Fluid Mechanics
Fluid Mechanics
8th Edition
ISBN: 9780073398273
Author: Frank M. White
Publisher: McGraw-Hill Education
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Chapter 2, Problem 2.95P
To determine

(a)

The specific gravity of the body if h=0.

Expert Solution
Check Mark

Answer to Problem 2.95P

SG=32

Explanation of Solution

Given information:

The width of uniform body is b. It’s in static equilibrium, when pivoted about point O.

Fluid Mechanics, Chapter 2, Problem 2.95P , additional homework tip  1

Assume that the horizontal force acting on the panel is F and can be given as below.

F=γAh1

In above equation,

γ - Specific weight of water

A - Area which the horizontal force is acting on

h1 - Distance between the top water level to the center of gravity of the panel (DG)

Assume that the vertical force acting on the panel is F and can be given as below:

F=γ(Vol)aboveOC

=γAh

In above equation,

γ - Specific weight of water

A - Area above the panel

h - Width of the panel

The below equation is used to find the line of action of horizontal force F

y=IAh1

In above equation,

y - Distance from the center of gravity (G) to the line of action.

I - Inertia of the plane which the horizontal force is acting on.

A - Area which the horizontal force is acting on

h1 - Distance between the top water level to the center of gravity of the panel (DG)

Calculation:

To find horizontal force:

According to the above-mentioned explanation:

F=γAh1=γ(Rb)(R2)=γ( R 2 b2)

Therefore, the horizontal component of the hydrostatic force = γ(R2b2)

To find the line of action,

y=IAh1

In this case, the inertia of the rectangular plane is equal to,

I=(112bh3)

Therefore,

y=IAh1=( 1 12 ( b ) ( R ) 3 )( Rb)( R 2 )=(R6)

The horizontal force is acting at a distance (R2)(R6)=(R3) from point O.

To find vertical force,

According to the above-mentioned explanation:

F=γ(Vol)aboveBC

=γAh

=γh(AreaofOCD)=γb( π ( R ) 2 4)=( π R 2 γb4)

The vertical force acts at a distance equals to 4R3π from left of point O.

Fluid Mechanics, Chapter 2, Problem 2.95P , additional homework tip  2

According to above diagram:

The body to be in static equilibrium, the moment about point O is equal to zero.

M O =0

F H ( R 3 )+ F V ( 4R 3π ) W A ( R 2 ) W B ( 4R 3π )=0

γ( R 2 b 2 )( R 3 )+( π R 2 γb 4 )( 4R 3π ) γ s ( Rhb )( R 2 ) γ s ( π R 2 b 4 )( 4R 3π )=0

γ( R 3 b 6 )+γ( R 3 b 3 ) γ s ( R 2 bh 2 ) γ s ( R 3 b 3 )=0

γ( R 3 b 2 )= γ s R 2 b( h 2 + R 3 )

( γ s γ )=( ( R 2 ) ( h 2 + R 3 ) )

( γ s γ )= ( h R + 2 3 ) 1

forh=0,SG=32

Conclusion:

The specific gravity of uniform body, when h=0 is equal to SG=32.

To determine

(b)

The specific gravity of the body if h=R.

Expert Solution
Check Mark

Answer to Problem 2.95P

SG=35

Explanation of Solution

Given information:

The width of uniform body is b. It’s in static equilibrium when pivoted about point O.

Fluid Mechanics, Chapter 2, Problem 2.95P , additional homework tip  3

Assume that the horizontal force acting on the panel is F and can be given as below:

F=γAh1

In above equation,

γ - Specific weight of water

A - Area which the horizontal force is acting on

h1 - Distance between the top water level to the center of gravity of the panel (DG)

Assume that the vertical force acting on the panel is F and can be given as below:

F=γ(Vol)aboveOC

=γAh

In above equation,

γ - Specific weight of water

A - Area above the panel

h - Width of the panel

The below equation is used to find the line of action of horizontal force F :

y=IAh1

In above equation:

y - Distance from the center of gravity (G) to the line of action.

I - Inertia of the plane which the horizontal force is acting on.

A - Area which the horizontal force is acting on

h1 - Distance between the top water level to the center of gravity of the panel (DG)

Calculation:

To find horizontal force,

According to the above-mentioned explanation:

F=γAh1=γ(Rb)(R2)=γ( R 2 b2)

Therefore, the horizontal component of the hydrostatic force = γ(R2b2)

To find the line of action:

y=IAh1

In this case, the inertia of the rectangular plane is equal to:

I=(112bh3)

Therefore,

y=IAh1=( 1 12 ( b ) ( R ) 3 )( Rb)( R 2 )=(R6)

The horizontal force is acting at a distance (R2)(R6)=(R3) from point O.

To find vertical force,

According to the above-mentioned explanation:

F=γ(Vol)aboveBC

=γAh

=γh(AreaofOCD)=γb( π ( R ) 2 4)=( π R 2 γb4)

The vertical force acts at a distance equals to 4R3π from left of point O.

Fluid Mechanics, Chapter 2, Problem 2.95P , additional homework tip  4

According to above diagram:

The body to be in static equilibrium, the moment about point O is equal to zero.

M O =0

F H ( R 3 )+ F V ( 4R 3π ) W A ( R 2 ) W B ( 4R 3π )=0

γ( R 2 b 2 )( R 3 )+( π R 2 γb 4 )( 4R 3π ) γ s ( Rhb )( R 2 ) γ s ( π R 2 b 4 )( 4R 3π )=0

γ( R 3 b 6 )+γ( R 3 b 3 ) γ s ( R 2 bh 2 ) γ s ( R 3 b 3 )=0

γ( R 3 b 2 )= γ s R 2 b( h 2 + R 3 )

( γ s γ )=( ( R 2 ) ( h 2 + R 3 ) )

( γ s γ )= ( h R + 2 3 ) 1

forh=R,SG=35

Conclusion:

The specific gravity of uniform body, when h=R is equal to SG=35.

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Chapter 2 Solutions

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