A Collection of Problems in Illustration of the Principles of Theoretical Hydrostatics and Hydrodynamics |
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Page 49
... displaced be equal to that of the solid , and that the line joining the centres of gravity of the solid and fluid dis- placed be vertical . The discovery of the principles of the equilibrium of floating bodies is due to Archimedes ; his ...
... displaced be equal to that of the solid , and that the line joining the centres of gravity of the solid and fluid dis- placed be vertical . The discovery of the principles of the equilibrium of floating bodies is due to Archimedes ; his ...
Page 50
... displaced fluid being equal to that of the lamina , we must have or and therefore Again , σ . PBCDQ = p . a3 , σ ( a3 − { xy ) = pa3 , - zy = 2 ( 1 - £ ) a xy σ • PE2 = x2 + AE2 · 2x AE cos - QE2 = y2 + AE ” - 2y AE cos and therefore ...
... displaced fluid being equal to that of the lamina , we must have or and therefore Again , σ . PBCDQ = p . a3 , σ ( a3 − { xy ) = pa3 , - zy = 2 ( 1 - £ ) a xy σ • PE2 = x2 + AE2 · 2x AE cos - QE2 = y2 + AE ” - 2y AE cos and therefore ...
Page 51
... displaced , AG = { AD , AH = { AE , and therefore AG : AD : AH : AE ; which shews that GH is parallel to DE . But GH must be vertical and therefore at right angles to B'C ' : hence DE is at right angles to B'C ' . Since △ DEB = L DEC ...
... displaced , AG = { AD , AH = { AE , and therefore AG : AD : AH : AE ; which shews that GH is parallel to DE . But GH must be vertical and therefore at right angles to B'C ' : hence DE is at right angles to B'C ' . Since △ DEB = L DEC ...
Page 54
... displaced , which will lie in a diameter VP through a point P , the tangent of which is horizontal . In order that the lamina may be at rest , HG must be vertical . Draw PE vertical to meet AD in E , and 54 EQUILIBRIUM OF SOLID BODIES ...
... displaced , which will lie in a diameter VP through a point P , the tangent of which is horizontal . In order that the lamina may be at rest , HG must be vertical . Draw PE vertical to meet AD in E , and 54 EQUILIBRIUM OF SOLID BODIES ...
Page 60
... displaced , 2pKgh = 3pKg . } h + 5pKgx , 2h = 3h + 5x 5x = h , x = ih . This result shews that half the axis of the cylinder is im- mersed in fluid , or that its middle point is in the surface of the upper fluid . 2. A circular lamina ...
... displaced , 2pKgh = 3pKg . } h + 5pKgx , 2h = 3h + 5x 5x = h , x = ih . This result shews that half the axis of the cylinder is im- mersed in fluid , or that its middle point is in the surface of the upper fluid . 2. A circular lamina ...
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Common terms and phrases
Albert Girard altitude angular velocity aperture axem axes axis vertical base Bernoulli body Bossut Bossut Traité Cambridge centre of gravity centre of pressure coinciding cone cos² cube Daniel Bernoulli denote the density denote the length depth determine distance dx dy Edition efflux elastic equal equation filled with fluid floating fluid displaced force free surface given hemisphere Hence humidum Hydrodynamique immersed vertically inclination incompressible fluid James Bernoulli lamina late Fellow latus rectum motion parabola parallel particles portion position of equilibrium position of rest problem quæ ratio required pressure resultant pressure right angles sesquialterum sin² small orifice solid cylinder specific gravity sphere supposed tangent Treatise triangle Trinity College unit of pressure University of Cambridge vertex vertical line vessel volume weight whole pressure
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