# Stability of object

#### foo9008

1. 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

2. Homework Equations

3. The Attempt at a Solution
ρ
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

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what is a ? what is b ?
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 $\frac 1{12}ab^3$.

#### foo9008

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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#### haruspex

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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?

#### foo9008

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

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how to know which axis is more danger of tipping ?
Is tipping more likely with a high metacentre or a low one?

#### foo9008

Is tipping more likely with a high metacentre or a low one?
low metccenter , right ?

#### foo9008

low metacenter means the M is below G , right ? so , the GM is negative , so the object is unstable ?

#### foo9008

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 ?

#### haruspex

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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?

#### foo9008

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

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the b is 1.2m , a is 0.9m , am i right ?
It depends how you are defining a and b. Is that using the a3b form or the ab3 form?

#### foo9008

It depends how you are defining a and b. Is that using the a3b form or the ab3 form?
I am using
ab3

#### haruspex

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I am using
ab3
Does a=.9, b=1.2 give a lower I than the other way around?

#### foo9008

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 cant be changed , right ?

#### haruspex

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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 cant 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.

#### foo9008

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 )

#### foo9008

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

#### haruspex

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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.
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.

#### foo9008

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.
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.

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#### foo9008

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.
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

#### foo9008

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.
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 wanna consider 2 axes for rotation of object?
Normally we consider only 1 axis of rotation of object, right?

#### SteamKing

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Why do we wanna 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.

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 cant 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.

"Stability of object"

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