Finding the density order of three floating bodies with different submersion depths

[brotherbobby]

Homework Statement

Rank in order, from largest to smallest, the $\mathbf{densities}$ of blocks a, b and c shown in figure below. Explain your reasoning.

Relevant Equations

Law of floatation : "The weight (or mass) of a floating body is equal to the weight (or mass) of the liquid displaced : $w_B = \Delta w_L$

The dimensions of the bodies given in the problem are visual. Clearly bodies a and b are cubical whereas c is not. One side of b is twice the side of a. Both a and b are submerged to the same depth but what is the depth of submersion of c? Arranging bodies (by copying and pasting a on b and c, as can be done using imaging programs), I paste the following diagram that show the depths and sides of the bodies shown. For simplicity, let us ignore the third dimension of the bodies or assume them to be equal, equivalently.

Please note the case for c . Also we have $y < x$.

Using $w_{\text{body}} = \Delta w_L$ and dropping the (constant) density of the liquid, we have for the weights of the three bodies :

$w_a = xy,\; w_b = 2xy,\; w_c = 2xy$.

As for the volumes of the three bodies,

$v_a = x^2,\; v_b = 4x^2,\; v_c = x(2x+y)$.

Dividing the first by the second and supressing the acceleration due to gravity $g$ from each, we get the densities of the three bodies,

$\rho_a = \frac{y}{x}, \; \rho_b = \frac{y}{2x}, \rho_c = \frac{2y}{2x+y}$.

Clearly, $\boxed{\rho_b < \rho_a}$.

Using simple algebra and remembering that $y<x$ yields the solution :

$\large{\boxed{\mathbf{\rho_a > \rho_c > \rho_b}}}$Of course I'd like to know if I am right. Even if I am, is there an easier and a more conceptual way to do this?

 

[Orodruin]

For simplicity, let us ignore the third dimension of the bodies or assume them to be equal, equivalently.

You can also ignore the width of the blocks as it is irrelevant. Only the percentage of each block that is submerged is going to matter for the computations, which also means that you do not need to assume equal extension in the third dimension either.

 

[brotherbobby]

"You can also ignore the width of the blocks as it is irrelevant. Only the percentage of each block that is submerged is going to matter for the computations".

Thank you, a point I hadn't realized. Though you might know the following well, I must carry it out if only to convince myself that the width of the block is irrelevant, only its depth of submersion relative to its height.

Let $l$ be the length and $b$ the width of the block.

Weight of the block : $w_B = \rho_B lb g$ and the weight of liquid displaced $\Delta w_L = \rho_L l' b g$, where $l'$ is the depth of immersion into liquid.

The two are equal due to the law of floatation.

Hence $\rho_B l = \rho_L l' \Rightarrow \frac{l'}{l} \propto \rho_B$ as the density of the liquid is a given.

Hence we only need to see the fraction (or percentage) $\frac{l'}{l}$ in my problem above for the three bodies.

Clearly we see that : $\rho_a > \rho_c > \rho_c$

 

[Merlin3189]

Just my opinion, but I don't think anyone (setting the question) would have intended you to do any maths here. It looks more to test understanding/ intuition.

 

[haruspex]

floatation

It's "flotation".