## 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 1848-1851Macmillan & Company, 1851 - 116 pages |

### From inside the book

Results 1-5 of 8

Page 45

...

...

**string**attached to the lower end of the rod and to a point in the wall . Find by geometrical construction the point in the wall to which the**string**must be attached . Let AB ( fig . 33 ) be the rod STATICS . 45. Page 46

...

...

**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 parallel ) in order that there may be equilibrium . We ... Page 54

...

...

**string**whose direction bisects the angle . The same**string**passes round all the pullies and is solicited by a certain force . Shew that the numbers of the**strings**between the pullies are as cos 4 : cos B : cos C. If n be the number of ... Page 55

Francis James Jameson. the

Francis James Jameson. the

**string**passing round all the pullies . The directions of these three tensions , by hypothesis , bisect the angles of the triangle , and therefore meet in a point . Let them meet in O ( fig . 43 ) . Then ... Page 110

...

...

**string**, ( 2 ) by means of a rigid rod ; what will be the conditions of equilibrium in the two cases , when given forces act in the direction of the**string**or rod , and perpen- dicular to it ? ( 4 ) . Equilibrium takes place on a ...### 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.