Here is the question: A pontoon of length L is ballasted so that the density in the bottom half is twice that in the top half.

enter image description here

a) Find the maximum value, smax, of the specific density, s, for which the pontoon will float in water.

b) Find the positions of the centre of buoyancy, CB and the centre of gravity, CG, when s=smax.

I have already completed part a and found that $S_{max} = \frac{2}{3}S$ using my diagram: enter image description here. $S = 1$ to make sure the pontoon is floating and $1$ to make sure $S$ will be the max.

Okay now part b), the center of bouayncy isnt too difficult, which is just the centroid of the displace volume, in this case, $\frac{H}{2}$

Now I am not sure how to find the center of gravity. Apparently, I have to take moment about the top of the body. Which results into the following.

enter image description here

Can someone please explain the solutions? Where are the forces? Why are we multiplying the mass by the height? Even if we have the weight instead of mass, wouldn't this be downwards?


2 Answers 2


enter image description here

If we consider the weight($W$) represents the force, it is embedded in the equation for calculating the centroid of the mixture:

$y_c = \dfrac {\sum A_iy_i}{\sum A_i}$

Since $A_1 = bh_1$ and $A_2 = bh_2$, for your case

$y_c = \dfrac {(bh_1)y_1 + (bh_2)y_2}{bh_1 + bh_2}$

Now, let's say the container has a length $L$, we can link $A_i$ to $W_i$ by consider

$W_i = \rho_i(bh_i)L$,

$bh_i = \dfrac{W_i}{\rho_i L}$

You can plug the result back (with the corresponding subscription) into the expanded equation for $y_c$, the result should remain the same with weight terms in it -

$bh_iy_i = \dfrac{W_iy_i}{\rho_i L}$, in which, $W_iy_i$ is the moment.

  • $\begingroup$ Hello, I am a bit confused about taking the moment about the top of the body part. The weight is pointing downwards, and the distance from the top of the body to the centre of weight of each block is also in the vertical direction. Then where is the perpendicular distance needed to find the moment? $\endgroup$
    – CountDOOKU
    Mar 31, 2022 at 5:24
  • $\begingroup$ ".. I am a bit confused about taking the moment about the top of the body". The multiplication is "A*arm" - A can be the area or height of each fluid. It is not a moment, but a step in finding the centroid through the "weighted average/mean" method. $\endgroup$
    – r13
    Mar 31, 2022 at 13:05

Centre of gravity is independent of the floating condition or the densities, it only depends on the mass distribution of the body (and technically the gravitational field, but that's irrelevant here). That is to say, as long as you have a (cuboidal) body of height H with the bottom half with density 2s and top half with density s, the centre of gravity will always be at a point 5H/12 from the bottom.

In your question (by symmetry), we know the CG of the top half is a fourth of the way down from the top and the CG of the bottom half is three fourths (the whole of the top part plus half of the bottom part down. It's only a matter of weighted addition of the two values, very little to do with moments.

If you have any doubts, just ask.


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