Question

You need to determine the density of a ceramic statue. If you suspend it from a spring scale, the scale reads $28.4\mathrm{N}$. If you then lower the statue into a tub of water, so that it is completely submerged, the scale reads $17.0\mathrm{N}$. What is the statue's density$?$

Short answer

Density of the statue is$2500\mathrm{kg}/{\mathrm{m}}^3$

Step-by-step solution

Expression for buoyant force 

When the statue is completely submerged it is in static equilibrium. Use the condition of static equilibrium and the expression for buoyant force to calculate the density of the statue.

The expression for buoyant force is,

$F_B={\rho }_{\mathrm{f}}{V}_{\mathrm{f}}g$

Here, ${\rho }_f$is density of the fluid, $V_f$displaced volume of the fluid (water), and $g$is acceleration due to gravity.

Free body diagram

The following figure represents the free body diagram of the statue when submerged in water.

In the figure, $T$represents the tension in the cable of the spring balance and $mg$is weight of the statue.

Calculation of FB

When the statue is submerged in water, the scale reading gives the tension $\left(T\right)$. When the statue is in air, the scale reading gives the original weight of the statue$\left(mg\right)$. So, mass of the statue is,

$mg=28.4\mathrm{N}$

$m=\frac{28.4\mathrm{N}}{g}$

Substitute $9.8m/s^2$for $g$.

$m=\frac{28.4\mathrm{N}}{g}$

$=2.9\mathrm{kg}$

The statue is in static equilibrium. So,

$\Sigma F_y=F_{\mathrm{B}}+T-mg$

$0=F_{\mathrm{B}}+T-mg$

$F_{\mathit{B}}=mg-T$

Substitute $17\mathrm{N}$for $T$and $28.4\mathrm{N}$for$\mathrm{mg}$.

$F_B=28.4\mathrm{N}-17\mathrm{N}$

$=11.4\mathrm{N}$

Calculation of density

Rearrange the expression $F_{\mathrm{B}}={\rho }_f{V}_{f}g$for $V_f$.

localid="1648145649485" $V_f=\frac{F_B}{{\rho }_{f}g}$

Substitute $1000kg/m^3$for ${\rho }_f$,$9.8m/s^2$for$g$,and $11.4\mathrm{N}$for $F_{\mathrm{B}}$.

$V_{\mathrm{f}}=\frac{(11.4\mathrm{N})}{\left(1000\mathrm{kg}/{\mathrm{m}}^3\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}$

$=0.00116{\mathrm{m}}^3$

This volume of the water that is displaced is equal to the volume of the statue. Thus, the density of the statue is,

$\rho =\frac{m}{V}$

Here, $\rho$is density of statue and $V$is volume of the statue.

Substitute $0.00116{\mathrm{m}}^3$for $V$and $2.9\mathrm{kg}$for $m$.

$p=\frac{2.9\mathrm{kg}}{0.00116{\mathrm{m}}^3}$

$\rho$$\rho$$=2500\mathrm{kg}/{\mathrm{m}}^3$