 Suppose we lower a cube of some material into a beaker of water:  We may have to hold it in place. Topic 2.2 Extended K – Fluid statics : Archimedes’

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Presentation transcript:

 Suppose we lower a cube of some material into a beaker of water:  We may have to hold it in place. Topic 2.2 Extended K – Fluid statics : Archimedes’ Principle  Note that all of the horizontal forces cancel each other out.  We can calculate the pressure on the bottom and the top of the cube: Question: Why is the lower force bigger than the upper force? h1h1 h2h2 p 1 =  gh 1 p 2 =  gh 2 F 1 = p 1 A F 2 = p 2 A F 1 =  gh 1 A F 2 =  gh 2 A  The buoyant force F b is the net upward force acting on the cube, and is the difference between the two forces: F b = F 2 - F 1 F1F1 F2F2 =  gh 2 A -  gh 1 A =  g(h 2 -h 1 )A F b =  gV f Archimedes' Principle "A body immersed wholly or partially in a fluid is buoyed up by a force equal in magnitude to the weight of the volume of liquid displaced." FYI: Archimedes lived from 287 to 212 BC. A RCHIMEDES' P RINCIPLE

A 20-N object is lowered into an unknown liquid in a graduated cylinder using a spring scale. While in the liquid the spring scale reads 16.5 N. The initial volume of the liquid is 50 mL. When the object is completely submerged the volume reads 125 mL. Topic 2.2 Extended K – Fluid statics : Archimedes’ Principle A RCHIMEDES' P RINCIPLE (a) What is the buoyant force acting on the object? F b = (b) What is volume of the object in m 3 ? V f = = 75 mL V f = 75 mL 1 cm 3 1 mL 1 m 100 cm 3 = 7.5  m 3 FYI: The volume of the object is the same as the volume of the liquid displaced. (c) What is density of the object?  = m/V = 2 kg/7.5  m 3 = kg/m 3 (d) What is density of the fluid? F b =  gV f   = F b /gV f = 3.5/(10)(7.5  ) = 4667 kg/m 3 = 3.5 N.