under a liquid or completely covered by a fluid

Partially submerged
Contrary to fully submerged partially submerged means that the considered surface or body is not
completely lying under the same fluid surface
o Therefore the main difference between fully and partially submerged surfaces can be identified as the
positioning of a considered surface with respect to a fluid surface. Fully and partially submerged
conditions of the same surface is illustrated in figure 06
17
(b)
(a)
Figure 06: Fully submerged (a) and partially submerged (b)
o However another point to be considered is that when a body is submerged hydrostatic forces act on its
submerged surface due to the hydrostatic pressure of the fluid. However hydrostatic force/pressure is
proportional to the depth of the fluid therefore the hydrostatic force increases proportionally as the
depth increase. Therefore when considering a fully and partially submerged surface of the same
surface area the hydrostatic force acting on the surfaces differ as seen in figure 07 below. A fully
submerged surface would have a larger hydrostatic force acting on it compared to a partially
submerged surface of the same area
(a)
(b)
Figure 07: Fully submerged (a) and partially submerged (b) with hydrostatic
forces
o
According to the Archimedes principle any body that is completely or partially submerged in a fluid at
rest is acted upon by an upward buoyant force which is equal to the weight of the fluid displaced by
the body therefore a partially submerged body would have a lesser buoyancy force acting on it
compared to a fully submerged body of the same dimensions.
o Based on the same principle another dissimilarity between partially submerged surface/body (floating
surface) and fully submerged surface/body is that a partially submerged surface would displace a fluid
volume such that its weight would be equal to the weight of the body whereas a fully submerged body
would displace a volume equal to the volume of the body
 Possible errors of the experiment and methods to improve the experiment
o When conducting the experiment water is poured to the tank using a beaker or a suitable vessel
therefore there is a possibility that while poring the water some of the water may accidently spill on
the balancing arm therefore when considering the moments around the pivot point an artificial weight
would be considered to balance the hydrostatic force acting on the vertical surface therefore this would
lead us to a erroneous hydrostatic force value. (Chapman, 2021) Therefore extra care should be taken
when adding water to the tank. Using a hose to add the water would be a more suitable method
18
o Furthermore the water that splashes o to the quadrant may also result in weighing it down therefore
this would again lead to an erroneous result this can also be avoided by following the precaution
mentioned above
o When adding water another error that could occur is creating water bubbles by a turbulent water
addition. The inclusion of air bubbles in the water would result in a decreased density of the water
therefore the calculated hydrostatic force would be erroneous. In order to avoid this if we are using a
hose we should make sure that the hose‘s open end would lie closer to the surface of water in the tank
o The weights used in the practical were never tested for their accuracy due to rust and accumulation of
dust the weights would not weigh as it is indicated on their surface. To prevent this weights that are to
be used should be tested for their accuracy and if needed be replaced.
o As the depth of water is read from the scale placed upon the quadrant there is a high chance of
encountering a parallax error therefore when taking the readings attention should be placed on the line
of sight the viewer
o The calibrated scale on the quadrant might not be effective in obtaining readings of high precision
therefore a difficulty in obtaining the accurate water depth measurement exist. Using a scale with a
higher sensitivity would be able to increase the precision of the readings
REFERENCES





Chapman, M., 2021. Chrome-extension mhjfbmdgcfjbbpaeojofohoefgiehjai index-material. [online]
Academia.edu. Available at:
<https://www.academia.edu/20123773/Chrome_extension_mhjfbmdgcfjbbpaeojofohoefgiehjai_index_
material> [Accessed 6 September 2021].
Claydon, J., 2015. Center of Pressure Apparatus. [image] Available at:
<http://www.jfccivilengineer.com/centre_of_pressure.htm> [Accessed 31 August 2021].
Greber, O., 2021. Density of water at an ambient pressure of 1,013 bar. [image] Available at:
<https://blog.wika.com/files/2017/12/temperaturerelateddensitychanges.jpg> [Accessed 6 September
2021].
Hamill, L., 2011. Understanding Hydraulics. 3rd ed. New York: Palgrave Macmillan, p.577.
Texas Education, 2021. Fig. 2.3 - Schematic diagram of laboratory apparatus. [image] Available at:
<https://www.caee.utexas.edu/prof/kinnas/319LAB/Lab/lab%202-HydroStatic%20Forces/IMG2009.GIF> [Accessed 4 September 2021].
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