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NEWTONS LAW OF MOTION

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FORCE (F)

Described as either push or pull that can cause a

mass (body) to accelerate (cause of motion)

A vector quantity that is the product of mass (m) &acceleration (a).

F = m a

In MKS it is in unit : Newtons (N) , 1 N = 1 kg-m/s2

In CGS it is in unit : dynes , 1 dyne = 1 g-cm/s2

In English Units : Pound ( or lbs) , Pound-force (lbf)

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CONVERSION :

1 dyne = 1x10-5 N = 10-5 N

1 lbs = 4.448 N

COMPONENTS

m

F = 5 N

m Fx = 5 N cos

Fy = 5 N sin

=

FORCES

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KINDS OF FORCES

1. Force due to Gravity / WeightFGRAV = W = mg, where g = 9.8 m/s

2 or 32 ft/s2

m

W = mg W = mg

Weight is ALWAYS directed TOWARDS (attractive to) the earth, even ifthe surface is at an angle.

Weight is ALWAYS acting on the body

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KINDS OF FORCES

2. Longitudinal Forces

Forces acting along the length of an object

Common in ropes, cables, solid cylinders

TENSION (T)

Pull Force on an object. Its end effect is

to STRETCH an object.

COMPRESSION (C)

Push Force on an object. Its end effect

is to FLATEN an object.- Compression usually is due to normal

forces between two objects in contact

C C

TT

Length (L)

Length (L)

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KINDS OF FORCES

3. Contact Forces

Forces due to interaction between different surfaces

FRICTIONAL FORCE /FRICTION (f)

Force that oppose motion of an object.

Always parallel to the contact surface& directed opposite the motion of the

object

NORMAL FORCE ( or N)

Reaction Force due to Weight of the

object(s) in contact

Always Perpendicular to the contact

surface

m

W = mgN

f

W = mg

f

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KINDS OF FORCES

3. Contact Forces

f N

f= N

Coefficient ofFriction

S Coefficient of StaticFriction

KCoefficient of KineticFriction

S = tan f, (This MUST be the Angle of friction or repose)

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KINDS OF FORCES

3. Contact Forces

Equations for NormalForce

m

W = mg W = mg

N

N

W = mg

N

F

N = W = mg N = Wy = W cos

N = mgcos

N = Wy Fy

N = mgcos Fsin

F

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KINDS OF FORCES

3. Contact Forces

Equations forFrictionalForce

m

W = mg W = mg

N

N

W = mg

N

F

N = W = mgN = Wy = W cos

N = mg cos

N = Wy Fy

N = mgcos F sin

F

f

f = N = mg

f

f

f = N = mg cosf = N

f = (mgcos F sin)

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NEWTONS FIRST LAW OF MOTION

A body acted on by NO net force either stays

motionless or moves, but with constant velocity

and zero acceleration


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NEWTONS SECOND LAW OF MOTION


A body requires a net force to accelerateThe

acceleration is directly proportional to the net force but

inversely proportional to the bodys massThe direction of the net force is the same as the

direction of the acceleration

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NEWTONS THIRD LAW OF MOTION

To every action there is always opposed an

equal reaction, same in magnitude but opposite

in direction.

m

W = mg

N

m

W = mg

T


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WRONG FBD CORRECT FBD

W = mg W = mg

f f

N

N

F

m

W = mgN

f

F

F

TIPS ON FBD

DONT Draw Vectors GOING towards the point mass (even though it mayshow in

the figure) draw it awayfrom the point mass, it does the same thing.

EXAMPLE 1

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EXAMPLE 2

W = mg

f

WRONG FBD * CORRECT FBD *

W = mg W = mg

f f

N

N

*Using the inclined surface as the x-axis

TIPS ON FBD

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PULLEYS & Weights

Pulleys are analyzed as frictionless & of negligible weight.

ForFCE : Tension of rope or cable passing through the pulleyis equal to the weights

HANGING from them.

T = W

2 kg

W = mg

T = W

T = W T = W

2 kg

W1 = mg

T = W1

T = W1

T = W1

T = W1

TIPS ON FBD

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Using the inclined surface as the x-axis (rotate of axis)

W = mg

f

1000 lbs

45

60

O

Must have at least 1 pair of perpendicular forces and 90

angle is visible.

ALLOWED since f & N are Perpendicular

NOTALLOWED :No perpendicular forces or

angles with respect to point O, thus not

practical to use any inclined as x-axis.

TIPS ON FBD

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TIPS ON FBD

If a system consists of two or more objects. Multiple FBDs maybe

required :

A

BP = ? A

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FIRST CONDITION OF EQUILIBRIUM (FCE)

Sample Problems :

1. A 5 kg block will start to slide down at

constant speed from a surface when

it is inclined at 40 with the

horizontal. Determine the Frictional

force, Normal force and the

coefficient of static friction. 40

Solution : Draw the forces acting

on the body

fN

Draw the FBD

W = mg

fN

40

40 W = mg

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Since we have perpendicular forces along the inclined (f & N) we can use theinclined as our x-axis

Re-draw the FBD

W = mg

f

N

40

Draw the component vectors of angled vectors

W

f

N

40

Wx

Wy


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W

f

N

40

Wx

Wy

Fx = 0 (+)

Fnet = 0

Wx + f = 0

f = Wx = Wsin

Fy = 0 (+)

+ N Wy = 0

N = Wy = Wcos

N = mg cos

N = (5kg)(9.8m/s2)(cos 40)

N = 37.54 N

Friction and Normal force and S


f = mg sin

f = (5kg)(9.8m/s2)(sin 40)

f = 31.5 N

f= N

= f/NS = f/N = 31.5 N/ 37.54 N S = 0.839

orS = tan f= tan 40 S = 0.839

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2. A woman at an airport is pulling a 15 kg suitcase (with wheels) at constant speed of 2 m/sby pulling on the handle attached to the bag (this makes an angle above the horizontal).

She pulls with a 56 N force, and the frictional force is 20N. What is the angle , the normal

force & the coefficient of kinetic friction?

F = 56 N

m = 15 kg

f = 20 N

N

W = mg

F = 56 N

f = 20 N

N

W = mg

Fy

f = 20 N

Fx

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Fx = 0 (+)

Fnet = 0

f + Fx

= 0

f = Fx

Fy = 0 (+)

+ N + Fy W = 0

N = W Fy

N = mg F sin

N = (15kg)(9.8m/s2) (56N)(sin 69.08)

N = 94.69 N

Solving for


f = F cos

20 N = 56 N(cos )

= 69.08

f= N

= f/NK

= f/N = 20 N/ 94.69 N K

= 0.211

N

W = mg

Fy

f = 20 N

Fx

,Normal force and K

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3

. A wet shirt weighs 4 N. It is hanged to dry on a metal clothesline. The shirt is placed atthe very center of the length of the clothesline, and the angle formed with respect to the

horizontal due to the weight of the shirt on either side are equal. What are the tensions

on each side of the clothesline?

W = 4 N

T2

T1

W = 4 N

T2y

T1y

T1x T2x

2 m

0.4 m

1 m

0.4 mtan = (0.4 m)/(1 m)

= 21.8

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Fx = 0 (+)

Fnet = 0

T1x

+ T2x

= 0

T2x = T1x

Fy = 0 (+)

+ T1y

+ T2y

W = 0

T1 sin + T2 sin = W

T1 sin + T1 sin = W

Solving for Tensions


T2 cos = T1cos

T2 = T1W = 4 N

T2y

T1y

T1x T2x

2(T1 sin) = W

T1 = W/ (2sin)

T1 = [(4N)/[2sin(21.8)]

T1 =5.385N T2 = 5.385N

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NEWTONS SECOND LAW OF MOTION


A body requires a net force to accelerateThe

acceleration is directly proportional to the net force but

inversely proportional to the bodys massThe direction of the net force is the same as the

direction of the acceleration

m

Body of mass m at rest on a

frictionless surface

m

Due to net force Fnet going to the left the

object will accelerate also to the left

Fnet

aFnet= ma

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mF

a

W = mgN

a)Fx= max & Fy= may

b)Fx= max & Fy= 0

c) Fx= 0 & Fy= may

If Fnet= ma

Possibilities :

Purely Horizontal Movement

Purely Vertical Movement

NEWTONS SECOND LAW OF MOTION (NSLM)

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NET FORCES

Net Force due to Contact Forces

m

W = mg W = mg

N

N

W = mg

N

F

N = W = mg N = Wy = W cos

N = mg cos

N = Wy

Fy

N = mgcos F sin

F

f

Fnet = F f

f

Fnet = Wx f

Fnet = Wsin f

Fnet = mg sin f

f

Fnet = Fx Wx f

Fnet = F cos W sin f

Fnet = F cos + mg sin f

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Sample Problems :

1. A 5 kg block slides down a plane

inclined at 40 to the horizontal. Find

the acceleration of the block

a) If the plane is frictionless

b) If the coefficient of kinetic friction is

0.20 40

Solution : Draw the forces acting

on the body

fN

Draw the FBD

W = mg

fN

40

40 W = mg

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Since we have perpendicular forces along the inclined (f & N) we can use theinclined as our x-axis

Re-draw the FBD

W = mg

f

N

40

a

Draw the component vectors of angled vectors

W

f

N

40

a

Wx

Wy

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Fx = max (+)

Fnet = ma

Wx + f = mama = Wx f

Fy = 0 (+)

+ N Wy = 0N = Wy = Wcos

N = mg cos ma = mg sin f

f = N

W

f

N

40

Wx

Wy

a

ma = mg sin - N

ma = mg sin mg cos

ma = mg (sin cos)


a = g (sin cos)

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1. A 5 kg block slides down a plane inclined at 40 to the horizontal. Find the

acceleration of the block

a) If the plane is frictionless

b) If the coefficient of kinetic friction is 0.20


(a) a = ? If f = 0, hence = 0

a = g [ (sin 40) ( cos 40) ]

a = (9.8 m/s2) [ (sin 40) 0 ( cos 40) ]

a = 6.3 m/s2

(b) a = ? If = 0.2

a = g [ (sin 40) ( cos 40) ]

a = (9.8 m/s2) [ (sin 40) 0.2 ( cos 40) ]

a = 4.8 m/s2

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2. A car (2,000 kg) is traveling at 28.7 m/s when the driver locks the breaks to stop thecar. What will be the shortest distance ( from the point where the breaks were

locked up to the full stopping point), if the coefficient of kinetic friction between

the tires and pavement (road) is 0.8?


VO = 28.7 m/s

s

VF = 0a

UsingK

inematics Eqn (3)

VF2 = VO

2 + 2as

s = (VF2 VO

2)/(2a)

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f

W = mg

N




af

W = mg

N

a

Using NSLM to determine the acceleration FBD :

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f

W = mg

N

a Fx = max (+)

Fnet = ma

f = + maFy = 0 (+)

+ N W = 0

N = W

N = mgf = N

N = +ma mg = +ma

g = +a

a = g = (0.8)(9.8 m/s2) = 7.84 m/s2

a = 7.84 m/s2

, deceleration

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VO = 28.7 m/s

s

VF = 0a

UsingK

inematics Eqn (3)

VF2 = VO

2 + 2as

s = (VF2 VO

2)/(2a)

s = [(02 (28.7m/s)2]/[(2)(7.84 m/s2)]

s = 52.53 m

a = 7.84 m/s2, deceleration

s = (823.69 m2/s2)/(15.68 m/s2)

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3. A 3.5 kg pail is dropped into a 15 m empty deep well, starting from rest at the top.The tension in the rope is constant at 14.8 N as the pail drops. What is the time to

reach the bottom of the well?

h = 15 m

VO = 0

Using Kinematics Eqn (2)

h = VOt + at2

Where a g

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Using NSLM to determine the acceleration FBD :

T = 14.8 N

W = mg

a

W = mg

T = 14.8 N

a

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W = mg

T = 14.8 N

a

Fnet = ma

Fx = 0 (+) Fy = may (+)

+ T W = ma

ma = mg T

a = ( mg T ) / m

a = [(3.5kg)(9.8 m/s2) 14.8 N]/(3.5kg)

a = 5.57 m/s2 , (downward)

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h = 15 m

VO = 0

Using Kinematics Eqn (2)

h = VOt + at2

Where a g

a = 5.57 m/s2 , (downward)

15m = (0)t + ( 5.57 m/s2)t2

15m = ( 2.785 m/s2)t2

t2 = (15m/ 2.785 m/s2)

t2 = 5.386s2

t = 2.32 s

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