Buoyancy
To calculate the vertical component of the resultant hydrostatic force, the body is considered to be divided into a number of elementary vertical prisms. The vertical forces acting on the two ends of such a prism of cross-section dAz (Fig. 5.4) are respectively
Therefore, the buoyant force (the net vertically upward force) acting on the elemental prism of volume is -
Hence the buoyant force FB on the entire submerged body is obtained as.20
(5.20)
Where is the total volume of the submerged body. The line of action of the force FB can be found by taking moment of the force with respect to z-axis. Thus
(5.21)
Substituting for dFB and FB from Eqs (5.19c) and (5.20) respectively into Eq. (5.21), the x coordinate of the center of the buoyancy is obtained as
(5.22)
which is the centroid of the displaced volume. It is found from Eq. (5.20) that the buoyant force FB equals to the weight of liquid displaced by the submerged body of volume . This phenomenon was discovered by Archimedes and is known as the Archimedes principle.
ARCHIMEDES PRINCIPLE
The buoyant force on a submerged body
The buoyant force on a partially immersed body
Stability of Unconstrained Submerged Bodies in Fluid
Let us suppose that a body is given a small angular displacement and then released. Then it will be said to be in
Stable Equilibrium
Consider a submerged body in equilibrium whose centre of gravity is located below the centre of buoyancy (Fig. 5.5a). If the body is tilted slightly in any direction, the buoyant force and the weight always produce a restoring couple trying to return the body to its original position (Fig. 5.5b, 5.5c).
Fig 5.5 A Submerged body in Stable Equilibrium
Unstable Equilibrium
On the other hand, if point G is above point B (Fig. 5.6a), any disturbance from the equilibrium position will create a destroying couple which will turn the body away from its original position (5.6b, 5.6c).
Fig 5.6 A Submerged body in Unstable Equilibrium
Neutral Equilibrium
When the centre of gravity G and centre of buoyancy B coincides, the body will always assume the same position in which it is placed (Fig 5.7) and hence it is in neutral equilibrium.
Fig 5.7 A Submerged body in Neutral Equilibrium
Therefore, it can be concluded that a submerged body will be in stable, unstable or neutral equilibrium if its centre of gravity is below, above or coincident with the center of buoyancy respectively (Fig. 5.8).
Fig 5.8 States of Equilibrium of a Submerged Body
(a) STABLE EQUILIBRIUM (B) UNSTABLE EQUILIBRIUM (C) NEUTRAL EQUILIBRIUM
Stability of Floating Bodies in Fluid
Fig 5.9 A Floating body in Stable equilibrium
Important points to note here are
Let the new line of action of the buoyant force (which is always vertical) through B' intersects the axis BG (the old vertical line containing the centre of gravity G and the old centre of buoyancy B) at M. For small values of qthe point M is practically constant in position and is known as metacentre. For the body shown in Fig. 5.9, M is above G, and the couple acting on the body in its displaced position is a restoring couple which tends to turn the body to its original position. If M were below G, the couple would be an overturning couple and the original equilibrium would have been unstable. When M coincides with G, the body will assume its new position without any further movement and thus will be in neutral equilibrium. Therefore, for a floating body, the stability is determined not simply by the relative position of B and G, rather by the relative position of M and G. The distance of metacentre above G along the line BG is known as metacentric height GM which can be written as
GM = BM -BG
Hence the condition of stable equilibrium for a floating body can be expressed in terms of metacentric height as follows:
GM > 0 (M is above G) Stable equilibrium
GM = 0 (M coinciding with G) Neutral equilibrium
GM < 0 (M is below G) Unstable equilibrium
The angular displacement of a boat or ship about its longitudinal axis is known as 'rolling' while that about its transverse axis is known as "pitching".
Floating Bodies Containing Liquid
If a floating body carrying liquid with a free surface undergoes an angular displacement, the liquid will also move to keep its free surface horizontal. Thus not only does the centre of buoyancy B move, but also the centre of gravity G of the floating body and its contents move in the same direction as the movement of B. Hence the stability of the body is reduced. For this reason, liquid which has to be carried in a ship is put into a number of separate compartments so as to minimize its movement within the ship.
Period of Oscillation
The restoring couple caused by the buoyant force and gravity force acting on a floating body displaced from its equilibrium placed from its equilibrium position is
W .GM sin θ (Fig. 5.9 ). Since the torque equals to mass moment of inertia (i.e., second moment of mass) multiplied by angular acceleration, it can be written
(5.23)
Where IM represents the mass moment of inertia of the body about its axis of rotation. The minus sign in the RHS of Eq. (5.23) arises since the torque is a retarding one and decreases the angular acceleration. If θ is small, sin θ=θ and hence Eq. (5.23) can be written as
(5.24)
Equation (5.24) represents a simple harmonic motion. The time period (i.e., the time of a complete oscillation from one side to the other and back again) equals to . The oscillation of the body results in a flow of the liquid around it and this flow has been disregarded here. In practice, of course, viscosity in the liquid introduces a damping action which quickly suppresses the oscillation unless further disturbances such as waves cause new angular displacements
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1. What is hydrostatic thrust on a submerged plane surface? |
2. How is the hydrostatic thrust calculated for a submerged plane surface? |
3. What factors affect the hydrostatic thrust on a submerged plane surface? |
4. How does the angle of inclination of a submerged surface affect the hydrostatic thrust? |
5. Can hydrostatic thrust be used to calculate the buoyant force on a submerged object? |
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