A Collection of Problems in Illustration of the Principles of Elementary Mechanics |
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Page xi
... IMPACT . 180 CHAPTER V. COLLISION OF SPHERES 185 I. Direct collision . ib . II . Oblique collision . 189 CHAPTER VI . IMPACT OF FALLING BODIES CHAPTER VII . IMPACT OF PROJECTILES 199 . 200 ib . . 214 I. II . Impact in the plane of motion ...
... IMPACT . 180 CHAPTER V. COLLISION OF SPHERES 185 I. Direct collision . ib . II . Oblique collision . 189 CHAPTER VI . IMPACT OF FALLING BODIES CHAPTER VII . IMPACT OF PROJECTILES 199 . 200 ib . . 214 I. II . Impact in the plane of motion ...
Page xii
William Walton. SECTION CHAPTER VIII . COLLISION OF PROJECTILES CHAPTER IX . IMPACT AND COLLISION OF BODIES MOVING UNDER CONSTRAINT I. Impact II . Collision CHAPTER X. DYNAMICAL UNITS Appendix . d2 ERRATUM . Page 73 , line 17 ...
William Walton. SECTION CHAPTER VIII . COLLISION OF PROJECTILES CHAPTER IX . IMPACT AND COLLISION OF BODIES MOVING UNDER CONSTRAINT I. Impact II . Collision CHAPTER X. DYNAMICAL UNITS Appendix . d2 ERRATUM . Page 73 , line 17 ...
Page 1
... a point be 120 ° , what is their resultant ? W. S. 1 Let CA , CB , fig . ( 2 ) RESULTANT OF TWO FORCES ACTING ON A POINT PAGE Geometrical method COLLISION OF PROJECTILES CHAPTER IMPACT AND COLLISION OF BODIES MOVING UNDER CONSTRAINT.
... a point be 120 ° , what is their resultant ? W. S. 1 Let CA , CB , fig . ( 2 ) RESULTANT OF TWO FORCES ACTING ON A POINT PAGE Geometrical method COLLISION OF PROJECTILES CHAPTER IMPACT AND COLLISION OF BODIES MOVING UNDER CONSTRAINT.
Page 12
... oppositely to the force 5 . In the latter case equilibrium is impossible , because 10 CHAPTER II Impact EQUILIBRIUM OF THREE FORCES ACTING ON A POINT Triangle of forces II II Collision CHAPTER DYNAMICAL UNITS Appendix.
... oppositely to the force 5 . In the latter case equilibrium is impossible , because 10 CHAPTER II Impact EQUILIBRIUM OF THREE FORCES ACTING ON A POINT Triangle of forces II II Collision CHAPTER DYNAMICAL UNITS Appendix.
Page 146
... impact , the projectile velocity Vcos a , parallel to the plane , must have been destroyed by the com- ponent of gravity g sin a in the time t : hence V cos e = 9 sin a.t From ( 1 ) and ( 2 ) it is plain that tan 0 = 1 cot a . 2 ...
... impact , the projectile velocity Vcos a , parallel to the plane , must have been destroyed by the com- ponent of gravity g sin a in the time t : hence V cos e = 9 sin a.t From ( 1 ) and ( 2 ) it is plain that tan 0 = 1 cot a . 2 ...
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Common terms and phrases
body is projected centre of gravity chord coefficient of friction collision component cos˛ 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 sin˛ sliding smooth sphere straight line string passing supposing suspended tanš tangent three forces uniform rod velocity of projection vertex vertical circle vertical plane
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