#### Transcript Document

```SECOND LAW OF MOTION
If there is a net force acting on an object, the object will
have an acceleration and the object’s velocity will
change.
Newton's Second Law of Motion states that for a
particular force, the acceleration of an object is directly
proportional to the net force and inversely proportional
to the mass of the object. The direction of the force is
the same as that of the acceleration. In equation form:
F
a
m
 F  ma
SECOND LAW OF MOTION
In the SI system, the unit for force is the newton (N):
A newton is that net force which, when applied to a 1-kg
mass, gives it an acceleration of 1 m/s2.
Acceleration and Force With Zero
Friction Forces
Pushing the cart with twice the force produces
twice the acceleration. Three times the force triples
the acceleration.
Acceleration and Mass Again With
Zero Friction
F
F
a/2
a
Pushing two carts with same force F
produces one-half the acceleration. The
acceleration varies inversely with the mass.
THIRD LAW OF MOTION
According to Newton's third law of motion, when one
body exerts a force on another body, the second body
exerts on the first an equal force in opposite direction.
The Third Law of Motion applies
to two different forces on two
different objects: " For every
action force, there must be an
equal and opposite reaction
force. "
Action and reaction forces never
balance out because they act on
different objects.
Newton’s Third Law
• Third Law: For every action force, there
must be an equal and opposite reaction
force. Forces occur in pairs.
Action
Reaction
Reaction
Action
Acting and Reacting Forces
• Use the words by and on to study
action/reaction forces below as they
relate to the hand and the bar:
Action
The action force is exerted by
bar
the hands
_____ on the _____.
The reaction force is exerted
by the _____
bar on the _____.
hands
Reaction
MASS
The property that a body has of resisting any change in
its state of rest or of uniform motion is called inertia.
The inertia of a body is related to the amount of matter it
contains. A quantitative measure of inertia is mass.
The unit of mass is the kilogram (kg).
WEIGHT (FG)
Weight (a vector quantity) is different from mass (a
scalar quantity). The weight of a body varies with its
location near the Earth (or other astronomical body),
whereas its mass is the same everywhere in the
universe.
The weight of a body is the force that causes it to be
accelerated downward with the acceleration of gravity g.
FG = mg
Units: Newtons (N)
g = acceleration due to gravity
= 9.8 m/s2
THE NORMAL FORCE
A normal force is a force exerted by one surface on
another in a direction perpendicular to the surface of
contact.
Note: The gravitational force and the normal force
are not an action-reaction pair.
Applying Newton’s Second Law
• Read the problem and draw a sketch.
• Write down the data in correct units:
m = kg
FG = N
• Draw a free-body diagram for each object.
• Choose x or y-axis along motion and choose
direction of motion as positive.
• Write Newton’s law for both axes:
SFx = m ax
SFy = m ay
• Solve for unknown quantities.
4.1 A 5.0-kg object is to be given an upward acceleration of
0.3 m/s2 by a rope pulling straight upward on it. What must be
the tension in the rope?
m = 5 kg
a = 0.3 m/s2
N2L
ΣFy = FT - FG = ma
FT
a (+)
FG
FT = m(a + g)
= 5(0.3 + 9.8)
= 50.5 N
5 kg
4.2 A cord passing over a frictionless pulley has a 7.0 kg mass
hanging from one end and a 9.0-kg mass hanging from the other.
(This arrangement is called Atwood's machine).
a. Find the acceleration of the masses.
m1 = 7 kg
m2 = 9 kg
N2L
FT
7kg
9 kg
FT
a (+)
a (+)
FG1
FG2
FT
FT
a (+)
a (+)
FG1
FG2
FG 2  FG1
a
mT
ΣF = FT - FG1 - FT + FG2 = mTOTALa
9(9.8)  7(9.8)
= 1.22 m/s2

97
FT
FT
a (+)
a (+)
FG1
FG2
b. Find the tension of the
cord
Using either side of the
pulley yields the same
answer!
FT – FG1 = m1a
FT = m1a + FG1
= m1 (a + g)
= 7(1.22 + 9.8)
= 77.1 N
APPARENT WEIGHT
The actual weight of a body is the gravitational force that
acts on it. The body's apparent weight is the force the
body exerts on whatever it rests on. Apparent weight
can be thought of as the reading on a scale a body is
placed on.
scale
4.3 What will a spring scale read for the weight of a 75 kg man in
an elevator that moves:
m = 75 kg
FG = 75(9.8) = 735 N
a. With constant upward speed of 5 m/s
N1L
m = 75 kg
FG = 735 N
FN
ΣFy = FN - FG = 0
F N = FG
= 735 N
FG
b. With constant downward speed of 5 m/s
FN = 735 N
FN
b. Going up with an
acceleration of
0.25 g
a = 2.45 m/s2
N2L
FG
ΣF = FN - FG = ma
FN = ma + FG
= 75 (2.45) + 735
= 919 N
FN
c. Going down with an
acceleration of 0.25 g
a = 2.45 m/s2
N2L
FG
ΣF = FG - FN = ma
FN = FG - ma
= 735 - 75 (2.45)
= 551 N
For free fall the only
force acting is FG
e. In free fall?
so FN = 0 N
N2L
FG
“Weightlessness”
More properly, this effect is called apparent
weightlessness, because the gravitational force still
exists. It can be experienced on Earth, but only briefly:
FRICTION: STATIC AND KINETIC FRICTION
Frictional forces act to oppose relative motion between
surfaces that are in contact. Such forces act parallel to
the surfaces.
Static friction occurs between surfaces at rest relative to
each other. When an increasing force is applied to a
book resting on a table, for instance, the force of static
friction at first increases as well to prevent motion. In a
given situation, static friction has a certain maximum
value called starting friction. When the force applied to
the book is greater than the starting friction, the book
begins to move across the table.
The origin of
friction: on a
microscopic scale,
most surfaces are
rough.
FRICTION: STATIC AND KINETIC FRICTION
The kinetic friction (or sliding friction) that occurs
afterward is usually less than the starting friction, so
less force is needed to keep the book moving than to
start it moving.
COEFFICIENT OF FRICTION
The frictional force between two surfaces depends on
the normal force FN pressing them together and on the
natures of the surfaces. The latter factor is expressed
quantitatively in the coefficient of friction  (mu) whose
value depends on the materials in contact. The frictional
force is experimentally found to be:
Ff   s FN Static friction
Ff   k FN Kinetic friction
4.4 A horizontal force of 140 N is needed to pull a 60.0 kg box
across the horizontal floor at constant speed. What is the
coefficient of friction between floor and box?
N1L
FN
Fa = 140 N
m = 60 kg
Fa
Ff
FG
ΣFx = Fa – Ff = 0
Fa = Ff = μFN
ΣFy = FN – FG = 0
FN = FG =mg
= 60(9.8)
= 588 N
Fa 140


FN 588
= 0.24
4.5 A 70-kg box is slid along the floor by a horizontal 400-N force.
Find the acceleration of the box if the value of the coefficient of
friction between the box and the floor is 0.50.
N2L
m = 70 kg
Fa = 400 N
μ = 0.5
ΣFy = FN – FG = 0
FN = FG =mg
= 70(9.8)
= 686 N
FN
Ff
Fa
FG
ΣFx = Fa – Ff = ma
Ff = μFN
= (0.5)(686)
= 343 N
m = 70 kg
Fa= 400 N
μ = 0.5
ΣFx = Fa – Ff = ma
a
FN
Ff
Fa
FG
Fa  F f
m
400  343

= 0.81 m/s2
70
4.6 A 70-kg box is pulled by a rope with a 400-N force at an angle of
30 to the horizontal. Find the acceleration of the box if the
coefficient of friction is 0.50
N2L
FN
Fa
Ff
FG
m = 70 kg
Fa = 400 N, 30
μ = 0.5
Fax = 400 cos 30 = 346.4 N
Fay = 400 sin 30 = 200 N
ΣFy = FN + Fay - Fg = 0
FN = FG - Fay
= 70(9.8) - 200
= 486 N
Ff = μFN
= (0.5)(486)
= 243 N
m = 70 kg
Fa = 400 N, 30
μ = 0.5
FN
Fa
Ff
ΣFx = Fax – Ff = ma
FG
a
Fax  F f
m
346 .4  243

= 1.47 m/s2
70
4.7 A box sits on an incline that makes an angle of 30˚ with the
horizontal. Find the acceleration of the box down the incline if the
coefficient of friction is 0.30.
N2L
FN
Ff
ΣFy = FN - FGy = 0
FN = FGy
ΣFx = FGx – Ff = ma
θ
FG
θ = 30
μ = 0.3
FGy
FGx
Ff = μFN
ΣFx = FGx –μ FGy = ma
FN
Ff
ΣFx = FGx – μ FGy = ma
mg sin θ - μ mg cos θ = ma
θ
FG
FGx
FGy
a  g (sin    cos  )
 9.8(sin 30   cos30 )
= 2.35 m/s2
4.8 Two blocks m1 (300 g) and m2 (500 g), are pushed by a force F.
If the coefficient of friction is 0.40.
a. What must be the value of F if the blocks are to have an
N2L
2
acceleration of 200 cm/s ?
m1 = 0.3 kg
m2 = 0.5 kg
μ = 0.4
a = 2 m/s2
FN1
Ff1
F
Ff2
FG1
FG1 = FN1
FG2 = FN2
FN2
FG2
ΣFx = F - Ff1 - Ff2 = mT a
F = mTa + Ff1 + Ff2
= mTa + μ FNT
= mTa + μ mT g
= 0.8(2) + (0.4) 0.8 (9.8)
= 4.7 N
b. How large a force does m1 then exert on m2?
m2 alone
FN2
Ff2
F12
FG2
ΣFx = m2a
F12 - Ff2 = m2a
F12 = Ff2 + m2a
= μ m2 g + m2a
= 0.4(0.5) (9.8) + 0.5 (2)
= 2.96 N
4.9 An object mA = 25 kg rests on a tabletop. A rope attached to it
passes over a light frictionless pulley and is attached to a mass
mB = 15 kg. The coefficient of friction between the table and block
A is 0.20. a. What is the acceleration of the 25 kg block?
FN
Ff
25 kg
FT
FGA
FT
15 kg
FGB
mA = 25 kg
mB = 15 kg
μ = 0.2
FN = FGA
= 25(9.8)
= 245 N
FfA = μFN
= 0.2 (245)
= 49 N
N2L
ΣF = FT - FfA + FGB - FT = mTa
FN
Ff
a
FT
FT
FGA
FGB
mA = 25 kg
mB = 15 kg
μ = 0.2
FGB  FfA
mT
15(9.8)  49

25  15
= 2.45 m/s2
b. What is the tension on the string?
ΣF = FT - Ff = mAa
FN
Ff
FT
FT
FGA
FGB
F T = m Aa + F f
= 25 (2.45) + 49
= 110.25 N
4.10 Blocks 1 and 2 of masses ml and m2, respectively, are
connected by a light string, as shown above. These blocks are
further connected to a block of mass M by another light string that
passes over a pulley of negligible mass and friction. Blocks l and
2 move with a constant velocity v down the inclined plane, which
makes an angle  with the horizontal. The kinetic frictional force on
block 1 is f and that on block 2 is 2f.
a. On the figure below, draw and label all the forces on block ml.
FN
FT
FG
Ff
Express your answers to each of the following in terms of
ml, m2, g, , and f.
b. Determine the coefficient of kinetic friction between the inclined
plane and block 1.
f  N
N  m1 g cos
f

m1 g cos 
c. Determine the value of the suspended mass M that allows
blocks 1 and 2 to move with constant velocity down the plane.
FN2
FN1
FT
N1L
2f
FT
FT
f
FT
m2 g
m1 g
Mg
m1g sin   f  m2 g sin   2 f  Mg  0
Mg  m1g sin  f  m2 g sin  2 f
3f
M  (m1  m2 ) sin  
g
d. The string between blocks 1 and 2 is now cut. Determine the
acceleration of block 1 while it is on the inclined plane.
m1 g sin   f  m1a
FN1
f
m1 g
N2L
f
a  g sin  
m1
Summary
Newton’s First Law: An object at rest or an object
in motion at constant speed will remain at rest or
at constant speed in the absence of a net force.
Newton’s Second Law: A net force produces an
acceleration in the direction of the force that is
directly proportional to the force and inversely
proportional to the mass.
Newton’s Third Law: For every action force,
there must be an equal and opposite reaction
force. Forces occur in pairs.
Summary: Procedure
FR  ma;
FR
a
m
N = (kg)(m/s2)
• Read and write data with units.
• Draw free-body diagram for each object.
• Choose x or y-axis along motion and choose
direction of motion as positive.
• Write Newton’s law for both axes:
SFx = m ax
SFy = m ay
• Solve for unknown quantities.
```