75_Problem CHAPTER 9

75_Problem CHAPTER 9 - PROBLEM 9.66 Show that the resultant of the hydrostatic forces acting on a submerged plane area A is a force P perpendicular

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PROBLEM 9.66 Show that the resultant of the hydrostatic forces acting on a submerged plane area A is a force P perpendicular to the area and of magnitude sin P Ay pA γ θ == , where is the specific weight of the liquid and p is the pressure at the centroid C of the area. Show that P is applied at a point , p C called the center of pressure, whose coordinates are / px y x IA y = and / yI A y = , where xy I xy dA = (see Sec. 9.8). Show also that the difference of ordinates p yy is equal to 2 / x ky and thus depends upon the depth at which the area is submerged. SOLUTION The pressure p at an arbitrary depth ( ) sin y is ( ) sin py = so that the hydrostatic force dP exerted on an infinitesimal area dA is ( ) sin dP y
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This note was uploaded on 07/11/2011 for the course EGM 2511 taught by Professor Jenkins during the Spring '08 term at University of Florida.

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