What is the tension of the string in FIGURE EX14.21?

Archimedes principle states that the upward force acts on the object when an object partly or full immersed in a fluid. This force is called buoyant force. Mathematically, buoyant force can be expressed as,

$F_n=\rho Vg$

Here,$\rho$is the density of fluid,$V$is volume of displaced fluid, and$g$is acceleration due to gravity.

The forces acting on the aluminum block are as shown in the below figure.

In this problem, an aluminum block is suspended by a cable and is fully submerged in a liquid. There are $3$forces acting on the block. The gravitational force on the block acts downward, tension and force acts on the block by the liquid are in upward direction.

According to Newton's second law of motion, if acceleration is zero, then,

$T+F_{\mathrm{B}}-\mathrm{w}=0$

Here, $T$is tension, role="math" localid="1648137802740" $F_B$is up thrust force, and $w$is weight.

Substitute $F_{\mathrm{B}}={\rho }_{fl}{V}_{Al}\mathrm{g}$in the above equation.

$T+{\rho }_{fl}{V}_{Al}g=m_{Al}g$

Here, $m$is the mass of the block; $V$is the volume of fluid, and $\rho$is the density of the liquid. Rearrange the above equation for $T$.

role="math" localid="1648138096737" $T=mg={\rho }_{fl}{V}_{Al}g$

The mass of the aluminum block can be expressed as,

$m_A={\rho }_{\mathrm{AI}}{V}_{AI}$

Now, the above equation changes as,

$T={\rho }_{\mathrm{A}I}{V}_{\mathrm{Al}}g-{\rho }_{fl}{V}_{\mathrm{Al}}g$

$=g{V}_{\mathrm{Al}}\left({\rho }_{\mathrm{Al}}-{\rho }_{fl}\right)$

Convert the unit of volume from ${\mathrm{cm}}^3$to ${\mathrm{m}}^3$

$V_{Al}=100{\mathrm{cm}}^3\left(\frac{{10}^{-6}{\mathrm{m}}^3}{{\mathrm{cm}}^3}\right)$

$=100\times {10}^{-6}{\mathrm{m}}^3$

Substitute $100\times {10}^{-6}{\mathrm{m}}^3$for $V_{Al}$, $9.8m/s^2$for $g$, $2700kg/m^3$for ${\rho }_{Al}$and $790\mathrm{kg}/{\mathrm{m}}^3$for ${\rho }_{fl}$in the equation.

$T=\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(100\times {10}^{-6}{\mathrm{m}}^3\right)\left(2700\mathrm{kg}/{\mathrm{m}}^3-790\mathrm{kg}/{\mathrm{m}}^3\right)$

$T$$=1.87\mathrm{N}$