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CLB 10102 : PHYSICS
CHAPTER 3
Newton’s Laws of Motion and Friction
Ahmad Azahari Hamzah
[email protected]
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Physical Quantities and Dimensions
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Forces Acting at a Point
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Newton’s Laws of Motion and Friction
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Work, Energy and Power
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Optics
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Simple Machines
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The Effects of Forces on Materials
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Heat Energy
Thermal Expansion
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
3.1 Equation of Motion in a Straight Line
Motion = A change of position.
Position and Displacement
 The most basic information you must have to describe the
motion of an object is its position and the time it was at
that position.
 The position of an object is always taken from some
reference point (which is usually "zero" on the scale).
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
 Displacement is a change from one position x1 to another
position x2. Displacement has two features: (1) its
magnitude; (2) its direction.
Δx = x2 - x1
 So it is a vector quantity.
 The value of the displacement of the object can be +ve /ve. It can move along a horizontal/vertical line.
 At reference point O, if the object moves 2m to the right,
it can be expressed as +2.0 m. then if the object moves
3m to the left :. – 3.0 m.
 Examples :
Δx = (12m) – (5m) = +7m
Δx = (1m) – (5m) = - 4m
Δx = (5m) – (5m) = 0
Example:
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
• An object P at the reference point O. Then it moves
along a path given as follows:
– 10 m to the east, then 5 m to the south, finally 15 m to the
west. Determine the displacement of the object P relative to
O when it arrives at the final position.
10 m
5m
p
15 m
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
Average Velocity
 Average velocity: the ratio of the displacement Dx that
occurs during a particular time interval Dt to that interval,
v =
Δx
=
Δt
x2 - x1
t 2 - t1
Δx = displacement (m)
Δt = time (s)
 Example 1: An athlete tried to run the 200m race to
break the world's record! Unfortunately, it takes him
25.73s to complete the run. Calculate his average
velocity.
v = Δx = 200m = 7.77 m/s
Δt
25.73s
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
 Example 2: A high speed train might be built between KL
Central and KLIA. It could travel between the two places at
an average velocity of +130km/h. The trip would take 45min.
Calculate how far apart KL Central and KLIA.
Solution:
v=
Δx
Δt
x = vt
= (130km/h) (0.75h)
= 97.5 km
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
Average Speed
 Please note that speed is not the same thing as velocity.
Average Speed is not a vector quantity and has only a
magnitude.
 It is calculated very much like average velocity except that
instead of calculating the total displacement, we calculate the
total distance a particle travels during an interval of time.
 So speed can be found with the formula:
s=
Total distance
Δt
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
Average Acceleration
• Acceleration is the change in velocity per unit time.
Δv
v 2 - v1
a =
=
Δt
t2 - t 1
• Example: A car accelerates along a straight road from rest to
75 km/h in 5.0 s. What is the magnitude of its average
acceleration?
v1 = 0 , v2 = 75 km/h , Δt = 5.0 s
75 km x 1000 m x 1 h = 21 m/s
h
1 km
3600 s
a = (21 – 0) m/s
5.0 s
= 4.2 m/s2
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
Motion at Constant Acceleration
Equations for motion at constant acceleration
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
1. Example: A brick of mass 1.0 kg is dropped from a height of
85 m. Neglect wind resistance. (a = g = 9.8 m/s2).
a) For how much time does the brick fall before hitting the
ground?
b) How fast is the brick travelling just before it hits the
ground?
Solution:
vo = 0 , a = 9.8 m/s2
85 m
CLB 10102 Physics
a)
x – xo = vo t + ½at2
2(x – xo) = 2vot + at2
t2 = 2(x – xo)
a
t = 2(x – xo)
a
= 2(85 m)
9.8 ms-2
= 4.16 s
CHAPTER 3 Newton’s Laws of Motion and Friction
b) v = vo + at
= 0 + 9.8 ms-2 ( 4.16 s)
= 40.77 ms-1
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
2. Example: Two long-separated friends, Jill and Bill, spot each
other at the National Park from a distance of D = 25 m apart.
Starting from rest, they cycle toward each other. Bill
accelerates from east at a constant rate of aB = 1.5 ms-2,
while Jill accelerates from the opposite direction at a
constant rate of aJ = 1.2 ms-2. How far from Jill’s initial
position do they meet?
Solution:
Bill
Jill
aJ = 1.2 ms-2
aB = -1.5 ms-2
xJ = xB
xoJ = 0
25 m
xoB = 25 m
CLB 10102 Physics
Jill:
xJ – xoJ = vot + ½aJt2
xJ = xoJ + vot + ½aJt2
CHAPTER 3 Newton’s Laws of Motion and Friction
Bill:
xB – xoB = vot + ½aBt2
xB = xoB + vot + ½aBt2
= 0 + 0 + ½ (1.2 ) t2
= 25 + 0 + ½ (-1.5 ) t2
= 0.6 t2 …..(1)
= 25 - 0.75 t2 …..(2)
xJ = xB
0.6 t2 = 25 – 0.75 t2
1.35 t2 = 25
t2 =
25 m
1.35 ms-2
t = 4.3 s
xJ = 0.6 t2
= 0.6 ms-2 (4.3 s)2
= 11 m
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
3. Example: A motorcycle slowing to a petrol station has an
average acceleration of -3.0 m/s2.[Note that a minus (-)
acceleration is commonly called deceleration, meaning that
the motorcycle is slowing down]. If its initial velocity is 20
m/s, how far does it travel in 4.0 s?
Solution:
a = -3.0 m/s2
t = 4.0 s
Δx = x – xo
vo = 20.0 m/s
Δx = ?
x – xo = vo t + ½at2
= 20 m/s (4.0 s) + ½(-3.0 m/s2)(4.0 s)2
= 80 m – 24 m
= 56 m
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
4. Example: A pitcher tosses a baseball up along a y axis, with
an initial speed of 12 m/s.
a) How far does the ball go?
b) How long is it in the air?
Solution:
a)
y = v 2 – vo 2
2a
= 0 – (12m/s)2
2(-9.8 m/s2)
= 7.3 m
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
b) v = vo+ at
t = v – vo
a
= (0 – 12) m/s
-9.8 m/s2
= 1.2 s
The time for the ball to reach its highest point is the same as
the time for it to fall into the pitcher’s hand.
2 x 1.2 s = 2.4 s
CLB 10102 Physics
3.2 Velocity-Time Graph
CHAPTER 3 Newton’s Laws of Motion and Friction
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
Its velocity increases in magnitude
until it reaches its cruising speed.
Its acceleration is greatest at the start
( from A to B), when its increase in
velocity is largest.
Velocity
Consider a bus moving out of the
station due east.
A
B
C
Time
Then the bus is moving at the relatively constant velocity
(from B to C), its acceleration is near zero.
Then as the bus approaches the first bus stop, its speed
decreases as it decelerates.
D
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Example 1:
Two rocks are dropped from rest from the top of a high
tower. Rock 1 is dropped exactly 1 s before rock 2 is
dropped. Assume negligible wind resistance throughout.
a) At the moment rock 2 gets dropped, what is the
distance between the two rocks?
b) On a single set of axes, sketch the velocity vs. time of
rock 1 and rock 2.
c) After the rocks have been falling for a few seconds,
does the distance between the two rocks increases,
decreases or stay the same as they fall?
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
Solution:
a) a = 9.8 ms-2 , vo = 0 , t = 1.0 s , xo = 0
x – xo = vo t+ ½at2
= 0 + 0 + ½(9.8 ms-2)(1.0 s)2
= 4.9 m
b) Both rocks accelerate at the same constant rate. So, both rock’s
v
graph are upward-sloped
straight lines with the same
Rock 1
slope.
Rock 2
1s
t
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
c) Since both rocks speed up, the slopes of their x vs. t graphs
increase. The upward curves indicate downward-moving rocks.
Both rocks are described by the same curve. Since rock 1 is
always moving faster than rock 2, so rock 1 gets farther and
farther ahead of rock 2.
The distance between the rocks keeps increasing.
x
Rock 1
Rock 2
1s
t
Example 2:
Amir is traveling down the interstate at 45.0 m/s. He looks
ahead and observes an accident which results in a pileup in
the middle of the road. By the time Amir slams on the
breaks, he is 50.0 m from the pileup. He slows down at a
rate of –15.0 m/s2.
a) Construct a velocity-time plot for Amir's motion.
b) Determine the distance which he would travel prior to
reaching a complete stop (if he does not collide with
the pileup). Will Amir hit the cars in the pileup?
a) The velocity-time graph for the motion is:
The distance traveled can be found by calculating the area
between the line on the graph and the axes.
b) Distance, x = area of triangle
= 0.5vt
= (0.5)(3.0 s)(45.0 m/s)
= 67.5 m
OR: we can use kinematic equation to solve the
problem.
v2 = vo2 + 2a∆x
(0 m/s)2 = (45.0 m/s)2 + 2(-15.0 m/s2)(∆x)
- 2025.0 m2/s2 = (-30.0 m/s2)(∆x)
∆x = 67.5 m
Since the accident pileup is only 50.0 m from Amir and
he needs 67.5 m to skid before coming to a complete
stop, Amir will indeed hit the pileup (unless, of course,
he veers aside).
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
3.3 Newton’s Laws of Motion
Newton’s First Law of Motion
Every object in a state of uniform motion tends to remain in
that state of motion unless an external force is applied to it.
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
 Newton’s First Law (Law of Inertia)
 Newton's First Law states that every object will
remain at rest or in uniform motion in a straight line
unless compelled to change its state by the action of
an external force.
 This is normally taken as the definition of inertia.
 The key point here is that if there is no net force acting
on an object (if all the external forces cancel each
other out) then the object will maintain a constant
velocity.
 If that velocity is zero, then the object remains at rest.
 If an external force is applied, the velocity will
change because of the force.
Newton’s Second Law of Motion
The acceleration of an object is directly proportional to the
net force acting on it and inversely proportional to its mass.
The relationship between an object's mass m, its acceleration
a, and the applied force F is F = ma.
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
 Newton’s Second Law (Law of Acceleration)
 The Second Law explains how the velocity of an object
changes when it is subjected to an external force.
 The law defines a force to be equal to change in
momentum (mass x velocity) per change in time.
 For an object with a constant mass m, the second law
states that the force F is the product of an object's mass
and its acceleration a:
F = ma
Newton’s Third Law of Motion
For every action there is an equal and opposite reaction.
F1 = -F2
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
 Newton’s Third Law (Law of Action & Reaction)
 The Third Law states that for every action (force) in
nature there is an equal and opposite reaction. In other
words, if object A exerts a force on object B, then
object B also exerts an equal force on object A.
 Notice that the forces are exerted on different objects.
 The third law can be used to explain the generation of
lift by a wing.
F1 = -F2
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
3.4 Coefficient of Friction
 Friction is typically characterized by a coefficient of friction
which is the ratio of the frictional resistance force to the
normal force which presses the surfaces together.
 In this case the normal force is the weight of the block.
 Typically there is a significant difference between the
coefficients of static friction and kinetic friction.
Ff = FN μf
where Ff = force of friction,
FN = force perpendicular to the contact
surface,
μf = the coefficient of friction
 The coefficient of friction is a scalar value used to calculate
the force of friction between two bodies.
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
 The coefficient of friction depends on the materials used.
 Example: ice on metal has a very low coefficient of friction
(they rub together very easily), while rubber on pavement
has a very high coefficient of friction (they do not rub
together easily).
 The force of friction is always exerted in a direction that
opposes movement. For example, a chair sliding to the right
across a floor experiences the force of friction in the left
direction.
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
 The coefficient of friction also depends on the type of
friction. There are two general types of friction:
 Static friction occurs when the two objects are not
moving relative to each other (like a desk on the
ground). The coefficient of static friction is typically
denoted as μs. The initial force to get an object moving
is often dominated by static friction. (Force between
objects at rest)
 Kinetic friction occurs when the two objects are moving
relative to each other and rub together (like a sled on the
ground). The coefficient of kinetic friction is typically
denoted as μk, and is usually less than the coefficient of
static friction.
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Kinetic Friction
Sliding Friction Solid surfaces slide over each other
Example: Ice skating
Rolling friction An object rolls over a surface
Example: Rollerblading, car's wheels on the ground
Fluid Friction An object moves through a fluid
Example: Submarine, swimmer
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
3.5 Application of Friction
1. The friction between the tires of your automobile and the
road determine your maximum acceleration, and more
importantly your minimum stopping distance. Without
friction, the tires could not push against the ground to
move the car forward and the brakes could not stop the
car. Without friction, a car is useless.
2. Friction between your pencil and your paper is necessary
for the pencil to leave a mark.
3. Without friction, you would just slip and fall when you
tried to walk.
4. To increase friction, make surface rougher, e.g: Sand
scattered on icy roads keeps cars from skidding.
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
3.6 Simple Problems on Mechanics Involving
Newton’s Laws and Friction
Example 1:
A force acts on a 2 kg mass and gives it an acceleration of 3 m/s2.
How strong is the force that does this? What acceleration would
the force produce on a 1.5 kg mass?
F = ma
= (2 kg)(3m/s2)
= 6 kgms-2 or 6 N
a = F/m
= 6 kgms-2
1.5kg
= 4 ms-2
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Example 2:
A 900 kg car is going 20 m/s along a level road. How large a
constant retarding force is required to stop it in a distance of 30 m?
v2 = v02+2a(x-x0)
a = v2-v02
2(x-x0)
= 0 – (20 m/s)2
2(30 m)
= -6.66 ms-2
F = ma
= (900 kg)(-6.66 ms-2)
= -5994 kgms-2 or -5994 N
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Example 3:
A block of mass m= 20 kg is hanging from a rope as shown. If
θ1 = 45o and θ2= 30o. What are the tensions, T1, T2, and T3 in the
two ropes?
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
Solution:
T3-mg = ma ……(1)
T2y
T1y
T1x
T2x
Block isn’t moving, so a = 0
T3-mg=0 …….(2)
T3 = mg
= (20 kg)(9.8 ms-2)
= 196 kgms-2 or N
Find the component for each vector:
For T1: T1x = -T1cos45 , T1y = T1sin45.
For T2: T2x = T2cos30 , T2y = T2sin30
For T3: T3x = 0
, T3y = 196 N
CHAPTER 3 Newton’s Laws of Motion and Friction
CLB 10102 Physics
T2y
T1y
T1x
T2x
T1x + T2x + T3x = 0
-T1cos 45 + T2 cos 30 + 0 = 0
T2cos 30 = T1cos 45
T1 = T2 cos 30
cos 45
T1 = 1.22 T2 …..(3)
T1y + T2y + T3y = 0
T1 sin 45 + T2 sin 30 - 196 N = 0
T1 sin 45 + T2 sin 30 = 196 N ….(4)
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Substitute (3) into (4):
1.22 T2 sin 45 + T2 sin 30 = 196 N
T2 (1.22 sin 45 + sin 30) = 196 N
T2 =
196 N
1.22 sin 45 + sin 30
= 143.83 N
T1 = 1.22 T2
= 1.22 (143.83 N)
= 175.47 N
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Example 4:
A block of mass, m= 20 kg is sitting on an inclined plane,
whose angle is θ = 30o as shown. Find the tension T in the rope,
which is holding the block from sliding down the incline.
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Solution:
Since the block doesn’t move,
-Fs + T = 0
T = Fs
Fs= mg sin θ
= (20 kg)(9.8 ms-2)(sin 30)
= 97.99 N
Therefore,
T = Fs
= 97.99 N
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Example 5:
Two blocks, one of mass m1= 10 kg and the other of mass m2 =
15 kg are an inclined plane as shown. The angle of the incline is
θ = 35o. Find the tension, T, in the rope that connects them
through the frictionless pulley, and the acceleration, a, of the
blocks.
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
Solution:
Firstly, we need to identify the
forces acting on each block.
For Block 1:
Fs= m1g sin θ ……(1)
For Block 2:
m2g - T = m2a ……(3)
-m1g sin θ + T = m1a
T = m1a + m1g sin θ ……(2)
m2g - (m1a + m1g sin θ) = m2a
Substitute (2) into (3): m2a + m1a = m2g – m1g sinθ
a(m2 + m1) = m2g – m1g sinθ
a = m2g – m1g sinθ
(m2 + m1)
……(4)
CLB 10102 Physics
CHAPTER 3 Newton’s Laws of Motion and Friction
a = m2g – m1g sinθ
(m2 + m1)
= (15 kg)(9.8 ms-2) – (10 kg)(9.8 ms-2)(sin 35)
(15 + 10) kg
= 3.63 ms-2
Substitute a into (2):
T = m1a + m1g sin θ
= (15 kg)(3.63 ms-2) + (10 kg)(9.8 ms-2)(sin 35)
= 110.66 N