Chapter 9, Problem 9WUE

(a)

To determine

The volume of the lead.

Answer to Problem 9WUE

The volume of the volume is $1.77\times {10}^{-3}{\text{m}}^{\text{3}}$.

Explanation of Solution

Given info: Mass of the lead is $20.0\text{kg}$ and density of the lead is $11.3\times {10}^3{\text{kg/m}}^{\text{3}}$.

The formula for the volume of the lead is,

$V_{\text{lead}}=\frac{M_{\text{lead}}}{{\rho }_{\text{lead}}}$

• $M_{\text{lead}}$ is mass of lead.
• ${\rho }_{\text{lead}}$ is density of lead.

Substitute $20.0\text{kg}$ for $M_{\text{lead}}$ and $11.3\times {10}^3{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{lead}}$ to find $V_{\text{lead}}$.

$\begin{array}{c}{V}_{\text{lead}}=\frac{20.0\text{kg}}{11.3\times {10}^3{\text{kg/m}}^{\text{3}}}\\ =1.77\times {10}^{-3}{\text{m}}^{\text{3}}\end{array}$

Thus, the volume of the volume is $1.77\times {10}^{-3}{\text{m}}^{\text{3}}$.

Conclusion:

Therefore, the volume of the volume is $1.77\times {10}^{-3}{\text{m}}^{\text{3}}$.

(b)

To determine

The buoyancy force on the lead.

Answer to Problem 9WUE

The buoyancy force on the lead is $17.3\text{N}$.

Explanation of Solution

The buoyancy force on the lead is $B=m_{\text{w}}g={\rho }_{\text{w}}{V}_{\text{lead}}g$ that is the force equal to the weight of the water displaced by the lead.

Given info: The density of water is $1.00\times {10}^3{\text{kg/m}}^{\text{3}}$, volume of the lead is $1.77\times {10}^{-3}{\text{m}}^{\text{3}}$, and acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$.

The formula for the buoyancy force on the lead is,

$B={\rho }_{\text{w}}{V}_{\text{lead}}g$

• ${\rho }_{\text{w}}$ is density of water.
• $V_{\text{lead}}$ is volume of the lead.
• $g$ is gravitational acceleration.

Substitute $1.00\times {10}^3{\text{kg/m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $1.77\times {10}^{-3}{\text{m}}^{\text{3}}$ is $V_{\text{lead}}$, and $9.80{\text{m/s}}^{\text{2}}$ for $g$ to find $B$.

$\begin{array}{c}B=\left(1.00\times {10}^3{\text{kg/m}}^{\text{3}}\right)\left(1.77\times {10}^{-3}{\text{m}}^{\text{3}}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\\ =17.3\text{N}\end{array}$

Thus, the buoyancy force on the lead is $17.3\text{N}$.

Conclusion:

Therefore, the buoyancy force on the lead is $17.3\text{N}$.

(c)

To determine

The weight of the lead.

Answer to Problem 9WUE

The weight of the lead is $196\text{N}$.

Explanation of Solution

Given info: The mass of the lead is $20.0\text{kg}$ and acceleration due to gravity is $9.80{\text{m/s}}^{\text{2}}$.

The formula for the weight of the lead is,

$W_{\text{lead}}=M_{\text{lead}}g$

• $M_{\text{lead}}$ is mass of the lead.
• $g$ is gravitational acceleration.

Substitute $20.0\text{kg}$ for $M_{\text{lead}}$ and $9.80{\text{m/s}}^{\text{2}}$ for $g$ to find $W_{\text{lead}}$.

$\begin{array}{c}{W}_{\text{lead}}=\left(20.0\text{kg}\right)\left(9.80{\text{m/s}}^{\text{2}}\right)\\ =196\text{N}\end{array}$

Thus, the weight of the lead is $196\text{N}$.

Conclusion:

Therefore, the weight of the lead is $196\text{N}$.

(d)

To determine

The normal force acting on the lead.

Answer to Problem 9WUE

The normal force acting on the lead is $179\text{N}$.

Explanation of Solution

The force acting on the lead is ${\displaystyle \sum F_y}=0=n+B-W_{\text{lead}}=0$ and the normal force on the lead is $n=W_{\text{lead}}-B$.

Given info: The weight of the lead is $196\text{N}$ and buoyancy force on the lead is $17.3\text{N}$.

The formula for the normal force on the lead is,

$n=W_{\text{lead}}-B$

• $W_{\text{lead}}$ is weight of lead.
• $B$ is buoyancy force.

Substitute $196\text{N}$ for $W_{\text{lead}}$ and $17.3\text{N}$ for $B$ to find $n$.

$\begin{array}{c}n=196\text{N}-17.3\text{N}\\ =179\text{N}\end{array}$

Thus, the normal force acting on the lead is $179\text{N}$.

Conclusion:

Therefore, the normal force acting on the lead is $179\text{N}$.