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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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