A Collection of Problems in Illustration of the Principles of Theoretical Hydrostatics and Hydrodynamics |
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Page 49
... fluid 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 ...
... fluid 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 ...
Page 50
... fluid . Then , the weight of the displaced fluid being equal to that of the lamina , we must have or and therefore Again , and therefore , since but σ . PBCDQ = p . a2 , - σ ( a2 − { xy ) = pa3 , xy = y - 2 ( 1-2 ) . α σ .. ( 1 ) . π ...
... fluid . Then , the weight of the displaced fluid being equal to that of the lamina , we must have or and therefore Again , and therefore , since but σ . PBCDQ = p . a2 , - σ ( a2 − { xy ) = pa3 , xy = y - 2 ( 1-2 ) . α σ .. ( 1 ) . π ...
Page 51
... fluid 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 DEBL DEC ...
... fluid 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 DEBL DEC ...
Page 54
... fluid , if the triangle be equilateral , provided that σ 7 be greater than ... fluid with its vertex A immersed , the plane of the lamina being vertical ... displaced , which will lie in a diameter VP through a point P , the tangent ...
... fluid , if the triangle be equilateral , provided that σ 7 be greater than ... fluid with its vertex A immersed , the plane of the lamina being vertical ... displaced , which will lie in a diameter VP through a point P , the tangent ...
Page 60
... fluid , and K the area of its section . Then , the weight of the cylinder being equal to that of the fluid displaced , 2pKgh = 3pKg . } h + 5pKgx , 2h = 3h + 5x 5x = { h , x = \ h . This result shews that half the axis of the cylinder ...
... fluid , and K the area of its section . Then , the weight of the cylinder being equal to that of the fluid displaced , 2pKgh = 3pKg . } h + 5pKgx , 2h = 3h + 5x 5x = { h , x = \ h . This result shews that half the axis of the cylinder ...
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Common terms and phrases
ABCD Albert Girard altitude angular velocity aperture axis axis of revolution axis vertical base Bernoulli Bossut Traité Cambridge centre of gravity centre of pressure cone constant quantity cos² Daniel Bernoulli denote the density denote the length denote the radius depth determine distance dx dy Edition elastic equal equation filled with fluid floating fluid displaced free surface given hemisphere Hence humidum inclination incompressible fluid integrating James Bernoulli John Bernoulli lamina latus rectum mass motion oscillation parabola paraboloid of revolution parallel particles Phoronomia portion position of equilibrium position of rest problem ratio represent right angles sesquialterum sin² small orifice solid cylinder specific gravity sphere supposed Traité d'Hydrodynamique Treatise triangle tube vertex volume w³y² weight whole pressure
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