Download Chapter 3

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3 .1 Force,weight,and gravitational mass
Forces
Friction force
m
a
s
s
Pushing force
Kinematics describes motion, but does not explain why
motion occurs. In order to explain the cause of motion,
we must learn about Statics and Dynamics … and FORCES.
What is a Force?
•a push or pull that one body exerts on
another
•Vector
•the force causes a change in velocity, or an
acceleration…
•There are two types of forces that we
can see
•Contact Forces
•At a distance forces
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Fkick
Fgrav
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Contact Force
Forces in which the two interacting objects are physically in contact with
each other.
EXAMPLES
Hitting - Pulling with a rope - Lifting weights - Pushing a couch
Frictional Force - Tensional Force - Normal Force- Air Resistance Force
Applied Force- Spring Force
At A Distance Force (field force)
•Forces in which the two interacting objects are not in physical
contact with each other, but are able to exert a push or pull despite
the physical separation.
Example:
Gravitational Force - Electrical Force - Magnetic Force- Nuclear
Forces
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Measuring Forces
• Forces are measured in newtons (kg . m/s2).
• Forces are measured using a spring scale.
Individual Forces
4N
Net Force
10 N
6N
3N
5N
36.9
4N
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What is meant by unbalanced force?
If the forces on an object are equal and opposite, they are said to
be balanced, and the object experiences no change in motion. If
they are not equal and opposite, then the forces are unbalanced
and the motion of the object changes.
• Are forces that results in an object’s motion
being changed.
+
velocity changes (object accelerates)
Fnet
Fpull
Ffriction
N
N
W
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Balanced Force
• A force that produces no
change in an object’s
motion because it is
balanced by an equal,
opposite force.
• no change in velocity
The object shown in the diagram must be at rest since ther
is no net force acting on it.
FALSE! A net force does not cause motion. A net force
causes a change in motion, or acceleration.
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Example
On a horizontal, frictionless surface, the blocks above are being acted upon by two
opposing horizontal forces, as shown. What is the magnitude of the net force acting
on the 3kg block?
A.
B.
C.
D.
E.
zero
2N
1.5 N
1N
More information is needed.
Example
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Weight
Weight = Mass x Gravity
W  m g
•The weight of an object is the gravitational force that the planet
exerts on the object. The weight always acts downward, toward
the center of the planet.
•SI Unit of Weight: Newton (N)
•depends on gravity
– m: mass of the body (units: kg)
– g: gravitational acceleration (9.8m/s2,
• As the mass of a body increases, its’ weight increases proportionally
•the mass of a body always the same
•A measure of the resistance of an object to changes in its motion due to a force
(describes how difficult it is to get an object moving)
•Scalar
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• what is the weight of a 2 kg mass?
• W = Fg = mxg = 2 kg x 9.8 m/s2
= 19.6 N
• What is the mass of a 1000 N person?
• W = Fg = mxg
m = Fg/g = 1000 N / 9.8 m/s2
=102 kg
• A girl weighs 745 N. What is his mass
m=F÷g
m = (745 N) ÷ (9.8 m/s2)
m = 76.0 kg
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3.3
Newton's first law
Every object continues in a state of rest , or of uniform motion in a
straight line , unless it is compelled to change that state by forces
acting upon it.
An equivalent statement of the first law is that :
An object at rest will stay at rest, and an object in
motion will stay in motion at constant velocity, unless acted
upon by an unbalanced force.
This, at first, does not seem obvious. Most things on earth tend
to slow down and stop. However, when we consider the
situation, we see that there are lots of forces tending to slow
the objects down such as friction and air resistance.
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The net force acting on an object is the
vector sum of all the forces acting on it.
Examples:
9 lN
8N
8 lN
6N
4N
7N
12 N
If an object is remaining at rest, it
is incorrect to assume that there
are no forces acting on the object.
We can only conclude that the
net force on the object is zero.
8 lN
?
5N
4N
4N
3 lN
7N
– “Law of Inertia”
• Inertia
– tendency of an object to resist any change in its motion
– increases as mass increases
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Inertia
• Inertia is a term used to measure the ability of an
object to resist a change in its state of motion.
• An object with a lot of inertia takes a lot of force to
start or stop; an object with a small amount of
inertia requires a small amount of force to start or
stop.
• The word “inertia” comes from the Latin word
inertus, which can be translated to mean “lazy.”
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3.5 Newton’s Third Law of Motion
– When one object exerts a force on a second object,
the second object exerts an equal but opposite force
on the first.
• Identifying Newton’s third law pairs
– Each force has the same magnitude
– Each force acts along the same line
but in opposite directions
– Each force acts at the same time
– Each force acts on a different object
– Each force is of the same type


F AB   F BA
All forces come in action-reaction pairs
Ex: feet push backward on floor, the floor pushes forward on feet
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• Problem:
 How can a horse
pull a cart if the cart
is pulling back on
the horse with an equal but opposite force?
 Aren’t these “balanced forces” resulting in no
acceleration?
NO!!!
 Explanation:
forces are equal and opposite but
act on different objects
– they are not “balanced forces”
– the movement of the horse
depends on the forces acting on
the horse
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• Action-Reaction Pairs
FG
 The rocket exerts a downward force
on the exhaust gases.
 The gases exert an equal but opposite
upward force on the rocket.
FR
Flying gracefully through the air,
birds depend on Newton’s third law
of motion. As the birds push down on
the air with their wings, the air
pushes their wings up and gives
them lift.
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Define the OBJECT (free body)
• Newton’s Law uses the forces
acting ON object
• N and Fg act on object
• N’ and Fg’ act on other objects
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Normal Force Is Not Always Equal to the Weight
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3.6 Newton’s Second Law
• Newton’s Second Law of Motion
• The force F needed to produce an acceleration a is
F = ma
– The acceleration of an object is directly proportional to the net force
acting on it and inversely proportional to its mass.
F: force (N)
m: mass (kg)
a:acceleration (m/s2)
1 N = 1 kg ·m/s2
a
 Action-Reaction Pairs
 Both objects accelerate.
F
m
F
m a
 The amount of acceleration depends on the mass of the object.
 Small mass  more acceleration
 Large mass  less acceleration
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• If you apply more
force to an object,
it accelerates at a
higher rate.
• If an object has more
mass it accelerates at
a lower rate because
mass has inertia.
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More about F = ma
If you double the mass, you double the force. If you
double the acceleration, you double the force.
What if you double the mass and the acceleration?
(2m)(2a) = 4F
Doubling the mass and the acceleration quadruples the force.
So . . . what if you decrease the mass by half? How much force
would the object have now?
Note that this is a vector equation, and should really be
worked in component form:
∑ Fx = max
∑ Fy = may .
We can now see that Newton’s First Law of Motion is
really just a special case of his Second Law of Motion.
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example
A railway engine pulls a wagon of mass 10 000 kg along a straight track at a
steady speed. The pull force in the couplings between the engine and
wagon is 1000 N.
A. What is the force opposing the motion of the wagon?
B .If the pull force is increased to 1200 N and the resistance to movement of
the wagon remains constant, what would be
The acceleration of the wagon?
Solution
a) When the speed is steady, by Newton’s first law, the resultant force must be zero.
The pull on the wagon must equal the resistance to motion. So the force resisting
motion is 1000 N.
b) The resultant force on the wagon is 1200 – 1000 = 200 N
By Equation
F  am
200  10000a
a  0.02m / s 2
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 What force would be required to accelerate a 40 kg mass by 4 m/s2?
F = ma = (40 kg)(4 m/s2) = 160 N
 A 4.0 kg shot-put is thrown with 30 N of force. What is its acceleration?
a = F ÷ m = (30 N) ÷ (4.0 kg)= 7.5 m/s2
F  ma  30000  1.5  45000 N  4.5  10 N
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Example :
a2
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Example :
F  (20)2  (15) 2
F  25N
a
25
 5m / s 2
5
  tan 1
F
x
(a)
15
 36.90
20
 7.5  20  27.5 N
F  30.4 N
  25.3o
a= 30.4 /5 = 6.08 m/s2
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example
How much force is needed to accelerate a 1,300 kg car
at a rate of 1.5 m/s2?
A- 867 N
F = m a or
b 1,950 N
2
=
1300Kg
x
1.5m/s
c 8,493 N
F = 1950 N
d 16,562 N
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A free-body-diagram is a diagram that represents the object
and the forces that act on it.
The net force in this case is:
275 N + 395 N – 560 N = +110 N
and is directed along the + x axis of the coordinate system.
•If the mass of the car is 1850 kg then, by Newton’s second law, the acceleration is
a

F
 110 N

 0.059 m s 2
m
1850 kg
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Equilibrium
• The condition of zero acceleration is
called equilibrium.
• In equilibrium, all forces cancel out
leaving zero net force.
• Objects that are standing still are in
equilibrium because their acceleration
is zero.
• Objects that are moving at constant
speed and direction are also in
equilibrium.
• A static problem usually means there
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is no motion.
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Balancing Forces (Statics)
There are often situations where a number of forces are acting on
something, and the object has no motion – it is STATIC or in
EQUILIBRIUM or at rest.
• This means the NET FORCE on the object is zero, or in other words
the forces balance each other out.
Objects in equilibrium (no net external
•ΣF = m a
a =0
force) also move at constant velocity
ΣF = 0
•
In a two-dimensional problem, we can separate this equation into
its two component, in the x and y directions:
Σ Fx = 0
unbalanced Forces (Dynamics)
Σ Fy = 0
•There are other situations where all the forces acting on something
do not cancel each other out completely.
•This means the NET FORCE on the object is not zero, the object will
change its motion and accelerate proportional to the object’s mass.
•ΣF = m a
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Example
a) Find the acceleration of a 20 kg crate along a horizontal
floor when it is pushed with a resultant force of 10 N parallel
to the floor.
b) How far will the crate move in 5s (starting from rest)?
Solution

F  ma
a  F / m  10 / 20  0.5m / s
Distance travelled
2
 x  v0 t  (1/2) a t 2
 x  0.5  0.5  5  6.25m
2
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Example
A 1kg stone fall freely from rest from a bridge.
a -What is the force causing it to accelerate?
b -What is its speed 4s later?
c -How far has it fallen in this time?
Solution
The force causing it to fall is its weight. As it is falling with acceleration due to
gravity
F  ma
F  1  9 .8m / s
v  v0  at
v  0  9.8  4
v  39.2 m / s
1
 x  v0 
at 2
2
1
x  0 
9 .8  ( 4 ) 2
2
x  78.4 m
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3 . 7 the significance of Newton's laws of motion
3 .7 some examples of Newton's laws
Make a diagram (conceptualize)
F 0
Categorize: no acceleration (at rest )
accelerating object:
 F  ma
Isolate each object and draw a free body diagram for each
object. Draw in all forces that act on the object.
Establish a convenient coordinate system.
Write Newton’s law for each body and each coordinate
component.  set of equations.
Finalize by checking answers.

F  0
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Example
• Three people are pulling on a wagon applying forces of 100 N,150
N, and 200 N.
• The wagon has a mass of 25 kilograms.
• Determine the acceleration and the direction the wagon moves.
(2m/s2 ) to the left
Example
• An airplane needs to accelerate at 5 m/s2 to reach take-off
speed before reaching the end of the runway.
• The mass of the airplane is 5,000 kilograms.
• How much force is needed from the engine? (F=25000N)
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Free body diagram
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Example :
A lift with its load has a mass of 2000 kg. It is supported by a
steel cable. Find the tension in the cable when it:
a -is at rest
b - accelerates upwards uniformly at 1m/s2
C - move upwards at a steady speed of 1 m/s
d - moves downwards at a steady speed of 1 m/s
e - accelerates downwards with uniform acceleration of 1 m/s2
a) When at rest we can use Newton’s first law which says
that the resultant force on the lift is zero.
Force acting down is the lifts weight, the force acting up is
the tension in the cable. These two must be equal and
opposite to give a resultant force of zero.
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b)
As the lift is accelerating upwards so T must exceed the weight mg. So
the resultant acceleration force
by Newton second law, F = ma, so


 F  ma
T  mg  ma
T  mg  ma
T  2000  19600  21600 N
c)
As in (a), by Newton’s first law, the resultant force on the lift must be zero, so
cons tan t

F  0
velocity  a  0
T  mg  0
T  mg  19600 N
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d) As in (c) the tension in the cable will still equal mg since
the change in direction of motion does not alter the fact
that there is no acceleration.
cons tan t velocity  a  0

F  0
T  mg  0
T  mg  19600 N
e ) If the lift accelerates downwards, then mg must exceed
the tension T. So the resultant accelerating force is


 F  ma
T  mg  ma
T  mg  ma
T  19600  2000  17600 N
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Example :
Two hanging objects connected by a light string passing over a
frictionless pulley
What are the acceleration of the objects and the tension on
( m1 = 4 kg
m2 = 7 kg )
the string?
m1 a1 = T - m1 g
- m2 a2 = T – m2 g
- m2 a = T – m2 g ⇒ m2 a = -T + m2 g
a = g (m2 – m1) / (m1 + m2)
= ( 7-4 ) x9.8 / ( 7+4 ) = 3x 9.8 / 11 =29.4 /11
a = 2 . 6 5 m/s2
m1 a = T - m1 g
T= m1a+ m1g
T = m1 ( a+ g)
T = 4( 2.65 + 9.8 )
N
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Example:
A body of mass 5kg lies on a smooth horizontal table. It is
connected by a light in extensible string, which passes over a
smooth pulley at the edge of the table, to another body of mass
3kg which is hanging freely. The system is released from rest. Find
the tension in the string and the acceleration.


F

m
a

T  m1 a
T 5a
(1 )
T  mg  ma
T  3g  3a
(2)
solving ( 1 ) and ( 2 )
5a - 3g  - 3a
5a  3a  3x9.8
a  3x9.8/ 8  29.4/8  3.68 m / s2
T  5  3.68  18.4 N
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Example:


F


a
m
1

T  m1 g   m1 a
T  9 g  9a


F

a
m
2

T  5a
5a  9 g  9a
a  9  9.8 / 14
a  6.3 m / s 2
T  5  6.3  31.5 N
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Example:
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example
• In Fig. a, a cord holds stationary a block of mass m = 15
kg, on a frictionless plane that is inclined at angle  = 27°.
(a) What are the magnitudes of the force T on the block
from the cord and the normal force N on the block from
the plane?
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  
T  N  Fg  0
T  0  mg sin   0
T  mg sin 
 (15 kg) (9.8 m / s 2 )(sin 27  )
 67 N
N  mg cos   0
N  mg cos 
 (15 kg) (9.8 m / s 2 )(cos 27 )
 131 N
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3-12
Friction
Friction results from relative motion between objects.
Frictional forces are forces that resist or oppose motion.
Sliding Friction:
• When two solid surfaces slide over each other.
• The amount of friction between two surfaces depends on two
factors.
1. The kinds of surfaces.
2. The force pressing the surfaces together (normal force).
• Static friction is the frictional force that prevents two surfaces from
sliding past each other.
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Normal force (N)
Friction force (N) Ff =  S N
Coefficient of friction
fs  f
max
s
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Friction force
To first order, we consider two effects on friction
1. Normal force of contact between the surfaces
2. Types of surfaces in contact (texture of surfaces
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The following empirical laws hold true about friction:
- Friction force, f, is proportional to normal force,N.
f s  s N
f k  k N
- s and k: coefficients of static and kinetic friction, respectively
- Direction of frictional force is opposite to direction of relative motion
- Values of s and k depend on nature of surface.
- s and k don’t depend on the area of contact.
- s and k don’t depend on speed.
- s, max is usually a bit larger than k.
- Range from about 0.003 (k for synovial joints in humans) to 1 (s for rubber on
concrete)..
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Table of friction coefficients
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Example
A 10 N force pushes down on a box that weighs 100 N.
As the box is pushed horizontally, the coefficient of
sliding friction is 0.25.
Determine the force of friction resisting the motion.
Fs = μ s x N
(10+100)x0.25 =27.5
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Example
The sled comes to a halt because the kinetic frictional force
opposes its motion and causes the sled to slow down.
Suppose the coefficient of kinetic friction is 0.05 and the total
mass is 40kg. What is the kinetic frictional force?
f k  k FN  k mg 


0.0540kg  9.80 m s 2  20kg
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Example
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example
A 1200 kg car moving at 100 km/h coasts to a stop in 25 seconds.
What is the value of the static coefficient of friction?
m = 1200 kg
vo = 100km/h = 27.8 m/s
t = 25 s
µ=F/N
where F = ma & N=mg
µ= ma/mg
=(m(|v – vo|)/t)/mg
vo
f
From v = vo + at,
a = (v-vo)/t
=(27.8m/s/25 s )/ 9.8 m/s2)
µ = 0.113
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A 10-kg box is being pulled across the table to the
right at a constant speed with a force of 50N.
a) Calculate the Force of Friction
b) Calculate the Force Normal
FN
Ff
Fa
Fa  F f  50 N
mg  N  (10)(9.8)  98 N
mg
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Suppose the same box is now pulled at an angle of 30 degrees
above the horizontal.
a) Calculate the Force of Friction
Fax  Fa cos   50 cos 30  43.3N
b) Calculate the Force Normal
F f  Fax  43.3N
FN
Ff
Fa
Fay
30
Fax
mg
N  mg
N  Fay  mg
N  mg  Fay  (10)( 9.8)  50 sin 30
N  73N
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Example
What is the weight of an object that is being pulled at a
constant velocity by a force of 25 N if the coefficient of sliding
friction between the object and the surface is 0.75?
∑ Fy = 0
N - W =0
∑ Fx = ma
(a =0
∑ Fx = 0
Fk= F = 25 N
Fk= μk N
N = Fk / μk
W = N = 25 / 0.75 = 33.3 N
W = 33.3 N
N=W
constant velocity )
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example
Suppose 35 kg crate was not moving at a constant speed,
but rather accelerating at 0.70 m/s2. Calculate the
applied force. The coefficient of kinetic friction is still
0.30.
N
Fa
Ff
mg
Example 3- 29
A block is at rest on an inclined plane the coefficient of static fraction
is µ
s
.what is the maximum possible angle of inclination θ ( max ) of the ssurface
for which the block will remain at rest
F s = w sin θ
N = w cos θ
tanθ

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


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Summary of Chapter 3
• Newton’s
first law: If the net force on an object is zero, it will remain
either at rest or moving in a straight line at constant speed.
• Newton’s second law:
• Newton’s third law:
• Weight is the gravitational force on an object W= m g.
• Free-body diagrams are essential for problem-solving. Do one object
at a time, make sure you have all the forces, pick a coordinate system
and find the force components, and apply Newton’s second law along
each axis.
•Kinetic friction:
Fk = μkN
• Static friction:
Fs ≤ μsN .
• Fk < Fs
μk < μs
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