MI High Percentile

41 47 50
MI 4chapters

Pl 10chapters

Difftinty
fromP
FORCES

Force Pull or Push

UNITS NEWTON N

15N

ION

Weight force of gravity on a mass

IN gravitational
acceleration
ON
Kg
gg

weight Mass

In
Inphysics 981
9
w
my

in Mt 10

mass 110kg 1110 lo 9

gps HOON

DIRECTION 90 towards ground


COMPONENTS OF FORCE

H FHcoso

F
AFsino Hsino

r
Fatso

gÉ µ
Hsing

10 T
155in40 Hcoso

150540
15N go

LIE R

INCLINED PLANE

1Weight 90 towardsground

É n 290 to inclined surface

ut 3 Parallel to inclined

p
so surface

Ino

10 Downward force Wcoso

IN

wcost I

n'sit

Of

To

7 30

Ya

1 Mr always stands
MI

at point
of contact

Mrm between object and surface

dm
2

up down MIMI Ifackkerenhestands

let
right
3 in complete MI we never

Fight take direction of forces

from our perspective

HDown

UP DOWN LEFT RIGHT

All forces are marked

from Mr M1 Pov

t
of
pyBToint
contact

Pcoso 01 Algap

Psind

D IN

Downward force Wt P since

P
I

ji

no

v Otetalgap Downwardforce Psind

IN

ring

inso it moves in direction of

130 biggerforce Cont Hence

I
AEfpointofcontent
the point contact is
of

made A
4N

way

12N

130


TENSION Force applied on an object
by a

string rope

4 SLACK NO TENSION

In I s Eg Taut Tension present

Direction AWAY FROM POINT OF OBSERVATION

Point observation is in

of given

the question

A B F is a fixed point

Box is the point of observation

TAT TTB

Ta and I are different

egg gq
Exception Ta T only

if length

A B R is a
Pitiable
ring

Box is the point of observation

TT TT

a removable
Tension is equal in both sides
Ii the length of

im
regardless of

rope on each side


PULLEY PEG Tension is always equal and

Towards the pulley

This is because we usually take object as point of observation

This is regardless D shape
ofdiagram

of 2 masses

of objects

3 Rest Motion

Td AT
At

TT
at

In M1 If they provide tan 0

34

and you need values since

DO NOT TAKE INVERSE AND TRY

of
TO FIND O
cost
7

III yo
Sino 3 cos0
14 4
4


EQUILIBRIUM

REST STATIONARY BALANCED

up down Left Right

D friction

Fsin50

Fsino
FIE

up down Left Right

Fs'm50 FSin20t12 Foos5otFcos20 G

Fsin50 Fsin20 12

F sin50 sin20 12 28.30 6550 0520 9

G 44 78

Sin50 sin20

F 28.30

0 fiction

Faso

Find

Left Right up down

Fast 7 4 Fsino

This a special simultaneous system Ps MI EYE's

F Sind 4 Foos 0 7

Step 1 Divide Step2 Square bothequation and add

find 4 F sin'd 42

Icoso I F cost 72

tand F'sin'O FOO 4772

F Sino 050 65
2
O tan 65
4 F

F 565

0 29.74

Fs in D
Recognize this

Faso

Fsino

Faso

left up down
Right

10 F cost Find 13

Fsino 13 FASO 10

Steps Steps
F'sing 132

Find 13

f cost TO F'cost 102

tano F'sin'd F 050 137102
If

F Sino 6507 269

O tan 1 3 F I 269

0 52.431 F 5269 16.40

 fiction

Left Right
r
Effigy

185in 30 P

P 9N

pissinso

rR R Left Right

185in30 Pcos30

thyiscosso P 185m30
v

pain poos 00530

Isinzo

D 10.41
y

ISN

Left Right

Toy 125in40 P

I D 7.71

v14

T t left
125in40
Right
Pcos40

TEX P 125in40

ago Y Cos40
Yasin
40

r D 16.069
12N

The forces on one side of pulley are independent

the other side

of forces on

Wisin40 Wasingo

e a
p Pz

jE go

M E TTz W

W T A
NTz W2

uptown Left Right up down up down

s W

T W q

Wi Wasinbo

Sinko

w 3.435 8 we twsbo 5
singing

Sin40

W 4 3969 Wz 3.2635

5.55in wsinfo O Wcoso

if
55 ÉÉÉ

s.IN FEW TEW

to

5.5 17.3

up down up down left Right up down

T 5.5 55cosotwsino 7.3 5 5sino WCoso Tz W

W 5.5Sino

5 5.5Sino 950

cos 0

5.5cost 5.5Sino 7.3

Coso

5.50050 5.550 7.3

cost

5.51650 sin o 7.3650

5 511 7.30050

Cos0 5.5

7 3

0 41.11

We 5.5Sino 5.5 Sin41.11 4.7996

cos 0 cosy

ÉÉation Tsionhay
I Tagg

Tasso

Th
Tsing

L Tim

Tcos20
8N

up down left Right

Tsin50 Tm20t8 Tosto To 20 X

Tinto Tsin 20 8 x T cos50 05201

Sino sinzo 8 4 18.86766550 0520

T 8 X 29.856

Sin50 sin20

T 18.867


Study

sleep Party

CONTACT

Force

FRICTIONAL COMPONENT F NORMAL COMPONENT R


FRICTION

F yr
Thisformula can
only be applied when
a object is about to move

EEEon FEELEY LIMITING EQUILIBRIUM

b Moving

COFFICIENT OF FRICTION NORMAL REACTION FORCE

U
Normal component of the R

contact force

Property ofSurface Direction Towards head of Mr M1

Smoothersurface yd

Roughersurface yt FORMULA
up down

0
This will not change even if
Smoothsurface 4 at rest
body is moving

or
Friction o

up down

R W

R lo n

losing up down

Rt 105in30 W

q
76530
R W losin30


down
up

25N Rt losin30 W 255in60

AR ION n

25560 I R W 255in60 105in30

losing

250560 10 0530

R
up down

R Wcoso

egg

Wsito

OC

IN

up down
R
R psin4o 120540

h Qr
7 1120540

To
post

Psi u
Isin40

r
12N

Psino

up down
P

É R Psino

R
aged

in

up down

Ian
n
30
125in
Rt 125in30 8

pointaunt t 130

q rooms

8N


12N

130

THOUGHTEXPERIMENT

In this example assume that maximum friction

between the box and ground is 15N

ion 10N

Rest

14.9N 149N

Rest

Friction
Limiting
Ypg 15N Equilibrium

Aboutto move

MayjFiction P 15N

More


LIMITING EQUILIBRIUM

BODY IS ABOUT TO MOVE SLIDE

STEM Mark direction

of friction against the

direction motion and label it F UR

of

STEPH Apply up down Left Right

and the unknown

find

1
I
F Isino Box is ABOUT TO SLIDE

In Tosa

YR

IN

2 Box is about to move

MR

Tasino

YO

3 R P P force is pulling the

object up the plane

Tincoso

Box is about to move

twsino we cannotdetermine direction


W of motion of box

Thr

TYRTywcoso
MRT Tywcoso

ursine trsino
yo yo

motion
R

Block B is about to move

BE I
AT

1dB

INA

Psino P AR
Block B is about to move
 01 O

C
Paso Pf F
t the

FB you cannot determine

a direction of motion

Ya

Eto e ur allowed
pointof
observation
LR u D

emotion
R

up down

R 12 15 sin 30
MR
50530
R 19.5
15sing

Left Right
156530

MR

15 cos 30 4 19.5

4 150530

19.5

Y 0.666 358
 W Yo

NOEMR
of
Point obser ID L R

You are allowed F Mr

af up down
Of

R Tsin to

Left Right
É

40 Tsing
40 50560 F

F 40 Tosto
in F
Y R

40 Tos 60 MLTsin60

40 Tos 60 0.7 Tsin 60

40 0.7 Tsin Go Tosto

40 T 0.75in 60 65607

T 40

0.75inGo cos60



INEQUALITIES OF REST AND MOTION

Moving stopping REST

E

forces forces

MOVE

Moving Stopping

forces forces

note incase
ofinequalitiesalways take maxFriction YR

3
I
sing
cos2 4

For Rup down

Inequality for Rest
R 12Sina 20
moving forces s stoppingforces
to

12sind R 27.2
14

j
q

12 Cosa E MR
n

I 4 136

48 I 136M

48 F M
136
 E Ey

4747

moves

motion

inequality for

moving forces Stopping forces

t
qt no

12 cosa 7 MR
12
E

won

four up down 48 64M

Rt 125in 2 20

4G y

Rt 12 20
31
u
I

ME

1 EQUILIBRIUM

2 LIMITINGEQUILIBRIUM

3 INEQUALITIES OF REST MOTION

4 RESUTANT OF FORCE


RESULTANT OF A FORCE

4N the and

1000540
dition of Resultant of

Y un these forces
401

losin40

ION

Horizontal Analysis Vertical Analysis Resultant Force

UN

409540
In Y Ry 4.34
glosinto
Rx 4.34 Op

Ry2.43

usePythagoras
2.432

magnitudeof resultant 4.974 RE4.342

Direction 29.22590 below trenaxis 12 4.974

Forangle

Imagine a grid at startofRx tano 2454

abaft O

vex axis
agent
re x axis tariff
belytay.EE beltwtvex axis

0 29.23


ADVANCED RESULTANT IS GIVEN ON THE DIAGRAM

METHOD REMOVE THE RESULTANT FROM MAIN

DIAGRAM AND MAKE SAME 3 COLUMNS

q
Psinbo

60
Pcos bottled

Horizontal
Analysis Vertical Analysis Resultant Force

12N
Q Psinbo a

Pasto Ry RTing

180

Rx 1200580

Rx d Pcos60

Now equate both Rx and both Ry with each other


Q Pos60 126580 Psinbo 125in80

0 126580 Paso P 12sin8o

0 120580 136460560 Sin60

0 8.907 P 13.646