A Collection of Problems in Illustration of the Principles of Elementary Mechanics |
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Page 1
... hence , by Euclid , the angle OAC , and therefore the angle AOB , that is , the angle between the two forces , must be a right angle . ( 2 ) If the angle between two equal forces acting on a point be 120 ° , what is their resultant ...
... hence , by Euclid , the angle OAC , and therefore the angle AOB , that is , the angle between the two forces , must be a right angle . ( 2 ) If the angle between two equal forces acting on a point be 120 ° , what is their resultant ...
Page 2
... hence but < CDA = 4 DCB : hence < DCA = < DCB , = and therefore , since ACB = 120 ° , DCA = 60 ° CDA : but the sum of the angles of the triangle ACD is 180 ° : hence also < CAD . = 60 ° , and the triangle CAD is equilateral . Thus CD is ...
... hence but < CDA = 4 DCB : hence < DCA = < DCB , = and therefore , since ACB = 120 ° , DCA = 60 ° CDA : but the sum of the angles of the triangle ACD is 180 ° : hence also < CAD . = 60 ° , and the triangle CAD is equilateral . Thus CD is ...
Page 3
... Hence the forces PT , PG , acting on P , are equivalent to the force 2PS acting on P in the direction PS . ( 5 ) Two forces , which act at a point of a curve , are repre- sented in magnitude and direction by chords passing through the ...
... Hence the forces PT , PG , acting on P , are equivalent to the force 2PS acting on P in the direction PS . ( 5 ) Two forces , which act at a point of a curve , are repre- sented in magnitude and direction by chords passing through the ...
Page 4
... Hence TF TE OE : OF : :: OB : OC :: AB : CD :: force along DC : force along AB . Hence the resultant of these two forces must pass through 0 . ( 7 ) The resultant of two equal forces , applied at a given point , is represented in ...
... Hence TF TE OE : OF : :: OB : OC :: AB : CD :: force along DC : force along AB . Hence the resultant of these two forces must pass through 0 . ( 7 ) The resultant of two equal forces , applied at a given point , is represented in ...
Page 6
... hence the pressure on the end C of the rod is equal to π + 2α 2P cos 4 ( 4 ) A string is wrapped round a regular polygon , the tension of the string being given : to find the sum of the pressures on the angles of the polygon , and ...
... hence the pressure on the end C of the rod is equal to π + 2α 2P cos 4 ( 4 ) A string is wrapped round a regular polygon , the tension of the string being given : to find the sum of the pressures on the angles of the polygon , and ...
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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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