The Principles of the Solution of the Senate-house 'riders,' Exemplified by the Solution of Those Proposed in the Earlier Parts of the Examinations of the Years 1848-1851 |
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Page 45
... rest , prove that their lines of action are either parallel or pass through a point ; and in both cases , shew that any two of the forces are inversely proportional to the perpendiculars drawn on their respective lines of action from ...
... rest , prove that their lines of action are either parallel or pass through a point ; and in both cases , shew that any two of the forces are inversely proportional to the perpendiculars drawn on their respective lines of action from ...
Page 46
... rest are and its weight acting along GW , tension of string along BP , reaction of wall along ARI NP . Now the principle of which ( 4 ) is the enunciation tells us that these three forces must meet in a point ( because they are not ...
... rest are and its weight acting along GW , tension of string along BP , reaction of wall along ARI NP . Now the principle of which ( 4 ) is the enunciation tells us that these three forces must meet in a point ( because they are not ...
Page 48
... rests in equi- librium with its base on a horizontal plane sufficiently rough to prevent all sliding . A force acts upon it in its own plane and in a given line drawn through the vertex and without the triangle : find by a geometrical ...
... rests in equi- librium with its base on a horizontal plane sufficiently rough to prevent all sliding . A force acts upon it in its own plane and in a given line drawn through the vertex and without the triangle : find by a geometrical ...
Page 52
... rest supported on it , find the least force along the plane requisite to drag the body up . In this case we have an extension of the problem ( 4 ) to the case where the plane is rough , the roughness being given by the fact that a body ...
... rest supported on it , find the least force along the plane requisite to drag the body up . In this case we have an extension of the problem ( 4 ) to the case where the plane is rough , the roughness being given by the fact that a body ...
Page 53
... rests on a smooth horizontal table , find the horizontal force which must act on the wedge to keep it at rest . From the equilibrium of the particle O ( fig . 41 ) we have , by the same method as is pursued in ( 4 ) , R = W.cosa ( 1 ) ...
... rests on a smooth horizontal table , find the horizontal force which must act on the wedge to keep it at rest . From the equilibrium of the particle O ( fig . 41 ) we have , by the same method as is pursued in ( 4 ) , R = W.cosa ( 1 ) ...
Other editions - View all
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2018 |
The Principles of the Solution of the Senate-House 'Riders: Exemplified by ... Francis J. Jameson No preview available - 2015 |
Common terms and phrases
AC² AN.NM Arithmetic arithmetical progression axis bisects body C₁ Cambridge centre of gravity chord CHURCHILL BABINGTON circle cloth cone Conic Sections conjugate hyperbola constant curvature curve cycloid describe diameter direction directrix distance drawn Edition ellipse equations equilibrium Fellow of St fluid focus geometrical given point Hence horizontal hyperbola inches inclined inscribed John's College joining latus-rectum least common multiple Lemma length locus meet mirror move number of seconds oscillation parabola parallel parallelogram particle perpendicular plane polygon pressure prop proportional proposition prove pullies quadrilateral quantity radius ratio rays rectangle refraction right angles sewed shew sides specific gravity spherical square straight line string surface tan² tangent triangle ABC Trinity College tube V₁ vary vertex vertical W₁ weight
Popular passages
Page 4 - To divide a given straight line into two parts, so that the rectangle contained by the whole and one of the parts, shall be equal to the square of the other part.
Page 6 - The angle at the centre of a circle is double of the angle at the circumference upon the same base, that is, upon the same part of the circumference.
Page 11 - AB is a diameter, and P any point in the circumference of a circle; AP and BP are joined and produced if necessary ; if from any point C of AB, a perpendicular be drawn to AB meeting AP and .BP in points D and E respectively, and the circumference of the circle in a point F, shew that CD is a third proportional of CE and CF.
Page 9 - IF the angle of a triangle be divided into two equal angles, by a straight line which also cuts the base; the segments of the base shall have the same ratio which the other sides of the triangle have to one another...
Page 4 - In any right-angled triangle, the square which is described upon the side subtending the right angle, is equal to the squares described upon the sides which contain the right angle.