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# Lesson 4 buoyancy ```HYDRAULICS
Engr. Michelle D. Enriquez
Lesson 4
Buoyancy and Stability of floating bodies
TOPICS
 Buoyancy
 Immersed and Floating Bodeis
 Stability of Floating Bodies
LEARNING OUTCOMES
At the end of the lesson, you should be able to:
 Understand why bodies immersed and float in liquid
 Analyse and design of dams
 Understand the principle of buoyancy
This lesson describes how to determine whether an object will tip over or remain in an upright position when placed in
a liquid. This topic is important for the design of objects such as ships and buoys. Although civil engineers are not boat
designers, they do have to deal with cases of buoyancy from time to time. Some typical examples are:
1. Buried gas pipelines in waterlogged ground
2. Exploration rigs used by oil or gas corporations
3. Towing large steel dock/lock gates by sea or river (assuming that the structure can float, of course)
Buoyancy
A fundamental principle affecting objects submerged in fluids was discovered by Greek mathematician and
natural philosopher Archimedes. Archimedes’ principle can be stated as follows:
“Any object completely or partially submerged in a fluid is buoyed up by a force
With magnitude equal to the weight of the fluid displaced by the object”
Everyone has experienced Archimedes’ principle. It’s relatively easy, for example, to lift someone if you’re
both standing in a swimming pool, whereas lifting that same individual on dry land may be a difficult task.
Water provides partial support to any object placed in it. We often say that an object placed in a fluid is
buoyed up by the fluid, so we call this upward force the buoyant force.
The buoyant force is not a mysterious new force that arises in fluids. In fact, the physical cause of the
buoyant force is the pressure difference between the upper and lower sides of the object, which can be
shown to be equal to the weight of the displaced fluid. The fluid inside the indicated sphere, colored darker
blue, is pressed on all sides by the surrounding fluid. Arrows indicate the forces arising from the pressure.
Because pressure increases with depth, the arrows on the underside are larger than those on top. Adding
them all up, the horizontal components cancel, but there is a net force upward. This force, due to differences
⃗ . The sphere of water neither rises nor falls, so the vector sum of the
in pressure, is the buoyant force 𝐵
buoyant force and the force of gravity on the sphere of fluid must be zero, and it follows that 𝐵 = 𝑀𝑔,
where 𝑀 is the mass of the fluid
HYDRAULICS
Engr. Michelle D. Enriquez
Replacing the shaded fluid with a cannon ball of the same volume, as in Figure b, changes only the mass on
which the pressure acts, so the buoyant force is the same: 𝐵 = 𝑀𝑔,, where M is the mass of the displaced
fluid, not the mass of the cannon ball. The force of gravity on the heavier ball is greater than it was on the
fluid, so the cannon ball sinks.
Archimedes’ principle can also be obtained, relating pressure and depth, using Horizontal forces from the
pressure cancel, but in the vertical direction 𝑃2 𝐴 acts upward on the bottom of the block of fluid, and 𝑃1 𝐴
and the gravity force on the fluid, Mg, act downward, giving:
𝑩 = 𝑷𝟐 𝑨 − 𝑷𝟏 𝑨 = 𝑴𝒈
where the buoyancy force has been identified as a difference in pressure equal in magnitude to the weight
of the displaced fluid. This buoyancy force remains the same regardless of the material occupying the
volume in question because it’s due to the surrounding fluid. Using the definition of density:
𝑩 = 𝝆𝒇𝒍𝒖𝒊𝒅 𝑽𝒇𝒍𝒖𝒊𝒅 𝒈
where𝝆𝒇𝒍𝒖𝒊𝒅 is the density of the fluid and 𝑽𝒇𝒍𝒖𝒊𝒅 is the volume of the displaced fluid. This result applies
equally to all shapes because any irregular shape can be approximated by a large number of infinitesimal
cubes.
A totally submerged object that is less dense than the fluid in which
it is submerged is acted upon by a net upward force. (b) A totally
submerged object that is denser than the fluid sinks.
HYDRAULICS
Engr. Michelle D. Enriquez
Case I: A Totally Submerged Object.
When an object is totally submerged in a fluid of density 𝝆𝒇𝒍𝒖𝒊𝒅, the
upward buoyant force acting on the object has a magnitude of
𝑩 = 𝝆𝒇𝒍𝒖𝒊𝒅𝑽𝒐𝒃𝒋 𝒈, where 𝑽𝒐𝒃𝒋 is the volume of the object. If the
object has density 𝝆𝒐𝒃𝒋 , the downward gravitational force acting on
the object has a magnitude equal to
𝑤 = 𝑚𝑔 =
𝝆𝒐𝒃𝒋 𝑽𝒐𝒃𝒋 𝒈, and the net force on it is 𝐵 − 𝑤 = (𝝆𝒇𝒍𝒖𝒊𝒅 −
𝝆𝒐𝒃𝒋 )𝑽𝒐𝒃𝒋 𝒈. Therefore, if the density of the object is less than the
density of the fluid, the net force exerted on the object is positive
(upward) and the object accelerates upward, as in figure a; whereas if
the density of the object is greater than the density of the fluid, as
Figure b, the net force is negative and the object accelerates
downward.
Case II: A Floating Object.
Now consider a partially submerged object in static equilibrium floating in a fluid, as in the figure above. In
this case, the upward buoyant force is balanced by the downward force of gravity acting on the object. If
𝑽𝒇𝒍𝒖𝒊𝒅 is the volume of the fluid displaced by the object (which corresponds to the volume of the part of the
object beneath the fluid level), then the magnitude of the buoyant force is given by
𝑩=
𝝆𝒇𝒍𝒖𝒊𝒅 𝑽𝒇𝒍𝒖𝒊𝒅 𝒈. Because the weight of the object is 𝑤 = 𝑚𝑔 = 𝝆𝒐𝒃𝒋 𝑽𝒐𝒃𝒋 𝒈, and because 𝑤 = 𝐵, it
follows that𝝆𝒇𝒍𝒖𝒊𝒅 𝑽𝒇𝒍𝒖𝒊𝒅𝒈 = 𝝆𝒐𝒃𝒋 𝑽𝒐𝒃𝒋 𝒈.
𝝆𝒐𝒃𝒋
𝑽𝒇𝒍𝒖𝒊𝒅
=
𝝆𝒇𝒍𝒖𝒊𝒅
𝑽𝒐𝒃𝒋
Immersed and Floating Bodies
Immersed Bodies
When a body is completely immersed in a liquid, its stability depends on the relative positions of the center of gravity of
the body and the centroid of the displaced volume of fluid, which is called the center of buoyancy. If the center of
buoyancy is above the center of gravity, such as in the figure shown below, any tipping of the body produces a righting
couple, and consequently, the body is stable. However, if the center of gravity is above the center of buoyancy, any
tipping produces an increasing overturning moment, thus causing the body to turn through 180&deg;. This is the condition
shown in c. Finally, if the center of buoyancy and center of gravity are coincident, the body is neutrally stable—that is, it
lacks a tendency for righting or for overturning, as shown in b.
Conditions of stability for immersed
bodies: (a) Stable. (b) Neutral. (c)
Unstable
HYDRAULICS
Engr. Michelle D. Enriquez
Floating Bodies
The question of stability is more involved for floating bodies
than for immersed bodies because the center of buoyancy
may take different positions with respect to the center of
gravity, depending on the shape of the body and the
position in which it is floating. For example, consider the
cross section of a ship shown in a. Here the center of
gravity G is above the center of buoyancy C. Therefore, at
first glance it would appear that the ship is unstable and
could flip over. However, notice the position of C and G
after the ship has taken a small
angle of heel.
Ship stability relations
As shown in b, the center of gravity is in the same position, but the center of buoyancy has moved outward of the
center of gravity, thus producing a righting moment. A ship having such characteristics is stable. The reason for the
change in the center of buoyancy for the ship is that part of the original buoyant volume, as shown by the wedge shape
AOB, is transferred to a new buoyant volume EOD. Because the buoyant center is at the centroid of the displaced
volume, it follows that for this case the buoyant center must move laterally to the right. The point of intersection of the
lines of action of the buoyant force before and after heel is called the metacenter M, and the distance GM is called the
metacentric height. If GM is positive—that is, if M is above G—the ship is stable; however, if GM is negative, the ship is
unstable. Quantitative relations
involving these basic principles of
stability are presented in the next
paragraph
(a) Plan view of ship at waterline. (b) Section A-A of ship
Consider the ship shown above, which has taken a small angle of heel 𝛼. First evaluate the lateral displacement of the
center of buoyancy, then it will be easy by simple trigonometry to solve for the metacentric height GM or to evaluate
the righting moment. Recall that the center of buoyancy is at the centroid of the displaced volume. Therefore, resort to
the fundamentals of centroids to evaluate the displacement. From the definition of the centroid of a volume,
𝑥̅ 𝑉 = ∑ 𝑥𝑖 ∆𝑉𝑖
Where:
𝑥̅ = 𝐶𝐶 ′ (𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑝𝑙𝑎𝑛𝑒 𝑎𝑏𝑜𝑢𝑡 𝑤ℎ𝑖𝑐ℎ 𝑚𝑜𝑚𝑒𝑛𝑡𝑠 𝑎𝑟𝑒 𝑡𝑜 𝑏𝑒 𝑡𝑎𝑘𝑒𝑛 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑟𝑜𝑖𝑑 𝑜𝑓 𝑉) w
𝑉 = 𝑡𝑜𝑡𝑎𝑙 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑑
∆𝑉𝑖 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡
𝑥𝑖 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑐𝑟𝑒𝑚𝑒𝑛𝑡 𝑣𝑜𝑙𝑢𝑚𝑒
HYDRAULICS
Engr. Michelle D. Enriquez
Take moments about the plane of symmetry of the ship. Recall from mechanics that volumes to the left produce
negative moments and volumes to the right produce positive moments. A convenient way to do this is to consider the
moment of the volume before heel, subtract the moment of the volume represented by the wedge AOB, and add the
moment represented by the wedge EOD. In a general way this is given by the following equation:
𝑥̅ 𝑉 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑉 𝑏𝑒𝑓𝑜𝑟𝑒 ℎ𝑒𝑒𝑙 − 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑉𝐴𝑂𝐵 + 𝑚𝑜𝑚𝑒𝑛𝑡 𝑉𝐸𝑂𝐷
Because the original buoyant volume is symmetrical with 𝑦 − 𝑦, the moment for the first term on the right is zero. Also,
the sign of the moment of 𝑉𝐴𝑂𝐵 is negative; therefore, when this negative moment is subtracted from the right-hand
side, the result is:
𝑥̅ 𝑉 = ∑ 𝑥𝑖 ∆𝑉𝑖𝐴𝑂𝐵 + ∆𝑉𝑖𝐸𝑂𝐷
In addition to the force of gravity or weight, all objects submerged in a fluid are acted on a force BF. The buoyant force
acts upward and is equal to the weight of the fluid displaced by the object. Buoyancy force can be simply defined as:
𝐵𝐹 = 𝛾𝑤 𝑉𝐷
Where:
𝐵𝐹 = 𝑏𝑢𝑜𝑦𝑎𝑛𝑡 𝑓𝑜𝑟𝑐𝑒
𝛾𝑤 = 𝑢𝑛𝑖𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 𝑜𝑓 𝑓𝑙𝑢𝑖𝑑
𝑉𝐷 = 𝑣𝑜𝑙𝑢𝑚𝑒 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒
For a freely floating object (with no external forces) the weight of the object is equal to the buoyant force on the object
(acting upward)
𝑊=𝐵𝐹
𝑊 = 𝛾𝑤 𝑉𝐷
Example 1:
A driver and his suit weight 890N. It requires 130N of lead to sink him in fresh water. If the
volume of the lead is 0.00117𝑚3 . What is the volume of the driver and his suit?
Solution:
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹1 + 𝐵𝐹1 = 𝑊1 + 𝑊2
𝛾𝑤1 𝑉𝐷1 + 𝛾𝑤2 𝑉𝐷2 = 890𝑁 + 130𝑁
𝛾𝑤 (𝑉𝐷1 + 11.3𝑉𝐷2 ) = 1020𝑁
𝑁
9810 3 (𝑉𝐷1 + 0.00117) = 1020𝑁
𝑚
𝑉𝐷1 = 0.1028𝑚3 .
HYDRAULICS
Engr. Michelle D. Enriquez
Example 2:
Cube A is 30cm long along edge and weighs 1445N. It is attached to the square prism B, which is 15cm x 15cm x 2.4m
𝑁
and weighs4700 𝑚3. What length of B will project above the liquid surface is the relative density of the liquids is 1.50.
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹1 + 𝐵𝐹1 = 𝑊1 + 𝑊2
1.5 (9810
𝑁
𝑁
𝑁
) (0.30)2 (1.0𝑚) + 1.5 (9810 3 ) (0.15)2 (2.4 − ℎ) = 1445𝑁 + 4700 3 (0.15m x 0.15m x 2.4m)
𝑚3
𝑚
𝑚
1324.5𝑁 + 1.5 (9810
𝑁
) (0.15)2 (2.4 − ℎ) = 1698.8𝑁
𝑚3
𝑁
) (0.15)2 (2.4 − ℎ) = 374.3𝑁
𝑚3
ℎ = 1.2694𝑚
1.5 (9810
Example 3:
A block of wood is 20mm thick floating in seawater. The specific gravity of wood is
0.65 while that of the seawater is 1.03. Find the area of a block of wood, which will
just support a woman weighing 80kg when the top of the surface is just at the water
surface.
Solution:
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹1 = 𝑊𝑤𝑜𝑜𝑑 + 𝑊𝑤𝑜𝑚𝑎𝑛
𝑆𝐺𝑠𝑤 𝛾𝑤 𝑉𝐷 = 𝑆𝐺𝑤 𝛾𝑤 𝑉 + 𝑊𝑤𝑜𝑚𝑎𝑛
(1.03) (9810
ℎ = 0.095𝑚3
𝑁
𝑚
3
) 𝐴(0.02𝑚) = 0.65 (9810
𝑁
𝑚
3
) (𝐴)(0.02𝑚) +
80𝑘𝑔
𝑚
(9.81 2 )
𝑠
Example 4:
The weight of certain crown in air is 14N and its weight in water is 12.7N. Assuming that
the crown is an alloy of gold (SG= 19.3) ad silver (SG = 10.5). Assume unit weight of
𝑘𝑁
water is 9.79 𝑚
3
a.
Compute the volume of the crown
b.
Compute the specific gravity of the brown
c.
Compute the fraction of silver in the crown.
HYDRAULICS
Engr. Michelle D. Enriquez
Solution:
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑊𝑎𝑖𝑟 − 𝑊𝑤𝑎𝑡𝑒𝑟
9,790
𝑘𝑁
𝑉
𝑚 3 𝑐𝑟𝑜𝑤𝑛
= 14𝑁 − 12.7𝑁
𝑽𝒄𝒓𝒐𝒘𝒏 = 𝟎. 𝟎𝟎𝟎𝟏𝟑𝟐𝟕𝒎𝟑
Specific gravity of the crown:
𝑊
𝛾𝑐𝑟𝑜𝑤𝑛 =
𝑉
14𝑁
𝛾𝑐𝑟𝑜𝑤𝑛 =
0.0001327𝑚3
𝑘𝑁
𝛾𝑐𝑟𝑜𝑤𝑛 = 105.422 3
𝑚
𝑘𝑁
105.422 3
𝛾𝑐𝑟𝑜𝑤𝑛
𝑚 = 10.768
𝑆𝐺𝑐𝑟𝑜𝑤𝑛= =
=
𝑘𝑁
𝛾𝑤
9.79 3
𝑚
Fraction of silver in the crown
Let 𝑥 = 𝑓𝑟𝑎𝑐𝑡𝑖𝑜𝑛 𝑜𝑓 𝑠𝑖𝑙𝑣𝑒𝑟 𝑖𝑛 𝑡ℎ𝑒 𝑐𝑟𝑜𝑤𝑛
10.5 𝑥 + (1 − 𝑥 )19.3 = 10.768
𝒙 = 𝟎. 𝟗𝟔𝟗𝟓
Example 5:
A wooden buoy of specific gravity 0.75 floats in a liquid with specific gravity of 0.85.
1. What is the percentage of the volume above the liquid surface to the total
volume of the buoy
2. If the volume above the liquid is 0.0145𝑚3, what is the weight of wooden
buoy?
3. What load will cause the buoy to be fully submerged?
Solution:
a. % of volume above the liquid surface
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑊
(0.75) (9.810
𝑘𝑁
𝑚3
) 𝑉 = (0.85) (9.810
𝑉 = 1.13333𝑉2
𝑉2 = 0.8824𝑉
𝑉1 = 𝑉 − 𝑉2
𝑉1 = 𝑉 − 0.8824𝑉
𝑉1 = 0.1177𝑉
𝑉1
= 0.1178
𝑉
𝑉1
= 11.78%
𝑉
𝑘𝑁
𝑚3
) 𝑉2
HYDRAULICS
Engr. Michelle D. Enriquez
b.
Weight of the wooden buoy
𝑊
𝛾𝑤𝑜𝑜𝑑 =
𝑉
𝑊 = 𝛾𝑤𝑜𝑜𝑑 𝑉
Since
𝑉1 = 0.1177𝑉 = 0.0145𝑚3 = 0.1177𝑉; 𝑉 = 0.1232𝑚3
𝑊 = (0.75) (9.810
𝑘𝑁
𝑚3
) (0.1232𝑚3 )
𝑊 = 0.9064𝑘𝑁
c.
Load that will cause the buoy to be fully submerged
𝑃 = 𝛾𝑏𝑢𝑜𝑦 𝑉1
𝑃 = (0.85) (9.810
𝑘𝑁
𝑚3
) (0.0145𝑚3 )
𝑃 = 0.1209 𝑘𝑁
Example 6:
A block of wood 0.60m x 0.60m x h meters in dimension was thrown into the water, it floats with
0.18m projecting above the water surface. The same block was thrown into a container of a
liquid having a specific gravity of 0.90 and it floats with 0.14cm projecting above the liquid
surface.
1. Determine the value of “h”
2. Determine the specific gravity of the block
3. Determine the weight of the block
Solution:
a. Value of h
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑊
𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝛾𝑏𝑙𝑜𝑐𝑘𝑠𝑢𝑏 𝑉
𝑘𝑁
𝑘𝑁
𝑆𝐺𝑏𝑙𝑜𝑐𝑘 (9.810 3 ) (0.60𝑚)(0.60𝑚)ℎ = (9.810 3 ) (0.60𝑚)(0.60𝑚)(ℎ − 0.18)
𝑚
𝑚
𝑆𝐺𝑏𝑙𝑜𝑐𝑘 ℎ = (ℎ − 0.18)
− 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 1
Thrown in other liquid:
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑊
𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝛾𝑏𝑙𝑜𝑐𝑘𝑠𝑢𝑏 𝑉
𝑘𝑁
𝑘𝑁
(9.810 3 ) (0.60𝑚)(0.60𝑚)ℎ = (0.90) (9.810 3 ) (0.60𝑚)(0.60𝑚)(ℎ − 0.14)
𝑚
𝑚
𝑆𝐺𝑏𝑙𝑜𝑐𝑘 ℎ = 0.90(ℎ − 0.14)
− 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 2
equating 1 and 2:
(ℎ − 0.18) = 0.90(ℎ − 0.14)
(ℎ − 0.18) = 0.90ℎ − 0.126
ℎ = 0.54𝑚
:
HYDRAULICS
Engr. Michelle D. Enriquez
b.
Specific gravity of block
From equation 1:
𝑆𝐺𝑏𝑙𝑜𝑐𝑘 ℎ = (ℎ − 0.18)
𝑆𝐺𝑏𝑙𝑜𝑐𝑘 (0.54) = (0.54 − 0.18)
𝑆𝐺𝑏𝑙𝑜𝑐𝑘 = 0.6667
c.
Weight of the block
𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝑊
𝑆𝐺𝑏𝑙𝑜𝑐𝑘 𝛾𝑏𝑙𝑜𝑐𝑘 𝑉 = 𝑊
𝑊 = (0.667) (9.810
𝑊 = 1.272𝑘𝑁
𝑘𝑁
) (0.60𝑚)(0.60𝑚)(0.54)
𝑚3
Stability of Floating Bodies
The stability of floating bodies is defined as its ability to return to its neutral position after the external force has been
applied and removed. The location of the metacentre is important in determining the stability. The metacentre is a point
on the vertical neutral axis through which the buoyant force always acts for small angles of tilt. For a large angle of tilt,
the metacentre may move along the neutral axis.
1. For stability to exist, the objects center of gravity must below its metacentre.
2. For objects submerged in fluids, the metacentre is located at its center of buoyancy.
3.
Where:
For partially objects the location of the metacentre is found from the following relation:
𝐴𝑀 = 𝐴𝐵𝑜 + 𝑀𝐵𝑜
𝐴𝑀 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡ℎ𝑒 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑚𝑒𝑡𝑎𝑐𝑒𝑛𝑡𝑒𝑟
𝐴𝐵𝑜 = 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑓𝑟𝑜𝑚 𝑡𝑒ℎ 𝑏𝑜𝑡𝑡𝑜𝑚 𝑜𝑓 𝑡𝑦ℎ𝑒 𝑜𝑏𝑗𝑒𝑐𝑡 𝑡𝑜 𝑡ℎ𝑒 𝑐𝑒𝑛𝑡𝑦𝑒𝑟 𝑜𝑓 𝑏𝑢𝑜𝑦𝑎𝑛𝑐𝑦
The location of the buoyancy 𝐵𝑜 is located at the geometric center of the displacement volume. The
magnitude and geometry of the displaced volume maybe found from the buoyancy equation. The distance from the
center of the buoyancy to the center of the metacentre
𝑀𝐵𝑜 =
𝐼𝑠
𝑉
Where:
𝐼𝑠 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑏𝑜𝑢𝑡 𝑡ℎ𝑒 𝑙𝑎𝑟𝑔𝑒𝑠𝑡 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑎𝑟𝑒𝑎 𝑝𝑟𝑜𝑑𝑢𝑐𝑒𝑑
HYDRAULICS
Engr. Michelle D. Enriquez
When the center of gravity is below the metacentre, there is a couple acting to return the vessel to its neutral
position but if the center of gravity is above the metacentre, the couple will continue to tip the vessel.
Vertical Stability of Floating Bodies:
𝑀𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑓𝑡𝑒𝑑 𝑤𝑒𝑑𝑔𝑒 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑑𝑢𝑒 𝑡𝑜 𝑡ℎ𝑒 𝐵𝐹
2
𝐵𝐹𝑥 = 𝐹 ( 𝐵)
3
𝐵 𝐻 (𝑦) 2
𝐵𝐹𝑥 =
( 𝐵) 𝛾𝑤
2 2
3
𝑦
tan 𝜃 =
𝐵
2
𝐵𝑡𝑎𝑛 𝜃
𝑦=
2
𝐵𝐹 = 𝛾𝑤 𝑉
𝑥 = 𝑀𝐵0 𝑠𝑖𝑛𝜃
𝐵𝑡𝑎𝑛 𝜃
𝐵𝐻( 2 ) 2
( 𝐵) 𝛾𝑤
2
2
3
𝐵 𝐻 (𝐵𝑡𝑎𝑛 𝜃) 𝐵
𝛾𝑤 𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 =
( ) 𝛾𝑤
2
2
3
3
𝐵 𝐻
𝛾𝑤 𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 = 12 𝑡𝑎𝑛 𝜃𝛾𝑤
𝐵3 𝐻
𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 =
𝑡𝑎𝑛 𝜃
12
𝛾𝑤 𝑉𝑀𝐵0 𝑠𝑖𝑛𝜃 =
But 𝑠𝑖𝑛𝜃 = tan 𝜃 ( 𝑓𝑜𝑟 𝑠𝑚𝑎𝑙𝑙 𝑎𝑛𝑔𝑙𝑒𝑠0
𝑉𝑀𝐵0 = 𝐼
𝐼
𝑀𝐵0 =
𝑉
Where:
𝐼 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑎𝑙𝑜𝑛𝑔 𝑎𝑥𝑖𝑠 𝑜𝑓 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛
𝑉 = 𝑚𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑠𝑢𝑏𝑚𝑒𝑟𝑔𝑒𝑑 𝑝𝑜𝑟𝑡𝑖𝑜𝑛
Metacentric Height:
𝑀𝐵0 =
𝐵2
𝑡𝑎𝑛2 𝜃
(1+
)
12𝐷
2
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
HYDRAULICS
Engr. Michelle D. Enriquez
Example 1:
A rectangular barge has dimensions of 30m long, 10m wide and 3m deth.
It weighs 4500kN when loaded, with its center of gravity along the vertical axis is
4m from the botoom of the barge. The barge is floating in seawater havinga a
secific gravity of 1.01
 Determine the draft of the barge in seawater
 Determine the metacentric height of the barge dur to rolling
 Determine the metacentruic height of the barge due to pitching
Solution:
a. Draft of the barge in seawater
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑊
𝛾𝑏𝑎𝑟𝑔𝑒 𝑉 = 𝑊
𝑆𝐺𝑏𝑎𝑟𝑔𝑒 𝛾𝑏𝑎𝑟𝑔𝑒 𝑉 = 𝑊
𝑘𝑁
(1.01) (9.810 3 ) (30𝑚)(10𝑚)(ℎ) = 4500
𝑚
𝐷 = 1.5139𝑚
b. Metacentric height if the barge is rolling
𝐼
𝑀𝐵0 = 𝑉
𝐵𝐻3
(30𝑚)(10𝑚)3
𝐼=
=
= 2500𝑚4
12
12
𝑉 = 𝐴ℎ = (30𝑚)(10𝑚)(1.5139𝑚) = 454.17𝑚3
𝑀𝐵0 =
𝐼
2500𝑚4
=
= 5.5045𝑚
𝑉
454.17𝑚3
Then the metacentric height:
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
𝑀𝐺 = 5.5045𝑚 − 3.245𝑚
𝑀𝐺 = 2.2595𝑚
HYDRAULICS
Engr. Michelle D. Enriquez
c. Metacentric height if the barge is pitching
𝐼
𝑉
𝐵𝐻3
(10𝑚)(30𝑚)3
𝐼=
=
= 22500𝑚4
12
12
𝑉 = 𝐴ℎ = (30𝑚)(10𝑚)(1.5139𝑚) = 454.17𝑚3
𝑀𝐵0 =
𝑀𝐵0 =
𝐼
22500𝑚4
=
= 49.5409𝑚
𝑉
454.17𝑚3
Then the metacentric height:
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
𝑀𝐺 = 49.5409 − 3.245𝑚
𝑀𝐺 = 46.2959𝑚
Example 2:
A rectangular boat 8m long by 6m wide and 2.4m high is sibmerged in water by 1.8m
a. Find the total draft after 𝑃 is placed at the edge of the boat without causing it
to sink. Assume water is fresh with density equa to 1000 𝑘𝑔/𝑚3
b. What is the maximum P that can be placed at the edge of the boat without
causing it to sink . assume water is fresh with density equal to 1000 𝑘𝑔/𝑚3
c. Solve the maximum P if the boat is submerged in sea water with a density at
1026 𝑘𝑔/𝑚3
Solution:
a.
Draft after P is placed at the edge of the boat
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑃
𝛾𝑤 𝑉 = 𝑃
𝑆𝐺𝜌𝑤 𝑉 = 𝑃
𝑘𝑔
𝑃 = (6.0𝑚)(8.0𝑚) (1000 3 ) (𝑥 )
𝑚
𝑃 = 48,000𝑥
𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑓𝑡𝑖𝑛𝑔 𝑤𝑒𝑑𝑔𝑒 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑛′ 𝑠 𝑤𝑒𝑖𝑔ℎ𝑡
(0.6 − 𝑥 )(3)
𝑘𝑔 2
𝑃 (3.0𝑚) =
(8.0𝑚) (1000 3 ) ( ) (6.0𝑚)
2
𝑚
3
(0.6 − 𝑥 )(3)
𝑘𝑔 2
48,000𝑥 (3.0𝑚) =
(8.0𝑚) (1000 3 ) ( ) (6.0𝑚)
2
𝑚
3
𝑘𝑔
48,000𝑥 (3.0𝑚) = (0.6 − 𝑥 )(12.0𝑚) (1000 3 ) (4.0𝑚)
𝑚
1
𝑥 = (0.6 − 𝑥 )( )
3
𝑥 = 0.15𝑚
𝑇𝑜𝑡𝑎𝑙 𝑑𝑟𝑎𝑓𝑡 𝑎𝑓𝑡𝑒𝑟 𝑃 𝑖𝑠 𝑙𝑜𝑎𝑑𝑒𝑑 = 1.8 + 0.15
𝑻𝒐𝒕𝒂𝒍 𝒅𝒓𝒂𝒇𝒕 𝒂𝒇𝒕𝒆𝒓 𝑷 𝒊𝒔 𝒍𝒐𝒂𝒅𝒆𝒅 = 𝟏. 𝟗𝟓𝒎
HYDRAULICS
Engr. Michelle D. Enriquez
b.
Maximum P at the edge of the boat
𝑃 = 48,000𝑥
𝑃 = 48,000(0.15)
𝑃 = 7200𝑘𝑔
c.
Maixumum P if the boat is submerged
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑊
𝑆𝐺𝜌𝑤 𝑉 = 𝑊
𝑊 = (6.0𝑚)(8.0𝑚) (1000
𝑘𝑔
) (1.8)
𝑚3
𝑊 = 86,400𝑘𝑔
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑊
𝑆𝐺𝜌𝑤 𝑉 = 𝑊
86400𝑘𝑔 = (6.0𝑚)(8.0𝑚) (1026
𝐷 = 1.7𝟓𝟒𝟒𝒎
∑ 𝐹𝑦 = 0 ↑ (+)
𝐵𝐹 = 𝑃
𝛾𝑤 𝑉 = 𝑃
𝑆𝐺𝜌𝑤 𝑉 = 𝑃
𝑃 = (6.0𝑚)(8.0𝑚) (1026
𝑃 = 49,248𝑥
𝑘𝑔
) (𝑥 )
𝑚3
𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑠ℎ𝑖𝑓𝑡𝑖𝑛𝑔 𝑤𝑒𝑑𝑔𝑒 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑚𝑎𝑛′ 𝑠 𝑤𝑒𝑖𝑔ℎ𝑡
(0.65 − 𝑥 )(3)
𝑘𝑔 2
𝑃 (3.0𝑚) =
(8.0𝑚) (1000 3 ) ( ) (6.0𝑚)
2
𝑚
3
(0.6 − 𝑥 )(3)
𝑘𝑔 2
49,248𝑥 (3.0𝑚) =
(8.0𝑚) (1000 3 ) ( ) (6.0𝑚)
2
𝑚
3
𝑘𝑔
49,248𝑥 (3.0𝑚) = (0.65 − 𝑥 )(12.0𝑚) (1000 3 ) (4.0𝑚)
𝑚
1
(
)
𝑥 = 0.65 − 𝑥 ( )
3
𝑥 = 0.1625𝑚
𝑃 = 49,248𝑥
𝑃 = 49,248(0.1625)
𝑃 = 8002.8𝑘𝑔
𝑘𝑔
) (𝐷 )
𝑚3
HYDRAULICS
Engr. Michelle D. Enriquez
Example 3:
A rectangualr scow 9.0m wide and 15.0m long and 3.6m high has a draft in seawater of 2.4m. its centyer of
gravity is 2.70m above the bottom of the scow.
a. Determine the initial metacentric height
b. If the scow is tilted until one end is just submerged in water, find the sidewise shjifting of the center of
buoyancy.
c. Determine the final metacentric height
Solution:
a. Initial metacentric height MG
𝜃=0
(15𝑚)(9𝑚)3
𝑏ℎ3
𝐼
911.25𝑚4
12
12
𝑀𝐵0 = =
=
= 2.8125𝑚
𝑉 𝑙𝑤ℎ
(15𝑚)(9.0𝑚)(2.4𝑚)
324𝑚3
𝐺𝐵0 = 2.70𝑚 − 1.20𝑚 = 1.50𝑚
Then the metacentric height:
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
𝑀𝐺 = 2.8125𝑚 − 1.50𝑚
𝑀𝐺 = 1.3125𝑚
b.
Sidewise shifting of the center of buoyancy
1.2
; 𝜃 = 14.93&deg;
4.5
𝑥𝑏
𝑠𝑖𝑛𝜃 =
𝑀𝐵𝑜
𝐼
𝑡𝑎𝑛2 𝜃
]
𝑀𝐵0 = [ 1 +
𝑉
2
𝑡𝑎𝑛2 14.93
]
𝑀𝐵0 = 2.8125 [ 1 +
2
𝑀𝐵0 = 2.9125
𝑥𝑏 = 2.9125 sin 14.93
𝑥𝑏 = 0.7497𝑚
𝑡𝑎𝑛𝜃 =
HYDRAULICS
Engr. Michelle D. Enriquez
c.
Final metacentric height
𝑀𝐵0 =
𝐵2
𝑡𝑎𝑛2 𝜃
(1+
)
12𝐷
2
𝑀𝐵0 =
(9)2
𝑡𝑎𝑛2 14.93
(1+
)
12(2.4)
2
𝑀𝐵0 = 2.9125
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
𝑀𝐺 = 2.9125𝑚 − 1.50𝑚
𝑀𝐺 = 1.4125𝑚
Example 4:
The center of gravity of the barge shown is on te mid sectios of 2.7m above the bottom of tyeh barge, the barge
has a dimesnion of 12m long, 4.5m wide and 3.0m high. The draft is 2.40m . If an outside force (wind or wave action)
hells the barge until point A is just the water surface.
a. Comoute the moment at the wedge shift in kg.m
b. What is the moment of the center of buoyancy shift?
c. Compute the value of rigting or upsetting moment.
Solution:
a. Moment of wedge shift
𝐵𝐹 = 𝛾𝑤 𝑉
𝐵𝐹 = (1000
𝑘𝑔
) (12𝑚)(4.5𝑚)(2.4𝑚)
𝑚3
𝐵𝐹 = 129,600 𝑘𝑔
𝑡𝑎𝑛𝜃 =
0.6
; 𝜃 = 14.93&deg;
2.25
𝑀𝐵0 =
𝐵2
𝑡𝑎𝑛2 𝜃
(1+
)
12𝐷
2
𝑀𝐵0 =
(4.5)2
𝑡𝑎𝑛2 14.93
(1+
)
12(2.4)
2
𝑀𝐵0 = 0.7281m
&lt; 𝐺𝐵0 = 1.5
HYDRAULICS
Engr. Michelle D. Enriquez
𝑀𝐺
𝑀𝐺
𝑀𝐺
𝑀𝐺
= 𝑀𝐵0 &plusmn; 𝐺𝐵0
= 2.9125𝑚 − 1.50𝑚
= −0.7719𝑚
= 0.7719𝑚 (𝑚𝑒𝑡𝑎𝑐𝑒𝑛𝑡𝑒𝑟 𝑖𝑠 𝑏𝑒𝑙𝑜𝑤 𝑡ℎ𝑒 𝑔𝑟𝑎𝑣𝑖𝑡𝑦)
𝑦 = 𝑀𝐵0 𝑠𝑖𝑛𝜃
𝑦 = 0.7281sin(14.93)
𝑦 = 0.1876𝑚
𝑥 = 𝑀𝐺𝑠𝑖𝑛𝜃
𝑥 = 0.7719sin(14.93)
𝑥 = 0.1989𝑚
(0.6𝑚)(2.25𝑚)
𝑘𝑔 2
(12.0𝑚) (1000 3 ) (4.5𝑚)
2
𝑚 3
𝑴 = 𝟐𝟒, 𝟑𝟎𝟎 𝒌𝒈𝒎
𝑀=
b.
Moment of center of buoyancy shift
𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑐𝑒𝑛𝑡𝑒𝑟 𝑜𝑓 𝑏𝑢𝑜𝑦𝑎𝑛𝑐𝑦 𝑠ℎ𝑖𝑓𝑡 = 𝐵𝐹𝑦
𝑀 = 129,600 𝑘𝑔(0.1876𝑚)
𝑴 = 𝟐𝟒, 𝟑𝟏𝟐. 𝟗𝟔𝒌𝒈𝒎
c.
Upsetting moment
𝑢𝑝𝑠𝑒𝑡𝑡𝑖𝑛𝑔 = 𝐵𝐹𝑥
𝑀 = 129,600 𝑘𝑔(0.1989𝑚)
𝑴 = 𝟐𝟓, 𝟕𝟕𝟕. 𝟒𝟒𝒌𝒈𝒎
HYDRAULICS
Engr. Michelle D. Enriquez
Example 5:
The barge shown is 10.0m wide by 30.0 m long when loade, the barge diaplecs
5MN and its center of gravity is 0.5m abiove the water line.
a. Find the value of 𝑀𝐵0
b. Determine the metacenter height above the center of gravity for a
roll angle of 10&deg;
c. Compute the righting moment
Solution:
a. Value of 𝑀𝐵0
𝜃=0
𝐵𝐹 = 𝑊
𝛾𝑤 𝑉 = 5 𝑥 106
𝑘𝑔
𝑚
(1000 3 ) (9.81 2 )(10𝑚)(30𝑚)ℎ = 5 𝑥 106
𝑚
𝑠
ℎ = 1.699𝑚
(30𝑚)(10𝑚)3
𝑏ℎ3
𝐼
12
12
𝑀𝐵0 = =
=
𝑉 𝑙𝑤ℎ
(10𝑚)(30𝑚)(1.699𝑚)
2500𝑚4
𝑀𝐵0 =
509.7𝑚3
𝑴𝑩𝟎 = 𝟒. 𝟗𝟎𝟒𝟖𝒎
b.
Metacenter height
𝐺𝐵0 = 0.5𝑚 + 0.85𝑚
𝐺𝐵0 = 1.35𝑚
Then the metacentric height:
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
𝑀𝐺 = 4.9048𝑚 − 1.35𝑚
𝑴𝑮 = 𝟑. 𝟓𝟓𝟒𝟖𝒎
HYDRAULICS
Engr. Michelle D. Enriquez
c.
Righting moment
𝑥 = 𝑀𝐺𝑠𝑖𝑛𝜃
𝑥 = 3.5548𝑚(10)
𝑥 = 0.6173𝑚
𝑅𝑀 = 5 𝑥 106 (0.6173𝑚)
𝑹𝑴 = 𝟑. 𝟎𝟖𝟔𝟓 𝒙 𝟏𝟎𝟔 𝑵𝒎
Example 6:
A cylindrical tank having a diamter of 40cm in diamter and 20cm in height
floats in mercutry ina vertical positio, its deth of immersion being 8.0cm .
a. Find the value of 𝑀𝐵0
b. Determine the metacenteric height
c. If the water is pured uinto the vessel over the mercury until the
cylinder is submerged; partly in mercury and partly in water, determine
the depth of immersion of mercury
Solution:
a. Value of 𝑀𝐵0
𝜃=0
𝜋(40𝑐𝑚)4
𝜋𝑑 4
𝐼
64
64
𝑀𝐵0 = =
=
𝜋(40𝑐𝑚)2
𝑉 𝜋𝑑 2
(8.0𝑐𝑚)
4 ℎ
4
40000𝜋𝑐𝑚 4
𝑀𝐵0 = 3200𝜋𝑐𝑚3
𝑴𝑩𝟎 = 𝟏𝟐. 𝟓𝟎𝒄𝒎
b.
Metacentric height
𝐺𝐵0 = 10𝑐𝑚 − 4𝑐𝑚
𝐺𝐵0 = 6𝑐𝑚
Then the metacentric height:
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
𝑀𝐺 = 12.50𝑐𝑚 − 6.0𝑐𝑚
𝑀𝐺 = 6.50𝑐𝑚
c.
Depth of immersion of mercury
𝐵𝐹 = 𝑊
𝛾𝑤 𝑉 + 𝛾𝐻𝑔 𝑉 = 𝑊
𝑔 𝜋(40𝑐𝑚)2
𝑔
𝜋(40𝑐𝑚)2
(20 − ℎ) + (1 3 ) (13.6)
(ℎ )
(1 3 )
𝑐𝑚
4
𝑐𝑚
4
𝑔
𝜋(40𝑐𝑚)2
(8.0𝑐𝑚)
= (1 3 ) (13.6)
𝑐𝑚
4
(20 − ℎ) + 13.6ℎ = 108.8𝑐𝑚
𝒉 = 𝟕. 𝟎𝟒𝟕𝟔𝒄𝒎
HYDRAULICS
Engr. Michelle D. Enriquez
Example 7:
For a ship with a waterline corss section as shown has adislacement of 5330kN
a. Determine the metacentric height to remain in stable
equilibrium
b. Determine the maximum distance that the center of gravity
may lie above the center of buoyancy if the ship is to remain
stable
c. Determine the horizontal displacement of the center of
buoyancy if the ship tilts an angle of 10&deg; without overturning
Solution:
a. metacentric height to ramin stable equilbirum
𝑀𝐺 = 0
b.
Maximum distance that the center of gravity may lie above the centerof buoyancy
𝐼=
𝑏ℎ 3
𝑏ℎ 3
+ 2 ( 36 )
(30𝑚0(7.5𝑚)3
7.5 ∗ 3.753
𝐼=
+ 2(
)
12
36
𝐼 = 1076.6602 𝑚4
12
𝛾𝑤 =
𝑉=
𝑊
𝑊
;𝑉 =
𝑉
𝛾𝑤
5350𝑘𝑁
= 545.3619𝑚3
𝑘𝑁
9.81 3
𝑚
𝐼
1076.6602 𝑚4
=
𝑉
545.3619𝑚3
𝑀𝐵0 = 1.9742𝑚
𝑀𝐵0 =
Then the metacentric height:
𝑀𝐺 = 𝑀𝐵0 − 𝐺𝐵0
𝑀𝐺 = 1.9742𝑚 − 0
𝑀𝐺 = 1.9742𝑚
c.
Horizontal displacement of the center of buoyancy if the ship tilts an angle of 10&deg; without overturning
𝑥 = 𝑀𝐺𝑠𝑖𝑛𝜃
𝑥 = 1.9742𝑚(10)
𝑥 = 0.3428𝑚
HYDRAULICS
Engr. Michelle D. Enriquez
Buoyancy and stability of floating bodies
1. A piece of wood floats in water with 50mm projecting above the water surface. When placed in glycerine
of specific gravity of 1.35, the block projects 75mm above the liquid surface
a. Find the height of the piece of wood
b. Find the specific gravity of the wood
c. Find the weight of wood if it has a cross sectional area of 200mm x 200mm
2. Two spheres each 1.20m in diameter are connected by means of a short
rope. One weighs 4kN and the other weighs 12kN when placed in water
a. Compute the tension in the rope
b. Compute the depth of the 4kN sphere
c. Compute the volume of sphere exposed above the water
surface
3. A concrete block with a volume of 0.023cu,m is tied to one end of a wooden post
𝑘𝑁
having dimensions of 200mm x 200mm by 3.0m long. Weight of wood is 6.4 3
𝑘𝑁
𝑚
and that of concrete is 23.5 3.
𝑚
a. Determine the length of the wooden post above the water surface
b. Determine the volume of additional concrete to be tied t the bottom of
the post to make its top flush with the water surface
c. Determine the total weight of concrete to make its top flush with the
water surface
4. A rectangular barge weighing 200,000kg is 14.0m long, 8.0m wide and 4.5m deep. It will transport to
𝑘𝑔
Manila the 20mm diameter 6.0m long steel reinforcing bars having a density of 7850 3. Density of salt
𝑘𝑔
water is 1026 3.
𝑚
a. Determine the draft of the barge on sea water before the bars as
b. If a draft to be maintained at 3.0m how many pieces of steel bars it
could carry
c. What is the draft of the barge when one half of its cargo is unloaded
𝑘𝑔
in fresh water (1000 3)
𝑚
d.
5. A concrete cube 0.50m on each side is to be held in equilibrium under water
by attaching a light foam buoy to its. Specific weight of concrete and foam are
𝑘𝑁
𝑘𝑁
𝑘𝑁
23.58 3. and 0.79 3. respectively. Assume weight of water 9.79 3.
𝑚
𝑚
𝑚
a. What is the minimum volume of foam required
b. What is the weight of the foam
c. What is the total weight of concrete and foam
𝑚
HYDRAULICS
Engr. Michelle D. Enriquez
6. A loaded scow has a draft of 1.8m in fresh water. The scow is 6.0m wide and 12.0m long and 2.4m
high. The center of gravity of the scow is 1.80m above the bottom along the vertical axis of symmetry
a. Determine the initial metacentric height
b. What is the maximum single weight that can be moved
transversely from the center of the unloaded scow over the
side without sinking the scow
c. If the maximum weight obtained in (b) is doubled, at what
distance from the center will the scow be on the verge of
submergence.
7. A cylindrical caisson having an outside diameter of 9.0m floats in sea water with its axis vertical and its
lower end submerged 9.0m below the water surface. If the center of gravity is on the vertical axis and is
3.6m above the bottom.
a. Determine the value of 𝑀𝐵0
b. Find the true height
c. Find the righting couple when the caisson is tipped
through an angle of 8&deg;
8. A scow having a length of 15m, 8.0 wide and 6.0m deep has a displacement of 164 metric tons. If the
scow tilts at an angle of 14&deg; without overturning
a. find the metacentric height
b. Determine the location of the center of gravity which lies on the
vertical axis measured from the bottom of symmetry
c. Determine the horizontal displacement of the center of
buoyancy
```