TABLE OF CONTENTNo Subject Page 1.1 Introduction 2 1.2 Objective 3 1.3 Learning Outcome 3 1.4 Theoretical Background 2.1 Problem Statement 8 3.1 Apparatus 9 3.2 Procedure 9-10 4.1 Result 10-12 4.2 Analysis 13-14 4.3 Discussion 15 4.4 Conclusion 16 4.5 Experimental precaution 16 4.6 References 16 4.7 Appendix 17 3-8 EXPERIMENT TITLE Determination of Metacentric height. 1.1 INTRODUCTION 1 G) must be equal to the buoyancy force which acts through the centre of buoyancy. W. The centre of gravity G is below the centroid B of the body. which is located at the centroid of the immersed cross-section. The activity is hope to slowly introduced and inculcate independent learning amongst students and prepare them for a much harder task of open-ended laboratory activities. 1. For stable equilibrium. For static equilibrium of the pontoon. the total weight. ( which acts through the centre of the gravity. by referring the distance from the centre of gravity (G). the metacentre M is identified as the point of intersection between the lines of action of the buoyancy force ( always vertical ) and BG extend.0: A floating body is stable if the body is (a). (b) The metacentre M is above G. In this laboratory activity students will be exposed to the equipment that used to measure the metacentric height of pontoon. However the answers to the assignment are left to the students to solve using the group creativity and innovativeness. 2 . When the pontoon heels through a small angle. M must be above G. B. Figure 5.2 OBJECTIVE To identify the position of the metacentre (M) of a floating body.Level 1 laboratory activity refers to condition where the problem and ways & means are guided and given to the students. (c) Unstable if M below G. F . iii.3 LEARNING OUTCOMES At the end of the laboratory activities. Analyse test data and present the solution to the open-ended problem. 3 .1. students would be able to: i. Work in a group to produce the relevant technical report 1. Determine the suitable laboratory tests to be conducted to address the given problem. B. the shape of the displaced liquid is also a rectangle with it centre at the geometrical centre namely the centre of buoyancy. G. W acting vertically down through the centre of gravity.1: A pontoon floating on even keel with W and F collinear.4 THEORETICAL BACKGROUND Pontoon is a term used to denote a flat bottomed vessel which is rectangular in cross section and in plan. Since the pontoon is floating in water with a constant depth immersion. it follows that there must be an equal force acting the opposing direction of the weight force. known as buoyancy force. Note that W and F act collinearly with G situated some distance above B. F acts upwards through B. we have the weight force. ii. The buoyancy force. which acts vertically up through the centre of gravity of the displaced water. of the pontoon. Since the pontoon is a simple rectangle. Figure 5. Considering Figure 2. What we now have is the situation depicted in Figure 5. but a couple of forces that return the pontoon to an even keel are formed. W of the pontoon and its load acts through the centre of gravity. When G becomes higher and the angle of tilt increases. This is because acts through the centre of gravity of the displaced liquid which is now trapezoidal in shape with its centre of gravity at B*. Figure 5. G which is relatively high. showing the overturning couple caused by W acting outside It a relatively tall piece of weight is placed on the pontoon as shown in Figure 4. As a result F and W are no longer collinear. showing the righting couple When a pontoon is tilted as shown in Figure 3. In this case the pontoon is capable of righting itself when tilted. the combined weight.3: A pontoon with a raised G and an imposed angle of tilt. F from B to B* is unlikely to be large enough to produce a righting couple. where the line of action 4 .2: A pontoon floating with an imposed angle of tilt. This is known as righting couple. This means that at the some point the movement of buoyancy force. W acts vertically down through G which maintained at the same position but F now acts through point B* instead of B.Figure 5. W acts further and turn further to the left. hence it is stable. iii. ii. Imagine line BG extends upwards and how consider the pontoon in tilted position as in Figure 6. The two forces F and W form an overturning couple. The distance of M above M constant. Figure 5. the centre of buoyancy moved from B to B*. The distance GM is called metacentric height of pontoon. the position of M can be found graphically. The vertical line through up B* through M. A line drawn vertically upwards through B* will intersect the line BG at the point labelled M in the diagram. Thus it is unstable. Considering the restoring moment that rights a rectangular pontoon to an even keel when it is tilted. it is usual to assume that the angle of tilt θ small. A line joining B to G would be as shown in Figure 4. that is vertical and at 90° to the deck of pontoon. i.4 : The position of metacentre A pontoon floating on an even keel has its center of buoyancy at B and its centre of gravity at G. the equation: BM = Iws / V 5 . This is necessary to simplify the theory by making the assumption that θ radians = sin θ = tan θ = θ radians. Provided the G does not move.of W is outside (nearer the edge of the pontoon) than the line along which F acts. Consequently if the location of B* can be calculated. Thus W is trying to overturn the pontoon. then for all relatively small angle of tilt. When considering the stability of floating body. This called the metacentre. Where: V = the volume of water displaced by the body Iws = the second moment of the area Figure 5. 6 . V. b. GM = BMBG. and with the pontoon floating on an even keel B is located at a height equal to half the depth of immersion (h/2) above the point O on the bottom of the pontoon. you should be able to see that BM = BG + GM or. B is the center of buoyancy. I and b.5: Plan of the pontoon where the tilt takes place about the longitudinal axis X-X LB 3 Iws = 12 It should be apparent that BM depends only upon: a. Now. then we can obtain GM and hence determine if the body is stable or unstable. the volume of displaced water which depends only upon the weight of the pontoon. the dimensions of the pontoon which govern the value of Iws . If we can calculate BG. Referring to Figure 7. The weight namely jockey weight (wj) is moved from the centreline known distance ( ) towards the side as shown in Figure 8. This is a simple procedure utilizing moveable weight positioned on the deck at approximately the middle of the longitudinal centreline and a pendulum hanging inside the vessel. wj x W GG* Where W is the total weight of the pontoon including the pontoon GG* GM tan x Combining both equations.6 : A pontoon showing the key points and dimensions It is common practice to carry out an experiment on vessel to assess its stability by calculating GM. The magnitude of GG* depends upon how far the jockey weight is moved and its size relatively to the total weight of the pontoon. 7 . Using the ratio of weight and x . wj dx W d GM It is important to remember that is in radian.Figure 5. This moves the centre of gravity of the pontoon from G on the centreline to a new position G* and causes the vessel to tilt at the angle of . 3.1 PROBLEM STATEMENT An important application of the buoyancy concept is the assessment of the stability of immersed and floating bodies when being place in a fluid.7: Movement of the jockey weight from the centreline 2. Metacentric Height Apparatus 8 . M location is vital and great importance in the design of ships and submarines. Students are required to perform a relevant experiment to fulfil the objective stated above using both method namely adjustable position traversed weight experiment and based upon geometry and depth of immersion. For computation purpose.1 APPARATUS a. the students are asked to find the equation from the literature or exiting manual for fluid and hydraulic laboratory. The body is said stable if M is above G and unstable if otherwise.Figure 5. Knowing metacentre. the pontoon is floated ensuring that the adjustable 5. This can be determined by using either a knife edge or by 4. The PVC plates provided is used to level the floating body and zero datum is 6. Plumb . Pontoon Body 2. The basin is filled with water.0 .Metacentric Height apparatus Figure 1. Linear Scale Figure 1. 9 . 3. Sliding Mass 6. The transverse adjustable mass is weighed.bar 3.1. 2.line 7. The pontoon is assembled and weighed. Mast 5. The sliding mass is positioned along the mast such that the center of gravity occurs at the top of the pontoon. Cross . mass is in its central position.2 PROCEDURE 1.Label of apparatus 3. The adjustable mass is moved to the right of centre in 5mm increments to the end of the scale. checked between plumb line and scale. suspending from a light string around the mast.1 . Adjustable Mass 4. nothing the angular displacement of the plumb line for each position. 343 0. 10.343×10-3 0.e.333×10-4 2.7.0335 0. for different centres of gravity. 8. 9. 4.0996 0. i. The value of dx for each sliding mass height is obtained.1 RESULTS Total weight of floating assembly (W) Weight of adjustable mass (ω) Weight of sliding mass (ω1) Breadth of pontoon (B) Length of pontoon (L) Second moment of area (I) Volume of water displaced (V) Height of metacentre above centre of buoyancy (BM) Depth of immersion of pontoon (IP) Depth of centre of buoyancy (CB) 2. The graph of lateral position of adjustable mass against angle of list for each sliding mass height is prepared. all the above is repeated for the sliding mass at different heights up the mast. adjustable 20 30 40 50 weight. GM and distance between the centre of buoyancy and the metacentre is calculated. All the reading is recorded in the result sheet.208 0. d the metacentric height. With the exception of weighing the adjustable weight and emptying and refilling the volumetric tank. y1 -50 -40 -30 -20 -10 (mm) 0 10 (mm) 0 54 42 30 20 10 0 -10 -20 -31 -42 -56 -66 50 - - - 62 32 0 -18 -46 -60 - - - 100 - - - - - 0 -50 -68 - - - - 10 60 .511 200 350 2.0167 kg kg kg mm mm m4 m3 m m m Height of Reading of list for adjustable weight lateral displacement from sail centre line. The adjustable mass is repeated for movement to the left centre. 7 7 Table 2: Angles for each height of adjustable weight 11 - .71 -3.3 60 1 8 - - 0 9.8 12.09 0 6.72 11.97 10.81 -1.Table 1: Lateral displacement of ropefor each height of adjustable weight from sail centre line In order to find the angle.46 12. y1 -50 (mm) 0 - -40 -30 -20 -10 -7.97 -5.2 50 0 - - - - -6.90 7. x1 adjustable (mm) 0 10 weight. Y=300 mm Height of Angles of list for adjustable weight lateral displacement from sail centre line.09 8.6 100 - - - 11.91 20 30 40 50 3.91 0 1.81 5. Where.4 7 1 - - - - - - 10. we need to use trigonometry rule. Height of adjustable -50 -40 -30 Metacentric Height (mm) -10 0 10 20 -20 30 40 50 60 weight.05 Table 2: Metacentric height for each height of adjustable weight 4.79 8.07 25.79 26.79 26.2 ANALYSIS 12 - .34 0 26.60 26.70 0 26.74 25.34 11.56 - - - 100 - - - - - 0 5.07 24. y1 (mm) 0 25.60 25.80 50 - - - 8.70 8.74 13.74 26.41 - - - 8. 25 Slope for 50mm dx/d = (10 – -20)/(6.81 – -3.64 i) Second Moment of Area.68) =1.46 – 0) dx/d = (20 – -20) / (3.Slope for 100mm Slope for 0mm dx/d = (10 – 0)/(9.09 – -11.06 = 5. I 13 .81) = 1. V W 103 2.333 10 4 m 4 I ii) Volume of water displaced. x 50.07 mm Metace ntric Height KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK KKKK 14 .LB 3 10 12 12 350 200 3 10 12 12 2.208 )( 50)005) 10.20 ( 2.343 103 2.343 103 m3 V iii) Metacentric Height.343 ) sin 1 iii) 25. 54 H mx m p sin ( 0. If M lies below G an overturning moment is produced. CB V 10 6 2 LB 2. Why the values of GM at lowest values of the angle (θ) are likely to be less accurate. ii. The effect of changing the position of G on the position of the metacentre (M).0335m IP vi) Depth of centre of buoyancy. 15 . IP V 10 6 LB (2. equilibrium is unstable and GM is regarded as negative.0996m BM v) Depth of immersion of pontoon. the body is in neutral equilibrium. If M coincides with G.343 10 3 10 6 2 350 200 0.0167m CB 4.3 DISCUSSION i.343 10 3 ) 10 6 350 200 0. Changing the position of G on the position of the metacentre (M) will cause change of stability of floating object.333 10 4 2.iv) Height of metacentre above centre of buoyancy. equilibrium is stable and GM is regarded as positive. BM I V 2.343 10 3 0. If M lies above G a righting moment is produced. 5 CONCLUSION Therefore. When the centre of buoyancy change. we can conclude that the objective of this experiment is achieved because we are able to identify the position of the metacentre (M) of a floating body. the angle occur when more of the hull on one side get deeper into the water.4 EXPERIMENTAL PRECAUTION 1) Make sure the water is in steady condition to prevent existence of large waves on the water before taking the reading. some precautions are also should be taken to increased the probability of our group results so that we can get the exact values and avoided more errors. Typically. the eyes must be perpendicular to the pontoon to get accurate reading. 4. 3) This experiment involves large volume of liquid. 16 . the eyes must be perpendicular to the pontoon to get accurate reading. ensure that the water flows accordingly in the container/apparatus to prevent flood occur in the laboratory. 4. when taking the reading of the angles for adjustable weight lateral displacement from sail centre line. 2) When taking the reading of the angles for adjustable weight lateral displacement from sail centre line. iii. We are also able to determine whether the pontoon is stable or unstable by getting the position of centre of gravity. The result is that the center of buoyancy shifts to the side where more water is displaced. by referring the distance of the adjustable weight from the centre of gravity (G). Besides. What is the most sensitive parameter that affects the accuracy of the results? Make sure the water is in steady condition to prevent existence of large waves on the water before taking the reading. this make the result GM less accurate and negative GM may be resulted.The angle(s) obtained during the experiment are directly related to GM. However. Thus. and the hull on the other side moves out of the water. while the center of gravity remains in the same place at the pontoon since the pontoon itself has not changed. 2006. Pearson. Sir Embam Ak Da’ud. 2nd Edition.codecogs.edu/~cfd/pdfs/57-020/stability.6 REFERENCES 1. Fluid Mechanics. Laboratory Manual of Hydraulics and Water Quality. Janusz M.Gasiorek.Jack .engineering.uiowa. Turahim ABd Hamid and Junaidah Ariffin. 2002 3.com/library/engineering/fluid_mechanics/floating_bodies/st ability-and-metacentric-height. Suhaimi Abdul Talib.php 5. Uitm Samarahan 2. 2. Faculty of Civil Engineering. Fluid Mechanics.4.F.Douglas. http://www. 5th Edition. http://www. and Lynne B. John A.Swaffield. 4.pdf APPENDIX 17 . John. Hamidon Ahmad.