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

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

NEWTON’S SECOND LAW : An object will accelerate in the direction of the net force applied to it. JP ©

A NET FORCE PRODUCES A CHANGE 7 FORCE 7 FORCE 7 FORCE 7 FORCE A NET FORCE PRODUCES A CHANGE IN A BODY’S VELOCITY JP ©

Fnet=kma so 1N = k x 1kg x 1m/s2 so k = 1 Fnet ma Fnet = kma THE NEWTON (force unit), IS DEFINED FOR k = 1 ONE NEWTON IS THE NET FORCE THAT CAUSES A MASS OF 1 kg TO ACCELERATE AT 1 m/s2 Fnet=kma so 1N = k x 1kg x 1m/s2 so k = 1 JP ©

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

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

GRAPH OF ACCELERATION VERSUS MASS acceleration / ms-2 mass / kg Fnet = 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 JP ©

(a) What is the force opposing the motion of the wagon? QUESTION A railway engine pulls a wagon of mass 10 tonnes along a level track at a constant velocity. The pull force in the couplings between the engine and wagon is 1000 N [E]. (a) What is the force opposing the motion of the wagon? The velocity is steady, so by Newton’s first law, the net force must be zero. The pull on the wagon must equal the resistance to motion. Answer is Ff = 1000 N [W] JP ©

a = Fnet ÷ m = 400 N ÷ 10 000 kg = 0.04 ms-2 [E] QUESTION A railway engine pulls a wagon of mass 10 tonnes along a level track at a constant velocity. The pull force in the couplings between the engine and wagon is 1000 N [E]. (b) If the pull force is increased to 1400 N [E] and the resistance to movement of the wagon remains constant, what would be the acceleration of the wagon? Fnet = 1400N – 1000N = 400 N [E] a = Fnet ÷ m = 400 N ÷ 10 000 kg = 0.04 ms-2 [E] 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 Upthrust [buoyancy] Thrust from engines SHIP Friction Weight Pull from tug Free body diagram for the ship JP ©

EXAMPLE 1 - AN ELEVATOR ACCELERATING UPWARDS If g = 9.81 ms-2, what “g force” does the passenger experience? The forces experienced by the passenger are her weight, mg and the normal reaction force FN. a = 20.0 ms-2 FN The resultant upward force which gives her the same acceleration as the lift is FN– mg. mg Apply F = ma FN – mg = ma Hence the forces she “feels”, FN = ma + mg The “g force” is the ratio of this force to her weight. JP ©

The force produced in moving the air downwards is given by: EXAMPLE 2 - A HOVERING HELICOPTER A helicopter hovers and supports its weight of 1000. kg by imparting a downward velocity,v, to all the air below its rotors. The rotors have a diameter of 6m. If the density of the air is 1.2 kg m-3 and g = 9.81 ms-2, find a value for v. The force produced in moving the air downwards has an equal and opposite reaction force, FN, which supports the weight of the helicopter, Fg. FN The force produced in moving the air downwards is given by: v As v is constant, so Fg Mass of air moved per second = Area(density)(v) = π x (3m)2 x 1.2kgm-3 x v 3m v = 18.1 m s-1 JP ©

Stop here. JP ©

Force x time = Area under the graph Force-Time Graphs The force applied to a body is rarely constant. Physicists tend to consider the force as a function of time and plot a graph of Force versus Time. force / N Time / s Force x time = Area under the graph = Change in velocity x mass  JP ©

“I’ve just crashed into a haystack!” Collision B Collision A “I’ve just crashed into a haystack!” “I’ve just crashed into a brick wall!” BOTH CARS HAD THE SAME MASS AND WERE TRAVELLING AT THE SAME SPEED force / N B A Time / s IDENTIFY WHICH COLLISIONS IS A AND WHICH IS B. COMMENT ON THE INFORMATION PROVIDED BY THE TWO GRAPHS JP ©