Class VIII Physics, Jan Session
Half-Yearly Term 2022
Handout-1
Chapter 1: Movement & Position
© Maple Leaf International School
Syllabus for Half-Yearly Exam 2022
Chapter 1: Movement & Position
Content Overview
• Distance & Displacement
• Speed – Instantaneous Speed & Average Speed
• Experiment to measure the average speed of an object
• Distance-Time Graphs
• Velocity
• Acceleration
• Velocity-Time Graphs
• Equation for uniformly accelerated motion
Distance & Displacement
Distance is the total length of the path of motion – it is a scalar quantity
Displacement is the linear distance between the initial and final point of an object – it is a vector quantity
Boy travels from A-B then B-C
Distance = 3m+4m = 7m
SI unit for both distance and displacement is meters
Displacement = 5m north east
symbol: m
Worked Example
Q. A student starts from point O and walks 4 meters east, 2 meters south, 4 meters west, and finally 2 meters north
as shown below. What is the distance walked by the student? What is his displacement?
Ans. Distance = 4m + 2m + 4m +2m = 12m
Displacement = 0m
-Since student is back to the point where he started from, initial point = final point. Therefore displacement is 0
-Displacement, being a vector quantity, must give attention to direction:
the 4 meters east cancels the 4 meters west; and the 2 meters south cancels the 2 meters north
- Distance is a scalar quantity - direction of motion ignored in determining overall distance
Speed
• Speed is the distance travelled per unit time:
𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑠𝑝𝑒𝑒𝑑 =
𝑡𝑖𝑚𝑒
• It is a measure of how fast an object is moving:
- higher speed means an object is moving faster
- lower speed means it is moving slower
- If an object is at rest or not moving at all,
it has zero speed
• Speed is a scalar quantity
• SI unit: m/s
symbol
𝑣=
𝑠
𝑡
Pg. 15 (Chapter Questions)
Q1 A sprinter runs 100 meters in 12.5 seconds.
Calculate the speed in m/s.
Ans
s=100m, t=12.5s
v=s/t
v=100m/12.5s
v=8m/s
Q2. A jet can travel at 350m/s. Calculate how far
it will travel at this speed in:
a. 30 seconds b. 5 minutes c. half an hour
v=s/t  s=vt
a. t=30s
s=350 x 30
b. t=5min=5x60=300s
s= 350m/s x 300s= 105000m (105km)
= 10,500m (10.5km)
c. t=half an hour=0.5h=0.5 x 60= 30min x 60= 1800s
s = 350m/s x 1800s = 630, 000m (630km)
1km= 1000m
Instantaneous Speed & Average Speed
1. Instantaneous speed - speed of an object at a particular moment (instant) in time
2. Average Speed =
Total Distance Covered
Total Time Taken
Example:
• A moving car does not have the same speed during its
travel. It stops at red lights then speeds up and again
slow down when approaching a zebra crossing.
• Sometimes it speeds up and sometimes it slows down.
• At a given instant of time what we read from the
speedometer is the instantaneous speed
• At the end of the trip, if we want to learn the average
speed of the car during the journey we divide total
distance covered by total time the trip takes.
Example
If a car travels from town A to town B, a distance of a 145 km
in 5h
Instantaneous speed-constantly changes throughout the
journey
Average speed =
𝑡𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒
𝑡𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒
=
145𝑘𝑚
5ℎ
= 29 km/h
Worked Example
Q. A car travelling from one city to another covers a total distance of 500 km in 5 hours. Calculate it’s average speed
during the journey
Ans. Average speed = total distance/total time
= 500km/5h
100/3.6 = 27.8 m/s (3 s.f)
=100km/h
Unit Conversion: km/h
m/s
i. km/h  m/s
Shortcut:
(𝑠𝑝𝑒𝑒𝑑 𝑖𝑛
𝑘𝑚
1000
𝑚
)𝑥
= (𝑠𝑝𝑒𝑒𝑑 𝑖𝑛 )
ℎ
3600
𝑠
Example : Convert 54km/h to m/s
Unitary method
In 1h distance covered = 54km=54000m [1km=1000m]
In 3600s distance covered = 54000m
In 1s distance covered = 54000/3600= 15m
Hence 15m is covered in 1s
Therefore 54km/h=15m/s
𝑠𝑝𝑒𝑒𝑑 𝑖𝑛 𝑘𝑚/ℎ
3.6
e.g. 54/3.6 = 15m/s
= speed in m/s
ii. m/s  km/h
Example
Q. A car travels a speed of 12.8m/s. What is it’s speed in km/h?
Unitary method:
In 1s car covers 12.8m
1hr=60x60=3600s
In 3600s car covers 12.8 x 3600 = 46080m
Shortcut:
1km=1000m
1000m=1km
1m=1/1000 km
46080m = 46080/1000 = 46.08km
Hence the car travels 46.08km in 1h
Therefore 12.8m/s=46.1km/h (3 s.f))
𝑚
3600
𝑘𝑚
(𝑠𝑝𝑒𝑒𝑑 𝑖𝑛 ) 𝑥
= (𝑠𝑝𝑒𝑒𝑑 𝑖𝑛
)
𝑠
1000
ℎ
(speed in m/s x 3.6) = speed in km/h
e.g. 12.8 x 3.6 = 46.1 km/h
Experiment to measure average speed of an object
Experimental setup
Pg. 7, Fig. 1.10
Apparatus:





toy car
wooden runway/track
clamp stand
meter rule
stop clock
Method
1. The track is securely clamped at height h using a clamp stand
2. The distance from A to B is measured using a meter rule
3. The car is released and a stop watch is turned on simultaneously
4. Stop watch is turned off when car passes point B, hence recording the time taken for car to move
through distance AB
Calculation
Distance moved from A to B =s meters
Time taken = t seconds
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑆𝑝𝑒𝑒𝑑, 𝑣 =
𝑇𝑜𝑡𝑎𝑙 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑡𝑟𝑎𝑣𝑒𝑙𝑙𝑒𝑑, 𝑠
𝑇𝑜𝑡𝑎𝑙 𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛, 𝑡
Sources of Error
1. When timing with a stop clock, human reaction time introduces measurement errors.
(Human reaction time is the amount of time it takes for a human to respond to a stimulus. Average reaction time for
humans is 0.25 seconds to a visual stimulus, 0.17s for an audio stimulus, and 0.15s for a touch stimulus.)
Reducing error:
i) Repeat the procedure to take several readings of time and use the average value to calculate average speed
Time/s
t1
t2
t3
Average time, t/s
Average Speed, v/ms-1
t= (t1 + t2 + t3) /3
v= distance AB/t
ii) Use electronically operated stop clock and electronic timing gates as shown below:
Pg 8. Fig 1.11
• The electronic stop clock will record the time elapsed (t) while the card strip attached to the moving car passes
through the timing gate.
• The length of the card strip (L) is measured using a ruler.
• The average speed of the car as it passes through the timing gate is calculated using v=L/t.
• Above method eliminates error introduced due to human reaction time
Precaution
Wear protective footwear
Finding the gradient/slope of a straight-line graph
• Gradient/slope of a graph refers to the steepness of the graph.
• The gradient of the line = (change in y-coordinate)/(change in x-coordinate) .
𝐺𝑟𝑎𝑑𝑖𝑒𝑛𝑡, 𝑚 =
𝑦2 − 𝑦1
𝑥2 − 𝑥1
To find the gradient of a straight-line graph, pick two points on the graph.
In the graph to the left, (x1,y1)=(6,6), (x2,y2)=(10,8)
gradient = (change in y-coordinate)/(change in x-coordinate)
m= (y2-y1)/(x2-x1 )
= (8-6)/(10-6) = 2/4 = 1/2
Gradient of a straight line can also be negative
 The more the gradient, steeper is the line
m= (y2-y1)/(x2-x1 )
here, (x1, y1) = (1,7) (x2,y2) = (4,1)
m=(1-7)/(4-1) = -6/3= -2
 The gradient of a straight line is constant
The gradient of a curve is not constant: changes from point to point
Gradient of a horizontal line is 0:
As x changes, y remains constant
m= (1-1)/(9-1) = 0/8 = 0
Distance-Time Graphs
• A distance-time graph shows how far an object has travelled in a given time
• Distance is plotted on the y-axis and time is plotted on the x-axis.
Time from start/s
0.0
0.5
1.0
1.5
2.0
2.5
Distance travelled from start/m
0.0
6.0
12.0
18.0
24.0
30.0
In a distance-time graph, 𝑮𝒓𝒂𝒅𝒊𝒆𝒏𝒕 =
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒅𝒊𝒔𝒕𝒂𝒏𝒄𝒆 (𝒔𝟐−𝒔𝟏)
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒊𝒎𝒆 (𝒕𝟐−𝒕𝟏)
= 𝑺𝒑𝒆𝒆𝒅
In Figure 1.6, to find gradient, select any two points on the line
e.g. (x1,y1)=(0.5,6.0), (x2,y2)=(2.5,30.0)
gradient, m= (y2-y1)/(x2-x1 )
= (30.0-6.0)/(2.5-0.5)
=12m/s
The steeper the line, more the gradient and hence the greater is the speed.
Pg. 6, Fig.1.7b
• Straight lines for both object A and B means gradients are constant hence
speed for both are constant
• Steeper line for object A than that of B indicates that graph of object A
has a higher gradient than that of object B. Hence A has a higher speed
than B
Remaining graphs in fig. 1.7
incorrect
• A distance time graph cannot have a line sloping downwards
• Speed is a scalar quantity – cannot be negative
• With time, distance can either remain unchanged or increase,
but it cannot decrease
*Displacement-time graph can have a downward sloping line (later)
Velocity
Pg 7, Example-1
Velocity is speed in a particular direction - it is a
vector quantity
e.g. 55 km/h to the right, 80m/s to the north,
72km/h south-west etc.
velocity=
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒
symbol
𝑠
𝑣=
𝑡
3𝑘𝑚
=
0.75ℎ
or
Δ𝑠
𝑣=
Δ𝑡
𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑑𝑖𝑠𝑝𝑎𝑙𝑐𝑒𝑚𝑒𝑛𝑡
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
**Recall:
Average Speed =
Total Distance Covered
Total Time Taken
Fig 1.9: Screen of a GPS
b.
The walker has to follow the
roads. So the distance walked is
greater than the straight line
distance between A and B (the
displacement). The walker’s
average speed calculated using
distance must be greater than
his average velocity (calculated
using displacement)
Speed v Velocity
• If a car travels at 50km/h around a bend, it’s speed is
constant
• But it’s velocity will be continuously changing as it turns the
corner
• Even though magnitude remains the same (50km/h) the
direction continuously changes as it turns the corner. Hence
velocity also changes continuously.
Displacement-Time Graph
• Shows how the displacement of a moving object
from a given point changes with time
• Slope or gradient of the line is equal to the
negative gradient = velocity in opposite direction
Object returns to starting point
∆𝑠
velocity of the object, m=
∆𝑡
=v
• Line with a negative gradient would indicate an
object going backwards
Pg. 7, Fig 1.8: Displacement is decreasing with time
Section of Graph
Description
O-A
Positive constant velocity; v=∆s/∆t=4/2=2m/s
A-B
Object stationary; v=(4-4)/(7-2) = 0/5= 0m/s
B-C
Negative constant velocity; v=(0-4)/(8-7) = -4/1 -4m/s
At C
Object has returned to starting position; s=0m
A
4m
-4m/s
2m/s
B
Acceleration
Defined as the rate of change of velocity i.e. change in velocity with respect to time.
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚
𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏, 𝒂 =
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒊𝒎𝒆
 Acceleration is a vector quantity
𝑎=
𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦−𝑖𝑛𝑖𝑡𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
𝑎=
𝑣−𝑢
𝑡
𝑎=
∆𝑣
∆𝑡
 SI unit for acceleration is m/s2
Example - Car moving with uniform acceleration
a=2m/s2
Every 1s the increase in velocity is constantUniform acceleration
Non uniform acceleration – Velocity does
not increase at a constant rate
When the velocity of an object increases with time,
we say that the object is accelerating
Pg. 9, Example-2
Q. A car is travelling at 20m/s. It accelerates steadily for 5s after which time it is travelling at 30m/s.
Calculate it’s acceleration
Ans u=20m/s, v=30m/s, t=5s
 When the velocity of an object
decreases with time, we say that the
object is decelerating
a=(v-u)/t
a=(30m/s-20m/s)/5s
a=2m/s2
t=0.02s
Pg. 10, Example-3
Q. An object hits the ground travelling at 40m/s. It is brought
to rest in 0.02s. What is it’s acceleration?
Ans
 Rate of decrease of velocity is known as
decelaration
 A decelerating object will have a smaller
velocity than it’s starting velocity
u=40m/s, v=0m/s (since brought to rest), t=0.02s
a = (v-u)/t =(0m/s-40m/s)/ (0.02s)
a = -2000m/s2
• Negative value for acceleration indicates that the object is decelerating
• Every 1s, the speed of the object would decrease by 2000m/s
 Direction of acceleration is opposite to
direction of motion (velocity)
– more later
Galileo's Experiment
 A ball made to roll down a slope and hit bells on it’s way
 Bells positioned in a way so that there is increasing distance
between successive bells
21
 Bells ring at equal time intervals as ball passes - Distance
travelled in equal time intervals increased.
15
9
3
12
27
Pg 10. Fig 1.13
Bell
1
2
3
4
5
Time/s
0.5
1.0
1.5
2.0
2.5
Average Velocity between successive bell:
Distance of
ball from
start/cm
3
12
27
48
75
• Starting Point - Bell 1: v1=3cm/0.5s = 6cm/s
Calculating ∆v: v2-v1=18-6=12cm/s
v3-v2= 30-18= 12cm/s
v4-v3=42-30= 12cm/s
• Bell 1- Bell 2: v2=(12-3)cm/(1-0.5)s = 9/0.5= 18cm/s
 Rate of increase of
velocity is uniform
 Uniform acceleration
• Bell 2- Bell 3: v3=(27-12)/(1.5-0.5)=15/0.5 = 30cm/s
• Bell 3 – Bell 4: v4=(48-27)/(2-1.5) = 21/0.5= 42cm/s
Velocity-Time Graphs
A velocity-time graph shows changes in velocity of a moving object over time.
Galileo’s Experiment calculation
Time/s
0.25
0.75
1.25
1.75
2.25
Avg. Velocity/cms-1
6
18
30
42
54
Pg 11, Fig. 1.14
Points plotted in the middle of each successive 0.5s time
interval since average velocity is calculated
e.g- Bell 1 - Bell 2, time interval window: 0.5-1: Avg. t = (0.5+1)/2 = 0.75
Bell 2 - Bell 3: time interval window: 1-1.5: Avg. t = (1+1.5)/2= 1.25
In a v-t graph, 𝑮𝒓𝒂𝒅𝒊𝒆𝒏𝒕 =
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚(𝒗𝟐−𝒗𝟏)
𝒄𝒉𝒂𝒏𝒈𝒆 𝒊𝒏 𝒕𝒊𝒎𝒆 (𝒕𝟐−𝒕𝟏)
= 𝒂𝒄𝒄𝒆𝒍𝒆𝒓𝒂𝒕𝒊𝒐𝒏
Pg. 13, Fig 1.17
• straight line
• constant gradient
• uniform acceleration
Here, (x1,y1)=(0.75,18), (x2,y2)=(1.75,42)
m= (y2-y1)/(x2-x1 )
= (42-18)/(1.75-0.75)= a = 24cm/s2
Modern Version of Galileo’s Experiment
Apparatus:
Air track, glider, card, light gates, electronic timer/data logger
𝜃
Pg. 12, Fig. 1.15
𝜃 1 = 1.50, 𝜃2 = 3.00
1. As the glider moves down the sloping track, time taken to get across
each light gate is measured electronically and recorded
2. Length of the card (L) is entered into the spreadsheet program
3. Velocity at each light gate is calculated by the spread sheet
program and a v-t graph is plotted
4. Experiment repeated for a different inclination of the track (𝜃)
Pg. 12, Fig 1.16
• Car moving with a constant velocity:
𝜃2 = 3.00
𝜃 1 = 1.50
Pg. 12, Fig 1.16
1.Straight lines mean acceleration is constant
for both cases
2. Steeper line for greater inclination implies
higher gradient and hence higher acceleration
Flat line  0 gradient  0 acceleration  constant velocity
Steeper the line, more is the gradient, hence greater is the acceleration
Some more shapes, Pg. 13 Fig 1.18
a. shallow gradient - low acceleration
c. horizontal (zero gradient) – no acceleration
b. steep gradient – high acceleration
d. negative gradient -negative acceleration
(deceleration)
Area under a velocity-time graph
Area under a v-t graph is equal to the distance travelled by the object in a given time interval
Pg. 14, Fig 1.19
Graph a shows a v-t graph for an object that travels with a constant velocity of 5m/s for 10s.
Total distance travelled, s = v x t = 5m/s x 10s = 50m
Area of the rectangle = length x breath = 10 x 5 = 50
Total distance travelled = Area Under Graph = 50m
Here, Total Distance = Total Displacement
50m
A
t=0s
B
t=10s
Graph b is a v-t graph for an object with a constant/uniform acceleration.
It’s average velocity during this time is given by
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑖𝑛𝑖𝑡𝑖𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑦 𝑢 + 𝑓𝑖𝑛𝑎𝑙 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 (𝑣)
2
0 + 10
Avg. vel. =
2
= 5m/s
𝐴𝑣𝑒𝑟𝑎𝑔𝑒 𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦 =
𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒 𝑖𝑛 𝑑𝑖𝑠𝑝𝑎𝑙𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒 𝑡𝑎𝑘𝑒𝑛
Increase in displacement = Avg. Vel x time = 5m/s x 4s = 20m
Here, Increase in Displacement = Total Distance travelled
a= (v-u)/t = (10-0)/4 = 2.5m/s2
20m
1
Area of the triangle = 2 𝑥 𝑏𝑎𝑠𝑒 𝑥 ℎ𝑒𝑖𝑔ℎ𝑡
=½ x 4 x 10
A
t=0s
= 20
Total distance travelled = Area Under Graph = 20m
B
t=4s
Pg. 16 (Chapter Questions)
Q8. A car is travelling at 20m/s. It accelerates uniformly at 3m/s2 for 5s
a. Sketch a velocity time graph for the car during the period that it is accelerating. Include numerical detail
on the axes of your graph
b. Distance travelled is equal to the area under the graph
Ans a. a=(v-u)/t  v =u + at
=20m/s+(3m/s2 x 5s)
v = 35m/s
Ans b. total area under curve= area of rectangle + area of triangle
= (l x b) + (0.5 x b x h)
= (5 x 20) + [0.5 x 5 x (35-20)]
= 137.5
therefore, distance travelled= 137.5m
Alternate way:
Distance = Area of trapezium
1
=2 𝑎 + 𝑏 (ℎ)
1
=2 20 + 35 (5)
=1/2 x 55 x 5
= 137.5
Distance travelled = 137.5m
Equation of uniformly accelerated motion
For an object moving with a constant/uniform acceleration, the following equation can be used:
(final speed)2= (initial speed)2+ (2 x acceleration x distance moved)
v2= u2 + 2as
Pg 18. Example-5
Q.A cylinder containing vaccine is dropped from a helicopter hovering at a height of 200m above the
ground. The acceleration due to gravity is 10m/s2. Calculate the speed at which the cylinder will hit
the ground.
u=0m/s
Ans:
We know, v2= u2 + 2as
u=0m/s (dropped from rest), a=g=10m/s2 , s=200m
v2
=
(0m/s)2
+ (2 x
10m/s2
g=10m/s2
200m
x 200m)
v2 = 4000m2/s2
therefore v= √(4000m2/s2)
= 63.25m/s
Both v= u + at & v2= u2 + 2as
Hold true only for uniform acceleration
i.e. constant a
End
Class VIII, Physics, Jan Session 2022
Handout-1
Class Rules & Syllabus
+
Chapter-2 Forces & Shape
© Maple Leaf International School
Class rules & How we will proceed:
1. Confirmation (from students) of transmission (audio and video) at start of class
2. Book: Edexcel International GCSE (9-1) PHYSICS Student Book
3. Presentation in slides
4. Lecture slides of each chapter will be uploaded as handouts
5. Lecture slides detailed enough to be used as study notes
6. Ask questions using the question panel in Portal. Relevant questions will be answered after class
7. Assignments will be given at the end of each chapter – must be completed and submitted within deadline
8. Full syllabus will be covered in online classes
Book Content (O’ Level Exam Syllabus)
1st Monthly Syllabus
1. Chapter 2: Forces and Shape
2. Chapter 18: Density &
Pressure
3. Chapter 19: Solids, Liquids &
Gases
Chapter 2: Forces and Shape
Content Overview
●
Defining Force
●
Types of Forces
●
SI Units
●
Scalar and Vector quantities
●
Calculating Resultant Force
●
Friction
●
Hooke’s law and Force-Extension Graph
●
●
Experiment to investigate how extension
varies with applied force force for different
materials
Elastic behavior of a material
What is a Force?
A force is a push or pull on an object due to it’s interaction with another object
Types of Forces
All forces between objects can be placed into two broad categories: Contact forces & Non-contact forces
Forces which can
be applied only
when objects are
in physical contact
Forces that are exerted even
when the objects are not in contact
(action-at-a-distance forces)
Examples of contact forces
Contact Forces
1. Frictional Force: Forces that oppose motion
2. Tension: Force exerted on an object by
a rope/cable/wire when pulled
crane
3. Normal Reaction Force:
- Force exerted by a surface on an object in contact.
-The force acts perpendicular (900) to the surface
example scenario: a book resting on a
table – here n.r.f is exerted by the table
on the book
4. Upthrust: Upward push exerted (by fluid)on an object immersed in the fluid.
-Acts opposite to the weight of the object
-Also called buoyant force
-This is why when swimming, a person ‘feels’ like
they have less weight.
-The person’s weight is being supported by the
upthrust.
5. Applied Force: Force exerted on an object by a person or another object
Other examples:
Examples of Non Contact forces:
1. Gravitational force : Force acting between masses
An apple falling
from a tree is
an example of a
gravitational
force
Weight of an object – downward
gravitational force exerted by the earth on
the object.
Gravitational forces are always attractive
another example
2. Electrostatic Force:
- Force between electric charges
- Can be attractive or repulsive depending on the
type of charge
A comb (after rubbed against hair)attracting
pieces of paper is an example of electrostatic
charge
Another example:
This happens because
of electrostatic force
between comb and hair
Pg 20, fig 2.4b
3. Magnetic Force:
- Force between magnets or between magnet and magnetic material (simple definition)
- Can be attractive or repulsive depending on the type of magnetic pole
Metallic paper clips attracted to a magnet is an
example of a magnetic force
Another example:
The compass needle, which is a magnet is affected by
the magnetic force between it and the other magnet
4. Nuclear Force:
-Force that acts between protons and neutrons in an atom
-Nuclear force is what binds the protons and neutrons together
in an atom
Atom
-An atom is the smallest piece of an element
-Everything is made of atoms
Structure of an atom
●
An atom consists of a center called the nucleus,
that contains the protons and neutrons
●
Electrons orbit in shells surrounding the nucleus
Pg 20, Fig. 2,4a
●
●
Positively charged protons tend to repel (push) away
one another.
But nuclear force is strong enough to overcome that
repulsion and hold the protons and neutrons together in
the nucleus
Units
In physics, units are standards for measurements of physical quantities
SI Units
•
International System of Units (SI Units) is a collection of units of measurement
•
Internationally agreed units of measurements used around the world
•
Allows scientists to use a single standard to present and communicate scientific data
without confusion
Few Examples of SI units
There are other collection of units
(system of measurement)
example CGS system where:
Length is measured in centimeter(cm)
Mass is measured in gram (g)
Time is measured in second (s)
-The SI unit of Force is measured in Newton, symbol N
-example: A 10N force is larger in size than a 5N force and vice versa
Scalars and Vectors
Scalars
– quantities that require only magnitude for specification
– magnitude implies “how much”
– examples: distance, speed, mass, temperature, time etc.
are scalar quantities
An object of mass 1kg/2kg
Vectors
– quantities that require both magnitude and direction for specification
– examples: displacement, velocity, acceleration, momentum, force etc.
are vector quantities
the car is moving with a velocity of 25m/s to the right
Body temperature of 98.70F
the car is moving with a velocity of 35m/s to the left
Force is a vector quantity – it has magnitude and direction
●
●
another example:
air resistance
A total weight of 1000N
(man+parachute) is acting
downwards
Air resistance of 400N is acting
on the parachute in the upward
direction
weight
thrust is the force
exerted by the engine
Resultant Force
1. Multiple forces can act on an object
2. All the forces can be replaced by a single force called the resultant force that has the same effect as
as all the original forces acting together
3. It is also called unbalanced force or net force
Calculating Resultant Force:
●
Forces acting in the same direction on an object are added:
Fig. b
Note: length of the arrow
corresponds to the magnitude
of the force: greater the magnitude
greater is the force
=
Fig.a
●
Resultant force = 3N + 6N
F = 9N (to the right)
Forces acting in opposite directions on an object are subtracted:
=
Fig. c
Fig d.
Resultant force = 6N – 3N
F= 3N to the left
Smaller force is subtracted from the
bigger force.
Direction of resultant force is in direction
of the bigger force
Fig e.
=
F = (6N + 1N) – 3N
= 7N – 3N
= 4N to the left
OR
Worked Example
Q. A small rocket has a weight
of 8000N. At take-off, the rocket
engine exerts an upward force
of 10,000N
What is the net upward force
acting on the rocket during take-off?
F = – 6N – 1N + 3N
= – 7N + 3N
= – 4N
Negative sign
implies net force
acting to the left
●
Forces to the right are positive
●
Forces to the left are negative
Ans. Net force = 10,000N – 8000N
F = 2000N upward
(or to the north)
If forces acting on opposite sides are equal, resultant force = 0N
Example: Tug of War
1. For much of the time, the rope does not move because the two forces exerted by the two teams are equal in
magnitude but opposite in direction, hence resultant force = 0N
2. Eventually one of the teams will become tired and its pull will be smaller than that of the other team
3. When the forces acting on the rope are unbalanced, the rope will start to move in direction of the greater force
●
Resultant force changes the speed and direction of an object.
●
If there is no unbalanced force, speed and direction of object stays the same
1. Forces perpendicular (900) to one another cannot be added or subtracted
2.Forces in the same plane (same line of action) can be added or subtracted
Resultant force in the horizontal plane (direction) = Thrust – Ground Friction – Air Resistance
Resultant force in the vertical plane (direction) = Weight – Normal Reaction Force
F = Thrust – N.R.F
or
F=Weight + Air Resistance
incorrect
●
●
N.R.F cannot be subtracted from thrust since they are at 900 to
one another
Weight cannot be added to air resistance as these forces
are perpendicular to one another
Friction
Force that opposes motion
Friction occurs when
1. Solid objects rub against (or move over) other solid objects – surface friction
2. Solid objects move through fluids (liquids and gases) – e.g. viscous drag, air resistance
Surface Friction
Fluid Friction
Ball dropped through liquid
air resistance
D= Drag
U = Upthrust
W= Weight
Air friction
weight
In some situation friction is not desired – we want to reduce friction as much as possible
●
✔
✔
Friction between different machine
parts cause degradation of
components.
Reducing friction means
machine work more efficiently
and do not wear out so quickly
Friction in machines:
●
In other cases, friction is desired
Sprint cyclists
wear suits of
special material
to reduce effect
of air friction
Aerodynamic design
of racing cars alters air flow
reducing drag
Parachutist
needs air
resistance
to slow them
down
Increasing friction can help make tyres that grip the
road better and prevent skidding
1. The friction on ice is very low.
2. Skater is able to glide for long distances
without having to do any work
Pg 22, Fig. 2.8a.
Friction generates heat
1. Friction causes particles on surfaces in contact to
gain kinetic energy and move faster
2. The more the K.E. of particles the greater is the
temperature (more later)
Rubbing your hands to
keep them warm is an
example of generating heat
due to friction
Heat generated by friction
ignites the material on tip of
the matchstick and it starts
burning
Investigating Friction (Pg. 22)
The apparatus shown in Fig. 2.9 can be used to
test different factors that may affect the size of the
friction force.
1. Weight is increased gradually until block starts to move.
This happens when pull of the suspended weight overcomes the
maximum friction force.
2.Friction between block and track has a maximum value
that depends on I) surfaces in contact
II) normal reaction force exerted by track on block
3, Experiment is repeated by adding more weights on the block:
More the weight greater the reaction force, hence greater the friction.
More weights needs to be hung to make the block move.
4. Experiment is repeated by replacing track with a rough surface:
Rougher the surface greater the friction. More weights need to be hung
to make the block move.
Factors affecting Friction
Factors affecting surface friction:
1. Roughness of the surfaces in contact: more the roughness greater is the friction
2. Weight of the object: more the weight, greater the normal reaction force hence greater is
the friction (friction is proportional to normal reaction force)
Factors affecting fluid friction (air resistance/drag):
1. Speed of the object: The more the speed of an object the greater is the air resistance/drag acting on it
2. Surface area of the object: More the surface area greater is the air resistance/drag
When parachute is not
open air resistance acting
is less
As parachute opens,
surface area increases
so air resistance
increases, parachutist
slows down
3. Viscosity of the fluid:
●
Viscosity: measure of resistance to fluid flow
●
More the viscosity of the fluid, greater is the viscous drag
Relation between mass and weight
Mass and weight are related by the following
equation:
symbol:
Or W=mg
g= 10N/kg on Earth
Example:
Q. An aircraft has a mass of 4500kg. Calculate
it’s weight.
Ans. W= 4500kg x 10N/kg
W= 45000N
What can forces do?
Forces can make
things start to
move
t=0s, v=5m/s
t=1s, v=10m/s
Forces can change the speed of an object
Forces can
change
direction of
an object
ball changes direction
when hit by batsman
compression
extension
Forces can change the shape of an object
Directly proportional - maths concept required to understand Hooke’s Law
1. Two quantities are said to be directly proportional if they always increase or decrease at the same rate
OR 2. the two quantities increase or decrease in the same ratio.
3. The symbol for directly proportional is
Ratio constant:
E1/H1 = 20/1 = 20
E2/H2 = 40/2 = 20
E3/H3 = 60/3 = 20
Constant ratio also called
Constant of proportionality
(symbol: k)
Earning = k x no. of hours worked
can be written as
E=kH
Example:
Q. You are paid $20 an hour. How much will you earn for working 8 hours? (here k=20)
Ans. E=kH
E = 20 x 8 = $160
y=5x
slope of line = 5 = k
(more later)
Hooke’s Law (Pg. 23)
Extension of a spring is directly proportional to applied force provided that the spring is within it’s
proportionality limit.
Force
extension
F=kx
Where k=proportionality constant
(also called spring constant)
In figure to the right (spring)
a. Force applied = 0N, Extension = 0cm
b. F = 5N, x=2.5cm, F/x = 5/2.5 = 2
c. F=10N, x=5cm, F/x = 10/5 = 2
●
Ratio F/x is a constant
●
F1/x1 = F2/x2 = F3/x3…
●
●
Hence F and x increases at the same rate
Or
F and x increases in the same ratio –
hence directly proportional
From graph:
1N/0.1m = 10
2N/0.2m = 10
3N/0.3m = 10
4N/0.4m = 10
hence ratio constant
Reminder:
When 2 variables
are directly
proportional to one
another, the graph
is a straight line
passing through the
origin
Proportionality limit:
Load (y axis) means force applied
● Point up to which F
and x remains directly
proportional
(point P in graph)
● Beyond P, F and x are
no longer proportional,
so graph no longer
straight but starts
curving
Pg 24, Fig 2.14
1.Hooke’s Law obeyed upto point P
2. Beyond P, as extension is no longer proportional to force, Hooke’s Law is no longer obeyed. Beyond P, extension
does increase with force applied but no longer in the same ratio
3. Point E is the elastic limit: point uptil which spring remains elastic, that is if the deforming force is removed,
spring goes back to it’s original length.
4. If stretched beyond point E, it no longer stays elastic – on removal of the applied force, spring does not go back
to it’s original length – there is now a permanent extension
Elastic – An object is said to be elastic in nature when
it can go back to it’s original shape and size after the
forces deforming them have been removed
Straight line passing through
origin – hence directly
proportional
Examples
1. Metal spring: as the force of 5N is
removed, spring goes back to original
length.
2. Rubber Band – the stretched band goes back to
original length as you let go of it.
Fig 18
Although a rubber band is elastic in nature
it does not obey Hooke’s Law.
1. As the force is removed from point X,
spring goes back to original length, hence
it is elastic
2. Although extension does increase with force,
It does not increase proportionally – graph is
not a straight line - Hence it does not obey
Hooke’s law.
load/
force
extension
Pg 25, fig 2.16
Opposite of elastic is plastic:
An object is said to be plastic in nature when
It changes shape permanently when a deforming
force is applied.
Plastic bottle changes
shape permanently
under a crushing force
Investigating Hooke’s Law
Apparatus – spring, known masses (1g, 2g, 3g…)
clamp stand, meter rule, set square
Method
1. The original length of the spring is measured using
the rule (l0)
2. A known mass is hung from spring (m 1)
3. New length of the spring is measured (l 1)
4. Procedure is repeated for a range of masses
and corresponding new length is measured each time
Tabulation
Pg 24, fig 2.13a
Pg 24, Fig 2.13a
Result
A graph of Force v Extension is drawn. A straight line passing through the origin
verifies that the spring is obeying Hooke’s law
Sources of error
1. Parallax error – error in reading due to incorrect position of eye
Pg 24, fig 2.13b
To prevent this error, ensure eye is leveled with scale marking when reading is taken
2. Ruler might not be vertical leading to incorrect measurement of height.
Solution: use a set square to keep ruler completely upright (90 0)
Precaution
1. Protective footwear to prevent feet injury
2. Safety glasses/eye-ware – spring might snap and hit eye
End of Chapter 2
Class VIII Physics, Jan Session
Half-Yearly Term 2022
Handout-2
Chapter 3: Forces & Movement
© Maple Leaf International School
Chapter 3: Forces & Movement
Content Overview
• Newton’s 1st Law of Motion
• Newton’s 2nd Law of Motion
• Weight & Free Fall
• Experiments to verify Newton’s 2nd Law
• Stopping distance
• Air Resistance/Drag & Terminal Velocity
Newton’s Laws of Motion
Three laws that describe the relationship between the motion of an object and the forces acting on it
Newton’s 1st Law of Motion
The law states that, an object at rest stays at rest and an object in motion stays in motion at a constant speed and
direction unless acted upon by an unbalanced force
Examples
1. A book resting on a table.
• Vertical Plane/Axis : downward weight=upward reaction force. Resultant force=0N
• Horizontal Plane/Axis: no forces are acting, so resultant force in this plane is also 0N
• The book is said to be in equilibrium and stays at rest on the table, hence obeying
Newton’s 1st law of motion.
 An object is said to be in equilibrium when the opposing forces are balanced such
that the resultant force acting on the object is zero and the object continues to be
in the state that it is in
 Objects obeying 1st law are said to be in a state of equilibrium
2. Car Moving at constant speed
1. Horizontal axis: engine thrust = total friction
2. Vertical axis: Weight = Normal Reaction Force
3. Resultant force (along both axes) is zero
4. The car is in equilibrium and continues to move at a
constant speed in the forward direction, hence obeying 1st law
In Summary:
1st Law in space:
 An astronaut who has his
screwdriver knocked into space
will see the screwdriver continue
on at the same speed and
direction forever
 With no forces (forward force or
air resistance), this object will
never stop.
Newton’s 2nd Law of Motion
For a resultant force acting on an object, the acceleration of the object is directly proportional to the
resultant force and inversely proportional to the mass of the object.
 2nd law describes behavior of objects for which all forces acting on an object are not balanced, that is,
forces do not cancel each other out as was the case in 1st law scenarios explained previously.
Meaning of “acceleration directly proportional to force”:
• As the resultant force acting on the object is increased, the acceleration of the object also increases, proportionately, 𝑎 ∝ 𝐹
1. Direction of motion of the object is the same as the direction of the resultant force – positive a – v increases
•
Different sized forces are applied to objects having
the same mass.
• The small force produces a smaller acceleration than
the large force
 The greater the resultant force, the higher the acceleration
and the quicker the speed increases
Pg. 29 Fig.3.2a
mass constant
2. Direction of motion of the object is opposite to the direction of the resultant force –
negative a (deceleration) – v decreases
 Car’s speed will decrease with time:
 It will slow down an eventually come to rest.
Resultant force, F
 The greater the resultant force, the higher the decelaration
and the quicker is the car brought to rest.
In both cases considered above, the direction of acceleration (positive or negative) is in the direction of resultant force
Meaning of “acceleration inversely proportional to mass”
1
• When the same force is applied to objects with different masses, the smaller mass will experience a greater acceleration 𝑎 ∝
𝑚
Fig.3.2b
resultant force constant
In Summary:
• When there is no resultant force there is no
acceleration
• Object remains in a state of equilibrium hence obeying
1st law
• 1st Law is a special case of 2nd Law
2nd Law Equation
𝑭 = 𝒎𝒂
Here, F=resultant/unbalanced force acting on the object
m=mass of the object
a=acceleration of the object
𝑎=
𝐹
𝑚
→𝑎=
1
𝐹
𝑚
𝐹
𝑎=
𝑚
If m is constant (i.e. for an object of constant mass)
For a constant resultant force
→ 𝑎 = 𝑘. 𝐹  a ∝ 𝑭
𝑎=
 So as F increases, a increases proportionately
(ratio of a/F is a constant: a1/F1 = a2/F2 =a3/F3……)
 A graph of acceleration against force gives a straight
line passing through the origin
𝑘
𝑚
𝑎=𝑘
1
𝑚
𝒂∝
𝟏
𝒎
 So as m increases, a decreases proportionately
(product of a1m1 = a2m2 = a3m3 …….. )
 A graph of acceleration against mass has the following
shape
10x4 = 5x8 = 40
Pg 30. Fig 3.4: Acceleration is proportional to force
Pg. 31, Fig 3.6 The graph of a v m is a curve
Q. Two forces are acting on a car of mass 1000kg:
a thrust force of 450N to the right and a frictional force
of 400N to the left. Calculate
a) resultant force acting on the car
b) acceleration of the car. State the direction of acceleration
Ans a. Resultant force, F=thrust force-air friction
F = 450N-400N
F =50N to the right
b. F=ma
a=F/m
a=50/1000
therefore a=0.05m/s2 to the right
400N
450N
Pg. 31, Ex-1
Weight
 Weight of an object is the gravitational force exerted by the Earth on the object
 Acts towards the center of the Earth
 The acceleration of a falling object because of gravitational force is called its acceleration
due to gravity and is denoted by g
 On Earth g=9.81m/s2 (in IGCSE O’ level syllabus, value of g used is 10m/s2)
 g is also known as gravitational field strength having unit N/kg. On Earth, g=10N/kg
We know, F=ma
If resultant force, F=Weight=W
then a=g=10m/s2
therefore, 𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑎𝑠𝑠 𝑥 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 𝑑𝑢𝑒 𝑡𝑜 𝑔𝑟𝑎𝑣𝑖𝑡𝑦
or
𝑊𝑒𝑖𝑔ℎ𝑡 = 𝑚𝑎𝑠𝑠 𝑥 𝑔𝑟𝑎𝑣𝑖𝑡𝑎𝑡𝑖𝑜𝑛𝑎𝑙 𝑓𝑖𝑒𝑙𝑑 𝑠𝑡𝑟𝑒𝑛𝑔𝑡ℎ
in symbols, 𝑾 = 𝒎 𝒙 𝒈
Pg. 34, Example-2
Mass (Earth) > Mass (Mars) > Mass (Moon)
 The gravitational field strength on the moon
is about 1.6N/kg, (or gmoon = 1.6m/s2)
 An object taken from the Earth to the Moon
will have less weight even though mass will
remain unchanged.
Free Fall
 An object is said to be in free fall when the only force acting on it is the force of gravity.
 In this situation the object will fall downwards with a constant acceleration = 10m/s2 = g
(irrespective of its mass or size)
• In the absence of air resistance, the only force acting
is the weight.
• Hence both the elephant and the feather accelerates
at the same rate (g=10m/s2) hence reaching the
ground at the same time.
**The situation is different in presence of
air resistance (more later)
𝐹
𝐹
𝑎 = 𝑚 = 10000N/1000kg
𝑎 = 𝑚 = 0.00008N/0.000008kg
a=10m/s2 = g
a = 10m/s2 = g
Experiments to verify Newton’s 2nd law
i. How acceleration of an object varies with resultant force applied on an object
Apparatus Required : trolley, nylon string, pulley, ramp/wooden runway, light gate and timer, stop clock, mass hanger and
known masses (e.g. 5g, 10g, 20g…..)
Procedure & Calculation
1. A known mass (m1) is hung on the mass hanger. Weight of the mass exerts a force on the trolley.
𝑎=
𝑣−𝑢
𝑡
2. A card is mounted on the top of a trolley. The length of the card (L) is measured.
3. As the trolley passes across the first light gate, the electronic timer measures the time taken
for card to cross the gate (t1)
4. Velocity at the position of first light gate is calculated: u= L ÷ t1
5. When passing by the second light gate, time is measured by electronic
timer (t2 ). Velocity is calculated by: v = L ÷ t2
6. The time taken for the trolley to travel from first to the second light gate is
measured using a stop watch (t3)
7. Acceleration is calculated by a=(v-u)/t3
8. The procedure is repeated by hanging different masses (m1, m2, m3…..mn) and consecutive acceleration (a1, a2 a3….an) is
calculated. Mass is kept constant during the investigation by using same trolley throughout
Tabulation & Graph
Mass/kg
m1
m2
m3
….
Force Applied, F=mg/N
F1
F2
F3
…..
Acceleration (m/s2)
a1
a2
a3
…..
A graph of (a V F) is drawn. A straight line passing through origin verifies that 𝑎 ∝ 𝐹, as predicted by the 2nd law
Precaution:
Footwear to avoid possible foot injury
ii. How acceleration varies with the mass of object for a given resultant force
Additional Apparatus Required:
Top Pan Balance
Experimental Setup
Procedure
1. A known mass is hung on the mass hanger. Weight of this mass exerts a force on the trolley. This force is kept constant
throughout the investigation.
2. As the trolley moves down the runway, combination of light gate, electronic timer and stop clock is used to
determine acceleration (as described previously)
3. Mass of the trolley is varied by adding known masses on top of the trolley.
(or use trolleys of different mass). Corresponding acceleration for each mass is calculated
Calculation & Graph
Mass of
trolley/kg
m1
m2
m3
….
Initial
u=L/t1
v=L/t2
Time taken from gate 1 to
gate 2 /s
t3
u1
u2
u3
….
v1
v2
v3
….
t3(1)
t3(2)
t3(3)
….
velocity/ms-1
Final
velocity/ms-1
A graph of (a V m) is drawn. The following shape, along
which product of (a x m) is constant would verify that
1
𝑎 ∝ 𝑚 as predicted by 2nd law
Fig. 3.6
a1
a2
a3
…..
A graph of (a V 1/m) can be drawn. A straight
line passing through origin would also verify
1
that 𝑎 ∝ 𝑚
m/kg
OR
Acceleration /ms-2
a= (v-u)/t3
𝟏
𝒎
/kg-1
a/ms-2
m1
1/m1
a1
m2
1/m2
a2
m3
1/m3
a3
..
…
Pg. 30, Fig 3.5
Stopping Distance
When a driver suddenly sees an object blocking the way ahead, he must bring the car to a stop in the shortest distance possible.
 Stopping distance is the distance the car moves from the moment the driver is aware of the need to stop to the point at which the vehicle
comes to a complete stop.
 Thinking distance is the distance the car travels in the time it takes for the driver to apply the brakes after realizing that the car needs to stop.
The time taken for the driver to respond to the new situation before applying the brakes is the reaction time.
 Braking Distance is the distance the car travels in the time after the driver has applied the brakes.
Factors affecting reaction time and hence thinking distance
•
•
•
•
tiredness of the driver: a tired driver will take more time to react
influence of drugs or alcohol: makes driver less aware and increases reaction time
visibility factors e.g. weather condition (fog), dirty windscreen
other distractions e.g. talking on phone while driving will increase reaction time
The longer the reaction time, the further the car will travel before breaking starts – that is, the longer will be the thinking distance
One other factor also affecting thinking distance is
•
car’s initial speed: higher the speed, the further the car will travel during this ‘thinking time’
Factors affecting Braking Distance:
•
poor road and weather conditions: e.g. wet icy roads means less friction between tyres and the
road hence increasing braking distance
•
poor vehicle conditions e.g. worn brakes or worn tyres - less friction between brakes and wheels.
•
Initial speed of the car: higher the car’s speed the greater is the time required to bring it to rest for a
given deceleration hence greater is the distance covered.
Stopping Distance in a velocity-time graph
Use the graph to find out:
a. how long the driver takes to react to seeing the obstacle (reaction time)
•
•
Area in purple represents thinking distance
Area in blue represents braking distance
Ans. 0.75s (the period during which the velocity of the car
is constant at 24m/s)
b. how far the car travels in this reaction time i.e. thinking distance
Ans. Thinking distance= area under graph for first 0.75s = 0.75s x24m/s = 18m
Pg 38, Chapter Questions Q6.
c. how long it takes to bring the car to a halt once the driver starts braking
Ans. This the period during which the velocity of the car decreases
from 24m/s to 0m/s: 3.25-0.75=2.5s
d. the total distance the car travels before stopping, i.e. stopping distance
Ans. Braking distance = triangle area= ½ x 2.5 x 24= 30m
Stopping distance=thinking distance + braking distance
= 18m + 30m = 48m
Drag/Air Resistance
•
•
An object moving through fluid (liquid or gas) experiences a force that opposes it’s movement. This frictional
force is called drag
Drag force exerted on objects moving through air is called air resistance
part a) -The ball has just been released
-Velocity is 0m/s.
-There is no drag force acting on it: FD. = 0
-The only force acting is the weight(W).
-Resultant force=W. Hence acceleration of the ball, a= g=10m/s2.
-Velocity of the ball increases as it falls.
part b) -Speed of the ball increases, drag force also increases.
-Resultant force= W- FD.
-As the resultant force decreases, acceleration of ball decreases: a<10m/s2.
- Speed increases at a lower rate
-As the ball travels further down, FD keeps increasing with speed
-Resultant downward force continue decreasing and so does acceleration
part c) -FD has increased to a point where it had become equal to the weight
-Resultant force=W- FD=0 .
-The ball has stopped accelerating
-It falls with this maximum constant velocity known as terminal velocity
Pg. 35. Fig 3.14
Terminal Velocity
Terminal velocity is the constant velocity that a falling object reaches when the upward drag force acting counterbalances the
downward weight and the object stops accelerating.
V-t graph for a falling object until it reaches terminal velocity
blue dotted line:
• With no air resistance, only force acting
is the weight. Hence a=g=10m/s2
• Straight line implies constant gradient and hence
constant acceleration
Graph below shows how acceleration of ball changes with time
Pg. 35, Fig 3.15
Elephant-Feather case in presence of air resistance
• Air resistance must be 10,000N to counterbalance weight
of elephant but only 0.00008N to counterbalance weight of
feather.
• Time taken for elephant to reach terminal velocity is much
greater than that for feather.
• Elephant accelerates for a longer period and reaches a
higher terminal velocity
• Elephant reaches ground first
• In this scenario, the weight is the dominant factor that
suppresses the effect the greater surface area of elephant
has on air resistance
Sky Divers and Terminal Velocity
When a skydiver jumps from a plane, he will accelerate for a time and eventually reach terminal velocity, in the same way
the ball reached terminal velocity in a previous discussion.
Diagram below describes the situation with a numerical calculation until he reaches terminal velocity.
• When he opens his parachute, surface area increases
and causes a sudden increase in drag force.
• At this velocity, the drag force acting is greater than
the weight of the sky diver.
• There is now an upward resultant force acting on the
parachutist.
• As the resultant force is opposite to direction of the
parachutist, it will cause him to decelerate i.e. the
downward speed will start to decrease.
• But as speed decreases, drag force also decreases. At
one point drag force again becomes equal to weight
and a new lower terminal velocity is reached.
• This new terminal velocity is low enough to allow the
parachutist to land safely.
the effect of opening the parachute is that it makes the
new terminal velocity lower
Velocity – Time graph for sky diver
Terminal Velocity in Liquids
• In general, objects have to accelerate to quite high speeds in air to reach
terminal velocity because of the low viscosity of air
• Objects falling through liquids experience a much larger drag force for a given
speed as liquids are more viscous
• Upthrust in liquids is also significant due to larger density of liquids and
partially counter balances the weight along with drag force
• Objects reach terminal velocity quicker
• Terminal velocity reached is also much lower than than that for objects falling
though air
End
Resultant force = W- (FD+U)
t=0s, R = 10 – (0 + 2) = 8N
t=1s, R = 10 – (4+2) = 4N
t=2s, R= 10 – (8+2) = 0N
Resultant force quickly becomes 0
Class VIII Physics, Jan Session
Half-Yearly Term 2022
Handout-3
Chapter 4: Momentum
© Maple Leaf International School
Chapter 4: Momentum
Content Overview
• Defining Momentum
• Newton’s 2nd Law in terms of Momentum
Before Explosion
• Newton’s 3rd Law of Motion
• Law of Conservation of Momentum
• Collisions and Explosions
• Momentum and Car Safety
After Explosion
Momentum
Use of the word ‘momentum’ in everyday life:
• We are often told that sports teams and political candidates have "a lot of
momentum".
• This implies that the team or candidate has had a lot of recent success and that
it would be difficult for an opponent to change their trajectory.
• This is also the essence of the meaning in physics, though in physics we need to
be much more precise: momentum can be defined as "mass in motion."
Momentum of an object is equal to the mass of the object times the velocity of the object
equation,
symbol,
𝑴𝒐𝒎𝒆𝒏𝒕𝒖𝒎 = 𝒎𝒂𝒔𝒔 × 𝒗𝒆𝒍𝒐𝒄𝒊𝒕𝒚
𝒑 = 𝒎𝒗
SI unit: kg m/s
Momentum is a vector quantity
e.g.
•
To fully describe, for example, the momentum of a 5kg bowling ball moving westward at 2 m/s,
the correct statement would be: the bowling ball has a momentum of 10kg m/s westward.
•
The direction of momentum is same as the direction of velocity
•
An object has a large momentum if both its mass and its velocity are large. Both variables are of equal
importance in determining the momentum of an object.
•
If we consider a truck and a roller skater moving down the street at the same speed, the considerably greater mass of the truck
gives it a considerably greater momentum.
•
Yet if the truck were at rest, then the momentum of the least massive skater would be greater. The momentum of
any object that is at rest is 0.
•
Objects at rest do not have momentum – they do not have any "mass in motion”
Worked Example
Determine the momentum in the following cases. Give units.
a. a 1000kg car moving northward at 20 m/s.
b. a ball of mass 400g travelling at 36km/h to the left
Ans. given m=1000kg, v=20m/s
p=mv=1000 x 20 = 20,000kg m/s
therefore, the car has a momentum of 20,000kg m/s to the north
Ans. (convert m & v to SI units first)
Newton’s Second Law in terms of Momentum
Recall from Chapter 3, equation for Newton’s 2nd Law:
Substituting equation (ii) into (i)
We get , 𝐹 =
F=
𝐹 = 𝑚𝑎 --- (i)
Recall from Chapter 1, the equation for acceleration:
𝑎 =
𝑣−𝑢
--- (ii)
𝑡
m=400/1000=0.4kg
v= (36 x 1000m)/3600s= 10m/s
p=m x v = 0.4kg x 10m/s=4kg m/s
therefore, the ball has a momentum of 4kg m/s to the left
•
•
𝑚 𝑣−𝑢
𝑡
𝑚𝑣−𝑚𝑢
𝑡
𝑚𝑣−𝑚𝑢
Here,
refers to rate of change of momentum
𝑡
Equation implies that the rate of change of momentum of an object is directly
proportional to the force applied on that object: this means that if you double
the force acting on an object, it’s momentum will change twice as quickly.
m=2000kg u= 200m/s, v=800m/s
If a thrust force of 300,000N is applied
F=
𝑚𝑣−𝑚𝑢
𝑡

300,000 =
2xF
2000(800−200)
𝑡
t= (2000x600)/300,000
= 1200,000/300,000
= 4s
It takes 4s for a momentum increase of 1200,000 kg m/s
** Momentum increase = mv-mu
= 2000*800 – 2000*200
=1200,000kgm/s
If a thrust force of 600,000N is applied
600,000 =
2000(800−200)
𝑡
t= (2000x600)/300,000
= 1200,000/600,000
= 2s
It takes 2s for a momentum increase of 1200,000 kg m/s
Another way to express the equation
We know, 𝐹 =
𝑚 𝑣−𝑢
𝑡
Therefore,
𝑭 𝒙 𝒕 = 𝒎 𝒗 − 𝒎𝒖
This arrangement of the equation
simply implies that a bigger force
applied to an object for a longer
time will result in a greater change
in the momentum of the object.
Here, t is the time for which the force has been applied.
mv – mu is change in momentum
b. If the rocket has a mass of 3000 tonnes,
what is the velocity of the rocket after the first stage
has completed it’s burn? (1tonne = 1000kg)
Ans. m= 3000 x 1000 = 3x106 kg
The rocket starts from rest, u=0; so initial momentum,
mu=0 kg m/s
Pg. 42, Ex-1:
Q. The first stage of rocket used in Moon missions provides an
unbalanced upward (away from the Earth) force of 30MN and
burns for 2.5minutes. [1MN = 106 N]
a. Calculate the increase in rockets momentum
that results.
F x t = mv-mu
4.5 x 109 = mv – 0
4.5 x 109 = 3 x 106 v
v= 1500m/s
Assumption: mass of the rocket stays the same.
Ans. (convert F and t to SI units)
F= 30 x 106 = 30,000,000 (3x107), t= 2.5min = 2.5 x 60 = 150s
F x t = mv-mu
mv-mu = 30,000,000 x 150 = 45,00000000 = 4.5x109 kg m/s
**In a real launch, fuel is used up so the mass of the rocket
would decrease and calculation would be harder
Newton’s 3rd Law of Motion
If object A exerts a force on object B, then object B exerts a force of equal magnitude but in opposite direction on object A
Every action has an equal and opposite reaction
• This law represents a certain symmetry in nature: forces always occur in pairs, and one body cannot exert a force on another
without experiencing a force itself.
• We sometimes refer to this law loosely as action-reaction, where the force exerted is the action and the force experienced as
a consequence is the reaction.
• The action-reaction forces are also called 3rd law pair forces.
Examples of 3rd Law pair forces:
Fish swimming through water :
• A fish uses its fins to push water backwards
• Following Newton’s 3rd Law, the water pushes the fish forwards, propelling the fish through the water
• The size of the force on the water equals the size of the force on the fish
• The direction of the force on the water (backwards) is opposite to the direction of the force on the fish (forwards)
push on water
by fish (action)
push on fish by water (reaction)
A bird flying upwards:
• A bird flies by use of its wings. The wings of a bird push air downwards (action)
• Following 3rd law, the air pushes the bird upwards (reaction)
• The size of the force on the air equals the size of the force on the bird
• The direction of the force on the air (downwards) is opposite to the direction of the force
on the bird (upwards)
**Helicopters can similarly move up by pushing air down, thereby experiencing an upward reaction force.
Rocket accelerating upwards
•
•
•
•
•
•
Rockets move forward by expelling gas backward at high velocity
This means the rocket exerts a large backward force on the gas in the rocket combustion chamber
Following 3rd law, the gas therefore exerts a large reaction force forward on the rocket
This reaction force is called thrust
When this upward push (or thrust) exceeds the weight of the rocket, there is an upward resultant force
Following Newton’s 2rd Law, the rocket accelerates in that direction (F=ma)
A book resting on a table:
• Book pushes down on the table (action force)
• Table pushes back on the book (reaction force) – also know as normal reaction force
Pg. 46, Fig 4.8
• Person X is clearly pushing on person Y but
it is not obvious that Y is pushing X back.
• When X and Y move as shown in part b), it is clear that X has
been affected by a force pushing him to the left.
• The force felt by X is the reaction force exerted by Y
Worked Example
Q. When a heavy football player (greater mass), and a light player
(smaller mass) run into each other (head-on collision), which player hits
the other with more force? Explain using Newton’s 3rd Law.
Ans. According to 3rd law, the action-reaction forces are equal in
magnitude. Hence the force exerted by the light football player on
his heavier counterpart is equal in magnitude. (Both players exert
and experience the same magnitude of force but opposite in direction)
So far all examples above of action-reaction forces are contact forces
Action-Reaction forces can also be non-contact in nature
e.g. Gravitational force acting between masses is an
example of 3rd law pair forces that are of non-contact type
• Earth exerts a downward gravitational
force on the apple.
• Apple, in turn exerts an upward gravitational
force, equal in magnitude on the earth
• Reason why apple accelerates down and
Earth does not move up is because of the
enormous difference in mass
Characteristics of action-reaction forces:
𝐹
• 𝑎 = 𝑚 . Although F is same for both,
mass of Earth >> mass of apple. Hence
aEarth << aapple
The forces:
 are equal in magnitude
 are opposite in direction
 act on two different bodies
 are of the same type e.g. if the action force is gravitational force, the reaction is also a
gravitational force.
 act along the same line of action
Law of Conservation of Momentum
In physics, the term conservation refers to something which doesn't change.
According to the principle of conservation of momentum:
Total momentum of a system of isolated objects remains constant
Isolated system:
• A system that is not acted on by
force external to the system
• Only forces acting are the action-reaction
forces between the objects within system
Momentum Conservation in collisions:
For a collision occurring between two objects, the total momentum of the objects before collision is equal to the total momentum of the
objects after collision provided that there is no external force acting on the system.
Collision case -1: Objects collide and move separately
-trolley A of mass m1 collides with trolley B having mass m2.
-Before collision, trolley A has initial velocity of u1 and trolley B has velocity u2.
-After collision, trolley A has velocity v1 and trolley B has velocity v2.
-In this scenario, the two trolleys together forms the system.
Total momentum before collision = m1u1+ m2u2
Total momentum after collision = m1v1 + m2v2
According to the conservation law:
Total momentum before collision = Total momentum after collision
Therefore, m1u1+ m2u2 = m1v1 + m2v2
Collision case -2: Objects collide and stick together, moving as a single body
illustrating with a numerical example:
Q. Trolley A of mass 4.0 kg moving at 3m/s along a horizontal frictionless surface, hits stationary trolley B having mass 2 kg.
After collision, the two trolleys stick together and move with a velocity of 2m/s.
a) Calculate momentum of system before collision (two
trolleys together forms the system)
Ans.
Momentum before collision = mAuA+ mBuB
= (4kg x 3m/s) + (2kg x0m/s)
= 12 kg m/s
b) Calculate momentum of system after collision
Ans. Momentum after collision = mAvA + mBvB
Common velocity, vA= vB=v =2m/s
Momentum after collision = mAv + mBv
= (mA + mB).v
= 6x2
= 12kgm/s
c) Show that this collision demonstrates law of conservation of momentum
From the above calculation,
Momentum before collision = Momentum after collision = 12kg m/s.
Hence momentum is being conserved in this collision.
• Only forces acting are the action-reaction forces; e.g. in a head-on collision
m Au A
Before collision:
mBuB
mAuA + mBuB = mAvA + mBvB
Total momentum before collision = Total Momentum after collision
During Collision
FAB
FBA
mAvA
After Collision:
mBvB
If there are external forces acting on the system
mAuA + mBuB ≠ mAvA + mBvB
Hence law of conservation of momentum will not hold
Momentum Conservation in Explosions
• In physics, an explosion involves a release of energy causing things to fly apart
Total momentum before explosion = Total momentum after explosion = 0 kg m/s
u=0
• Although, destructive bomb explosions are the first thing that comes to mind, other
simpler or more common scenarios in everyday life can be described as explosion.
• For example, if two trolleys with a compressed spring in between are held together
with a string, when the string is cut the two trolleys move in opposite direction:
the event of two trolleys moving in opposite direction can be regarded as an explosion
Before Explosion
After Explosion
Mu = m1v1 + m2v2 = 0
Where M= m1+ m2
Total momentum before explosion= mAu1+ mBu2
Both trolleys are at rest, therefore, u1=u2=0
Total momentum before explosion
= (2kgx 0m/s) + (1kg x 0m/s)
= 0 kg m/s
Total momentum after explosion
= mAv1 + mBv2
= (2kg x -1m/s) + (1kg x 2m/s)
= -2kg m/s + 2kg m/s
= 0kg m/s
• In any explosion, the object/part with lower mass will have a greater velocity in order to conserve momentum
• For example, in the above scenario, trolley B with mass 1kg moves with velocity 2m/s which is greater than velocity of
trolley A with double the mass
Another example of controlled explosion: Rocket propulsion
• Rocket motors use the principle of conservation of momentum to propel spacecraft through space.
• They produce a continuous, controlled explosion, that forces large amounts of fast-moving gases (produced by the
fuel burning) out of the back of the rocket at a certain velocity; ejected gas gains momentum downwards
• The spacecraft gains an equal amount of momentum, in the opposite direction to that of the moving exhaust gases.
• Total momentum of system before and after explosion = 0kgm/s
Momentum of rocket
=
Momentum of expelled gas
Momentum & Car Safety
• In a car crash, the force exerted on a person can be very large causing severe injuries; can also be fatal
• Example below illustrates this situation
Pg. 45, Ex - 3
Q. A car travelling at 20m/s collides with a stationary lorry and is brought to rest in just 0.02s. A woman in the car has a
mass of 50kg. She experiences the same deceleration when she comes in contact with a hard surface in the car
(such as the dashboard or the windscreen). What force does the person experience?
Ans. F =
F=
𝑚𝑣−𝑚𝑢
𝑡
50 𝑥 0 − 50 𝑥 20
0.02
F = - 50,000N
 As the woman is brought to rest in a very short time (0.02s), her momentum is reduced to 0 very fast.
 Rate of change of momentum (
𝑚v −mu
)
t
is high. Since 𝐹 =
𝑚v −mu
,
t
a large force is exerted on the passenger
Safety Features
• Cars are now designed with various safety features that increases the time over which the car’s momentum changes
in an accident.
1. Crumple Zones
 Crumple zones are areas of a vehicle that are designed to crush in a controlled way in a collision.
 They increase the time taken for the vehicle to slow down in an impact
 In 𝑭 =
𝒎𝐯 −𝐦𝐮
, as t increases, momentum of the passenger changes over
𝒕
a longer time.
 Rate of change of momentum decreases, hence force exerted on passenger also decreases.
 In the previous worked example above, If the deceleration time is increased from 0.02s to 1s,
F=

50 𝑥 0 − 50 𝑥 20
1
= -1000N
•
•
1000N << 50,000N
Greatly reduces probability of severe injury or fatality
Alternative: F x t = mv-mu; since t increases, lower F is needed to bring about the same change in momentum:
mv-mu = 50 x 0 - 50 x 20 = -1000kgm/s
2. Airbag
•
On the occasion of an accident, if the passenger hits on an airbag instead of direct
impact with hard surface such as a steering wheel/dashboard/windscreen, the time taken
𝒎𝒗 −𝒎𝒖
for speed of passenger’s head to decrease from maximum to 0 is more (𝑭 =
)
𝒕
•
Rate of change of momentum decreases. Hence force exerted on passenger’s head also
decreases.
Fig: air bag test demonstration
3. Seat Belt
•
During an accident, the passenger’s body continue moving forward.
•
The passenger moves against the seat belt exerting a force on it.
•
Following 3rd law the seat belt exerts a force back on the person.
•
This force exerted by the seat belt is opposite to the direction the
person is moving, hence causing the person to decelerate and come to rest.
•
The belt, as it stretches, allows the body to decelerate over a longer period of
time resulting in force reduction
•
It is also designed to distribute the force on the strongest parts of the body,
shoulder and hip, hence protecting the spine and head
Differentiating between balanced forces (1st Law) and action-reaction forces (3rd Law)
• It can sometimes be difficult to sort out balanced forces from action-reaction forces
balanced forces
force exerted by table on book (N.R.F)
downward gravitational pull
by Earth on book (weight)
similarity: both balanced forces and action-reaction forces are opposite in direction.
action-reaction forces
force exerted by table on book (N.R.F)
forces exerted
on table by
book
downward
gravitational
pull by
Earth on book
upward
gravitational
pull by book
on Earth
differences: i. Balanced forces act on the same body; action-reaction forces act on different bodies
ii. Balanced forces usually of different types; action reaction forces always of the same type
iii. Balanced forces – 1st Law; action reaction forces - 3rd Law
End
Class VIII Physics, Jan Session
Half-Yearly Term 2022
Handout-4
Chapter 5: The Turning Effect of Forces
© Maple Leaf International School
Chapter 5: The Turning Effect of Forces
Content Overview
• Moment of a Force
• Center of Gravity
• Conditions for an object to be in equilibrium
-Principle of Moments
• Forces on a beam
Moment of a Force
• A force acting on an object may cause the object to change shape, to start or stop moving, to accelerate or decelerate
• A force can also cause an object to turn around a fixed point
In physics, the turning effect of a force about a fixed point is called moment
The fixed point is also referred to as the pivot
We use turning effect of forces all the time; examples:
Opening a door:
• You push on the door handle and the door rotates around
its hinges (hinges in this case is the pivot)
• You exerted a force that caused the door to rotate – the
rotation was the result of the moment of your pushing
force
Removing a lid off a tin:
Using a spanner to rotate a nut
• Screwdriver is used to exert a force about lid edge (pivot)
• Lid comes off
Swing
-A person pushing a swing will make the swing
rotate about the pivot (A)
Body joints
-forces of muscles make parts of our body turn around joints
like elbows or knees.
Equation
Moment of a force is calculated using the following equation:
Moment of a force = force  perpendicular distance from pivot
Symbol:
M=Fd
SI unit: Nm
Moment is a vector quantity
𝑀
𝐹=
𝑑
d=
𝑀
𝐹
The direction of the moment can be either clockwise or anti-clockwise
• Clockwise is the direction in which the hands of a clock move.
Anti-clockwise is the opposite direction
Since moment is a product of force and perpendicular distance
The larger the force and the greater the perpendicular distance, the greater will be the turning effect of the force
Opening a door
- Man is applying a force of 10N, 20cm from hinge
- Woman pulls door knob that is at a distance of 80cm
with a force of only 2.5N and prevents man from
closing door
- Woman has to apply a much smaller force to
create same turning effect/moment:
- Man: 10N x 0.2m = 2Nm Woman: 2.5N x 0.8m= 2Nm
Worked Example
Calculate the moment in each cases:
a.
Pg. 49, Fig 5.2: Distance of force from pivot is crucial
b.
Ans.
moment = F x d
About pivot shown, F=5N, d=0.5m
M = 5N x 0.5m
= 2.5Nm anti-clockwise
Ans. moment = F x d
About pivot (bolt), F=50N, d= 80cm=0.8m
moment = 50N x 0.8m
= 40Nm clockwise
Center of Gravity
Center of gravity is the point through which the weight of a body seems to act
e.g. For a uniform ruler, the mass of which is equally spread throughout it’s length,
the center of gravity lies at the center
The center of gravity is also sometimes called the center of mass
Ruler in Balance
• If center of the ruler is placed on our finger tip, i.e. the finger is placed at the center of gravity
the moment produced by the weight of the ruler is 0 since perpendicular distance between
weight and pivot (finger tip) is 0m (fig 5.6, also fig a. below)
• Hence the ruler’s weight produces 0 moment (or no turning effect) and balance on finger tip
• If we try to balance the ruler at any point along the ruler other than the center, there will be a
finite perpendicular distance between weight of the ruler and pivot (finger tip) (fig. b below)
• Weight of the ruler produces a moment about pivot and the ruler will tip over
d
G
fig. a
fig. b
Pg. 51, Fig 5.6
Objects with a uniform mass
uniform mass – mass evenly distributed throughout the object
axis of symmetry - line that divides a
shape into 2 equal halves
Pg. 52, Fig. 5.8: The center of gravity for these regular shapes is where the axes of symmetry cross. For the rectangle for
example, it lies at the intersection of the two diagonals.
Uniform beam:
-mass uniformly/evenly distributed throughout the beam
-center of gravity lies in the middle of the beam
- when a uniform beam is not pivoted trough its center of gravity,
the weight of the object will produce a turning effect
Objects with non-uniform mass
Pg. 52, Fig. 5.9: the weight of the beam causes a clockwise moment about
the pivot
- Mass not evenly spread out – some parts have more mass than other parts
- Center of gravity not at the center of the object
- G will be closer to the heavy end of the object where most of its mass
is concentrated
Center of gravity and Stability
• The position of G of an object affects it’s stability
• A stable object is one that is difficult to push over – when pushed and released, it tends to return to it’s original position
Fig. 5.12: stable and unstable object
Fig 5.12a.
• the broad base of the bottle, means more of it’s mass is
concentrated at the bottom.
• so center of gravity (black dot) is closer to it’s base. There
is less chance of the bottle toppling over.
Fig. 5.12b.
• the vase is more massive at the top resulting in
center of mass shifting upwards
• this makes the vase less stable and it has greater
chances of toppling over
Principle of Moments
An object will be in balance (i.e. it will not turn about a pivot point) if
sum of clockwise moment = sum of anticlockwise moment
**If they are not equal, the object will rotate
in the direction of the greater moment
 This is the 1st condition for an object to be in equilibrium
a.
b.
Pg. 50, fig 5.4
Girl: Clockwise moment = 300N  1.5m = 450Nm
Girl: Clockwise moment = 300N  1.0m = 300Nm
Boy : Anticlockwise moment = 200N  1.5m = 300Nm
Boy: Anticlockwise moment = 200N  1.5m = 300Nm
Clockwise moment > Anticlockwise moment
Clockwise moment = Anticlockwise moment
see-saw is balanced
see-saw rotates clockwise about pivot and lean to the
right
10
Pg. 50, Ex-1
Q. Look again at Fig 5.2. Person A pushes the door with a force of 200N at a distance 20cm from the hinge (the pivot in
this example) Person B opens the door by pulling on the handle, that is 80cm from the hinge. What is the minimum
pulling force that the woman must exert to open the door?
Ans.
• For the door to open, moment produced by woman
must be greater than that produced by man
• Moments are equal when force applied by woman is 50N
• Therefore, the woman must exert a force greater
than 50N to open the door (final answer)
11
Pg. 52, Ex-2:
Q. Diagram shows a uniform wooden meter ruler with a mass of 0.12kg. A pivot at the 40cm mark supports
the rule. When a brass weight is placed at the 20cm mark, the ruler just balances in the horizontal position.
Take moments about the pivot to calculate mass of the brass weight.
Ans. About pivot, clockwise moment = anticlockwise moment
0.12g x 10 = mg x 20
1.2 = 20m
m= 0.06kg
Fig. 5.10
100cm
Crane & Principle of Moment
 The pivot point is the crosspiece of the crane supported by the crane tower
 The weight of the long load arm and the load must be counterbalanced by large concrete blocks at
the end of the short arm that projects out behind the crane controller’s cabin.
Fig. 5.6
 Perpendicular distance between counterbalance weights and pivot point is less than that between
load and pivot
 Hence the counterbalance weights must be large so that ACM=CM and the crane is balanced.
 If ACM ≠ CM, turning forces about pivot could cause crane to rotate and collapse
2nd condition for an object to be in equilibrium
 In addition to the clockwise and anticlockwise moment being equal about the pivot point, for an object to be in balance,
there is a second condition
The total upward forces acting on the object must be equal to the total downward forces acting
 In other words, the resultant force acting on the object in the vertical axis/plane must be zero.
Example:
• Two masses hung from a beam exerting downward forces of 1.5N and 2N
R
• Since the beam is in balance: CM= ACM
CM= 2N x 0.3m=0.6N, ACM= 1.5N x 0.40m = 0.6Nm
• And, total downward force must be counterbalanced by total upward
force
• The upward reaction force exerted by the pivot on the beam = total
downward weight: R=1.5N + 2N
Assumption: weight of the beam is assumed to
be negligible
Conditions for an object to be in equilibrium:
 about pivot, sum of clockwise moment = sum of anticlockwise moment
 total upward forces acting on the object must be equal to the total downward forces acting
Forces on a Beam
a. Man standing on a beam at the center (G):
a.
• The weight of the man is 400N
• The beam is supported at it’s two ends
• The beam is 100m long
• Total Downward force = Total Upward Forces
400N = 200N + 200N
• Taking moment about G:
CM produced by force at A = 200  50 =10000Nm
ACM produced by force at B = 200  50 = 10000Nm
• So clockwise moment = anticlockwise moment
Forces acting on the beam:
i. downward push of 400N on the beam by the man
ii. supports at two ends exert upward forces on the beam
of 200N each
• In this case, weight of man does not produce any
moment as he is standing at G (perpend. dist. = 0)
b. If the man moves 25m to the left of G:
• His weight produces an anticlockwise moment
= 400 x 25 = 10000Nm
• Shift of the man causes the upward force at A to increase to
300N and at B to decrease to 100N
• The change in forces is to ensure clockwise moment remains
equal to anti-clockwise moment so that the beam stays in
equilibrium
• About G,
In Summary:
ACM by upward force at B: 100 x 50 = 5000Nm
Total ACM about G = 10000Nm (by man) + 5000Nm(by B) =15000Nm
CM by force at A : 300 x 50 = 15000Nm
• Total clockwise moment = Total anticlockwise moment
Total downward force equal to total upward forces: 400N= 300N +100N
• Hence the beam stays in equilibriums and do not turn over
End
• Forces exerted by supports on the beam
changes as the man walks along it to
ensure total CM stays equal to total
ACM and the beam stays balanced
• Change in magnitude of the reaction
forces at the supports depend on where
the load is applied