Question

In Figure 14-38, a cube of edge length $\mathbf{L}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{600}\mathbf{}\mathbf{m}$and mass $\mathbf{450}\mathbf{}\mathbf{kg}$is suspended by a rope in an open tank of liquid of density $\mathbf{1030}\mathbf{}\raisebox{1ex}{$\mathbf{k}\mathbf{g}$}\!\left/ \!\raisebox{-1ex}{${\mathbf{m}}^{\mathbf{3}}$}\right.$. (a)Find the magnitude of the total downward force on the top of the cube from the liquid and the atmosphere, assuming atmospheric pressure is $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{atm}$, (b) Find the magnitude of the total upward force on the bottom of the cube, and (c)Find the tension in the rope. (d) Calculate the magnitude of the buoyant force on the cube using Archimedes’ principle. What relation exists among all these quantities?

Short answer

a) The total force on the top of the block is$1.422\times {10}^3\mathrm{N}$

b) The total force on the bottom of the block is$3.54\times {10}^3\mathrm{N}$

c) The tension in the rope is$4.41\times {10}^6\mathrm{N}$

d) The buoyant force is$2.118\times {10}^3\mathrm{N}$

Step-by-step solution

The given data

• Density of air,$\mathrm{\rho }=1.3\mathrm{kg}/{\mathrm{m}}^3$

• Acceleration due to gravity,$\mathrm{g}=9.8\mathrm{m}/{\mathrm{s}}^2$

• Length of the side of the cube,$\mathrm{L}=0.600\mathrm{m}$

• Distance from the surface of the water to the top of the cube,$\frac{\mathrm{L}}{2}-0.300\mathrm{m}$

Understanding the concept of Archimedes’ Principle

We can find the pressure using density, acceleration, and height. Since pressure is force per unit area, we can find the force using the given area. Force of buoyancy can be found using the Archimedes Principle. As the volume displaced and density is known, we can find the force of buoyancy.

Formula:

Total pressure on the top of the cube, ${\mathrm{\rho }}_{\text{top}}={\mathrm{p}}_{\mathrm{atm}}+\mathrm{\rho gh}$(i)

Total pressure on the bottom of the cube, ${\mathrm{\rho }}_{\text{bottom}}={\mathrm{p}}_{\mathrm{atm}}+\mathrm{\rho gh}$(ii)

Pressure on a body through force $\mathrm{F}$on area $\mathrm{A}$,$\mathrm{p}=\frac{\mathrm{F}}{\mathrm{A}}$(iii)

a) Calculation of total force on top of the block

We can write the equation (i) for the pressure as:

${\mathrm{p}}_{\text{top}}={\mathrm{p}}_{\mathrm{atm}}+\mathrm{\rho g}\frac{\mathrm{L}}{2}$ (The top of cube is$\raisebox{1ex}{$\mathrm{L}$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$under the water)
$=1.01\times {10}^3+9.8\times 1000\times 0.3$

$=3.95\times {10}^3\mathrm{pa}$

The area of the block on which the pressure acts is as follows:

$\mathrm{A}={\mathrm{L}}^2$

$=0.6^2$

$=0.36{\mathrm{m}}^2$

So, the force on the top using equation (iii) is given as:

${\mathrm{F}}_{\mathrm{top}}=3.95\times {10}^3\times 0.36$

$=1.422\times {10}^3\mathrm{N}$

Hence, the force on the top of the block is $1.422\times {10}^3\mathrm{N}$$1.422\times {10}^3\mathrm{N}$

b) Calculation of force on the bottom of the block

We can write the equation (ii) as follows:

${\mathrm{p}}_{\text{bottom}}={\mathrm{p}}_{\text{Atm}}+\mathrm{\rho g}\left(\mathrm{L}+\frac{\mathrm{L}}{2}\right)$

$=1.01\times {10}^3+9.8\times 1000\times 0.9$

$=9.83\times {10}^5\mathrm{Pa}$

Using equation (iii) and the value from equation (a), the force on the bottom is as follows:

${\mathrm{F}}_{\text{Dolfoin}}=9.83\times {10}^3\times 0.36$

$=3.54\times {10}^3\mathrm{N}$

Hence, the force on the bottom of the cube is$3.54\times {10}^3\mathrm{N}$$3.54\times {10}^3\mathrm{N}$

c) Calculation of the tension in the rope

Difference in the force at the bottom and at the top would give us the force of buoyancy. Tension is in the upward direction, and weight is in the downward direction. So we can write,

$\mathrm{T}+\left({\mathrm{F}}_{\text{botiom}}-{\mathrm{F}}_{\mathrm{lo\rho }}\right)-\mathrm{mg}=0$ (from figure)

$\mathrm{T}=\mathrm{mg}-\left({\mathrm{F}}_{\text{bottom}}-{\mathrm{F}}_{\text{fop}}\right)$

$=450\times 9.8-\left(3.54\times {10}^3-1.422\times {10}^3\right)$

$=4.41\times {10}^6\mathrm{N}$

Hence, the tension in the rope is$4.41\times {10}^6\mathrm{N}$

d) Calculation of buoyant force

The buoyant force can be given as the net force applied on the block, which is given as:

${\mathrm{F}}_{\mathrm{b}}=\left({\mathrm{F}}_{\text{batom}}-{\mathrm{F}}_{\mathrm{t}00}\right)$

$=\left(3.54\times {10}^3-1.422\times {10}^3\right)$

$=2.118\times {10}^3\mathrm{N}$

Hence, the value of buoyant force is$2.118\times {10}^3\mathrm{N}$