Chapter 9, Problem 3SP

A copper block with a density of 8960 kg/m³ is suspended from a string in a beaker of water so that the block is completely submerged but not resting on the bottom. The block is a cube with sides of 4 cm (0.04 m).

1. a. What is the volume of the block in cubic meters?
2. b. What is the mass of the block?
3. c. What is the weight of the block?
4. d. What is the buoyant force acting on the block?
5. e. What tension in the string is needed to hold the block in place?

To determine

Volume of copper block.

Answer to Problem 3SP

Volume of copper block is $6.4\times {10}^{-5}{\text{m}}^3$

Explanation of Solution

Given info: Density of copper is $8960\text{kg}/{\text{m}}^{\text{3}}$ , side of cube is $4\text{cm}$

Write the equation to find the volume of a cube.

$V=a^3$

Here,

$V$ is the volume

$a$ is the side of cube

Substitute $4\text{cm}$ for $a$ in equation to find the volume.

$\begin{array}{c}V={\left(4\times {10}^{-2}\text{m}\right)}^3\\ =6.4\times {10}^{-5}{\text{m}}^3\end{array}$

Conclusion:

Volume of copper block is $6.4\times {10}^{-5}{\text{m}}^3$

To determine

Mass of block.

Answer to Problem 3SP

Mass of the copper block is $0.57\text{kg}$

Explanation of Solution

Write the equation to find the mass.

$M=\rho V$

Here,

$M$ is the mass

$\rho$ is the density

$V$ is the volume

Substitute $8960\text{kg}/{\text{m}}^{\text{3}}$ for $\rho$ and $6.4\times {10}^{-5}{\text{m}}^3$ for $V$ in equation to find $M$

$\begin{array}{c}M=\left(8960\text{kg}/{\text{m}}^{\text{3}}\right)\left(6.4\times {10}^{-5}{\text{m}}^3\right)\\ =0.57\text{kg}\end{array}$

Conclusion:

Mass of the copper block is $0.57\text{kg}$

To determine

Weight of the block.

Answer to Problem 3SP

Weight of the copper block is $5.62\text{N}$

Explanation of Solution

Write the equation to find the weight.

$W=Mg$

Here,

$W$ is the weight

$M$ is the mass

$g$ is the acceleration due to gravity

Substitute $0.57\text{kg}$ for $M$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation to get $W$

$\begin{array}{c}W=0.57\text{kg}\times \text{9}\text{.8}\text{m}/{\text{s}}^{\text{2}}\\ =5.62\text{N}\end{array}$

Conclusion:

Weight of the copper block is $5.62\text{N}$

To determine

The buoyant force acting on block.

Answer to Problem 3SP

The buoyant force is $0.627\text{N}$

Explanation of Solution

Buoyant force acting on the block is equal to the weight of the water displaced.

Write the equation to find the buoyant force.

$\begin{array}{c}{F}_{bouyant}=m_{displaced}g\\ ={\rho }_{displaced}\times V_{water}\times g\end{array}$

Here,

$F_{bouyant}$ is the buoyant force

$\rho$ is the density of water

$V_{water}$ is the volume of water displaced

$g$ is the acceleration due to gravity

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{displaced}$ ,$6.4\times {10}^{-5}{\text{m}}^3$ for $V_{water}$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ in equation to find $F_{bouyant}$

$\begin{array}{c}{F}_{bouyant}=1000\text{kg}/{\text{m}}^{\text{3}}\times 6.4\times {10}^{-5}{\text{m}}^3\times 9.8\text{m}/{\text{s}}^{\text{2}}\\ =0.627\text{N}\end{array}$

Conclusion:

The buoyant force is $0.627\text{N}$

To determine

The tension in the string to hold the block.

Answer to Problem 3SP

The tension is $4.99\text{N}$

Explanation of Solution

The string in the fluid is acted upon by two forces. They are weight of block down wards and buoyant force upwards.

By Newton’s first law of motion the sum of forces acting on a body must be zero.

$T=weight-F_{\text{buoyant}}$

Here,

$T$ is the tension

$Weight$ is the weight of cube

$F_{\text{bouyant}}$ is the buoyant force

Substitute $0.627\text{N}$ for $F_{bouyant}$ and $5.62\text{N}$ for $weight$ in equation to get $T$

$\begin{array}{c}T=5.62\text{N}-0.627\text{N}\\ \text{=4}\text{.99}\text{N}\end{array}$

Conclusion:

The tension is $4.99\text{N}$