## Transcript

PAMANTASAN NG LUNGSOD NG MAYNILA (University of the City of Manila) Intramuros, Manila COLLEGE OF ENGINEERING AND TECHNOLOGY
1. ARCHIMEDES STORY AND PRINCIPLE
2. BOUYANCY (Concept, Principle & Examples) a. APPLICATION AND EXPERIMENTS 3. STABILITY OF FLOATING BODIES a. APPLICATION AND EXPERIMENTS
BSCE-5
ENGR.
September 12, 2014
I.
ARCHIMEDES STORY AND PRINCIPLE
Archimedes Story Archimedes's tale takes place some 2,200 years ago when King Heron II of Syracuse in Sicily gave a jeweller a bar of gold and ordered him to make it into a crown. The king, however, suspected that the jeweller had substituted some of the gold for cheaper metal like silver, while pocketing the leftover gold. The king had no way of proving his suspicions, so he asked Archimedes – a Greek mathematician, engineer, inventor, and astronomer – to find a definitive answer. Archimedes had spent a long time trying to figure out the answer, which came to him when he noticed how water would splash out of his bath tub the moment he stepped into it, and the more he stepped into the tub, even more water got displaced. At the time, Archimedes had known that gold was denser than silver, so if a certain weight of silver had been substituted for the same weight of gold, the crown would occupy a larger space than an identical one of pure gold. So to find the crown’s volume, all Archimedes had to do was essentially immerse the crown and exact measurement of pure gold in a tub filled with water to the brim, measure the spillage, and compare the volume of spillages – if the jeweller had indeed made a crown of pure gold the volume should be the same. Archimedes was said to be so thrilled with this discovery that he immediately hopped out of the bath and ran onto the streets naked shouting 'Eureka!' 'Eureka!'. And in case you were wondering, the jeweller was indeed cheating the king.
Archimedes Principle
II.
BOUYANCY (Concept, Principle & Examples)
Buoyancy Concept and Principle
Buoyancy is an upward force exerted by a fluid that opposes the weight of an immersed object. The principle of buoyancy holds that the buoyant or lifting force of an object submerged in a fluid is equal to the weight of the fluid it has displaced. The concept is also known as Archimedes' principle, but Archimedes’ principle does not consider the surface tension (capillarity) acting on the body, but this additional force modifies only the amount of fluid displaced, so the principle that Buoyancy = weight of displaced fluid remains valid. “Any body, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight displaced by the volume” – Archimedes Principle
Buoyancy Concept
Buoyant Force = weight of object = (unit weight of water x volume displaced) Examples
Icebergs Water is an unusual substance in a number of regards; not least its behavior as it freezes. Close to the freezing point, water thickens up, but once it turns to ice, it becomes less dense. This is why ice cubes and icebergs float. However, their low density in comparison to the water around them means that only part of an iceberg stays atop the surface. Submarines A submarine uses ballast as a means of descending and ascending underwater: when the submarine captain orders the crew to take the craft down, the craft is allowed to take water into its ballast tanks. If, on the other hand, the command is given to rise toward the surface, a valve will be opened to release compressed air into the tanks. The air pushes out the water, and causes the craft to ascend.
Ships at Sea A steel ball dropped into the water will sink straight to the bottom, because it is a lowvolume, high-density object that outweighs the water it displaced. 3 3 But how can ships made out of steel, with a density of 487 lb/ft (77,363 N/m ), float on a salt-water ocean with an average density of only about one-eighth that amount? The answer lies in the design of the ship's hull. If the ship were flat like a raft, or if all the steel in it were compressed into a ball, it would indeed sink. Instead, however, the hollow hull displaces a volume of water heavier than the ships own weight: once again, volume has been maximized, and density minimized. For a ship to be seaworthy, it must maintain a delicate balance between buoyancy and stability. Balloons There are only three gases practical for lifting a balloon: hydrogen, helium, and hot air. Each is much less than dense than ordinary air, and this gives them their buoyancy. In fact, hydrogen is the lightest gas known, and because it is cheap to produce, it would be ideal — except for the fact that it is extremely flammable. After the 1937 crash of the airship Hindenburg, the era of hydrogen use for lighter-than-air transport effectively ended. Helium, on the other hand, is perfectly safe and only slightly less buoyant than hydrogen. This makes it ideal for balloons of the sort that children enjoy at parties; but helium is expensive, and therefore impractical for large balloons. Hence, hot air — specifically, air heated to a temperature of about 570°F (299°C), is the only truly viable option.
Sample Problem:
The iceberg that hit titanic is having a specific gravity of 0.92 floats in salt water having a 3 specific gravity of 1.09. If the volume of ice above the surface is 700 yd. what is the total volume of the iceberg?
= +
=700+
=
= ℎ
0.9262.4 = 1.0362.4 0.92700 +
= 1.0362.4
= 5854.55
= 700 + 5854. 55 = 6554.55
IIa.
APPLICATION AND EXPERIMENTS
Sinking Orange Experiment
Do oranges float or sink in water? Materials
For the Sinking Orange experiment you will need the following materials:
An orange
A deep bowl or container
Water
Procedures
1. Fill the bowl with water. 2. Put the orange in the water and watch what happens. 3. Peel the rind from the orange and try the experiment again, what happens this time? Discussion
The first time you put the orange in the bowl of water it probably floated on the surface, after you removed the rind however, it probably sunk to the bottom, why? The rind of an orange is full of tiny air pockets which help give it a lower density than water, making it float to the surface. Removing the rind (and all the air pockets) from the orange increases its density higher than that of water, making it sink. Density is the mass of an object relative to its volume. Objects with a lot of matter in a certain volume have a high density, while objects with a small amount of matter in the same volume have a low density.
Floating Golf Ball Experiment
The Floating Golf Ball Experiment explains why materials (such as an golf ball) float more in salt water than in dishwashing liquid. Materials
For the Floating Golf Ball Experiment you will need the following materials:
Table salt
Two containers
Tablespoon
Dishwashing Liquid
Two golf balls
Procedures
1. Fill the first container with tap water. 2. Add about 6 tablespoons of salt in one container and stir it well with a tablespoon until the salt has completely dissolved in the water. 3. Place a golf ball in the first container. 4. Fill the second container with dishwashing liquid. 5. Place the other golf ball in the second container. 6. Observe which one of the golf balls float in the container and which one sinks. 7. Pour some dishwashing liquid in the container of saltwater. 8. Observe what will happen to the golf ball. Discussion
The explanation behind this phenomenon is simple - DENSITY! In the Floating Golf Ball experiment, you have observed that the golf ball placed in saltwater floated and the one in dishwashing liquid didn't. Because saltwater is denser than dishwashing liquid, the golf ball does not end up sinking like it usually does!
Tin Foil Boat Experiment
In this easy science experiment you will be designing a tin foil boat that will hold the greatest number of coins. Get together with some friends and have a contest to see who can design a cargo boat that will hold the most number of coins. Materials
For the Tin Foil Boat experiment you will need the following materials:
Tin foil
Bowl
Scissors
Coins
Water
Procedures
1. Cut a piece of tin foil 5 x 6 inches. 2. Fold up the sides of the boat so it will not sink and hold a cargo of coins. 3. Place the boat in the bowl of water. Begin adding coins for the boat's cargo. 4. See how many coins your boat can carry before it sinks. 5. Have some friends over and try this experiment with them. See which one of you can create the boat that will carry the greatest amount o f cargo. 6. Be sure to dry the coins before you begin adding them as cargo because remember water has weight! 7. Have the person who created the boat begin adding coins to their boat while another person counts the number of coins as they are added to the boat. 8. Try different ways to distribute the weight of the coins on your barge so you can carry the maximum number. Discussion
There are two primary forces acting on this science experiment. The first force is gravity. Gravity is trying to pull the tin foil and coins downward. The force of buoyancy is pushing the boat toward the surface. The gravitational force is determined by the weight of the tin foil and the weight of the coins in the boat. The force of buoyancy is the weight of the water displaced by the boat. Your boat will continue to float as long as the force of buoyancy is greater than the force of gravity and you do not overload the boat so it will tip over or leak.
III.
STABILITY OF FLOATING BODIES
Stability
A floating body reaches to an equilibrium state, if a. its weight = the buoyancy b. the line of action of these two forces become collinear . The equilibrium: stable, or unstable or neutrally stable .
Stable equilibrium: if it is slightly displaced from its equilibrium position and will return to that position. Unstable equilibrium : if it is slightly displaced form its equilibrium position and tends to move farther away from this position. Neutral equilibrium : if it is displaced slightly from this position and will remain in the new position.
Motion of a Ship: 6 degree of freedom - Surge - Sway - Heave - Roll - Pitch - Yaw
Righting & Heeling Moments
A ship is designed to float in the upright position . •
Righting Moment : exists at any angle of inclination where the forces of weight and buoyancy act to move the ship toward the upright position. • Heeling Moment : exists at any angle of inclination where the forces of weight and buoyancy act to move the ship away from the upright position. For a displacement ship,
G---Center of Gravity, B---Center of Buoyancy M--- Transverse Metacenter *If M is above G, we will have a righting moment, and *If M is below G, then we have a heeling moment.
Upsetting Forces (overturning moments)
• • • • •
Beam wind, wave & current pressure Lifting a weight (when the ship is loading or unloading in the harbor.) Offside weight (C.G is no longer at the center line) The loss of part of buoyancy due to damage (partially flooded, C.B. is no longer at the center line) Turning
Static Stability
1. 2.
The initial stability (aka stability at small inclination) and, The stability at large inclinations .
The initial (or small angle) stability: studies the right moments or right arm at small inclination angles. The stability at large inclination (angle): computes the right moments (or right arms) as function of the inclination angle, up to a limit angle at which the ship may lose its stability (capsizes).
Initial stability
•
•
Righting Arm: A symmetric ship is inclined at a small angle dΦ. C.B has moved off the ship’s centerline as the result of the inclination. The distance between the action of buoyancy and weight, GZ , is called righting arm. Transverse Metacenter: A vertical line through the C.B intersects the original vertical centerline at point, M .
Location of the Transverse Metacenter Transverse metacentric height : the distance between the C.G. and M ( GM ). It is important as an index of transverse stability at small angles of inclination. GZ is positive, if the moment is righting moment. M should be above C.G, if GZ >0.
If we know the location of M , we may find GM , and thus the righting arm GZ or righting moment can be determined given a small angle dΦ. How to determine the location of M ?
WoLo – Waterline (W.L) at upright position W1L1 – Inclined W.L Bo – C.B. at upright position, B1 – C.B. at inclined position - The displacement (volume) of the ship v1 , v2 – The volume of the emerged and immersed g1, g2 – C.G. of the emerged and immersed wedge, respectively
Equivolume Inclination
( v1 =v2 )
If the ship is wall-sided with the range of inclinations of a small angle dΦ, then the volume v1 and v2 , of the two wedges between the two waterlines will be same. Thus, the displacements under the waterlines WoLo and W1L 1 will be same. This inclination is called equivolume inclination. Thus, the intersection of WoLo, and W1 L1 is at the longitudinal midsection . For most ships, while they may be wall-sided in the vicinity of WL near their midship section, they are not wall-sided near their sterns and bows . However, at a small angle of inclination, we may still approximately treat them as equivolume inclination .
STABILITY OF FLOATING BODIES
Any floating body is subjected by two opposing vertical forces. One is the body's weight W which is downward, and the other is the buoyant force BF which is upward. The weight is acting at the center of gravity G and the buoyant force is acting at the center of buoyancy BO. W and BF are always equal and if these forces are collinear, the body will be in upright position as shown below.
The body may tilt from many causes like wind or wave action causing the center of buoyancy to shift to a new position as shown below.
Point is the intersection of the axis of the body and the line of action of the buoyant force, it is called metacenter . If M is above G, BF and W will produce a righting moment RM which causes the body to return to its neutral position, thus the body is stable. If M is below G, the body becomes unstable because of the overturning moment OM made by W and BF. If M coincides with G, the body is said to be just stable which simply means critical. The value of righting moment or overturning moment is given by
The distance
is called metacentric height .
Metacentric height, Use (-) if G is above BO and (+) if G is below BO. Note that M is always above BO. Value of MBO
Assume that the body is rectangular at the top view and measures B by L at the waterline when in upright position. The moment due to the shifting of the buoyant force is equal to the moment due to shifting of wedge.
For small value of θ, tan θ ≡ sin θ and note that 1/12 LB3 = I, thus,
The formula above can be applied to any section. Where W = weight of the body BF = buoyant force M = metacenter G = center of gravity of the body BO = center of buoyancy in upright position BO' = center of buoyancy in tilted position MG = metacentric height or the distance from M to G MBO = distance from M to BO GO = distance from G to BO v = volume of the wedge either immersion or emersion s = horizontal distance between the center of gravity of the wedges θ = angle of tilting I = moment of inertia of the waterline section of th e body RM = righting moment OM = overturning moment For rectangular section
Name: ___________________________ Yr. & Blk. : ______
Score: __________
EXPERIMENT NO. ___ STABILITY OF FLOATING BODY OBJECTIVE: To determine the stability of floating body (scow) through the center of gravity at various heights. Materials: Scow, Rectangular container whose dimensions is greater than the scow, weights, clamp, ruler and water.
Angular Scale
Horizontal Weights
Pole Plumb Line
Vertical Weights Draft Scale
Fig. 1 Scow Procedure: Metacentric Height: 1. Set up the apparatus. Fill the container with w ater of about two-thirds. 2. Check the scow if the weights are in balance positions. The horizontal weights should be in middle of bar and the vertical weights should rest on the platform. 3. Measure the outside dimension of the rectangular scow. Compu te the volume of the scow afterwards. 4. Place the scow in the container with water. Let it rest for a while until it is stable. 5. Compute the moment of inertia of the waterline section with the scow. 6. Determine the distance of metacenter to the center of buoyancy.
= .
7. Determine the distance of center of gravity to the center of buoyancy.
= + , where G is the distance of the center of gravity from the waterline and Bo is the distance of the center of buoyancy to the waterline. 8. Calculate the metacentric height:
= 9. Record the results.
Righting/Overturning Moment 1. 2. 3. 4. 5.
Now, adjust the horizontal weights to the right side end. Also, adjust the vertical weights to a height of 3 .0cm. Let it rest for a while until it is settle. Measure the angle from the plumb line to the pole. Determine the distance of metacenter to the center of buoyancy.
=
6. Determine the distance of center of gravity to the center of buoyancy.
= + , where G is the distance of the center of gravity from the waterline and Bo is the distance of the center of buoyancy to the waterline. 7. Calculate the metacentric height:
= 8. Compute for the righting moment: R.M. = W x GM 9. Repeat step 2 to 8 by raising incrementally the vertical weights. 10. Record the data and analyze.
Data Table: A.
Initial Metacentric Height Moment of Inertia, I
Volume, V
MBo
GBo
GM
B. Righting Moment
Height of weight (cm) 3.0 4.0 5.0 6.0 7.0 Computation:
Draft Height (cm)
MBo
GBo
GM
R.M.