1. S I R I S A A C N E WTON
( )
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2. WHEN FORCES ARE NOT BALANCED A NET FORCE CHANGES A BODY’S VELOCITY
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3. NEWTON’S SECOND LAW :
An object will accelerate in the direction of the net force applied to it.
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4. A NET FORCE PRODUCES A CHANGE7
FORCE 7
FORCE 7
FORCE 7
FORCE
A NET FORCE PRODUCES A CHANGE
IN A BODY’S VELOCITY
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5. 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
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6. u n i t s NEWTON’S SECOND LAW Fnet in Newtons m in kilograms
a in metres per second2
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7. ACCELERATION IS DIRECTLY PROPORTIONAL TO THE NET FORCE
acceleration / ms-2
force / N
N.B. - STRAIGHT LINE THROUGH THE ORIGIN
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8. GRAPH OF ACCELERATION VERSUS MASS
acceleration / ms-2
mass / kg
Fnet = m a
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9. ACCELERATION IS INVERSELY PROPORTIONAL TO MASS OF THE BODY
acceleration / ms-2
N.B. - STRAIGHT LINE THROUGH THE ORIGIN
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10. NEWTON’S SECOND LAW IS USED IN TWO FORMS
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11. (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]
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12. 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 ÷ kg = 0.04 ms-2 [E]
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13. 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.
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14. 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
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15. 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.
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16. The force produced in moving the air downwards is given by:
EXAMPLE 2 - A HOVERING HELICOPTER
A helicopter hovers and supports its weight of 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
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17. Stop here.
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18. 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 
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19. “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
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