Chapter 11, Problem 64QAP

To determine

(a)

Derivation of the difference between top and bottom of the cube

Answer to Problem 64QAP

The pressure difference between the top and bottom of the cube is $\text{s}\times \text{d}\times \text{g}$

Explanation of Solution

Given:

Side of the cube

  = s

Liquid

  = fresh water

Amount of cube submerged

  = completely

Formula used:

  $\begin{array}{l}\text{F = Thrust = mg}\\ \text{Density = }\frac{\text{mass}}{\text{volume}}\\ \text{Volume = s}\times \text{ s }\times \text{ s}\\ \text{Area = s}\times \text{s}\\ \text{Pressure = }\frac{\text{Thrust}}{\text{Area}}\end{array}$

Calculation:

When a cube is completely submerged in water then the amount of thrust acting will be equal to the weight of the cube.

  $\begin{array}{l}\text{F = Thrust = m}\times \text{g}\\ \text{Density = }\frac{\text{mass}}{\text{volume}}\\ \text{mass = Density}\times \text{volume}\\ \text{So,}\\ \text{Thrust = d}\times \text{V}\times \text{g}\\ \text{Now,}\\ \text{Volume = s}\times \text{ s }\times \text{ s}\\ \text{Thrust = d}\times {\text{s}}^3\times \text{g}\\ \text{Also,}\\ \text{Area = s}\times \text{s}\\ \text{Pressure = }\frac{\text{Thrust}}{\text{Area}}\\ \text{Pressure = }\frac{\text{d}\times {\text{s}}^3\times \text{g}}{{\text{s}}^2}\\ \text{Pressure (}\Delta \text{P) = s}\times \text{d}\times \text{g}\end{array}$

Conclusion:

The pressure difference between the top and bottom of the cube is $\text{s}\times \text{d}\times \text{g}$

To determine

(b)

Using free-body diagram derive an expression for net force acting on the cube.

Answer to Problem 64QAP

Net force acting on the cube will be the product of volume, density and acceleration due to gravity, i.e. Vdg.

Explanation of Solution

Given:

Side of the cube

  = s

Liquid

  = fresh water

Amount of cube submerged

  = completely

Formula used:

  $\begin{array}{l}\text{F = Thrust = mg}\\ \text{Density = }\frac{\text{mass}}{\text{volume}}\\ \text{Volume = s}\times \text{ s }\times \text{ s}\\ \text{Area = s}\times \text{s}\\ \text{Pressure(P) = }\frac{\text{Thrust (F)}}{\text{Area (A)}}\end{array}$

Calculation:

Substituting the values,

  $\begin{array}{l}{\text{F}}_{\text{net}}{\text{ = F}}_2-{\text{F}}_1\\ {\text{F}}_{\text{net}}={\text{h}}_2\text{dgA}-{\text{h}}_1\text{dgA}\\ {\text{F}}_{\text{net}}={\text{A(h}}_2-{\text{h}}_1)\text{dg}\\ {\text{Volume(V) = A(h}}_2-{\text{h}}_1)\\ {\text{F}}_{\text{net}}\text{ =Vdg}\end{array}$

Conclusion:

Net force acting on the cube will be the product of volume, density and acceleration due to gravity, i.e. Vdg.

To determine

(c)

Weight of water displaced by the cube

Answer to Problem 64QAP

Weight of water displaced by the cube is $\text{d }\times \text{V}\times \text{g}$

Explanation of Solution

Given:

Side of the cube

  = s

Liquid

  = fresh water

Amount of cube submerged

  = completely

Formula used:

  $\begin{array}{l}\text{F = W = mg}\\ \text{Density = }\frac{\text{mass}}{\text{volume}}\\ \text{Volume = L}\times \text{ B }\times \text{ H}\end{array}$

From Archimedes' principle,

Weight of the block = weight of the water displaced by it = buoyant force

Calculation:

Substituting the values,

  $\begin{array}{l}\text{Volume = L}\times \text{ B }\times \text{ H}\\ \text{Volume = s }\times \text{ s }\times \text{ s }\\ {\text{Volume = s}}^3\\ \text{Density = }\frac{\text{mass}}{\text{volume}}\\ \text{mass = Denisty }\times \text{ volume}\\ \text{mass = d }\times {\text{s}}^3\\ \text{F = W = mg}\\ \text{W = d }\times {\text{s}}^3\times \text{g}\\ \text{W = d }\times \text{V}\times \text{g}\end{array}$

Conclusion:

Weight of water displaced by the cube is $\text{d }\times \text{V}\times \text{g}$