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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
NEWTONS LAW OF MOTION
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NEWTONS SECOND LAW OF MOTION
NEWTONS 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
NEWTONS LAW OF MOTION
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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
FIRST CONDITION OF EQUILIBRIUM (FCE)
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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
FIRST CONDITION OF EQUILIBRIUM (FCE)
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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FIRST CONDITION OF EQUILIBRIUM (FCE)
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
FIRST CONDITION OF EQUILIBRIUM (FCE)
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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FIRST CONDITION OF EQUILIBRIUM (FCE)
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
FIRST CONDITION OF EQUILIBRIUM (FCE)
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
NEWTONS 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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NEWTONS SECOND LAW OF MOTION (NSLM)
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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NEWTONS SECOND LAW OF MOTION (NSLM)
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)
NEWTONS SECOND LAW OF MOTION (NSLM)
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
NEWTONS SECOND LAW OF MOTION (NSLM)
(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?
NEWTONS SECOND LAW OF MOTION (NSLM)
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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NEWTONS SECOND LAW OF MOTION (NSLM)
f
W = mg
N
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?
af
W = mg
N
a
Using NSLM to determine the acceleration FBD :
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NEWTONS SECOND LAW OF MOTION (NSLM)
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?
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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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?
NEWTONS SECOND LAW OF MOTION (NSLM)
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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NEWTONS SECOND LAW OF MOTION (NSLM)
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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NEWTONS SECOND LAW OF MOTION (NSLM)
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?
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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NEWTONS SECOND LAW OF MOTION (NSLM)
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?
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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NEWTONS SECOND LAW OF MOTION (NSLM)
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
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