Solved Examples on Deformable Bodies and Elastic Deformation:-

Question 1:-The tension in a string holding a solid block below the surface of a liquid (of density greater than the solid) is

T_{0}when the containing vessel in the below figure is at rest.Show that the tension

T, when the vessel has an upward vertical accelerationa, is given byT_{0}(1+a/g).

Concept:-The various forces acting on the block when it is inside the water are the weight, the buoyant force and the tension in the string. When the vessel is at rest, there is no net force acting on the block. The sum of the three forces is equal to zero when the vessel is at rest.

When the vessel moves upward with an acceleration, the buoyant force and the tension in the string will be differ. There is a net force acting vessel. Thus, the sum of the three forces is equal to the net force acting on the block.

Solution:-The weight of the block having mass

mwhich is acting downward is

W=mgHere,

Acceleration due to gravity at the point of observation is

gThe magnitude of the buoyant which acts upwards is

F_{b}=VρgHere,

Volume of the water displaced by the block is

VDensity of the water is

ρThe various forces acting on the block which is placed inside the vessel is shown below:

As the vessel is rest, the three forces are in equilibrium. Thus, the sum of the three forces is equal to zero.

Thus,

∑

F= 0This makes

W-F_{b}-T_{0}= 0Here,

Tension in the string when the vessel is at rest is

T_{0}.Insert the values of the various terms involved in the above equation gives

F_{b}-W-T_{0}= 0

Vρg-mg-T_{0}= 0

T_{0}=Vρg-mgThis represents the tension in the string when the vessel is at rest.

When the vessel is moving with an upward vertical acceleration

a, there is net forces acting on the block. So, there is change in the buoyant force and the tension in the string.The buoyant forces acting on the block is

F_{b,a}=Vρ(g+a)Thus, the net forces is given as

∑

F=maIt is equal to the sum of the buoyant force, the weight of the block and tension (

T).Thus,

F_{b,a}-W-T=maSubstitute the values of

F_{b,a}andWin the above equation gives

F_{b,a}-W-T=ma

Vρ(a+g) –mg-T=ma

T = Vρa+ Vρg -mg-maTo obtain the tension in the string, substitute

T_{0}=Vρg-mgin the above equation.

T = Vρg - mg+Vρa -ma=

T_{0}+a(Vρ-m)Dividing both numerator and denominator in the second part in the right hand side of the equation by

ggives

T=T_{0}+[(a/g) (Vρg–mg)]=

T_{0}+[(a/g)T_{0}]=

T_{0}[1+ (a/g)]This represents the tension in the string when the vessel is accelerating upward.

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Question 2:-Using a soap solution for which the surface tension is 0.025N/m, a child blows a soap bubble of radius 1.40 cm. How much energy is expended in stretching the soap surface?

Concept:-Surface tension is the surface potential energy per unit area of surface.

The surface area of the soap bubble blown by the child having radius r is

A= 4πr^{2}The surface tension of the soap bubble is

γ=U/2AThe factor 2 is due to the presence of two surfaces of the soap bubble.

Substitute

A=4πr^{2}in the above equation, the surface tension becomes

γ=U/2A=

U/[2(4πr^{2})]=

U/8πr^{2}

Solution:-From the above equation, the energy stored in the surface of the soap bubble is

U= 8γπr^{2}This much of energy is used in stretching the soap surface.

To obtain required energy, substitute 1.40 cm for the radius

rof the soap bubble, 0.025 N/m for the surface tension of the soap bubbleγin the equationU= 8γπr^{2}

U= 8γπr^{2}

^{ }= 8(0.025 N/m) (3.14) (1.40 cm)^{2}(10^{-2}m/1 cm)^{2}= (1.2308810

^{-4 }N.m) (1 J/1 N.m)= 1.2308810

^{-4 }JRounding off to three significant figures, the energy expend in stretching the soap surface is 1.23×10

^{-4}J._____________________________________________________________________________________

Question 3:-In 1654 Otto von Guericke, Burgermeister of Magdeburg and inventor of the air pump, gave a demonstration before the Imperial Diet in which two teams of horses could not pull apart two evacuated brass hemispheres.

(a)Show that the forceFrequired to pull apart the hemispheres isF= πR^{2}Δp, whereRis the (outside) radius of the hemispheres and Δpis the difference in pressure outside and inside the sphere in the below figure.(b)TakingRequal to 0.305 m and the inside pressure as 0.100 atm, what force would the team of horses have had to exert to pull apart the hemispheres?(c)Why were two teams of horses used? Would not one team prove the point just as well?

Concept:-The two brass hemispheres with an open flat can be replaced with two hemispheres with a closed flat end.

The force required to pull apart the hemispheres is equal to pressure times the surface area.

Solution:-(a) It is given that the pressure difference between outside and inside the sphere is Δ

p. The radius of the circle formed by the half hemispheres which face each other isR.Now, the surface area of the hemisphere is

A= πR^{2}The force required to pull the apart the brass hemisphere is

F= ΔpA

=Δp(πR^{2})= π

R^{2}ΔpTherefore, the force required to pull apart the two hemispheres is π

R^{2}Δp.(b) It is given that the inside pressure is 0.11 atm.Thus, the pressure difference between the inside and outside of the hemispheres is

Δ

p= 1.00 atm – 0.11 atm= 0.89 atm

Substitute 0.305 m for

Rand 0.89 atm for Δpin the equationF= πR^{2}Δp,F = π

R^{2}Δp

=(3.14) (0.305 m)^{2}(1.01×10^{5}Pa/1 atm) (1 kg/m.s^{2}/1 Pa)= (2.5997×10

^{4}kg.m/s^{2}) (1 N/1 kg.m/s^{2})= 2.5997×10

^{4 }NRounding off to three significant figures, the force required to pull apart the two hemispheres by the team of horses is 2.60×10

^{4}N.(c) The two teams of horses were used so as to pull apart the brass hemispheres. They could not do so as the pressure difference is too high. For such pressure difference, large amount of force is required.