Find Meta Center for Object Stability

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Homework Statement


is it necessary to find the meta center for this question ? the formula of meta center is I /Vd , the I ( moemtn of inertia ) for rectangle is (ab^3) / 12 , but what is a ? what is b ? I'm confused now

Homework Equations

The Attempt at a Solution


ρ[/B] gVsubmerged = weight(579N)
1025(9.81)Vsubmerged = weight(579N)
Vsubmerged = 0.058m^3

0.058m^3 = (1.2x0.9x hsubmerged)
hsubmerged =0.054m

center of buoyancy = 0.054/2 = 0.027m
metacenter = I / Vd = [(ab^3) / 12 ] /Vd
 

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haruspex said:
For a rectangle measuring a by b, the second moment of area about an axis through the mid points of the sides of length b is ##\frac 1{12}ab^3##.
what is the value of a and b ? is the second moment about x -axis(Ixx) , right ? IMO , a is 0.9m , b is 1.2m , am i correct ?
 
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foo9008 said:
what is the value of a and b ? is the second moment about x -axis(Ixx) , right ? IMO , a is 0.9m , b is 1.2m , am i correct ?
In principle, you should consider each axis as a potential axis for tipping. However, it is obvious that one of ab3, a3b is the larger. Which gives the greater danger of tipping, the larger or the smaller?
 
haruspex said:
In principle, you should consider each axis as a potential axis for tipping. However, it is obvious that one of ab3, a3b is the larger. Which gives the greater danger of tipping, the larger or the smaller?
how to know which axis is more danger of tipping ?

the hydrostatic pressure act on the top and the bottom surface of the object , so i am consdiering the moment about x of tipping
 
haruspex said:
Is tipping more likely with a high metacentre or a low one?
low metccenter , right ?
 
foo9008 said:
in the link , it shows that the high metacenter of object enable the object to have higher stability against overturning, so the low metecanter of the object less stable , right ?
Yes.
To get a low metacentre, does "I" need to be large or small? If a>b, which of ab3 and a3b will give that less stable i?
 
haruspex said:
Yes.
To get a low metacentre, does "I" need to be large or small? If a>b, which of ab3 and a3b will give that less stable i?
to get low metacenter , the I has to be small. since the water pressure act on the bottom of the box , the b is 1.2m , a is 0.9m , am i right ?
 
haruspex said:
It depends how you are defining a and b. Is that using the a3b form or the ab3 form?
I am using
ab3
 
haruspex said:
Does a=.9, b=1.2 give a lower I than the other way around?
a=1.2 , b = 0.3 , i would get lower metacenter , but the formula is yp = yc + Ixx / (ycA) , Ixx is the second moment of inertia about x-axis . Ir can't be changed , right ?
 
foo9008 said:
a=1.2 , b = 0.3 , i would get lower metacenter , but the formula is yp = yc + Ixx / (ycA) , Ixx is the second moment of inertia about x-axis . Ir can't be changed , right ?
The object in question could tip about either of two axes. If you fix one direction as x and apply that formula you will only be considering one axis of tipping.
 
haruspex said:
The object in question could tip about either of two axes. If you fix one direction as x and apply that formula you will only be considering one axis of tipping.
why not 3 axes? since the object is 3d ( as shown in the diagram )
 
haruspex said:
Does a=.9, b=1.2 give a lower I than the other way around?
for a = 1.2 , b = 0.3 , i will get metacenter = 1.25m , but , which value to choose ? 1.25m or 2.22m( a = 0.3, b = 1.2 ) ?
if metacenter = 1.25m or 2.22m , G located at 0.86m +0.3m =1.16 m , object is stable
 
foo9008 said:
for a = 1.2 , b = 0.3 , i will get metacenter = 1.25m , but , which value to choose ? 1.25m or 2.22m( a = 0.3, b = 1.2 ) ?
if metacenter = 1.25m or 2.22m , G located at 0.86m +0.3m =1.16 m , object is stable
Yes, it's 3D, and you are right that in principle it could tip over any axis. But it is fairly obvious that the most likely tip is about the natural axes. Feel free to analyse the general case.:wink:
To check for stability, consider the worst case, i.e. the least metacentric height.
The metacentric height you calculated, 1.25m, is the height from the centre of buoyancy, not from the base.
But that only makes it even more stable than you concluded.
 
haruspex said:
Yes, it's 3D, and you are right that in principle it could tip over any axis. But it is fairly obvious that the most likely tip is about the natural axes. Feel free to analyse the general case.:wink:
To check for stability, consider the worst case, i.e. the least metacentric height.
The metacentric height you calculated, 1.25m, is the height from the centre of buoyancy, not from the base.
But that only makes it even more stable than you concluded.
 
Last edited:
haruspex said:
Yes, it's 3D, and you are right that in principle it could tip over any axis. But it is fairly obvious that the most likely tip is about the natural axes. Feel free to analyse the general case.:wink:
To check for stability, consider the worst case, i.e. the least metacentric height.
The metacentric height you calculated, 1.25m, is the height from the centre of buoyancy, not from the base.
But that only makes it even more stable than you concluded.
i thought when we consider the center of pressure , we only need to consider the Ixx ( second moment of inertia about x -axis) , y -axis is the vertical axis ... this is the first question i ever encountered that we need to find the second moment of inertia about 2 axes
 
haruspex said:
Yes, it's 3D, and you are right that in principle it could tip over any axis. But it is fairly obvious that the most likely tip is about the natural axes. Feel free to analyse the general case.:wink:
To check for stability, consider the worst case, i.e. the least metacentric height.
The metacentric height you calculated, 1.25m, is the height from the centre of buoyancy, not from the base.
But that only makes it even more stable than you concluded.
Why do we want to consider 2 axes for rotation of object?
Normally we consider only 1 axis of rotation of object, right?
 
foo9008 said:
Why do we want to consider 2 axes for rotation of object?
Normally we consider only 1 axis of rotation of object, right?
You're getting confused here.

We're talking about the stability of this platform. The center of pressure, etc. has no bearing on this analysis.

foo9008 said:
a=1.2 , b = 0.3 , i would get lower metacenter , but the formula is yp = yc + Ixx / (ycA) , Ixx is the second moment of inertia about x-axis . Ir can't be changed , right ?

The location of the metacenter M depends on the inertia of the waterplane cut by the hull of the platform. Using b = 0.3 would make no sense, since this dimension represents the depth of the platform's hull.

The Ixx used in the equation for yp represents something else and is not applicable here.

For the stability of this platform, BM = I / V, but I can be calculated about an axis which either runs parallel to the length of the platform or parallel to the breadth of the platform, which ever choice leads to the least value of I. V = volume of displacement and remains the same in any event.
 
SteamKing said:
You're getting confused here.

We're talking about the stability of this platform. The center of pressure, etc. has no bearing on this analysis.
The location of the metacenter M depends on the inertia of the waterplane cut by the hull of the platform. Using b = 0.3 would make no sense, since this dimension represents the depth of the platform's hull.

The Ixx used in the equation for yp represents something else and is not applicable here.

For the stability of this platform, BM = I / V, but I can be calculated about an axis which either runs parallel to the length of the platform or parallel to the breadth of the platform, which ever choice leads to the least value of I. V = volume of displacement and remains the same in any event.
thanks for pointing out that i mixed up the Ixx in the equation of center of pressure with the second moment of inertia of MB
 
SteamKing said:
You're getting confused here.

We're talking about the stability of this platform. The center of pressure, etc. has no bearing on this analysis.
The location of the metacenter M depends on the inertia of the waterplane cut by the hull of the platform. Using b = 0.3 would make no sense, since this dimension represents the depth of the platform's hull.

The Ixx used in the equation for yp represents something else and is not applicable here.

For the stability of this platform, BM = I / V, but I can be calculated about an axis which either runs parallel to the length of the platform or parallel to the breadth of the platform, which ever choice leads to the least value of I. V = volume of displacement and remains the same in any event.
the platform has 6 faces ( 3 different types) of surface , how do we determine which surfaces should to consider into the calculation ?
 
foo9008 said:
the platform has 6 faces ( 3 different types) of surface , how do we determine which surfaces should to consider into the calculation ?
You want to consider the surfaces which cut through the water when the platform is floating. The waterplane thus created will be parallel and coincident with the surface of the water. It is the moment of inertia of this waterplane which is the key parameter in determining the value of the distance BM.
 
SteamKing said:
You want to consider the surfaces which cut through the water when the platform is floating. The waterplane thus created will be parallel and coincident with the surface of the water. It is the moment of inertia of this waterplane which is the key parameter in determining the value of the distance BM.
what do you mean by cut thru the water ? when the object is floating , the bottom part will surely sink into the water , whereas the other 4 surface planes is partly submerged with water . so, IMO , there are 5 planes will cut thru water

edit : i think i understand with the previous diagram .
 

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