A Collection of Problems in Illustration of the Principles of Elementary Mechanics |
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Page 1
... former forces . Let O , fig . ( 1 ) , be the point , OA , OB , the directions of the forces 6 , 8 , respectively : take OA = 6 units of length , and OB = 8 units of length : complete the parallelogram OACB , and join OC . Then OC must ...
... former forces . Let O , fig . ( 1 ) , be the point , OA , OB , the directions of the forces 6 , 8 , respectively : take OA = 6 units of length , and OB = 8 units of length : complete the parallelogram OACB , and join OC . Then OC must ...
Page 9
... of 90 ° : to compare the magnitudes of the resultants in the two cases . The ratio of the former to the latter resultant is equal to W. S. ( 1 + 1 / 2 ) 2 ( 22 ) Two equal forces P , P , RESULTANT OF TWO FORCES ACTING ON A POINT . 9.
... of 90 ° : to compare the magnitudes of the resultants in the two cases . The ratio of the former to the latter resultant is equal to W. S. ( 1 + 1 / 2 ) 2 ( 22 ) Two equal forces P , P , RESULTANT OF TWO FORCES ACTING ON A POINT . 9.
Page 12
... former case , equilibrium is possible , the two forces 2 and 3 acting in a straight line oppositely to the force 5 . In the latter case equilibrium is impossible , because 10 CHAPTER II Impact EQUILIBRIUM OF THREE FORCES ACTING ON A ...
... former case , equilibrium is possible , the two forces 2 and 3 acting in a straight line oppositely to the force 5 . In the latter case equilibrium is impossible , because 10 CHAPTER II Impact EQUILIBRIUM OF THREE FORCES ACTING ON A ...
Page 17
... former case is to that in the latter as 3 to 4 . ( 8 ) One end of a string AOP is attached to a fixed point A , and to the other end is fixed a weight P : a string BO , shorter than AB , is attached to a point B in the same horizontal ...
... former case is to that in the latter as 3 to 4 . ( 8 ) One end of a string AOP is attached to a fixed point A , and to the other end is fixed a weight P : a string BO , shorter than AB , is attached to a point B in the same horizontal ...
Page 23
... former force being inclined at an angle of 60 ° , and that of the latter at an angle of 45 ° to a given straight line passing through the point : to find the mag- nitude and direction of the resultant . = Let Rthe magnitude of the ...
... former force being inclined at an angle of 60 ° , and that of the latter at an angle of 45 ° to a given straight line passing through the point : to find the mag- nitude and direction of the resultant . = Let Rthe magnitude of the ...
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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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