Chapter 14, Problem 71AP

To determine

The equilibrium readings of both upper and lower scales.

Answer to Problem 71AP

The lower scale reading is $\underset{\_}{31.7\text{N}}$ and upper scale reading is $\underset{\_}{17.3\text{N}}$.

Explanation of Solution

The weight of the iron block is balanced by the sum of tension on spring and the buoyant force exerted on iron block by the oil when viewed from the upper part of the scale.

    $T+B=F_{g,\text{iron}}$                                                                                                       (I)

Here, $T$ is the tension force on the spring scale, $B$ is the buoyant force on iron block, and $F_{g,\text{iron}}$ is the force of gravity on the iron block.

Write the expression for density of iron block.

    ${\rho }_{\text{iron}}=\frac{m_{\text{iron}}}{V_{\text{iron}}}$                                                                                                     (II)

Here, ${\rho }_{\text{iron}}$ is the density of iron block, $m_{\text{iron}}$ is the mass of iron block, and $V_{\text{iron}}$ is the volume of block.

Rearrange equation (II) to find $V_{\text{iron}}$.

    $V_{\text{iron}}=\frac{m_{\text{iron}}}{{\rho }_{\text{iron}}}$                                                                                                        (III)

By Archimedes law, volume of iron block dipped in oil is equal to the volume of oil displaced from the jar.

    $V_{\text{displaced oil}}=\frac{m_{\text{iron}}}{{\rho }_{\text{iron}}}$                                                                                           (IV)

Here, $V_{\text{displaced oil}}$ is the volume of the displaced oil.

Write the expression for the buoyant force exerted by the oil on the iron block.

    $B={\rho }_{\text{oil}}{V}_{\text{iron}}g$                                                                                                     (V)

Here, ${\rho }_{\text{oil}}$ is the density of oil, $V_{\text{iron}}$ is the volume of iron block, and $g$ is the acceleration due to gravity.

Rearrange equation (I) to find $T$.

    $T=F_{g,\text{iron}}-B$                                                                                              (VI)

Use expression (V) in (VI) to find $T$.

    $T=F_{g,\text{iron}}-{\rho }_{\text{oil}}{V}_{\text{iron}}g$                                                                                 (VII)

Write the expression for force of gravity on iron block.

    $F_{g,\text{iron}}=m_{\text{iron}}g$                                                                                               (VIII)

Here, $m_{\text{iron}}$ is the mass of the iron block.

Use expression (VIII) in (VII).

    $T=m_{\text{iron}}g-{\rho }_{\text{oil}}{V}_{\text{iron}}g$                                                                                      (IX)

Use expression (III) in (IX) to find $T$.

    $\begin{array}{c}T=m_{\text{iron}}g-{\rho }_{\text{oil}}\left(\frac{m_{\text{iron}}}{{\rho }_{\text{iron}}}\right)g\\ =\left(1-\frac{{\rho }_{\text{oil}}}{{\rho }_{\text{iron}}}\right)m_{\text{iron}}g\end{array}$                                                                           (X)

Now observe the system from the bottom side of scale. Let $n$ be the upward normal force acting on the system.

Write the sum of all the vertical forces acting on the system.

    ${\displaystyle \sum F_y}=T+n-F_{g,\text{beaker}}-F_{g,\text{oil}}-F_{g,\text{iron}}$                                                         (XI)

Here, ${\displaystyle \sum F_y}$ is the sum of vertical forces acting on the system, $F_{g,\text{beaker}}$ is the force of gravity on beaker.

At equilibrium the sum of all vertical forces is equal to zero.

    $T+n-F_{g,\text{beaker}}-F_{g,\text{oil}}-F_{g,\text{iron}}=0$                                                                 (XII)

Write the expression for $F_{g\text{,beaker}}$.

    $F_{g\text{,beaker}}=m_{\text{beaker}}g$                                                                                      (XIII)

Here, $m_{\text{beaker}}$ is the mass of the beaker.

Write the expression for $F_{g\text{,oil}}$.

    $F_{g\text{,oil}}=m_{\text{oil}}g$                                                                                               (XIV)

Here, $m_{\text{oil}}$ is the mass of the oil.

Use expressions (XIV), (XIII), and (VIII) in expression (XII) and solve for $n$.

    $\begin{array}{c}T+n-m_{\text{beaker}}g-m_{\text{oil}}g-m_{\text{iron}}g=0\\ n=\left(m_{\text{beaker}}+m_{\text{oil}}+m_{\text{iron}}\right)g-T\end{array}$                       (XV)

Conclusion:

Substitute $916{\text{kg/m}}^3$ for ${\rho }_{\text{oil}}$, $7860{\text{kg/m}}^3$ for ${\rho }_{\text{iron}}$, $2.00\text{kg}$ for $m_{\text{iron}}$, and $9.80{\text{m/s}}^2$ for $g$ in equation (X) to find $T$.

    $\begin{array}{c}T=\left(1-\frac{916{\text{kg/m}}^3}{7860{\text{kg/m}}^3}\right)\left(2.00\text{kg}\right)\left(9.80{\text{m/s}}^2\right)\\ =17.3\text{N}\end{array}$

Substitute $1.00\text{kg}$ for $m_{\text{beaker}}$, $2.00\text{kg}$ for $m_{\text{oil}}$, $2.00\text{kg}$ for $m_{\text{iron}}$, $9.80{\text{m/s}}^2$ for $g$, and $17.3\text{N}$ for $T$ in equation (XV) to find $n$.

    $\begin{array}{c}n=\left(1.00\text{kg}+2.00\text{kg}+2.00\text{kg}\right)\left(9.80{\text{m/s}}^2\right)-17.3\text{N}\\ \text{=31}\text{.7}\text{N}\end{array}$

Therefore, the lower scale reading is $\underset{\_}{31.7\text{N}}$ and upper scale reading is $\underset{\_}{17.3\text{N}}$.