Please help me to determine how to solve following problems in case of they will come on test
A cubical block of metal weighs $6$ pounds. How much will another cube of the same metal weigh if its sides are twice as long?
My interest is if it is related to surface area or volume? For example let us take cube with side=$2$,then it's surface are is
$6*2*2=24$
and volume is equal to $2^3=8$;
now let us double it's side length so we have side=$4$; then it's surface area is
$6*4*4=96$ and volume is $4*4*4=64$; if we divide new surface area by old one we get $96/24=4$ and if we divide new volume by old one we get $64/8=8$; as we see surface area is increased by $4$ and volume by $8$, so I have to multiply $6$ by $4$ or $6$ by $8$?or weight of adjusted cube is $24$ or $48$? Thanks in advance
Best Answer
Although you derivation takes on account both the Surface area and the Volume, I'd like to put this solution in prospective of the second term.
Since I do not know the value of the sides $\ell_{\mathrm{box}}$ an the volume $\ell_{\mathrm{box}}^3$, but I know his weight, then the only factor that correlates them is the Density, which I'll call $\varrho_x$, expressed by: $$ \varrho_x=\frac{m_{\mathrm{box}}}{\ell_{\mathrm{box}}^3} $$ Now, if the metal is the same, and so $\varrho_x$ doesn't change, then I could write a simple equation: $$ \varrho_x=\frac{m_{\mathrm{box}(i)}}{\ell_{\mathrm{box}(i)}^3}=\frac{m_{\mathrm{box}(f)}}{(2\ell_{\mathrm{box}(i)})^3} $$ The only unknown is $m_{\mathrm{box}(f)}$, so: $$ m_{\mathrm{box}(f)}=\frac{m_{\mathrm{box}(i)}\cdot 8\ell_{\mathrm{box}}^3}{\ell_{\mathrm{box}(i)}^3}=m_{\mathrm{box}(i)}\cdot 8 $$ As you can see, the difference between the surface area is related to the difference of the volume, so you have to multiply by 8, as this simple $2^3$ times the side of the original lenght.
Let's take a generalized example.
If you imagine one simple cube, with lenght $a$, then the volume and the surface area are: $$ S=6\cdot a^2,\quad V=a^3 $$ It's true that they have different factors, but they do not mean the same thing: if you imagine the increasing of weight, it's only given by the infinite variation of the Volume, as a simple integral: $$ V_f=V_i+\left(\int_{a}^{2a}\mathrm{d}a\right)^3 $$ Above it can be observed that one single infintesimal change into the third dimension it's different from a single change of the surface; because the expansion is also inside the cube, which is not counted in the "$4$" factor, as two dimensional.
As you measure only the external surface, the volume takes the "internal expansion" into account, with either the relation of the density and the factor of $2$ "cubed" into the third dimension.