 Ch 2. Fluid Statics Multimedia Engineering Fluids PressureVariation PressureMeasurement HydrostaticForce(Plane) HydrostaticForce(Curved) Buoyancy
 Chapter 1. Basics 2. Fluid Statics 3. Kinematics 4. Laws (Integral) 5. Laws (Diff.) 6. Modeling/Similitude 7. Inviscid 8. Viscous 9. External Flow 10. Open-Channel Appendix Basic Math Units Basic Equations Water/Air Tables Sections Search eBooks Dynamics Fluids Math Mechanics Statics Thermodynamics Author(s): Chean Chin Ngo Kurt Gramoll ©Kurt Gramoll FLUID MECHANICS - CASE STUDY SOLUTION Force Equilibrium Submerged Water Line The total weight of the canoe and the paddlers is      W = 120 lb + 630 lb = 750 lb For equilibrium, the total weight must be balanced by the buoyant force. That is,      W = FB where the buoyant force is the weight of the liquid displaced by the volume and is given by      FB = ρgVdisplaced The submerged waterline is denoted as h and the displaced volume for each design will be expressed in terms of h as follows: For Design Layout A: Displaced Volume of Design Layout A The displaced volume of layout A is given by      Vdisplaced = 2 (10 + h) h Applying force equilibrium yields      750 = (1.94) (32.2) (2) (10 + h)h      h2 + 10h - 6 = 0 Using the Quadratic Equation gives,      h = 0.5678 ft The submerged height h (0.568 ft) is less than the height of the canoe (1 ft), hence this canoe will float. For Design Layout B: Displaced Volume of Design Layout B The displaced volume of layout B is given by      Vdisplaced = 2 (11 + h) h Applying force equilibrium yields      750 = (1.94) (32.2) (2) (11 + h)h      h2 + 11h - 6 = 0      h = 0.5208 ft The submerged height h (0.521 ft) is greater than the height of the canoe (0.5 ft), hence this canoe will sink.

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