Chapter 15, Problem 25P

A 10.0-kg block of metal measuring 12.0 cm by 10.0 cm by 10.0 cm is suspended from a scale and immersed in water as shown in Figure P15.24b. The 12.0-cm dimension is vertical, and the top of the block is 5.00 cm below the surface of the water. (a) What are the magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water? (b) What is the reading of the spring scale? (c) Show that the buoyant force equals the difference between the forces at the top and bottom of the block.

To determine

The magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water.

Answer to Problem 25P

The magnitude of the force acting on the top of the block due to the surrounding water is $1.0179\times {10}^3\text{ N}$ and that on the bottom of the block is $1.0297\times {10}^3\text{ N}$.

Explanation of Solution

Write the equation for the pressure at a depth.

  $P=P_0+\rho gh$        (I)

Here, $P$ is the pressure at the bottom of the fluid, $P_0$ is the pressure at the surface of the fluid, $\rho$ is the density of the fluid, $g$ is the acceleration due to gravity and $h$ is the depth from the surface of the fluid.

Write the equation for pressure.

  $P=\frac{F}{A}$

Here, $P$ is the pressure and $A$ is the area over which the force acts.

Rewrite the above equation for $F$.

  $F=PA$        (II)

Use equation (II) to write the expression for the force at the top of the block.

  $F_{\text{top}}=P_{\text{top}}A$        (III)

Here, $F_{\text{top}}$ is the force on the top of the block, $P_{\text{top}}$ is the pressure at the top of the block and $A$ is the area of the block.

Use equation (II) to write the expression for the force on the bottom of the block.

  $F_{\text{bot}}=P_{\text{bot}}A$        (IV)

Here, $F_{\text{bot}}$ is the force on the bottom of the block and $P_{\text{bot}}$ is the pressure at the bottom of the block.

Conclusion:

The value of $P_0$ is $1.013\times {10}^5\text{ Pa}$ , the value of $g$ is $9.80{\text{ m/s}}^2$ and the density of water is $1000{\text{ kg/m}}^3$.

Substitute $1.013\times {10}^5\text{ Pa}$ for $P_0$ , $1000{\text{ kg/m}}^3$ for $\rho$ , $5.00\text{ cm}$ for $h$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (I) to find the pressure at the top of the block.

  $\begin{array}{c}{P}_{\text{top}}=1.013\times {10}^5\text{ Pa}+\left(1000{\text{ kg/m}}^3\right)\left(5.00\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)\left(9.80{\text{ m/s}}^2\right)\\ =1.0179\times {10}^5{\text{ N/m}}^2\end{array}$

Since the vertical dimension of the block is $12.0\text{ cm}$ and the block is $5.00\text{ cm}$ below the surface of the water, the depth for the bottom of the block will be $17.0\text{ cm}$.

Substitute $1.013\times {10}^5\text{ Pa}$ for $P_0$ , $1000{\text{ kg/m}}^3$ for $\rho$ , $17.0\text{ cm}$ for $h$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (I) to find the pressure at the bottom of the block.

  $\begin{array}{c}{P}_{\text{bot}}=1.013\times {10}^5\text{ Pa}+\left(1000{\text{ kg/m}}^3\right)\left(17.0\text{ cm}\frac{1\text{ m}}{100\text{ cm}}\right)\left(9.80{\text{ m/s}}^2\right)\\ =1.0297\times {10}^5{\text{ N/m}}^2\end{array}$

Substitute $1.0179\times {10}^5{\text{ N/m}}^2$ for $P_{\text{top}}$ and $\left(10.0\text{ cm}\times 10.0\text{ cm}\right)$ for $A$ in equation (III) to find $F_{\text{top}}$.

  $\begin{array}{c}{F}_{\text{top}}=\left(1.0179\times {10}^5{\text{ N/m}}^2\right)\left(10.0\text{ cm}\times 10.0\text{ cm}\frac{1{\text{ m}}^2}{{10}^4{\text{ cm}}^2}\right)\\ =1.0179\times {10}^3\text{ N}\end{array}$

Substitute $1.0297\times {10}^5{\text{ N/m}}^2$ for $P_{\text{bot}}$ and $\left(10.0\text{ cm}\times 10.0\text{ cm}\right)$ for $A$ in equation (IV) to find $F_{\text{bot}}$.

  $\begin{array}{c}{F}_{\text{bot}}=\left(1.0297\times {10}^5{\text{ N/m}}^2\right)\left(10.0\text{ cm}\times 10.0\text{ cm}\frac{1{\text{ m}}^2}{{10}^4{\text{ cm}}^2}\right)\\ =1.0297\times {10}^3\text{ N}\end{array}$

Therefore, the magnitude of the force acting on the top of the block due to the surrounding water is $1.0179\times {10}^3\text{ N}$ and that on the bottom of the block is $1.0297\times {10}^3\text{ N}$.

To determine

The reading of the spring scale.

Answer to Problem 25P

The reading of the spring scale is $86.2\text{ N}$.

Explanation of Solution

The free-body diagram of the system is shown below.

The tension in the string is the spring scale reading.

The tension in the string is balanced by the vector sum of the weight of the block and the buoyant force.

Write the equation for the tension in the string.

  $T=F_g-B$        (V)

Here, $T$ is the tension in the string, $F_g$ is the weight of the block and $B$ is the buoyant force.

Write the equation for the weight of the block.

  $F_g=mg$        (VI)

Here, $m$ is the mass of the block and $g$ is the acceleration due to gravity.

Write the equation for the buoyant force.

  $B=\rho gV$        (VII)

Here, $V$ is the volume of the block.

Put equations (VI) and (VII) in equation (V).

  $\begin{array}{c}T=mg-\rho gV\\ =\left(m-\rho V\right)g\end{array}$        (VIII)

Conclusion:

Substitute $10.0\text{ kg}$ for $m$ , $1000{\text{ kg/m}}^3$ for $\rho$ , $\left(12.0\text{ cm}\times 10.0\text{ cm}\times 10.0\text{ cm}\right)$ for $V$ and $9.80{\text{ m/s}}^2$ for $g$ in equation (VIII) to find $T$.

  $\begin{array}{c}T=\left(10.0\text{ kg}-\left(1000{\text{ kg/m}}^3\right)\left(12.0\text{ cm}\times 10.0\text{ cm}\times 10.0\text{ cm}\frac{1{\text{ m}}^3}{{10}^6{\text{ cm}}^3}\right)\right)\left(9.80{\text{ m/s}}^2\right)\\ =86.2\text{ N}\end{array}$

Therefore, the reading of the spring scale is $86.2\text{ N}$.

To determine

To show that the buoyant force equals the difference between the forces at the top and bottom of the block.

Answer to Problem 25P

It is showed that the buoyant force equals the difference between the forces at the top and bottom of the block.

Explanation of Solution

Write the equation for the difference between the forces at the top and bottom of the block.

  $F_{\text{dif}}=F_{\text{bot}}-F_{\text{top}}$        (IX)

Here, $F_{\text{dif}}$ is the difference between the forces at the top and bottom of the block.

Conclusion:

Substitute $1.0297\times {10}^3\text{ N}$ for $F_{\text{bot}}$ and $1.0179\times {10}^3\text{ N}$ for $F_{\text{top}}$ in equation (IX) to find $F_{\text{dif}}$ .

  $\begin{array}{c}{F}_{\text{dif}}=1.0297\times {10}^3\text{ N}-1.0179\times {10}^3\text{ N}\\ =11.8\text{ N}\end{array}$

Substitute $1000{\text{ kg/m}}^3$ for $\rho$ , $9.80{\text{ m/s}}^2$ for $g$ and $\left(12.0\text{ cm}\times 10.0\text{ cm}\times 10.0\text{ cm}\right)$ for $V$ in equation (VII) to find $B$ .

  $\begin{array}{c}B=\left(1000{\text{ kg/m}}^3\right)\left(9.80{\text{ m/s}}^2\right)\left(12.0\text{ cm}\times 10.0\text{ cm}\times 10.0\text{ cm}\frac{1{\text{ m}}^3}{{10}^6{\text{ cm}}^3}\right)\\ =11.8\text{ N}\end{array}$

The value of the difference between the forces at the top and bottom of the block and the magnitude of the buoyant force are equal.

Thus, it is showed that the buoyant force equals the difference between the forces at the top and bottom of the block.