Question

What is the tension in the string in the figure?

In the figure there is a block hanging off a string, completely immersed in ethyl alcohol.
it is 100 cm3 of aluminum (the block), and 2700 kg/m3 is the density of the aluminum.

Answer 1

Concepts and reason

The concepts required to solve this problem are Archimedes’ principle, and Equilibrium force condition.

Initially, use the density equation to solve for the mass in terms of volume. Then use the mass to solve for the force of gravity. Finally, use the equilibrium force condition to calculate the tension force.

Fundamentals

According to Archimedes’ principle, the upward buoyant force on any object immersed in a fluid is equal to the weight of liquid displaced by it. That is, the buoyant force is,

${F_{\rm{B}}} = {\rho _{\rm{w}}}gV$

Here, ${\rho _{\rm{w}}}$ is the density of the fluid, $g$ is acceleration due to gravity, and $V$ is the volume of liquid displaced.

The equilibrium force condition is applied when the system is static or at rest and acceleration is zero. The equilibrium force condition is,

$\sum {F = 0}$

Here, $\sum F$ is the sum of all the forces on the object.

The force of gravity is given as,

${F_{\rm{g}}} = mg$

Here, $m$ is mass, $g$ is acceleration due to gravity.

The density is given as,

$\rho = \frac{m}{V}$

Here, $m$ is mass, and $V$ is volume.

Rearrange the density equation $\rho = \frac{m}{V}$ to solve for mass.

$\begin{array}{c}\\\rho = \frac{m}{V}\\\\m = \rho V\\\end{array}$

Use the force of gravity equation.

Substitute $\rho V$ for $m$ in the equation ${F_{\rm{g}}} = mg$.

${F_{\rm{g}}} = \rho Vg$

Use the equilibrium force condition.

Substitute $T - {F_{\rm{g}}} + {F_{\rm{B}}}$ for $\sum F$ in the equation $\sum F = 0$ and solve for Tension $T$.

$\begin{array}{c}\\T - {F_{\rm{g}}} + {F_{\rm{B}}} = 0\\\\T = {F_{\rm{g}}} - {F_{\rm{B}}}\\\end{array}$

Substitute $\rho Vg$ for ${F_{\rm{g}}}$, and ${\rho _{\rm{w}}}gV$ for ${F_{\rm{B}}}$ in the equation $T = {F_{\rm{g}}} - {F_{\rm{B}}}$.

$\begin{array}{c}\\T = \rho Vg - {\rho _{\rm{w}}}Vg\\\\ = \left( {\rho - {\rho _{\rm{w}}}} \right)Vg\\\end{array}$

Here, $\rho$ is the density of aluminum and ${\rho _{\rm{w}}}$ is the density of ethyl alcohol.

Substitute $2700{\rm{ kg/}}{{\rm{m}}^3}$ for $\rho$, $790{\rm{ kg/}}{{\rm{m}}^3}$ for ${\rho _{\rm{w}}}$, $100{\rm{ c}}{{\rm{m}}^3}$ for $V$, and $9.8{\rm{ m/}}{{\rm{s}}^2}$ for $g$ in the equation $T = \left( {\rho - {\rho _{\rm{w}}}} \right)Vg$.

$\begin{array}{c}\\T = \left( {2700{\rm{ kg/}}{{\rm{m}}^3} - 790{\rm{ kg/}}{{\rm{m}}^3}} \right)\left( {100\;{\rm{c}}{{\rm{m}}^3}} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = \left( {{\rm{1910 kg/}}{{\rm{m}}^3}} \right)\left( {100\;{\rm{c}}{{\rm{m}}^3}\left( {\frac{{{{10}^{ - 6}}{\rm{ }}{{\rm{m}}^3}}}{{1{\rm{ c}}{{\rm{m}}^3}}}} \right)} \right)\left( {9.8{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 1.87{\rm{ N}}\\\end{array}$

Ans:

The tension in the string is 1.87 N.