Tata AIG General Insurance Company - Insurer Innovation Award 2024
Forces & Eqm
1. FORCES & EQUILIBRIUM
1. Revise the effects of balanced and unbalanced forces on the motion of an object.
2. Recognise that an object accelerates in response to the resultant force and that F
= ma (Newton’s second law)
3. Recognise the resultant force on an object as being the sum of the individual
forces acting or the sum of the components
4. Know the definition of torque and use it to calculate torque in a variety of
everyday situations.
5. Explain that for a body to be in rotational equilibrium, the sum of the torques
acting around any point equals zero and solve problems using this principle.
6. Explain that for a body to be in translational equilibrium, the sum of the forces
acting in any direction equals zero and solve problems using this principle.
7. Explain that for a body to be in equilibrium then the conditions for both rotational
and translational equilibrium coexist and be able to solve problems based on
everyday situations, using these principles.
Embedded SLO’s
1. Explain the difference between mass and weight
2. Recognise that forces are vectors and as such
they can be resolved into components
Read p.105 to 117
Monday, 24 May 2010
2. Revision FORCE AND ACCELERATION
First Law
An object is reluctant to change its state of motion. An object will remain at rest
or continue to travel at a steady speed unless acted upon by an unbalanced
force.
Second Law
An object will accelerate in the direction of an unbalanced force according to the
equation:
F = ma
Third Law
For every action there is an equal and opposite reaction
Examples
1. Use Newton’s laws to explain why during heavy braking (in a car) the seat belt
digs into your chest.
Monday, 24 May 2010
3. 2. A sky rocket of mass 0.1 kg accelerates upwards at 10 ms-2. What is the
unbalanced force that causes this acceleration.
3. A cricketer throws a ball with initial acceleration of 90 ms-2. The force exerted is
30 N. Calculate the mass of the ball.
4. A force of 60N is exerted on a car of mass 1000kg. If the frictional forces acting on
the car are 10N then calculate the car’s acceleration.
Monday, 24 May 2010
4. RESULTANT FORCE
• Force is a vector quantity
• The resultant force is the sum of all the force vectors acting on a body.
• The resultant force is the single force which has the same effect as the
combination of forces.
Examples
1. Find the resultant force in each of the following situations
2. In each instance draw a vector diagram that shows how the resultant force has
been determined:
1 2 5N
8N
12 N
10 N
Monday, 24 May 2010
5. 3 Calculate the resultant force in the following situation:
50000N
80000N 10000N
40000N
Calculate the force that is effectively pushing the car down the slope Ex.9A Q.1 & 2
4 (hint: this is a component of one of the forces)
Support force 50N
Force due to gravity 55N
25o
Draw a vector triangle showing the relationship between these three forces.
Monday, 24 May 2010
6. 5 A 450 g trolley rests on a rough table. A horizontal force is applied by the
hanging masses and as a , the trolley travels 0.60 m from rest and reaches a
final speed of 2.07 ms-1. [adapted from CR p54 Q.2 & 3]
a Pulley
m2 = 450 g Trolley
a
Hanging masses m1 = 250 g
(a) Use vf2 = vi2 + 2ad to calculate the acceleration of the mass.
(b) Label the horizontal force on the diagram and calculate its size. (This is the
tension force and causes the acceleration of the trolley.
(c) Calculate the size of the force that causes m1’s acceleration. (Hint: Use a vector
diagram to show the forces acting on m1.
Ex.9A Q.3 to 5
Monday, 24 May 2010
7. RESULTANT AS THE SUM OF THE COMPONENT FORCES
A force of 1000 N acts at an angle of 30o to the direction of a cart’s motion. What is
the force that acts in the direction of the cart’s motion?
Direction of cart’s motion
20 Kg
30o (axis)
1000 N
30o
1000 N
Monday, 24 May 2010
8. RESULTANT AS THE SUM OF THE COMPONENT FORCES
A force of 1000 N acts at an angle of 30o to the direction of a cart’s motion. What is
the force that acts in the direction of the cart’s motion?
Direction of cart’s motion
20 Kg
30o (axis)
1000 N
The component of the 1000 N force that acts in the
direction in which the cart moves is shown below:
30o
1000 N
Monday, 24 May 2010
9. RESULTANT AS THE SUM OF THE COMPONENT FORCES
A force of 1000 N acts at an angle of 30o to the direction of a cart’s motion. What is
the force that acts in the direction of the cart’s motion?
Direction of cart’s motion
20 Kg
30o (axis)
1000 N
The component of the 1000 N force that acts in the
direction in which the cart moves is shown below:
1000.cos30o
30o
1000 N
Monday, 24 May 2010
10. RESULTANT AS THE SUM OF THE COMPONENT FORCES
A force of 1000 N acts at an angle of 30o to the direction of a cart’s motion. What is
the force that acts in the direction of the cart’s motion?
Direction of cart’s motion
20 Kg
30o (axis)
1000 N
The component of the 1000 N force that acts in the
direction in which the cart moves is shown below:
The trolley accelerates
in response to this force
1000.cos30o
30o
1000 N
Monday, 24 May 2010
11. RESULTANT AS THE SUM OF THE COMPONENT FORCES
A force of 1000 N acts at an angle of 30o to the direction of a cart’s motion. What is
the force that acts in the direction of the cart’s motion?
Direction of cart’s motion
20 Kg
30o (axis)
1000 N
The component of the 1000 N force that acts in the
direction in which the cart moves is shown below:
The trolley accelerates
in response to this force
1000.cos30o
30o
1000 N
Calculating the trolley’s acceleration:
Monday, 24 May 2010
12. RESULTANT AS THE SUM OF THE COMPONENT FORCES
A force of 1000 N acts at an angle of 30o to the direction of a cart’s motion. What is
the force that acts in the direction of the cart’s motion?
Direction of cart’s motion
20 Kg
30o (axis)
1000 N
The component of the 1000 N force that acts in the
direction in which the cart moves is shown below:
The trolley accelerates
in response to this force
1000.cos30o
30o
1000 N
Calculating the trolley’s acceleration: a = F = 1000.cos30o = 43.3 ms-2
m 20
Monday, 24 May 2010
13. RESULTANT AS THE SUM OF THE COMPONENT FORCES
A force of 1000 N acts at an angle of 30o to the direction of a cart’s motion. What is
the force that acts in the direction of the cart’s motion?
Direction of cart’s motion
20 Kg
30o (axis)
1000 N
Remember that a vector can be expressed as the sum of two components that are
drawn at right angles to each other.
The component of the 1000 N force that acts in the
direction in which the cart moves is shown below:
The trolley accelerates
in response to this force
1000.cos30o
30o
1000 N
Calculating the trolley’s acceleration: a = F = 1000.cos30o = 43.3 ms-2
m 20
Monday, 24 May 2010
14. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
Monday, 24 May 2010
15. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
Monday, 24 May 2010
16. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
1000 N
30o
Monday, 24 May 2010
17. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
1000 N
30o 1000.cos30o
Monday, 24 May 2010
18. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
1000 N
30o 1000.cos30o
30o
1000 N
Monday, 24 May 2010
19. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
1000 N
30o 1000.cos30o
30o
1000.cos30o
1000 N
Monday, 24 May 2010
20. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
1000 N
30o 1000.cos30o
30o
1000.cos30o
The trolley accelerates
in response to both of
1000 N
these forces
Monday, 24 May 2010
21. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
1000 N
30o 1000.cos30o
30o
1000.cos30o
The trolley accelerates
in response to both of
1000 N
these forces
Calculating the trolley’s acceleration:
Monday, 24 May 2010
22. Now consider that a second force of 1000 N is introduced and that this force also acts
at an angle of 30o to the direction of the cart’s motion. What is the force that acts in
the direction of the cart’s motion?
1000 N
30o Direction of cart’s motion
20 Kg
30o (axis)
1000 N
There are now two components which
act in the direction of the cart’s motion:
1000 N
30o 1000.cos30o
30o
1000.cos30o
The trolley accelerates
in response to both of
1000 N
these forces
Calculating the trolley’s acceleration: a = F = 2 x1000.cos30o = 86.6 ms-2
m 20
Monday, 24 May 2010
23. Examples
1. A large container ship is being towed out to sea by two tug boats. Each boat exerts
a force of 150000 N and pulls on the ship at an angle of 20o. The ship has a mass
of 1000 tonnes. (1 tonne = 1000 kg).
20o
20o Tug boats
The ship is initially at rest. Calculate the acceleration of the ship.
Monday, 24 May 2010
24. 2. A climber hangs off a bar with both arms. Each arm is at an angle of 45o to the
vertical. The mass of the climber is 65 kg. Calculate the tension force in each arm
when the climber is simply hanging stationery.
Exercises: “Vectors & Components”
Monday, 24 May 2010
25. WHAT IS TORQUE?
Torque is a force which causes rotation (i.e. it is a turning force)
Examples
Because a force is applied that causes rotation, a torque is being used in each case
Exercises: “Talking about torque”
Monday, 24 May 2010
26. WHAT IS TORQUE?
Torque is a force which causes rotation (i.e. it is a turning force)
Examples
Because a force is applied that causes rotation, a torque is being used in each case
Torque = force applied x distance between the point of application and the pivot.
= F.r⊥
Exercises: “Talking about torque”
Monday, 24 May 2010
27. WHAT IS TORQUE?
Torque is a force which causes rotation (i.e. it is a turning force)
Examples
Because a force is applied that causes rotation, a torque is being used in each case
Torque = force applied x distance between the point of application and the pivot.
= Torque in Newton-metres
= F.r⊥ Where (Nm)
F = Force applied (N)
r⊥ = perpendicular distance
Exercises: “Talking about torque”
Monday, 24 May 2010
28. WHAT IS TORQUE?
Torque is a force which causes rotation (i.e. it is a turning force)
Examples
Because a force is applied that causes rotation, a torque is being used in each case
Torque = force applied x distance between the point of application and the pivot.
= Torque in Newton-metres
= F.r⊥ Where (Nm)
F = Force applied (N)
r⊥ = perpendicular distance
r⊥ implies that the force and the distance must be perpendicular to each other.
Exercises: “Talking about torque”
Monday, 24 May 2010
29. Exercises TALKING ABOUT TORQUE
A
Torque in B is greater than torque in A
because
B
C
Torque in C is greater than torque in B because
Demo: Torque wrench vs spanner
Monday, 24 May 2010
30. Demo: The pattern of wear on chainrings can show where
Examples in the pedal cycle maximum torque is applied.
1. The diagram below shows a bicycle crank and chain. On the crank the force and the distance
between the point of application of the force and the pivot are perpendicular to each other.
Calculate the torque applied in this situation:
80 kg
An 80 kg cyclist stands on
the pedal whilst the crank arm is in the
position shown
175 mm
F
2. The crank arm turns through an angle of 30o below the horizontal and the cyclist is still in a
standing position.
(i) Draw the situation that exists now.
(ii) Calculate the torque applied in this situation. (Hint: You could find the component of the
force that is perpendicular to the crank arm first)
Monday, 24 May 2010
31. ROTATIONAL EQUILIBRIUM
Definition
A system that is not accelerating in rotational motion is said to be in rotational
equilibrium. For a system to be in rotational equilibrium the following situation needs
to exist:
Σ any direction = 0 where = torque
“The clockwise and anti-clockwise torques must be equal in size to each other”
Σ clockwise =Σ anti-clockwise
Bicycle crank - opposing torques Observation
10 N
10 N
http://www.walter-fendt.de/ph14e/lever.htm
Monday, 24 May 2010
32. Example
1. A mother and her daughter were playing on a see-say at their local park. Mum has a mass of
55 kg and the daughter is much lighter at 35 kg. The see-saw measures 2.5 m from the pivot
to each end. The daughter sits at one end of the see-saw. How far from the other side of the
pivot must the mother sit in order to achieve balance?
55 kg
35 kg
d 2.5 m
2. A 70 kg painter stands on a 10 kg plank that is supported by two saw-horses. The plank is
4 m long and is supported 500 mm from each end. The painter stands 1 m from saw-horse A
as shown. The weight of the plank acts at the centre of mass of the plank.
There are 2 support forces, F1 and F2. Calculate the size
F1 70 kg of F2.
F2
A B
Fgrav
Ex.9B Q.4 to 6 Plank pbms - in finder
Monday, 24 May 2010
33. TRANSLATIONAL EQUILIBRIUM
Definition
A system that is not accelerating (at rest or travelling at a constant speed) is said to be in
translational equilibrium. For a system to be in translational equilibrium, the sum of the
forces along any axis must equal zero.
ΣFany direction = 0 When vectors add to zero there is no resultant
~ ~
Demo - Ping Pong ball/Hair dryer
There are 3 forces acting on the ball as shown by the free - body force diagram:
Lift
Ping pong ball Thrust
Hair dryer
Gravity
Adding these three vectors gives a zero
resultant:
Monday, 24 May 2010
34. TRANSLATIONAL EQUILIBRIUM
Definition
A system that is not accelerating (at rest or travelling at a constant speed) is said to be in
translational equilibrium. For a system to be in translational equilibrium, the sum of the
forces along any axis must equal zero.
ΣFany direction = 0 When vectors add to zero there is no resultant
~ ~
Demo - Ping Pong ball/Hair dryer
There are 3 forces acting on the ball as shown by the free - body force diagram:
Lift
Ping pong ball Thrust
Hair dryer
Gravity
Adding these three vectors gives a zero
resultant:
Monday, 24 May 2010
35. TRANSLATIONAL EQUILIBRIUM
Definition
A system that is not accelerating (at rest or travelling at a constant speed) is said to be in
translational equilibrium. For a system to be in translational equilibrium, the sum of the
forces along any axis must equal zero.
ΣFany direction = 0 When vectors add to zero there is no resultant
~ ~
Demo - Ping Pong ball/Hair dryer
There are 3 forces acting on the ball as shown by the free - body force diagram:
Lift
Ping pong ball Thrust
Hair dryer
Gravity
Adding these three vectors gives a zero
resultant:
Monday, 24 May 2010
36. TRANSLATIONAL EQUILIBRIUM
Definition
A system that is not accelerating (at rest or travelling at a constant speed) is said to be in
translational equilibrium. For a system to be in translational equilibrium, the sum of the
forces along any axis must equal zero.
ΣFany direction = 0 When vectors add to zero there is no resultant
~ ~
Demo - Ping Pong ball/Hair dryer
There are 3 forces acting on the ball as shown by the free - body force diagram:
Lift
Ping pong ball Thrust
Hair dryer
Gravity
Adding these three vectors gives a zero
resultant:
Monday, 24 May 2010
37. Examples
1. The diagram below shows a birds eye view of a sling shot which is ready to launch a
stone contained in its pouch.
26.6 N
20o
50 N 20o
26.6
Show that the stone is in translational equilibrium using a vector diagram, drawn
to scale.
2. The diagram below shows a painter who is standing still on a plank that is supported
at two points A and B. (a) Calculate the weight of the painter (Fgrav).
70 kg
500 N (b) Calculate the force acting at point B
A B
Fgrav (c) Name this force.
Ex.9B Q.1 to 3
Monday, 24 May 2010
38. Since we often deal with horizontal and vertical forces:
Sum of Forces acting upwards = Sum of Forces acting downwards
ΣFup = ΣFdown
Sum of Forces acting to the left = Sum of Forces acting to the right
ΣFleft = ΣFright
3. A mass is suspended by two strings that are 20o 20o
fixed to the ceiling. Use a vector diagram to 70 N 70 N
calculate the weight of the mass, W.
Mass
W
Monday, 24 May 2010
39. A BODY IN EQUILIBRIUM IS IN BOTH TRANSLATIONAL AND ROTATIONAL
EQUILIBRIUM
Examples
1. Below is a picture of two children on a see saw which is in equilibrium. One child has a mass of
30 kg, whilst the other child has an unknown mass. The distances from the pivot (or fulcrum)
are shown in the picture.
30 kg m=?
1.5 m 2.5 m
(a) In terms of the children and the see saw, explain what is meant by the term “rotational
equilibrium”
__________________________________________________________________________
__________________________________________________________________________
(b) Calculate the mass, m of the child at the far end of the see saw.
__________________________________________________________________________
__________________________________________________________________________
2. A wheelbarrow filled with soil has a mass of 30 kg (this can be considered to act at the centre
of mass). The dimensions of the barrow are shown in the diagram. Calculate the effort
required to lift the barrow.
____________________________________ Centre of Mass .
____________________________________
____________________________________ 0.7 m
____________________________________ 1.3 m
Ex.9B Q.7 to 12
Monday, 24 May 2010
40. 3. A Y12 physics experiment is set up as shown in the diagram. A student uses masses as
described in the information below and marks angles on the paper as indicated in the
diagram. The knot that connects the three strings is not moving. Show using a vector diagram
drawn to scale that tension forces T1, T2 and T3 are in translational equilibrium.
m1 = 0.1000 kg T1 Knot T3
m2 = 0.1200 kg
m3 = 0.0833 kg 25o 20o
let g = 10 ms-2
T2
m1 m3
m2
Sheet of paper
Monday, 24 May 2010
41. MASS AND WEIGHT
The difference between the two
• Mass is a measure of the amount of matter in an object. (unit: kilogram, kg)
• Weight is the force due to gravity acting on that object. (unit: Newton, N)
Weight, Fw or Fgrav, is proportional to the mass, m of the object. (Fw α m)
Fw = mg where g is a constant (gravitational acceleration)
where g = 10 ms-2 (usually considered to be negative when direction is relevant)
Note
g is often referred to as gravitational acceleration since any free-falling object falls
with acceleration equivalent to g (usually rounded to 10 ms-2)
It is sometimes convenient to refer to g as the gravitational constant as it is also the
force per unit mass on an object. (in fact g = 10 ms-2 = 10 Nkg-1)
g depends on how strong the gravitational field is around the planet/moon. The
gravitational field strength depends on the size of the planet/moon. Different
planets/moons have different values of g. For example the Earth’s moon has a much
smaller value of g than Earth.
Example
On Earth a 50 kg mass will have a weight force of 500 N acting on it.
Monday, 24 May 2010
42. Exercises VECTORS & COMPONENTS
y
Any vector can be expressed
as the sum of 2 vectors at right angles x
to each other.
The diagram shows the components of a vector in the x and y axes.
Components could also be vertical or horizontal. IN FACT components
could be determined in any axis or any direction
The two components add to give the original vector, v. v is also could also
be called the resultant of the vector addition.
Calculate the components of the vectors shown along the axes/lines given (drawn as
dotted lines and draw them on the diagram given. Show your calculation below the
diagram.
100 m
1 -1
60 ms 2
20o 75o
Horizontal
__________________________ __________________________
__________________________ __________________________
__________________________ __________________________
Monday, 24 May 2010
43. 25 ms-1
2 km
50 ms-1
Lakewood Drive
6. The ceiling beam: A ceiling beam of mass 250 kg and length 7 m is pitched at
an angle of 20o and supported at each end by timber framing.
(a) Draw a diagram of the situation in the space provided:
(b) Mark the point at which we can consider the force due gravity acts and
calculate this force. ________________________________________
(c) Explain why this beam has a tendency to slide “downhill” and state what
must be done to prevent this from happening.
______________________________________________________________
(d) Calculate the size of this “downhill” force.
______________________________________________________________
Monday, 24 May 2010
55. 12 PHYSICS FORCES ASSIGNMENT Name
1. Coming into the summer sports season the ground staff have been requested to prepare the
cricket wicket for the forthcoming games. The following diagram represents the efforts of the
ground staff in pushing the roller. The force exerted down the handle of the roller is 200 N.
200 N
30°
Use the information in the diagram to answer the question that follows.
(a)Draw a labelled force diagram to show the four forces acting on the roller.
(b)Show that the horizontal component of the push on the roller is 173 N.
Monday, 24 May 2010
56. (c) If the roller is being pushed at a constant speed along the wicket, determine the magnitude of
the force opposing the motion of the roller and explain your reasoning.
_______________________________________________________________________
_______________________________________________________________________
_______________________________________________________________________
(d) The vertical component of the push on the roller is 100 N. The mass of the roller is 100 kg.
Calculate the value of the reaction force on the roller.
___________________________________________________________________________
___________________________________________________________________________
___________________________________________________________________________
(e) The length of the wicket between the stumps is 20 m. Calculate the work done in pushing the
roller eight times along the pitch.
___________________________________________________________________________
___________________________________________________________________________
__________________________________________________________________________
2. A wheelbarrow has a mass of 15 kg and contains 50 kg of soil. The centre of mass
of the load is shown on the diagram and the pivot is also illustrated. Load and effort
forces are drawn and labelled on the diagram. Calculate the effort required to lift
the load. ___________________________________________
___________________________________________
___________________________________________
___________________________________________
___________________________________________
Monday, 24 May 2010
57. 3. A body-building machine is drawn below. Load, effort and pivot are shown.
Calculate the effort required to lift the 25 kg mass. The bar has mass of 5 kg. The
centre of mass of the bar and masses (combined) is shown on the diagram.
Calculate the effort needed to lift the mass.
317 mm C of M 483 mm
Effort
25 kg
5 kg bar
Load
___________________________________________________________________
___________________________________________________________________
___________________________________________________________________
4. A uniform girder is 3 m long and weighs 700 N. It is carried horizontally on the
shoulders of two men who support it at points 30 cm and 60 cm from each end.
Find the load carried by each man.
___________________________________________________________________
___________________________________________________________________
Monday, 24 May 2010
58. [p144 “Physics in Action”]
5. Find the centre of mass of a set of barbells, set up with 10 kg on one end and 30 kg
on the other. The central bar is of negligible mass and is 4.0 m long. To solve this,
we could assign a centre of mass, x metres from one end and use the fact that the
barbells would balance at the centre of mass.
• Bars or Dumbells balance at
their centre of mass.
• The weight of an object acts at is
centre of mass
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
____________________________________________________________________
Monday, 24 May 2010
