Chapter 15, Problem 33PQ

A rectangular block of Styrofoam 25.0 cm in length, 15.0 cm in width, and 12.0 cm in height is placed in a large tub of water. Assume the density of Styrofoam is 3.00 × 10² kg/m³. a. What volume of the block is submerged? b. A copper block is now placed atop the Styrofoam block so that the top of the Styrofoam block is level with the surface of the water. What is the mass of the copper block?

To determine

The submerged volume of the block.

Answer to Problem 33PQ

The submerged volume of the block is $\underset{\_}{1.35\times {10}^{-3}{\text{m}}^{\text{3}}}$.

Explanation of Solution

The submerged volume of the object is equal to the volume of the water displaced.

Write the equation of force at equilibrium.

  $F_B=F_g$                                                                                                                 (I)

Here, $F_B$ is the buoyant force and $F_g$ is the gravity force.

Write the expression for the buoyant force.

  $F_B={\rho }_f{V}_{\text{displaced}}g$                                                                                                  (II)

Here, ${\rho }_f$ is the density of fluid, $V_{\text{displaced}}$ is the volume of displaced water and $g$ is the gravitational acceleration.

Write the expression for the gravity force.

  $F_g={\rho }_s{V}_{b}g$                                                                                                        (III)

Here, ${\rho }_s$ is the density of Styrofoam and $V_b$ is the volume of block.

Write the expression of the volume of the block.

  $V_b=lwh$                                                                                                            (IV)

Here, $l$ is length, $w$ is width and $h$ is height.

Rewrite the expression from equation (I) to express displaced volume by using (II), (III) and (IV).

  $\begin{array}{l}{\rho }_f{V}_{\text{displaced}}={\rho }_s\left(lwh\right)\\ V_{\text{displaced}}=\frac{{\rho }_s\left(lwh\right)}{{\rho }_f}\end{array}$                                                                                          (V)

Conclusion:

Substitute $3.00\times {10}^2{\text{kg/m}}^{\text{3}}$ for ${\rho }_s$, $1000{\text{kg/m}}^{\text{3}}$ for ${\rho }_f$, $25.0\text{cm}$ for $l$, $15.0\text{cm}$ for $w$ and $12.0\text{cm}$ for $l$ in equation (V) to find $V_{\text{displaced}}$.

  $\begin{array}{c}{V}_{\text{displaced}}=\frac{\left(3.00\times {10}^2{\text{kg/m}}^{\text{3}}\right)\left[\begin{array}{l}\left\{\left(25.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}\left\{\left(15.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}\\ \left\{\left(12.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}\end{array}\right]}{\left(1000{\text{kg/m}}^{\text{3}}\right)}\\ =\frac{\left(3.00\times {10}^2{\text{kg/m}}^{\text{3}}\right)\left[\left(25.0\times {10}^{-2}\text{m}\right)\left(15.0\times {10}^{-2}\text{m}\right)\left(12.0\times {10}^{-2}\text{m}\right)\right]}{\left(1000{\text{kg/m}}^{\text{3}}\right)}\\ =1.35\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Thus, the submerged volume of the block is $\underset{\_}{1.35\times {10}^{-3}{\text{m}}^{\text{3}}}$.

To determine

The mass of the copper block.

Answer to Problem 33PQ

The mass of the copper block is $\underset{\_}{3.15\text{kg}}$.

Explanation of Solution

Write the equation of force at equilibrium.

  $F_B=F_g+M_{\text{copper}}g$                                                                                         (VI)

Here, $F_B$ is the buoyant force, $F_g$ is the gravity force, $M_{\text{copper}}$ is the mass of copper block and $g$ is the gravitational acceleration.

Write the expression for the buoyant force.

  $F_B={\rho }_f{V}_{b}g$                                                                                                    (VII)

Here, ${\rho }_f$ is the density of fluid, $V_b$ is the volume of block.

Write the expression for the gravity force.

  $F_g={\rho }_s{V}_{b}g$                                                                                                    (VIII)

Here, ${\rho }_s$ is the density of Styrofoam, $V_b$ is the volume of block.

Write the expression of the volume of the block.

  $V_b=lwh$                                                                                                          (IX)

Here, $l$ is length, $w$ is width and $h$ is height.

Rewrite the expression from equation (VI) to express mass of copper block by using (VII), (VIII) and (IX).

  $M_{copper}=\left({\rho }_f-{\rho }_s\right)\left(lwh\right)$                                                                                  (X)

Conclusion:

Substitute $3.00\times {10}^2{\text{kg/m}}^{\text{3}}$ for ${\rho }_s$, $1000{\text{kg/m}}^{\text{3}}$ for ${\rho }_f$, $25.0\text{cm}$ for $l$, $15.0\text{cm}$ for $w$ and $12.0\text{cm}$ for $l$ in equation (X) to find $M_{\text{copper}}$.

  $\begin{array}{c}{V}_{\text{displaced}}=\left[\left(1000{\text{kg/m}}^{\text{3}}\right)-\left(3.00\times {10}^2{\text{kg/m}}^{\text{3}}\right)\right]\left[\begin{array}{l}\left\{\left(25.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}\left\{\left(15.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}\\ \left\{\left(12.0\text{cm}\right)\left(\frac{1\times {10}^{-2}\text{m}}{1\text{cm}}\right)\right\}\end{array}\right]\\ =\left[\left(1000{\text{kg/m}}^{\text{3}}\right)-\left(3.00\times {10}^2{\text{kg/m}}^{\text{3}}\right)\right]\left[\left(25.0\times {10}^{-2}\text{m}\right)\left(15.0\times {10}^{-2}\text{m}\right)\left(12.0\times {10}^{-2}\text{m}\right)\right]\\ =3.15\text{kg}\end{array}$

Thus, the mass of the copper block is $\underset{\_}{3.15\text{kg}}$.