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 , P Q R sin Q , R : sin R , P : sin P , Q. But 4Q , R = 4BOC = π B + C 2 = π A + and therefore sin Q , R = cos B similarly , sin R , P = cos 2 EQUILIBRIUM ...
... intersect in the point 0 , fig . ( 15 ) . Then , the forces being in equilibrium , P Q R sin Q , R : sin R , P : sin P , Q. But 4Q , R = 4BOC = π B + C 2 = π A + and therefore sin Q , R = cos B similarly , sin R , P = cos 2 EQUILIBRIUM ...
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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accelerating attached balance ball beam body centre of gravity chord circle collision component cylinder denote descent described determine direction distance draw drawn Edition elasticity equal equation equilibrium extremity falling feet fixed forces act former friction fulcrum given point greatest ground hangs heavy hence horizontal plane inches inclined plane intersection join latter length lower magnitude mass meet middle point motion moving opposite parallel particle passes placed portion position of equilibrium pounds pressure produce projection proportional prove pully radius ratio reaction represented respectively rest resultant right angles ring rough seconds SECT shew sides sliding smooth space sphere square straight line string supported supposing suspended tension three forces triangle uniform unit velocity vertical vertical plane wall weight wheel whole
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Page 145 - OUT of childhood into manhood Now had grown my Hiawatha, Skilled in all the craft of hunters, Learned in all the lore of old men, In all youthful sports and pastimes, In all manly arts and labors. Swift of foot was Hiawatha ; He could shoot an arrow from him, And run forward with such fleetness, That the arrow fell behind him...