Forces & Motion

Forces & Newton’s Laws GRADE 11 K WARNE

Newton’s First Law
” A body will................................ of ...... or ................. .................... in a straight line, unless acted on by a ......................................"
This law describes - ................
IMPORTANT POINTS: (A moving object…)
Continues in its state
Rest or uniform motion in a straight line
Unless acted upon by (external) forces

Newton’s First Law
” A body continues in its state of rest or of uniform motion in a straight line , unless acted on by a resultant force ."
This law describes - Inertia.
IMPORTANT POINTS: (A moving object…)
Continues in its state
Rest or uniform motion in a straight line
Unless acted upon by (external) forces

Newton’s Laws Second Law
A …………………. on an object causes it to ……………… in the direction of that force.
F res = …………… and F res = ………….. F res = …………….. And F g = ………… These are the two most basic scenarios - any of the ………… can be calculated if ……………. are known. F m a F m T W

Newton’s Laws Second Law
A resultant force on an object causes it to accelerate in the direction of that force. The magnitude of the acceleration will be directly proportional to the force and inversely proportional to the mass of the object.
F res = F + f and F res = m x a F res = F g + T F res = m x a And F g = mg These are the two most basic scenarios - any of the variables can be calculated if all the others are known. F m a F m T F g

Inertia
A body has a natural tendency to resist any changes to its state of motion.
This resistance is known as Inertia .
Examples:
What will happen if the card in the picture is flicked ? Explain why.
The moon moves around the sun in a circular orbit . Explain why.
Every object in a state of uniform motion tends to remain in that state of motion unless an external (unbalanced)force is applied to it. Gravity Motion Earth 1. 2. Motion Inertia Friction

Inertia
A body has a ............................................ any changes to its state of motion.
This resistance is known as ....................... .
If the card in the picture is flicked ................. ....................................... . Inertia keeps the peg stationary when the card is moved quickly .
The peg’s Inertia is overcomes .................. ................ which try to keep it’s position on the card.
The moon was moving past the earth in a straight line but became ................. by the ............................
Gravity does not act against the direction of motion (90 o ) so the motion continues because ............................................... to .................... .
(The question is who threw it in the first place!)

Inertia
A body has a natural tendency to resist any changes to its state of motion.
This resistance is known as Inertia .
If the card in the picture is flicked the peg should fall into the glass . Inertia keeps the peg stationary when the card is moved quickly .
The peg’s Inertia is overcomes the friction forces which try to keep it’s position on the card.
The moon was moving past the earth in a straight line but became trapped by the earth’s gravity.
Gravity does not act against the direction of motion (90 o ) so the motion continues because there is no unbalanced force to stop it .
(The question is who threw it in the first place!)

The Normal or Reaction Force
This is the Reaction Force (Newton …..) exerted by a surface on any object that rests (exerts a force on) that surface.
Always EQUAL in magnitude and to the initial force and at 90 0 to the SURFACE.
Calculate the normal force in each example here if the ball has a 2 kg mass. Weight (F g ) F ┴ N = ………. Normal Reaction Force (N) Weight (F g ) Normal Reaction Force (……) N = ……..

The Normal or Reaction Force
This is the Reaction Force (Newton 3) exerted by a surface on any object that rests (exerts a force on) that surface.
Always EQUAL in magnitude and to the initial force and at 90 0 to the SURFACE.
Calculate the normal force in each example here if the ball has a 2 kg mass. Weight (F g ) F ┴ N = - F ┴ Normal Reaction Force (N) Weight (F g ) Normal Reaction Force (N) N = - F g

FRICTION
Friction forces exists between two surfaces in contact.
There are two types …………… FRICTION and ………………(moving) FRICTION.
Static friction is ………………. than dynamic friction.
The maximum force exerted on a stationary object …………..it begins to move ( accelerate ) is equal to the maximum static frictional force.
The force required to ………. an object moving with ……………… velocity on a surface is equal to the dynamic friction.
Max force Static friction V = 0 Force dynamic friction constant velocity

FRICTION
Friction forces exists between two surfaces in contact.
There are two types STATIC FRICTION and DYNAMIC (moving) FRICTION.
Static friction is GREATER than dynamic friction.
The maximum force exerted on a stationary object before it begins to move ( accelerate ) is equal to the maximum static frictional force.
The force required to keep an object moving with constant velocity on a surface is equal to the dynamic friction.
Max force Static friction V = 0 Force dynamic friction constant velocity

Coefficient of FRICTION
STATIC FRICTION and
Can vary from zero to a ……………... of
F s = …………
DYNAMIC (moving) FRICTION.
Always ………………… value given by
F d = ………..
Max force Static friction V = 0 Force Dynamic friction constant velocity N = -F g N = -F g

Coefficient of FRICTION
STATIC FRICTION and
Can vary from zero to a MAX IMUM of
F s = μ s N
μ s = coefficient of static friction
DYNAMIC (moving) FRICTION.
Always constant value given by
F d = μ d N
Max force Static friction V = 0 Force Dynamic friction constant velocity N = -F g N = -F g

F N (N) F f N μ s = = gradient F f (N) N F f N μ s = = gradient DRY surface WET surface static kinetic Graphs of Frictional force against normal force. 0 1 2 3 0 1 2 3 F f (N) static kinetic

Friction Examples 5kg F A A block is held against a vertical surface by a horizontal force. What is the coefficient of friction if the minimum applied force is 20 N?

Friction Examples Object on an inclined plane. 2 kg  What is the maximum angle that the slope can have before the block begins to slide if the coefficient of friction is 0.8? 2 kg

Exam Question
The diagram shows a truck with a crate immediately behind the cab in position A.
The truck suddenly accelerates forward and the crate slides towards the back of the truck and comes to rest in position (B)
Draw a force diagram, indication and labeling the horizontal force(s) acting on the crate while the truck accelerates. (2)
Briefly explain why the crate slides towards the back of the truck. (4)
Name and state the law or principle that you have applied in order to reach an answer in (b). (4)
WCED Nov 2001 TOTAL [10]
A

Exam Question
The diagram shows a truck with a crate immediately behind the cab in position A.The truck suddenly accelerates forward and the crate slides towards the back of the truck and comes to rest in position (B)
Draw a force diagram, indication and labeling the horizontal force(s) acting on the crate while the truck accelerates.
(2)
Friction between the crate and truck A B

Exam Question
Briefly explain why the crate slides towards the back of the truck.
According to Newtons 1st law  the crate will tend to maintain its state of rest (in its position)  until it experiences a resultant force – the friction between the truck and the box is not enough to accelerate the box at the same rate as the truck - so as the truck moves forwards the crate stays where  it is but appears to move towards the back of the truck  ). (4)
c) Name and state the law or principle that you have applied in order to reach an answer in (b). Newton 1  A body will continue…    (4)
WCED Nov 2001 TOTAL [10]
Friction between the crate and truck A B

Newton’s 3rd Law
"For every action there is an ………… and ………………… reaction .”
The gasses experience a force ………………… out of the rocket, this has an equal but opposite …………………. which drives the rocket forward ! The force exerted by the hand on the head is the same …………………. as the force exerted by the head on the hand! The head hits the hand just as hard as the hand hits the head!

Newton’s 3rd Law
"For every action there is an equal and opposite reaction .”
The gasses experience a force backwards out of the rocket, this has an equal but opposite reaction force which drives the rocket forward ! The force exerted by the hand on the head is the same magnitude as the force exerted by the head on the hand! The head hits the hand just as hard as the hand hits the head!

An apple on a table.
Non contact forces Gravity
Exerted by the ……… on the apple
Reaction by the ……….. on the earth.
Contact forces: Table Apple exerts a force ……… on the table. Table exerts a …………. or …………. force on the apple. •
All these forces are …………. so there is no …………… force.
Forces on apple: gravity (down) & reaction (up)

An apple on a table.
Non contact forces Gravity
Exerted by the earth on the apple
Reaction by the apple on the earth.
Contact forces: Table Apple exerts a force down on the table. Table exerts a Normal or Reaction force on the apple. •
All these forces are equal so there is no resultant force.
Forces on apple: gravity (down) & reaction (up)

Walking & Pushing
Draw sketches to show and explain all the reaction pairs of forces that occur when
a person is walking
A person is pushing a car.

Person Walking
The person pushes ……………. on the ground .
The ground pushes …………….. on the person.
The person has an …………. force so moves in the direction of that force.
(The earth…………
movement Forces on person

Person Walking
The person pushes backwards on the ground .
The ground pushes forwards on the person.
The person has an UNBALANCED force so moves in the direction of that force.
(The earth…………
movement Forces on person

Pushing a car
The person pushes backwards on the ground.
The ground pushes forwards on the person.
The person pushes forward on the car.
The car pushes back on the person.
The forces on the car are less than the forces on the ground.
The person has an unbalanced force so moves in the direction of that force.
Movement Forces on person

Newton’s 3rd BOB (who isn't very bright) is tired of pushing his van to the petrol station. He gets an idea -- "Hey! I go much faster when I am roller-blading than when I am walking. Why don't I wear my roller-blades!" So BOB gets his roller blades from the back of his van, and starts pushing. Will this work? Explain using Newton's laws what will happen. (Calculate the acceleration and displacement each will experience. BOB weighs 50 kg, and the van weighs 2,000 kg.)

Newton’s 3rd ANS : Since every action has an equal and opposite reaction, when BOB pushes the van, the van pushes BOB with the same size force but opposite direction. Assume he pushes for one second. BOB weighs 50 kg, and the van weighs 2,000 kg. Taking right as positive:

Newton’s 3rd ANS : Since every action has an equal and opposite reaction, when BOB pushes the van, the van pushes BOB with the same size force but opposite direction. BOB weighs 50 kg, and the van weighs 2,000 kg. Taking right as positive: a bob = F/m =(-100/50)= -2m/s 2 s bob = ut + 1 / 2 at 2 =0(1)+1/2(-2)(1)= -1m a van = F/m =(100/2000)=0.05m/s 2 s van = ut+ 1 / 2 at 2 =0(1)+1/2(0.05)(1)= +0.025m

Pairs of forces

Pairs of forces
Action-reaction pairs
Man against ground - ground against man – greater for …. than for ……….
Man on rope – rope on man – all ………….. but less than ……... ground forces and greater than ….. ground forces.
A and B both experience a ………………………….
2. 1. A B

Pairs of forces
Action-reaction pairs
Man against ground - ground against man – greater for B than for A.
Man on rope – rope on man – all equal but less than B’s ground forces and greater than A’s ground forces.
A and B both experience a nett force to the RIGHT.
The rope is part of the system and does not accelerate in the system – only as part of the system.
2. 1. A B

Pulley system
The system shown is in equilibrium. What is the magnitude and direction of the friction force acting on the block?
50kg

Pulley system
The system shown is in equilibrium. What is the magnitude and direction of the friction force acting on the block?
W 10kg 250N 30 o 250N 250N 30 o 30 o 250N 100N F f = ? F ┴ F || F ||

Eg - Hanging Weight
A weight of 30 N hangs on a rope fron a ceiling. A horizontal force pulls it to the side. The angle between the rope and ceiling is 60 0 .
Determine: the tension in the rope (T) and the horizontal force.
30N 60 o Force Diagram 30N T Fg T F 60 o

Newton’s - Connected objects
In both these cases the tension gets ………………… as more blocks are connected I.e.
T 1 ….. T 2 …. T 3
The systems can be treated as a …….. …………:
F … = …………………………..
Where
F … = ………. or F … = ………..
So
…………………………………… ..
To find the tension between blocks you must consider them ……………. I.e. ………………………………………………. T 1 T 2 T 3 m 1 m 2 m 3 f m 3 m 2 m 1 T 3 T 2 T 1 Fg

Newton’s - Connected objects
In both these cases the tension gets larger as more blocks are connected I.e.
T 1 < T 2 < T 3
The systems can be treated as a single unit:
F res = m total a = (m 1 +m 2 +m 3 )a
Where
F res = T 3 + f or F res = T 3 + F g
So
T 3 + f( or F g ) = (m 1 +m 2 +m 3 )a
To find the tension between blocks you must consider them individually masses I.e. T 2 + F g = (m 1 +m 2 )a and T 1 + F g = m 1 a T 1 T 2 T 3 m 1 m 2 m 3 f m 3 m 2 m 1 T 3 T 2 T 1 F g

Newton’s Laws - Pulley systems The accelerations of these systems can be easily found by considering them as a …………….: F res = ………… where m total = ………… And F res = ………………….. (always opposed) To find tension; ……………………………….. Individual masses or tensions can be found by considering masses individually: …………………………………… ... Tension in any one string is the same throughout! F g1 F g2 m 2 m 1 T T m 1 m 2 m 3 T 1 T 2 F g1 F g2

Newton’s Laws - Pulley systems The accelerations of these systems can be easily found by considering them as a unit: F res = m total a where m total = m 1 +m 2 … And F res = F g1 - F g2 (always opposed) To find tension; T + F g1 = m 1 a or T + F g2 = m 2 a Individual masses or tensions can be found by considering masses individually: T 1 + F g1 = m 1 a or T 2 – F g2 = m 2 a Tension in any one string is the same throughout! F g1 F g2 m 2 m 1 T T m 1 m 2 m 3 T 1 T 2 F g1 F g2

Examples Newton 1 & 2
1. A 25kg container is acted upon by a horizontal force of 65N. The frictional force between the container and the floor is 15N. Calculate the acceleration of the container.
(a = 2m.s -2 )

A ball with a mass of 18kg is suspended by a string.
Calculate the tension in the string if the ball is
stationery,
accelerates upwards at 3m.s -2 ;
accelerates downwards at 3m.s -2 .
T = 176.4N upwards T = 230.4 upwards T = 122.4N upwards

Two blocks rest on a frictionless horizontal surface connected by a string. A force of 80N is exerted on the 7kg block. Calculate:
the acceleration of the blocks
the force exerted by the string on the 13 kg block.

4. A constant force of 120N is used to lift two masses of 4 kg and 6kg which are attached to each other with string. Ignore any mass of the string and calculate the acceleration of the system and the tension in the string.
4kg 6kg 120 N

Examples Newton 1 & 2 &3
5. Calculate: The frictional force on the system and the force that the 4kg block exerts on the 6kg block.
4kg 6kg a = 2m.s -2 (Ff=10N (6N of 6kg block 4N on 4kg block) )

Examples Newton 1 & 2 &3
5.
4kg 6kg a = 2m.s -2

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Forces & Motion

by Keith Warne on May 12, 2010

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