 | Sir Richard Phillips - Physics - 1832 - 286 pages
...other. Observation 1. The centre of gravity of a circle, sphere, or any regular body, is its centre. The centre of gravity of a parallelogram, is at the intersection of its diagonals. 2. To find the centre of gravity of a triangle, draw lines from any two angles to bisect... | |
 | John Francis Twisden - 1860 - 380 pages
...and DE DC AB : BC .'. (Ex aequali) DG DC GA : BC But DC=4 BC .'. DG = i GA=iDA. Ex. 191. -Show that the centre of gravity of a parallelogram is at the intersection of the diagonals. 47. Centre of gravity of solids. — The above method can easily be extended to the case... | |
 | John Francis Twisden - 1863 - 390 pages
...DEO is similar to ABO and EDC to ABC ; therefore and therefore (ex sequali) Ex. 232. — Show that the centre of gravity of a parallelogram is at the intersection of the diagonals. 65. Centre of Gravity of Solids. — The above method can easily be extended to the case... | |
 | John Francis Twisden - 1863 - 412 pages
...DEO is similar to ABO and EDC to ABC ; therefore and therefore (ex sequali) Ex. 232. — Show that the centre of gravity of a parallelogram is at the intersection of the diagonals. 65. Centre of Gravity of Solids. — The above method can easily be extended to the case... | |
 | Isaac Todhunter - Mechanics - 1867 - 368 pages
...direction of R makes with AC and CB are given. 6. See Art. 38, and Euclid, HI. 21, 22. 8. The point must be at the intersection of the straight lines which join the middle points of opposite sides. 9. The forces 1 and «y3 are at right angles; the forces 2 and 1 at 120°. 11. Let CD be the resultant... | |
 | John Francis Twisden - Mechanics - 1868 - 364 pages
...: BC therefore (ex sequali) DO : DC :: OA : BC But DC = JBC.'. DG = £G\=IDA. Ex. 262.—Show that the centre of gravity of a parallelogram is at the intersection of the diagonals. 65. Centre of Gravity of Solids.—The above method can easily be extended to the case of... | |
 | William Guy Peck - Mechanics - 1870 - 322 pages
...similar reason, the centre of gravity is on it ; it is, therefore, at G, their point of intersection. Hence, the centre of gravity of a parallelogram is at the intersection of two straight lines joining the middle points of the opposite sides. The diagonals of a parallelogram... | |
 | William Chauvenet - Geometry - 1871 - 380 pages
...parallelogram whose perimeter is equal to the sum of the diagonals of the quadrilateral (I. 122). 24. The intersection of the straight lines which join the middle points of opposite sides of any quadrilateral, is the middle point of the straight line which joins the middle points of the... | |
 | William Chauvenet - Geometry - 1872 - 382 pages
...parallelogram whose perimeter is equal to the sum of the diagonals of the quadrilateral (I. 122). 24. The intersection of the straight lines which join the middle points of opposite sides of any quadrilateral, is the middle point of the straight line which joins the middle points of the... | |
 | George Wightwick - 1875 - 394 pages
...it at L, so that AL may be to LB as the weight at B is to that at A. A Parallelogram. — The centre of a parallelogram is at the intersection of the straight...lines which join the middle points of opposite sides, or of the diagonals. A Triangle. — The centre of gravity of a triangle is found by following rule... | |
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The intersection of the straight lines which join the middle points of opposite sides of any quadrilateral, is the middle point of the straight line which joins the middle points of the diagonals (I. 
