Forces & Motion

For FULL presentation click HERE >> www.warnescience.net

Forces & Newton’s Laws

K WARNE

http://www.warnescience.net/


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


SAMPLE ONLY SAMPLE ONLY SAMPLE ONLY


Inertia

1. 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.

2. 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 (90o) so the motion continues because there is no unbalanced force to stop it.

• (The question is who threw it in the first place!)

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

MotionEarth

1.

2.

Motion

Inertia

Friction




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 900 to

the SURFACE.

Weight (Fg)

F┴

N = - F┴

Normal Reaction

Force (N)

Weight (Fg)

Normal Reaction

Force (N)

N = - Fg

Calculate the normal force in each example here if the ball

has a 2 kg mass.

30o

F┴ = Fg cos

= 19.6 cos 30

= 16.97 N

FN = 16.97 N




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




= =

Coefficient of FRICTIONSTATIC FRICTION and

• Can vary from zero to a MAX IMUM of

Fs = μs........

μs = coefficient of .......... Friction

DYNAMIC (moving) FRICTION.

• Always constant value given by

Fd = μdN

μd = coefficient of .......... friction

....... force

Static friction

V = ....

Constant Force

Dynamic friction

constant velocity

N = -Fg

N = -Fg

Ff

N

FA

Fg




FN (N)

Ff

Nμs = = gradient

Ff

(N)

Ff

Nμs = = gradient

DRY surface WET surface

static kinetic

Graphs of Frictional force against normal force.

0 1 2 3 0 1 2 3

Ff

(N)

static kinetic

The force of friction Ff is directly proportional to the Normal force (weight)

The gradients of these graphs would be EQUAL to the coefficients of friction.

FN (N)




Friction Examples

5kgFA

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.

What is the maximum

angle that the slope can

have before the block

begins to slide if the

coefficient of friction is

0.8?




A B

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)

a) 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




Newton’s Laws Second LawA ..............................on an object causes it to ...................in the direction of

that force. The magnitude of the acceleration will be directly proportional to the ...................and inversely proportional to the ..............of the object.

Fres = F + f

and

Fres = m x a

Fm

a = ?

Fm=1kg

T = ?

Fg

Fres = Fg + T

Fres = m x a

And

Fg = mg

These are the two most basic scenarios

- any of the variables can be calculated

if all the others are known.

a=2m.s-2




Newton’s - Connected objectsIn both these cases the tension gets

………………… as more blocks

are connected I.e.

T1 ….. T2 …. T3

The systems can be treated as a ……..

…………:

F… = …………………………..

Where

F … = ………. or F … = ………..

So

……………………………………..

T1 T2 T3m1 m2 m3f

m3

m2

m1

T3

T2

T1

Fg

To find the tension between blocks you must consider them

……………. I.e. ……………………………………………….




Newton’s Laws - Pulley systemsThe accelerations of these systems can be easily

found by considering them as a …………….:

Fres = ………… where mtotal = …………

And Fres = ………………….. (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!

m2

m1

T T

Fg1

Fg2

m1

m2

m3

T1T2

Fg1Fg2




Examples Newton 1 & 2 &3

5. The frictional force on the system is 10 N (6 N & 4

N). Calculate the force that the 4kg block exerts on

the 6kg block.

4kg6kga = 2m.s-2

(Ff=10N) (6N of 6kg block 4N on 4kg block)

Consider 4 kg block: Fnett = ma = (4)*(2) = 8 N

Fnett = Fa + Ff .: 8 = Fa + (-4)

Fa = 12 N

.: F exerted by 4kg on 6kg = 12 N

Consider whole system: Fnett = ma = (10)*(2) = 20 N

Fnett = F + Ff .: 20 = F + (-10)

F = 30 N

F




Examples Newton 1 & 2

2. A ball with a mass of 18kg is suspended by a string.

a) Calculate the tension in the string if the ball is

i) stationery,

ii) accelerates upwards at 3m.s-2;

iii) accelerates downwards at 3m.s-2.

T = 176.4N upwards

T = 230.4 upwards

T = 122.4N upwards

ii) Fres = ma = (18)*(3) = 54 N

Fres = T + fg .: 54 = T + fg

but fg = mg = (18)*(-9.8) = -176.4 N

.: 54 = T + (-176.4)

.: T = 54 + 176.4 = 230.4 N upwards




Examples Newton 1 & 23. Two blocks rest on a frictionless horizontal surface

connected by a string. A force of 80N is exerted on the

7kg block. Calculate:

a) the acceleration of the blocks

b) the force exerted by the string on the 13 kg block.

a) Fres = ma .: 80 = (13+7)*a .: a = 80/20 = 4 m.s-2 right

13 kg 7kg 80 N

13 kg Tb)

Fres = ma .: T = (13)*4 .: T = 52 N right




Examples Newton 1 & 2

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



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!




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)




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 calculate the acceleration experianced by both Bob and the

van as well as the distance moved by both.

abob= F/m =(-100/50)= -2m/s2 sbob = ut + 1 /2 at2 =0(1)+1/2(-2)(1)= -1m

avan= F/m =(100/2000)=0.05m/s2 svan = ut+1/2at2 =0(1)+1/2(0.05)(1)= +0.025m




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




2.

Pairs of forces

Action-reaction pairs

1. Man against ground - ground against man – greater for …. than for ……….

2. Man on rope – rope on man – all ………….. but less than ……... ground forces and greater than ….. ground forces.

3. A and B both experience a ………………………….

1.

A B




Hi -

This is a SAMPLE presentation only.

My FULL presentations, which contain a lot more more slides and other resources, are freely

available on my resource sharing website:

www.warnescience.net(click on link or logo)

Have a look and enjoy!

WarneScience





Education

Forces & Motion