A Collection of Problems in Illustration of the Principles of Elementary Mechanics |
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Page 3
... same parts , and are respectively proportional to these sides : to prove that the resultant will pass through the intersection of the diagonals . Let ABCD , fig . ( 6 ) , be RESULTANT OF TWO FORCES ACTING ON A POINT . 3.
... same parts , and are respectively proportional to these sides : to prove that the resultant will pass through the intersection of the diagonals . Let ABCD , fig . ( 6 ) , be RESULTANT OF TWO FORCES ACTING ON A POINT . 3.
Page 5
... intersection of the two other sides . SECT . 2. Trigonometrical Method . Let two forces P , Q , act upon a point , and let a be the angle between their directions : then , if R be their resultant , and if R's direction be inclined to ...
... intersection of the two other sides . SECT . 2. Trigonometrical Method . Let two forces P , Q , act upon a point , and let a be the angle between their directions : then , if R be their resultant , and if R's direction be inclined to ...
Page 12
... intersection of W's direction . with the horizontal line through any point A of the plane , draw EF at right angles to the plane . Then , since P , W , R , produce equilibrium , they must be proportional to the sides FO , OE , EF ...
... intersection of W's direction . with the horizontal line through any point A of the plane , draw EF at right angles to the plane . Then , since P , W , R , produce equilibrium , they must be proportional to the sides FO , OE , EF ...
Page 15
... intersect in the point 0 , fig . ( 15 ) . Then , the forces being in equilibrium , PQR sin Q , R : sin R , P : sin P , Q. B + C π A But < Q , R = < BOC = π 2 and therefore sin Q , R = cos = + 2 2 B similarly , sin R , P = cos and ...
... intersect in the point 0 , fig . ( 15 ) . Then , the forces being in equilibrium , PQR sin Q , R : sin R , P : sin P , Q. B + C π A But < Q , R = < BOC = π 2 and therefore sin Q , R = cos = + 2 2 B similarly , sin R , P = cos and ...
Page 19
... intersection of lines drawn from the angles of a triangle ABC to bisect opposite sides , to prove that the point will be in equilibrium . Let P , Q , R , be the points of bisection of the sides : then the resultant of the forces OA , OB ...
... intersection of lines drawn from the angles of a triangle ABC to bisect opposite sides , to prove that the point will be in equilibrium . Let P , Q , R , be the points of bisection of the sides : then the resultant of the forces OA , OB ...
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body is projected centre of gravity chord coefficient of friction collision component cos² cylinder denote described determine directrix distance equal forces equation feet find the position find the pressure fixed point forces act fulcrum geometrical progression given circle given point hangs hence highest point horizontal plane impact impinges inclined plane intersection join latus rectum lever line of quickest lowest point magnitude and direction mass middle point motion moveable parabola parallel parallelogram particle perfectly elastic ball perpendicular point of projection position of equilibrium pound weight pounds prove pully quickest descent radii radius ratio reaction respectively rest resultant right angles SECT shew sides sin² sliding smooth sphere straight line string passing supposing suspended tan¹ tangent three forces uniform rod velocity of projection vertex vertical circle vertical plane
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