## Saturday, January 24, 2009

### Will the block topple?

This post will be vulnerable to the assumptions i have made. If you find one too far stretched or simply wrong, please comment. And of course you are free to comment otherwise also :-)

If you place a cuboidal block in water it will float if it is lighter than water, but in what position. If you lay a block flat it will float like that, but if it is put standing, it may float, or may fall to a flatter position. This much is observation, or just intution, but I wanted to get there mathematically. I wanted to find an expression which would tell me if the block will fall, or not.

So I have a block lighter than the liquid it is in. I try to find an expression which will tell me if the position it is put in is stable or not. So fr
om the position, i deflect it by a small angle θ and check if a restoring couple is created by the pair of forces (weight and buoyancy) or if the couple helps topple the block.

Approach:
Simple, find the center of buoyancy and the center of gravity after tilting the block by a small angle.
Check the points at which the buoyancy and weight act, to see the direction of the couple formed.

At this point of time, my guesses. High density of liquid will make it difficult for the block to float in a erect position, that is on a smaller face. And of course larger the vertical dimension with
respect to the flat one, difficult it should be to float.

To start the so
lution, here is the diagram. All the required dimensions are marked. The block has been tilted by an angle θ.
The axes X and Y are stuck to the blocks flat position, thus tilted to vertical and horizontal by θ.
Since the density o
f the liquid is ρ the displaced volume of the solid so that it is in equilibrium is ρHB (third dimension is 1, say). Thus the average depth in water is ρH.
Another parameter η has been used which just defines the shape of the block. η=B/H.

Next is to find the buoyancy point. That would be the center of mass of the submerged portion which is trapezoidal. We find that the distance of the center from the origin along both axes are:

Using simple geometry, the horizontal distance of this point from the origin is [ΔXsinθ-ΔYcosθ]. This is to be compared with the location of the center of mass of the block, since the weight acts through this point. In a position with B horizontal, the distance of the center from the bottom base is H/2,
while the base is at a depth of below the water level. Thus the distance from the origin is (ρH-B/2), and hence the horizontal distance is (ρH-B/2)sinθ.

The latter should be greater for the restoring couple to act, so that as seen in the diagram the couple formed is favorable for restoring the position. Thus using (ρH-B/2)sinθ>ΔXsinθ-ΔYcosθ we get

After coming this far, now i do not know how correct this is and more importantly i had fun doing all this but what do i do with it now!
:-(

Ninja a.k.a. Talli said...

wtf
kya tha be ye sab/??!!!!
:-o

Mithun Pandey said...

what is tan theta bhaiya and theta ko likhte kaise hai keyboard se bhaiya

Voice said...

1. i am not good in mechanics
2. you are too good in mechanics

having said that...
i dont think that the rod/block will be floating erect.
if u throw a block (light one) from a tall building it will start rotating as well. this probably happens due to
1. because it is not a concentrated mass as we assume in physical problems
2. other forces, like air drag, low pressure zone created due to velocity of air.
so i assume the block will be a bit tilted from the very beginning.

now if u give a tap (say theta) and it starts oscillating then it may cross the vertical position and then force will start acting in totally different direction.

----------------
oh u had fun that is well understood or else aisa blog kaun likhega :D

x said...

not as much fun as i had reading it, thats for sure. math blows holes in my brain and that was a lot of match vocabulary for a visual observation.