Lecture Note
Stability: Jeans Mass and Spiral Structure
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University:
Massachusetts Institute of Technology -
Course:
8.902 | Astrophysics II Academic year:
2022
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Views:
469
Pages:
13
Author:
prediguemzhv
RECITATION #2
STABILITY OF A FLOATING BODY
2D body floating on the surface of a fluid
The sketch above shows a 2D body (l into paper > band a) floating on the surface of a fluid of density ρ. The draft of the floating body is d and its width is 2b. The body is assumed to be symmetrical around the vertical z-axis.
W: weight of the floating body
G: center of gravity of the floating body
Vg: volume displaced by body = 2bd
B: center of gravity of displaced fluid
We know from hydrostatics and force equilibrium that:
[Equation relating weight, volume, and fluid density]
Fa: buoyancy force is vertically upwards (Fa = ρg2bd), passes through B, and has a magnitude = weight of displaced fluid (2)
We: weight of the floating body acts vertically downward (We = ρg2bd), passes through G, and has a magnitude = Fa (3)
Both B and G are located on the vertical z-axis, i.e. at x=0, because of symmetry, and we have the coordinates (0, 2b) for G, (0, 2d) for B (4)
If 2d > 2b and we turn the body a small angle around G (but retain the forces We and Fa), it would appear that the body would continue to turn, i.e., it would be unstable and roll over due to the moment of Fa (M_Fa = (2d - 2b)Fa) (5)
Is this really true?
Problem #2
Same 2-D body as in problem #1, but rotated 6° around the centerline of the waterline, O. (cosθ = 0.995, sinθ = tanθ = 0.1)
[Diagram of a 2D body floating on the surface of a fluid, rotated 6°]
Determine the net horizontal force on the rotated floating body.
Determine the uplift (buoyancy) force on the rotated floating body.
Is the rotated floating body in equilibrium?
RECITATION #2
Problem #1
a)
Body is in equilibrium. Weight of body = Ws = ρg(a+d)l.
Buoyancy force = weight of displaced fluid = Fb = ρgdl.
ρs = ρa(d/(a+d)) = 0.75ρ = 750 kg/m^3 (Wood!)
b)
Rectangle CG is right smack in the middle.
xG = l/2 = 2m
zG = (a+d)/2 = 1/2 = 0.5m
d)
Same thing, except only submerged portion of floating body is considered.
xG = l/2 = 2m
zG = d/2 = 0.75/2 = 0.375m
Notice: Net horizontal force is zero. On the left part of body, ρgd^2 pushes towards the right, and is balanced by ρgd^2 acting towards the left from the right part of body.
Notice also: Lines of action of vertical forces We & Fb coincide (xG = xB) and so do the lines of action of horizontal forces (zHor,far = 1/3 d).
Problem No: 2
a)
Horizontal force (to the right) = hydrostatic force (to the right) on body's projection onto a vertical plane. This vertical plane extends from the free surface down to the deepest corner, which is a distance of dmax = (d + tanθ)cosθ + d/2 = 0.75 + 0.2 = 0.95m
Same way for force to the left: same result.
Net Horizontal force = 0
b)
Compared to original θ=0 position, an extra triangular column (l/2k)tanθ has been submerged (dashed ---). But an exact replica (dashed ===) is no longer submerged. So the volume displaced by the rotated body is identical to that of the body when θ=0. Same buoyancy force as before, i.e., Fa = We (Vertical Force Equilibrium!)
c)
Horizontal forces balance! Vertical forces balance! Looks as if body is in equilibrium.
But wait: There is more to equilibrium than forces! How about Moment balance?
This is the topic referred to as STABILITY OF FLOATING BODIES
To answer this question, let us look at the forces acting on the floating body when this is rotated a small angle, δθ.
[Diagram of a 2D body floating on the surface of a fluid, rotated a small angle δθ]
For the floating body's position shown in the sketch above [note: the floating angle δθ is assumed small, but shown as "large" for clarity], we note that the location of the waterline W has moved to a new location, W'.
On the original x-axis, simple geometry tells us that (in the original (x, z)-coordinates):
Xw - Xw' = -(zG - Zw)sin(δθ)
Zw - Zw' = -(zG - Zw)(cos(δθ) - 1)
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