Chapter 9, Problem 29P

(a)

To determine

Whether $1.0\text{kg}$ of lead or $1.0\text{kg}$ of aluminum has larger buoyant force acting on it when immersed in water.

Answer to Problem 29P

Aluminum has larger buoyant force when it is immersed in water, since for the same mass aluminum is less dense and has larger volume than lead.

Explanation of Solution

Buoyant force is the force acting on object that is partially or completely immersed in water. Buoyant force is equal to weight of the water displaced by the object. Water displaced by the object depends on the volume of the object. Therefore, buoyant force is large for an object having larger volume than small volume.

Conclusion:

Lead is denser than aluminum. Therefore, for same mass lead has small volume than aluminum. Aluminum displaces more water than lead. Therefore, buoyant force will be larger for aluminum than lead.

Therefore, Aluminum has larger buoyant force when it is immersed in water, since for the same mass aluminum is less dense and has larger volume than lead.

(b)

To determine

Whether $1.0\text{kg}$ of steel that is sinking to the bottom of the lake or $1.0\text{kg}$ of wood with a density of $500\text{kg}/{\text{m}}^3$ that is floating on the lake has greater buoyant force acting on it.

Answer to Problem 29P

Steel is denser than wood. Since their masses are same volume is larger for wood than steel. Since floating wood displaces more water than sinking steel. Buoyant force is larger force wood that is floating than steel that is sinking.

Explanation of Solution

Buoyant force is the force acting on object that is partially or completely immersed in water. Buoyant force is equal to weight of the water displaced by the object. Water displaced by the object depends on the volume of the object. Therefore, buoyant force is large for an object having larger volume than small volume.

Conclusion:

Steel is denser than wood. Since both have same mass, volume of wood is larger compared to steel. Even though the wood is floating, it displaces more water than does steel. Buoyant force is equal to weight of the water displaced. Therefore, wood has larger buoyant force than steel, since water displaced by wood that is floating is larger compared to steel that is sinking.

Therefore, steel is denser than wood. Since their masses are same volume is larger for wood than steel. Since floating wood displaces more water than sinking steel. Buoyant force is larger force wood that is floating than steel that is sinking.

(c)

To determine

Find quantitative answers for part (a) and part (b).

Explanation of Solution

Buoyant force is equal to weight of the water displaced by object.

Write the expression for buoyant force.

$F_{\text{b}}={\rho }_{\text{w}}g{V}_{\text{Pb}}$

Here, $F_{\text{b}}$ is the buoyant force, ${\rho }_{\text{w}}$ is the density of water, $g$ is the acceleration due to gravity and $V_{\text{pb}}$ is the volume of lead.

Substitute $\frac{m_{\text{Pb}}}{{\rho }_{\text{Pb}}}$ for $V_{\text{Pb}}$ in above equation to get $F_{\text{b}}$.

$F_{\text{b, Pb}}={\rho }_{\text{w}}g\left(\frac{m_{\text{Pb}}}{{\rho }_{\text{Pb}}}\right)$ (I)

Here, $F_{\text{b, Pb}}$ is the buoyant force acting on lead ,$m_{\text{Pb}}$ is the mass of lead and ${\rho }_{\text{Pb}}$ is the density of lead.

Similarly, write the expression for buoyant force acting on Aluminum.

$F_{\text{b, Al}}={\rho }_{\text{w}}g\frac{m_{\text{Al}}}{{\rho }_{\text{Al}}}$ (II)

Here, $F_{\text{b, Al}}$ is the buoyant force acting on Aluminum,${\rho }_{\text{w}}$ is the density of water, $g$ is the acceleration due to gravity,$m_{\text{Al}}$ is the mass of Aluminum and ${\rho }_{\text{Al}}$ is the density of Aluminum.

Similarly, write the expression for buoyant force acting on steel.

$F_{\text{b, steel}}={\rho }_{\text{w}}g\frac{m_{\text{steel}}}{{\rho }_{\text{steel}}}$ (III)

Here, $F_{\text{b, steel}}$ is the buoyant force acting on steel,${\rho }_{\text{w}}$ is the density of water, $g$ is the acceleration due to gravity,$m_{\text{steel}}$ is the mass of steel and ${\rho }_{\text{steel}}$ is the density of steel.

Since wood is floating, its buoyant force is equal to its weight.

Calculate weight of wood.

$F_{\text{b,wood}}=m_{\text{wood}}g$ (IV)

Here, $F_{\text{b,wood}}$ is the buoyant force of wood and $m_{\text{wood}}$ is the mass of wood.

Conclusion:

Substitute $1.00\times {10}^3\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{w}}$, $9.80\text{m}/{\text{s}}^2$ for $g$, $1.0\text{kg}$ for $m_{\text{Pb}}$ and $11300\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{Pb}}$ in equation (I) to get buoyant force acting on lead.

$\begin{array}{c}{F}_{\text{b,Pb}}=\left(1.00\times {10}^3\text{kg}/{\text{m}}^3\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(\frac{1.0\text{kg}}{11300\text{kg}/{\text{m}}^3}\right)\\ =0.87\text{N}\end{array}$

Substitute $1.00\times {10}^3\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{w}}$, $9.80\text{m}/{\text{s}}^2$ for $g$, $1.0\text{kg}$ for $m_{\text{Al}}$ and $2702\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{Al}}$ in equation (II) to get buoyant force acting on Aluminum.

$\begin{array}{c}{F}_{\text{b,Al}}=\left(1.00\times {10}^3\text{kg}/{\text{m}}^3\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(\frac{1.0\text{kg}}{2702\text{kg}/{\text{m}}^3}\right)\\ =3.6\text{N}\end{array}$

Substitute $1.00\times {10}^3\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{w}}$, $9.80\text{m}/{\text{s}}^2$ for $g$, $1.0\text{kg}$ for $m_{\text{steel}}$ and $7860\text{kg}/{\text{m}}^3$ for ${\rho }_{\text{Al}}$ in equation (III) to get buoyant force acting on Aluminum.

$\begin{array}{c}{F}_{\text{b,steel}}=\left(1.00\times {10}^3\text{kg}/{\text{m}}^3\right)\left(9.80\text{m}/{\text{s}}^2\right)\left(\frac{1.0\text{kg}}{7860\text{kg}/{\text{m}}^3}\right)\\ =1.2\text{N}\end{array}$

Substitute $1.0\text{kg}$ for $m_{\text{wood}}$ and $9.80\text{m}/{\text{s}}^2$ for $g$ in equation (IV) to get $F_{\text{b,steel}}$.

$\begin{array}{c}{F}_{\text{b,wood}}=\left(1.0\text{kg}\right)\left(9.80\text{m}/{\text{s}}^2\right)\\ =9.8\text{N}\end{array}$

From above equations, $F_{\text{wood}}>F_{\text{Al}}>F_{\text{Steel}}>F_{\text{Pb}}$.

Therefore, buoyant force is larger force wood that is floating than steel that is sinking and Aluminum has larger buoyant force than lead.