S I R I S A A C N E WTON (1647 - 1727) JP

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(1686) Philosophiae Naturalis Principia Mathematica In this work, he proposed three “laws” of motion: JP

NEWTON’S FIRST LAW : “Every body continues in a state of rest or uniform motion in a straight line unless impressed forces act upon it.” JP

“Every body continues in a state of rest or parked up NEWTON’S FIRST LAW : “Every body continues in a state of rest or uniform motion in a straight line unless impressed forces act upon it.” JP

AAAAAAAH STATE OF REST UNLESS ……. ! JP

YOU UNDERSTAND WHAT I AM SAYING NEWTON’s FIRST LAW … constant velocity, unless …….. YOU UNDERSTAND WHAT I AM SAYING JP

THE PRACTICAL MEASURE OF INERTIA IS NEWTON’s FIRST LAW INTRODUCES THE IDEA OF INERTIA THE RELUCTANCE OF A BODY AT REST TO MOVE (OR OF A MOVING BODY TO CHANGE ITS STATE OF MOTION) MASS THE PRACTICAL MEASURE OF INERTIA IS JP

The greater the mass of a body, the greater its INERTIA Make me JP

Car stops but inertia carries pendulum forward CAR SEAT BELT locking rod pivots seat belt Car stops but inertia carries pendulum forward ratchet wheel Direction of motion JP

WHEN FORCES ARE BALANCED OBJECTS IN MOTION, STAY IN MOTION [SAME SPEED & DIRECTION] OBJECTS AT REST, STAY AT REST JP

A body is in a state of equilibrium if the forces acting on it are in balance 2 forces balanced 8 N 8N 1200 1200 1200 3 forces balanced JP

THERE ARE TWO CONDITIONS FOR A BODY TO BE IN EQUILIBRIUM: THE SUM OF THE FORCES IN ANY DIRECTION IS ZERO THE SUM OF THE MOMENTS ABOUT ANY POINT IS ZERO JP

CAN YOU EXPLAIN WHAT HAPPENS BELOW? JP

WHEN FORCES ARE NOT BALANCED A RESULTANT FORCE CHANGES A BODY’S VELOCITY JP

NEWTON’S SECOND LAW : “THE RATE OF CHANGE OF MOMENTUM OF A BODY IS DIRECTLY PROPORTIONAL TO THE RESULTANT EXTERNAL FORCES ACTING UPON IT, AND TAKES PLACE IN THE DIRECTION OF THAT FORCE” JP

A RESULTANT FORCE PRODUCES A CHANGE 7 FORCE 7 FORCE 7 FORCE 7 FORCE A RESULTANT FORCE PRODUCES A CHANGE IN A BODY’S MOMENTUM A RESULTANT FORCE AN ACCELERATION JP

F = m a JP

u n i t s NEWTON’S SECOND LAW F in Newtons m in kilograms a in metres per second2 JP

ACCELERATION IS DIRECTLY PROPORTIONAL TO THE APPLIED FORCE acceleration / ms-2 force / N N.B. - STRAIGHT LINE THROUGH THE ORIGIN JP

GRAPH OF ACCELERATION VERSUS MASS acceleration / ms-2 mass / kg F = m a JP

ACCELERATION IS INVERSELY PROPORTIONAL TO MASS OF THE BODY acceleration / ms-2 N.B. - STRAIGHT LINE THROUGH THE ORIGIN JP

NEWTON’S SECOND LAW IS USED IN TWO FORMS Where F is the RESULTANT FORCE JP

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“ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE” NEWTON’S THIRD LAW : “ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE” “IF A BODY A EXERTS A FORCE ON BODY B, THEN B EXERTS AN EQUAL AND OPPOSITELY DIRECTED FORCE ON A” JP

WELL ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE!! devishly clever I’LL PULL HIM WELL ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE!! JP

WELL ACTION AND REACTION ARE ALWAYS EQUAL AND OPPOSITE!! SO WHY DOES THE GIRL MOVE FASTER? JP

NEWTON’S THIRD LAW PAIRS THEY ARE EQUAL IN MAGNITUDE THEY ARE OPPOSITE IN DIRECTION THEY ACT ON DIFFERENT BODIES JP

NEWTON’S THIRD LAW PAIRS SIMILARITIES DIFFERENCES The 2 forces act for the same length of time The 2 forces act on different bodies The 2 forces are in opposite directions The 2 forces are the same size The 2 forces act along the same line Both forces are of the same type JP

THE CLUB EXERTS A FORCE F ON THE BALL THE BALL EXERTS A N EQUAL AND OPPOSITE FORCE F ON THE CLUB F F JP

Drawing Free-Body Diagrams A free-body diagram singles out a body from its neighbours and shows the forces, including reactive forces acting on it. Free-body diagrams are used to show the relative magnitude and direction of all forces acting upon an object in a given situation. The size of an arrow in a free-body diagram is reflective of the magnitude of the force. The direction of the arrow reveals the direction in which the force is acting. Each force arrow in the diagram is labeled to indicate the exact type of force. JP

Free body diagram for the ship Tug assisting a ship Free body diagram for the ship Upthrust [buoyancy] Thrust from engines SHIP Pull from tug weight Friction JP

EXAMPLE 1 - A LIFT ACCELERATING UPWARDS If g = 10 ms-2, what “g force” does the passenger experience? The forces experienced by the passenger are her weight, mg and the normal reaction force R. a = 20 ms-2 R The resultant upward force which gives her the same acceleration as the lift is R – mg. mg Apply F = ma R – mg = ma Hence the forces she “feels”, R = ma + mg The “g force” is the ratio of this force to her weight. JP ©