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FLUID MECHANICS - CASE STUDY SOLUTION


X-Component of the Resultant Force

 

In order to determine the magnitude of the resultant force exerted on the curved surface AB, first find its force components FRx and FRy.

The horizontal projection of the curved surface AB is the plane area AC. The x-component of the resultant force is given by the normal force acting on this plane area. That is,

     FRx = ρghc AAC
           = (1000 kg/m3) (9.8 m/s2) (4.5 m) (3 m) (8 m)
           = 1,058 kN

Note that hc is the vertical distance to the centroid of plane area AC.

     

Y-Component of the Resultant Force
 

The y-component of the resultant force is the weight of the water directly above the curved surface (i.e., imaginary volume ABEF).

     FRy = ρg VolABEF = ρg (VolADEF + VolABD)
           = (1000 kg/m3) (9.8 m/s2) [(3 m) (3 m) (8 m)
                 + (π 32/4) m2 (8 m) ]
           = 1,260 kN

Hence, the resultant force is given by

     FR = (FRx2 + FRy2)0.5
          = [(1,058 kN)2 + (1,260 kN)2]0.5
          = 1,645 kN

     

Resultant Force and its Line of Action
 

And the angle θ is given by

     θ = tan -1 (FRy / FRx)
        = tan -1 (1,260 kN / 1,058 kN)
        = 50o

Also, the resultant force FR has to pass through point D since all the pressure forces are perpendicular to the curved surface.

     
   
 
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