Explanation Given information: The arbitrary depth d = y sin θ . Calculation: Calculate the moment of inertia about x axis as follows: I x = ∫ y 2 d A Calculate the moment of inertia about y axis as follows: I y = ∫ x y d A Find the pressure ( p ) at an arbitrary depth as shown below: p = γ ( y sin θ ) Find the hydrostatic force ( d P ) exerted on infinitesimal area as follows: d P = ( γ y sin θ ) d A Find the magnitude P resulting force acting on the plane as follows. P = ∫ d P Substitute ( γ y sin θ ) d A for d P . P = ∫ ( γ y sin θ ) d A = γ y ¯ sin θ ∫ d A = γ A y ¯ sin θ p ¯ = γ y ¯ sin θ P = p ¯ A Thus, the resultant of hydrostatic force acting on submerged plane is P = p ¯ A _ Apply the Equivalence of the force and couple and the system of infinitesimal hydrostatic force requires as follows. Show that the coordinate y p as shown below: Summation of moment about x axis. ∑ M x = 0 − y ¯ p P = − ∫ ( y d P ) Substitute ( γ y sin θ ) d A for dP . ∫ ( y d P ) = ∫ y ( γ y sin θ ) d A = γ sin θ ∫ y 2 d A = ( γ sin θ ) I x y p P = ( γ sin θ ) I x y p = ( γ sin θ ) I x γ sin θ ( y ¯ A ) y P = I x A y ¯ Thus, the center of pressure co-ordinate is y P = I x A y ¯ _ . Show that the coordinate x p as shown below: ∑ M y = 0 x p P = ∫ x d P ∫ x d P = ∫ x ( γ y sin θ ) d A = γ sin θ ∫ x y d A Substitute I x y for ∫ x y d A

Loose Leaf for Vector Mechanics for Engineers: Statics and Dynamics

12th Edition

ISBN: 9781259977206

Author: BEER, Ferdinand P., Johnston Jr., E. Russell, Mazurek, David, Cornwell, Phillip J., SELF, Brian

Publisher: McGraw-Hill Education

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Chapter 9.2, Problem 9.66P

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Show the resultant of hydrostatic force acting on a submerged plane.

Show the force P is applied at a point CP called the center of pressure whose coordinates are xP=Ixy/Ay¯ and yP=Ix/Ay¯

Show that the difference of coordinates yP−y is equal to k¯2x′/y¯.

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Chapter 9.2, Problem 9.65P

Chapter 9.3, Problem 9.67P

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Loose Leaf for Vector Mechanics for Engineers: Statics and Dynamics

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Show the resultant of hydrostatic force acting on a submerged plane. Show the force P is applied at a point C P called the center of pressure whose coordinates are x P = I x y / A y ¯ and y P = I x / A y ¯ Show that the difference of coordinates y P − y is equal to k ¯ 2 x ′ / y ¯ .

Solution Summary: The author shows the resultant of hydrostatic force acting on a submerged plane.

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Show the resultant of hydrostatic force acting on a submerged plane.

Show the force P is applied at a point CP called the center of pressure whose coordinates are xP=Ixy/Ay¯ and yP=Ix/Ay¯

Show that the difference of coordinates yP−y is equal to k¯2x′/y¯.

Want to see the full answer?

Chapter 9 Solutions

Show the resultant of hydrostatic force acting on a submerged plane. Show the force P is applied at a point C P called the center of pressure whose coordinates are x P = I x y / A y ¯ and y P = I x / A y ¯ Show that the difference of coordinates y P − y is equal to k ¯ 2 x ′ / y ¯ .

Solution Summary: The author shows the resultant of hydrostatic force acting on a submerged plane.

Concept explainers

Videos

Show the resultant of hydrostatic force acting on a submerged plane. Show the force P is applied at a point CP called the center of pressure whose coordinates are xP=Ixy/Ay¯ and yP=Ix/Ay¯ Show that the difference of coordinates yP−y is equal to k¯2x′/y¯.

Want to see the full answer?

Chapter 9 Solutions

Show the resultant of hydrostatic force acting on a submerged plane.

Show the force P is applied at a point CP called the center of pressure whose coordinates are xP=Ixy/Ay¯ and yP=Ix/Ay¯

Show that the difference of coordinates yP−y is equal to k¯2x′/y¯.