Chapter 14, Problem 14.27P

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 P14.11b. 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.

Figure P14.11 Problems 11 and 12.

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 14.27P

The magnitudes of the forces acting on the top and on the bottom of the block due to the surrounding water is $1018.15\text{N}$ and $1029.91\text{N}$ respectively.

Explanation of Solution

The mass of the block is $10.0\text{kg}$, the height of the block is $12.0\text{cm}$, the length of the block is $10.0\text{cm}$, the breadth of the block is $10.0\text{cm}$ and the top of the block is $5.0\text{cm}$ below the surface of the water.

Formula to calculate the absolute pressure at the top of the block is,

    $P_{\text{top}}=P_0+{\rho }_{\text{w}}g{h}_{\text{top}}$

Here, $P_{\text{top}}$ is the absolute pressure at the top of the block, ${\rho }_{\text{w}}$ is the density of water, $g$ is the acceleration due to gravity and $h_{\text{top}}$ is the depth of the top of the block in the water.

Formula to calculate the force acting on the top of the block is,

    $\begin{array}{c}{F}_{\text{top}}=P_{\text{top}}A\\ =P_{\text{top}}\times \left(b\times l\right)\end{array}$

Here, $F_{\text{top}}$ is the force acting on the top of the block, $A$ is the area of the top face of the block, $l$ is the length of the block and $b$ is the breadth of the block.

Substitute $P_0+{\rho }_{\text{w}}g{h}_{\text{top}}$ for $P_{\text{top}}$ in the above equation.

    $F_{\text{top}}=\left(P_0+{\rho }_{\text{w}}g{h}_{\text{top}}\right)\left(l\times b\right)$

Substitute $1.01325\times {10}^5\text{Pa}$ for $P_0$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $5.0\text{cm}$ for $h_{\text{top}}$, $10.0\text{cm}$ for $l$ and $10.0\text{cm}$ for $b$ to find $F_{\text{top}}$.

    $\begin{array}{c}{F}_{\text{top}}=\left(1.01325\times {10}^5\text{Pa}+1000\text{kg}/{\text{m}}^{\text{3}}\times 9.8\text{m}/{\text{s}}^{\text{2}}\times \left(5.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ \times \left(10.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\times 10.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =101815\text{Pa}\times 0.01{\text{m}}^{\text{2}}\\ =1018.15\text{N}\end{array}$

Formula to calculate the absolute pressure at the bottom of the block is,

    $P_{\text{bottom}}=P_0+{\rho }_{\text{w}}g{h}_{\text{bottom}}$

Here, $P_{\text{bottom}}$ is the absolute pressure at the bottom of the block and $h_{\text{bottom}}$ is the depth of the bottom of the block in the water.

Formula to calculate the force acting on the bottom of the block is,

    $\begin{array}{c}{F}_{\text{bottom}}=P_{\text{bottom}}A\\ =P_{\text{bottom}}\times \left(b\times l\right)\end{array}$

Here, $F_{\text{bottom}}$ is the force acting on the bottom of the block and $A$ is the area of the bottom face of the block.

Substitute $P_0+{\rho }_{\text{w}}g{h}_{\text{bottom}}$ for $P_{\text{bottom}}$ in the above equation.

    $F_{\text{bottom}}=\left(P_0+{\rho }_{\text{w}}g{h}_{\text{bottom}}\right)\left(l\times b\right)$

Substitute $1.01325\times {10}^5\text{Pa}$ for $P_0$, $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$, $\left(5.0\text{cm}+12.0\text{cm}\right)$ for $h_{\text{bottom}}$, $10.0\text{cm}$ for $l$ and $10.0\text{cm}$ for $b$ to find $F_{\text{bottom}}$.

    $\begin{array}{c}{F}_{\text{top}}=\left(1.01325\times {10}^5\text{Pa}+1000\text{kg}/{\text{m}}^{\text{3}}\times 9.8\text{m}/{\text{s}}^{\text{2}}\times \left(\left(5.0\text{cm}+12.0\text{cm}\right)\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)\\ \times \left(10.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\times 10.0\text{cm}\times \frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\\ =102991\text{Pa}\times 0.01{\text{m}}^{\text{2}}\\ =1029.91\text{N}\end{array}$

Conclusion:

Therefore, the magnitudes of the force acting on the bottom of the block due to the surrounding water is $1029.91\text{N}$.

To determine

The reading of the spring scale.

Answer to Problem 14.27P

The reading in the spring scale is $86.24\text{N}$.

Explanation of Solution

The scale reading when the block is immersed in water is equal to the tension in the chord supporting the block.

Apply the equilibrium condition.

    $\begin{array}{c}{\displaystyle \sum F_{\text{y}}}=0\\ T+F_{\text{bottom}}-F_{\text{top}}-mg=0\end{array}$

Here, $T$ is the tension in the chord and $m$ is the mass of the block.

Rearrange the above equation for $T$.

    $T=F_{\text{top}}+mg-F_{\text{bottom}}$

Substitute $1029.91\text{N}$ for $F_{\text{bottom}}$, $10.0\text{kg}$ for $m$, $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ and $1018.15\text{N}$ for $F_{\text{top}}$ to find $T$.

    $\begin{array}{c}T=1018.15\text{N}+10.0\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}-1029.91\text{N}\\ =86.24\text{N}\end{array}$

Conclusion:

Therefore, the reading in the spring scale is $86.24\text{N}$.

To determine

That the buoyant force equals the difference between the force at the top and bottom of the block.

Explanation of Solution

When an object is immersed in a liquid, then the upward force exerted on it by fluid is called buoyant force and its magnitude equals the weight of the liquid displaced by the object.

Formula to calculate the buoyant force is,

    $B={\rho }_{\text{w}}Vg$

Here, $B$ is the buoyant force, ${\rho }_{\text{w}}$ is the density of water, $V$ is the volume of block and $g$ is the acceleration due to gravity.

Substitute $1000\text{kg}/{\text{m}}^{\text{3}}$ for ${\rho }_{\text{w}}$, $\left(12.0\text{cm}\times 10.0\text{cm}\times 10.0\text{cm}\right)$ for $V$ and $9.8\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $B$.

    $\begin{array}{c}B=1000\text{kg}/{\text{m}}^{\text{3}}\times \left(\left(12.0\text{cm}\times 10.0\text{cm}\times 10.0\text{cm}\right)\times \frac{{10}^{-6}{\text{m}}^3}{1{\text{cm}}^{\text{3}}}\right)\times 9.8\text{m}/{\text{s}}^{\text{2}}\\ =11.76\text{N}\end{array}$

From part (a), the difference between the force at the top and bottom of the block is,

    $\begin{array}{c}{F}_{\text{bottom}}-F_{\text{top}}=1029.91\text{N}-1018.15\text{N}\\ =11.76\text{N}\end{array}$

From the above calculations it is clear that,

    $B=F_{\text{bottom}}-F_{\text{top}}$

Conclusion:

Therefore, the buoyant force equals the difference between the force at the top and bottom of the block.