§2.7 Buoyancy force and stability of snorkeling

2.7.1 Buoyancy force and three states of submerged body

1．Buoyancy force and buoyancy center of a submerged body

The obj ect floating on the surface of the water or submerged below the water surface also bears hydrostatic pressure, which is the sum of hydrostatic pressure at each point on the surface of the obj ect.

As illustrated in Fig.2-30, vertical tangents (AA＇, B B＇, …)of the submerged body were drawn.These tangents are the generatrix of vertical column tangent to the surface of the submerged body.The intersection of the column surface and the surface of the submerged body decompose the surface of the submerged body into two parts:AD B and ACB.The vertical component Fx1 of the total hydrostatic pressure acting on the surface of the submerged body above the intersection is equal to the weight of the pressure prism above the curved surface AD B in magnitude.The direction is downward.The vertical component Fx2 of the total hydrostatic pressure acting on the surface of the submerged body below the intersection is equal to the weight of the pressure prism above the curved surface ACB, and the direction is upward.

The vertical component Fz acting on the whole surface of the submerged body is:

where VACBDA is the volume of the liquid displaced by the submerged body.

Equation(2-30)indicates the vertical force acting on the submerged body is equal to the weight of the displaced fluid.

In the same way, the submerged body can be divided into two parts:CAD (left) and CBD(right).The horizontal component Fx acting on the body is the sum of F1 and F2.F1 and F2 are equal to the hydrostatic pressure acting the vertical proj ection of the curved surface CAD and CBD in magnitude.The proj ection areas of these two parts on the vertical plane are equal.The positions of the proj ection are the same.Thus, F1and F2 are equal in magnitude, but opposite in direction.These two forces counteract to each other.As a result, the horizontal component acting on the submerged body is zero.

In summary, the total hydrostatic pressure acting on the obj ect in liquid only has vertical component, which is equal to the weight of the liquid displaced by the obj ect in magnitude.This is Archimedes principle.

Because Fz tends to push the obj ect to the surface of liquid, Fz is called buoyancy. The point through which the buoyant force acts is called the center of buoyancy.The center of buoyancy overlaps with the centroid of the displaced volume.

2．Three states of submerged body

When the obj ect immersed in liquid is not supported by other obj ects, it is only subj ected to gravity G and buoyant force Fz.According to the relative magnitude of gravity and buoyancy, the immersed obj ect has three states:

(1)Sinking body

When G>Fz, the obj ect sinks to the bottom.

(2)Submerged body

When G=Fz, the obj ect is in suspended state.

(3)Floating body

When G<Fz, the obj ect surfaces and keeps balance till the weight of the liquid displaced by the part below the liquid surface is equal to the weight of the obj ect.The obj ect is in a floating state.

2.7.2 The balance and stability of submerged body

There is a submerged obj ect with the weight of G.The barycenter and center of buoyancy of this obj ect locate at point C and D, respectively.According to the relative position of the center of gravity and buoyancy in the same vertical line, the equilibrium stability, i.e.the ability to restore to its original equilibrium state when the submerged body is tilted due to external disturbances, can be categorized as follows.

(1)Indifferent equilibrium

The center D of buoyancy overlaps with the center C of gravity (see Fig.2-31 a). The orientation of the obj ect in a liquid is arbitrary, i.e.the submerged body is balanced at any position.

(2)Stable equilibrium

For the completely submerged body, which has a center D of buoyancy is above the center C of gravity(see Fig.2-31 b), a rotation from its equilibrium position will create a restoring couple formed by the gravity G and buoyant force, Fz, which causes the body rotate back to its original position.If the external disturbances are displaced, it automatically returns to its equilibrium position.It is to be noted that as long as the center of gravity falls below the center of buoyancy, this will always be true; that is, the body is in a stable equilibrium position.

(3)Unstable equilibrium

If the center D of buoyancy is below the center C of gravity(see Fig.2-31 c), the resulting couple formed by the gravity G and buoyancy Fz will cause the submerged body to overturn.Even if the external disturbances are displaced, it moves to a new equilibrium position.Thus, a completely submerged body with its center of gravity above its center of buoyancy is in an unstable equilibrium position.

Therefore, the center D of buoyancy must be above the center C of gravity in order to keep the balance of the submerged body.

2.7.3 The balance and stability of floating bodies

The conditions for the equilibrium of the floating bodies and that of submerged bodies are the same.The equilibrium conditions are:(1)Gravity G and buoyancy Fz are equal in magnitude, but opposite in direction; (2)Gravity G and buoyancy Fz act on the same vertical line.However, the conditions for stability of them are different.For floating obj ects, it can be stable even though the center C of gravity lies above the center D of buoyancy.This is true since as the body rotates the buoyant force, Fz, shifts to pass through the centroid of the newly formed displaced volume and combines with the gravity G to form a couple which will cause the body to return to its original equilibrium position.However, for the relatively tall, slender body, a small rotational displacement can cause the buoyant force and the gravity to form an overturning couple.

1．Metacenter radius and eccentricity

As shown in Fig.2-32a, there is a symmetrical floating body.After it tilts, its center of gravity keeps unchanged.The center of buoyancy moves from D to D＇due to the shape change of the part immersed in water as described in Fig.2-32b.There are a few new concepts.The line H-H passes through the center D of buoyancy and the center C of gravity is called floating axle.The crossover point M of the action line of buoyancy F＇zpassing through D＇and the floating axle is called metacenter.The distance from metacenter M to the original center D of buoyancy is metacenter radium, denoted asρ.The distance between the center C of gravity and the center D of buoyance is called eccentricity, denoted as e.

2．Balance of floating body

Whether the floating bodies can restore to their original equilibrium position after they tilt depends on the relative position of the center C of gravity and metacenter M.

It can be seen from Fig.2-32:if the metacenter M of the floating body is above the center C of gravity, i.e.ρ>e, the joint effect of gravity G and buoyancy F＇zacting on the floating body after inclination generates a rotational moment which helps the floating body restore to the original equilibrium position.The floating body is in stable equilibrium.On the contrary, if the metacenter M of the floating body is below the center C of gravity, i.e.ρ<e, the joint effect of gravity G and buoyancy F＇zgenerates a rotational moment which makes the floating body tend to be more inclined.If point M and point C overlap, i.e.ρ=e, the joint effect of gravity M and buoyance Fz does not generate moment of force.The floating body is in indifferent equilibrium.

3．Stability of floating body

Therefore, the conditions for floating body to keep balance are that the center C of gravity must be below the metacenter M, i.e.the metacenter radiusρis larger than the eccentricity e.

For a symmetrical floating body with a fixed center of gravity, the eccentricity e can be determined if the shape and weight of the floating body are definite.Thus, whether the floating body is stable depends on the magnitude of metacenter radius.

For the floating body with a small inclination angle (α< 10°), we can write

where I0is the moment of inertia of the floatation plane with respect to the central vertical axle, and V is the volume of the liquid displaced by the floating body.

The intersection plane between the floating body and water surface is called floatation plane.

Equation(2-31)indicates the magnitude of metacenter radium ρ is related to the moment of inertia of the floatation plane on the central vertical axle and the volume of the liquid displaced by the floating body.If ρ is larger than the eccentricity e, the floating body is stable, otherwise it is unstable.The larger the metacenter radius, the more stable the floating body.

Questions

1．What kind of forces are acting on the real fluid and ideal fluid at rest or in motion, respectively?

2．How much is the unit mass force of the fluid at rest, with which mass force only includes gravity? (The coordinate axis z is vertically upward)

3．Are the differential balance equations applicable to the fluid at rest? Why?

4．What is equipressure surface? What are the conditions for an equipressure surface to have?

5．Is the equipressure surface of the fluid at relative balance horizontal plane? Why? What are the conditions for an equipressure surface to become a horizontal plane?

6．Is the pressure measured by pressure gauge and manometer absolute pressure or relative pressure?

7．Why is mercury usually used as working fluid of piezometer tube in traditional experiments?

8．How much is the pressure on the wall of the container when an open container with liquid is in free fall?

9．What is the limit of diving depth for a diver if the maximum pressure which a person can bear is 1.274 MPa (relative pressure)?

10.What is minimum length of a piezometer tube when the piezometric head is 1 m and pressure head is 1.5 m?

11.Is the piezometric head at any point equal to each other in the fluid at rest or in flowing fluid?

12.There are two kinds of liquid (ρ1<ρ2)in a same container, as shown in Fig. 2-33.Two piezometer tubes were set on the side wall of the container.Is the water level in piezometer tubes marked in the figure below correct?

13.There are two points A and B in a tank with water at rest, as shown in Fig. 2-34.Their water depth difference is h.In the U-shaped manometer which connects pint A and B, the height difference of the liquid level is hm.Among the three values of hm, which one is correct?

14.There are three planar objects submerged in the water, with different shape and one side blocking water as shown in the Fig.2-35.The area of these three obj ects is same.The water depth at their centroids is same.Is the total hydrostatic pressure acting on these three obj ects equal? Which center of pressure is the deepest among these three obj ects?

15.The rectangular gate with one side blocking water is submerged in the water (See the figure of the question above).If it rotatesαaround the horizontal axis passing through its centroid C, do the magnitude, direction and acting point of the total hydrostatic pressure change? Why?

16.If graphic method and analytic method are used to calculate the total hydrostatic pressure acting a surface, is there any limit to the shape of the compressed surface? Why?

17.What are the conditions of equilibrium and stability for floating body and submerged body? How many types of equilibrium do they have?

Exercises

1．On a square with side length of 0.02 m in a horizontal plane xOy, uniform force acting on the square is F=4i+3j+9k and the unit is N.Calculate:(1)the vertical component and tangential component; (2)pressure and shearing stress.

2．Calculate the absolute pressure and relative pressure at the location 2 m below the free surface of freshwater (Absolute pressure on the free surface is one engineering atmosphere).

3．The tube is filled with oil (ρg=8.5 kN/m3)at rest (see Fig.2-36).Calculate the pressure with mH2O as unit at point A and B.

4．Determine the height of piezometer tube and piezometric head at point A, B, and C in the Fig.2-37.

5．Multi-tube mercury pressure gauge is used to measure pressure.The unit of the elevation marked in the Fig.2-38.Calculate the pressure p0on the water surface.

6．There is a narrow crack inside mountain massif.The vertical depth is 1200 m and the crack is filled with seeping water at rest.Calculate the horizontal hydrostatic thrust acting on the mountain massif per square meter in the deepest of the crack.

7．A cylindrical container with variable cross section is filled with water (see Fig.2-39).Calculate the vertical component acting on the tapered segment ABCD.

8．In order to measure the acceleration of an obj ect, a small-diameter U-shaped tube with liquid inside moves along with the object (see Fig.2-40).If the height difference of liquid surface is h and the horizontal length of U-shaped tube is l, calculate expression equation of the acceleration a.

9．An open container filled with water accelerates.Calculate the hydrostatic pressure distribution inside the container in the following two cases:(1)Free fall;(2)Upward movement with constant acceleration a.

10.There is a rectangular metal plate gate which is supported by two I-steel beams. The height h and width b of the gate is 3 and 1 m, respectively.The water retaining surface is on an even height with the top of the gate as shown in Fig.2-41.If the force acting on the two beams is equal, how much are L1and L2?

11.Calculate the unit width pulling force required to open the gate of sluice (see Fig.2-42).The weight of the gate and shaft friction can be ignored.

12.The radius of gate is L.How high does upstream water depth h rise so that the overflow will be automatically opened (see Fig.2-43)? The resistance is ignored.

13.There is a rectangular automatic gate as shown in Fig.2-44.If the given upstream and downstream water depth gate can automatically open, calculate the position where the hinge M locates.

14.Try to plot the pressure prism of each curved surface ABC in Fig.2-45.

15.Calculate the moment M applied on the hinge of the cylindrical gate (see Fig. 2-46).The length of the gate is 2 m.The weight is 5000 N.The friction of the hinge is neglected.

16.Determine (1)the magnitude and line of action of the horizontal component acting on the radial gate; (2)the magnitude and line of action of the vertical component acting on the radial gate; (3)the magnitude and the direction of the resultant force acting on the radial gate (Fig.2-47).