Solved Examples on Fluid Mechanics

Problem 1:-

The tension in a string holding a solid block below the surface of a liquid (of density greater than the solid) is T0 when the containing vessel (see below figure) is at rest. Show that the tension T, when the vessel has an upward vertical acceleration a, is given by T0 (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 m which is acting downward is

W = mg

Here,

Acceleration due to gravity at the point of observation is g.

The magnitude of the buoyant which acts upwards is

Fb = Vρg

Here,

Volume of the water displaced by the block is V

Density of the water is ρ.

The 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,

ΣF = 0

This makes

W – Fb – T0 = 0

Here,

Tension in the string when the vessel is at rest is T0.

Insert the values of the various terms involved in the above equation gives

Fb – W – T0 = 0

Vρg – mg – T= 0

So, T0 = Vρg – mg

This 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 figure which shows the forces acting on the block when the vessel is accelerating upward is

The buoyant forces acting on the block is

Fb,a = Vρ (g+a)

Thus, the net forces is given as

ΣF = ma

It is equal to the sum of the buoyant force, the weight of the block and tension (T).

Thus,

Fb,a – W -T = Ma

Substitute the values of Fb,a and W  in the above equation gives

Fb,a – W – T = ma

Vρ (g+a) – mg – T = ma

So, T = Vρa +Vρg – mg – ma

To obtain the tension in the string, substitute T0 = Vρg – mg in the above equation

T = Vρg – mg + Vρa – ma = T0 + a (Vρ – m)

Dividing both numerator and denominator in the second part in the right hand side of the equation by g gives

T = T0 + [a/g (Vρg – mg)]

  = T0 +[(a/g) T0] = T0 (1+a/g)

This represents the tension in the string when the vessel is accelerating upward.

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Problem 2:-

A cubic object of dimensions L = 0.608 m on a side and weight W = 4450 N in a vacuum is suspended by a wire in an open tank of liquid of density ρ = 944 kg/m3, as shown in below figure.

(a) Find the total downward force exerted by the liquid and the atmosphere on the top of the object. (b) Find the total upward force on the bottom of the object. (c)  Find the tension in the wire. (d) Calculate the buoyant force on the object using Archimedes principle. What relation exists among all these quantities?

Concept:-

The pressure acting on the top of the object is

pT = p0 +ρg (L/2) 

Here, atmospheric pressure is p0, density of the liquid is ρ, acceleration due to gravity is g and length of the side of the object is L.

The total downward force exerted by the liquid and the atmosphere on the top of the object is equal to pressure times the surface area of the object.

Thus,

FT = pTL= [p0 + ρg (L/2)] L2

The pressure acting on the bottom of the object is

pB = p0 +ρg (3L/2) 

The total upward force acting on the object is

FB = pB L= [p0 +ρg (3L/2)]L2 

According to Archimedes’ principle the buoyant force acting on the object is

FBuoyant = L3ρg

The tension in the wire is

T = W – FT + FB

Here, weight of the object is W.

Solution:-

(a) One atmospheric pressure is equal to 1.01×105 Pa

To obtain the force acting on the bottom of the object, substitute  1.01×105 Pa for p0, 944 kg/m3 for ρ  , 9.81 m/s2 for g and 0.608 m for L in the equation FT = [p0 + ρg (L/2)]L2,

FT = [p0 + ρg (L/2)]L2 

    = [1.01×105 Pa + (944 kg/m3) ( 9.81 m/s2) (0.608 m/2)] (0.608 m)2

    = (3.8376×104 kg.m/s2) [1 N/(1 kg.m/s2)] = 3.8376×104 N

Rounding off to three significant figures, the total downward force exerted by the liquid on the object is 3.8376×104 N.

(b) To obtain the total downward force exerted by the liquid, substitute  1.01×105 Pa for p0, 944 kg/m3 for ρ, 9.81 m/s2 for g , and 0.608 m for L in the equation FB = [p0 +ρg (3L/2)]L2,

FB = [p0 +ρg (3L/2)]L2

    =  [1.01×105 Pa + (944 kg/m3) ( 9.81 m/s2) (3(0.608 m)/2)] (0.608 m)2

   = (4.0455×104 kg.m/s2) [1 N/(1 kg.m/s2)] = 4.0455×104 N

Rounding off to three significant figures, the total downward force exerted by the liquid on the object is 4.0455×104 N.

(c) To obtain the buoyant force acting on the object, substitute 0.608 m for L, 944 kg/m3 for ρ, 9.81 m/s2 for g  in the equation FBuoyant = L3ρg,

FBuoyant = L3ρg

            = (0.608 m)3 (944 kg/m3) (9.81 m/s2)

           = (2.0793×103 kg.m/s2) [1 N/(1 kg.m/s2)] = 2.0793×103 N

Rounding off to three significant figures, the buoyant force acting on the object due to the liquid is 2.08×103 N .

(d) The various force acting on the object in the liquid is shown below:

To obtain the tension in the wire, substitute  4450 N for W, 3.84×10N for FT and 4.05×104 N for FB in the equation T = W – FT + FB,

T = W – FT + FB

  = 4450 N – 3.84×10N + 4.05×104 N = 2350 N

Rounding off to three significant figures, the tension in the wire is 2350 N.

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Problem 3:-

A cylindrical barrel has a narrow tube fixed to the top, as shown with dimensions in below figure. The vessel is filled with water to the top of the tube. Calculate the ratio of the hydrostatic force exerted on the bottom of the barrel to the weight of the water contained inside. Why is the ratio not equal to one? (Ignore the presence of the atmosphere.)

Concept:-

The pressure at the bottom of the cylindrical barrel is

p = ρgh

Here, density of the liquid is ρ, acceleration due to gravity is g and height of the cylindrical barrel is h.

The hydrostatic force acting at the bottom of the barrel is

F = PA

 Here, surface area of the barrel at the bottom is A.

The weight of the liquid is

W = ρgV

Here, total volume of the cylindrical barrel is V.

Solution:-

Now, the total height of the cylindrical barrel is

h = 1.8 m + 1.8 m

From above figure, substitute 3.6 m for h in the equation p = ρgh gives,

p = ρgh = ρg (3.6 m)

This gives the pressure at the bottom of the barrel.

The volume of the thin barrel at the top is

V1 = (4.6 cm2) (1.8 m)

    = (4.6 cm2) (10-2 m/1 cm)2 (1.8 m) = 8.24×10-4 m3

The radius of the barrel at the lower part is

r = 1.2 m/2 = 0.6 m

The volume of the barrel at the lower part is

V2 = πr2 (1.8 m)

    = (3.14) (0.6 m)2 (1.8 m) = 2.03472 m3

Now, the total volume of the cylindrical barrel is

V = V1 + V2

  = 8.24×10-4 m3 +  2.03472 m3 = 2.035548 m3

The surface area of the barrel at the bottom is

A =  πr2 = (3.14) (0.6 m)= 1.1304 m2

To obtain the hydrostatic force exerted by the liquid at the bottom, substitute  ρg (3.6 m) for p and 1.1304 m2 for A in the equation F = pA gives,

 F = pA

    = ρg (3.6 m) (1.1304 m2= ρg (4.06944 m3)

To obtain the weight of the liquid in the cylindrical barrel, substitute 2.035548 m3  for V in the equation W = ρgV gives,

 W = ρgV = ρg ( 2.035548 m3)

Now, the ratio of the hydrostatic force to the weight of the liquid is

F/W = [ρg (4.06944 m3)]/[ρg ( 2.035548 m3)] = 1.9992

Rounding off to two significant figures, the ratio of the hydrostatic force to the weight of the liquid is 2.0.

The hydrostatic pressure is depending on the height of the liquid column. So, the same amount of liquid when taken in different volume of containers having different heights will not be the same. Weight is the volume times the density of the liquid. Therefore, the ratio is not equal to one.

 

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