AQA Physics Forces in Equilibrium.pptx

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2015
6 Forces in Equilibrium
3.4.1 Force, Energy and
Momentum
7407 / 7408 PHYSICS KS5
6.1 Vectors and Scalars
6.2 Balanced Forces
6.3 The Principles of Moments
6.4 More on Moments
6.5 Stability
6.6 Equilibrium Rules

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2015
Interactive Links
Title What to do Site Type URL

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Introduction
This chapter guides students towards an understanding of the conditions for
static equilibrium of a body. They will learn to interpret and manipulate
vectors and to analyse three forces in equilibrium. They will work with
moments and centre of mass and apply the principle of moments. Finally,
they will practise solving problems involving objects in equilibrium by
considering both forces and moments. Opportunities for practical work
involve the measurement of forces and moments in equilibrium, and some of
the experiments require students to draw and interpret graphs and to discuss
errors and uncertainties. The principle mathematical techniques used in this
chapter are vector skills and trigonometry.
Prior knowledge
1. Students will need to build on their knowledge and skills, from Key Stage
4 Science and Mathematics, of:
2. changing the subject of an equation
3. vector quantities: quantities which have magnitude and direction
4. objects acted on by two equal and opposite forces are at rest or moving
at constant velocity.

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3.4.1.1 Scalars and vectors / 6.1& 6.2
Learning Objectives...
• Define a vector quantity.
• Describe how we represent vectors.
• Explain how we add and resolve vectors.
• Explain why we have to consider the direction in which a force acts.
• Demonstrate when two (or more) forces have no overall effect on a point object.
• Explain the parallelogram of forces.
Skills:
This topic provides opportunities to practise the following mathematical skills:
MS 0.6 Use calculators to handle sin x, cos x, and tan x when x is expressed in
degrees or radians
MS 3.2 Plot two variables from experimental or other data
MS 3.3 Understand that y = mx + c represents a linear relationship
MS 3.4 Determine the slope and intercept of a linear graph
MS 4.2 Visualise and represent 2D and 3D forms including 2D representations of 3D
objects
MS 4.4 Use Pythagoras’ theorem and the angle sum of a triangle
MS 4.5 Use sin, cos, and tan in physical problems.

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Mr D Powell
2018
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6.1 Vectors and Scalars – what are they?
C
CHALLENGE
What is a vector or scalar quantity
and can I identify one?
How do we represent vectors/ Addition of
vectors using a scale diagram
Addition of two perpendicular vectors using a
calculator or resolving a vector into two
perpendicular components
What happens when you are over 90?
A
B

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A Definition.... M
Scalars Vectors
mass, temperature, time,
length, speed, energy
displacement, force, velocity,
acceleration, momentum
TASK: Make a note of each quantity, then look at
the units and give a reason why they have been
sorted as such..

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2015
Displacement and velocity
A runner completes one lap of an athletics
track.
400m
What is her final displacement?
If she ends up exactly where she started, her
displacement from her starting position is zero.
What is her average velocity for the lap, and
how does it compare to her average speed?
What distance has she run?

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What did you decide?
Despite the trip moving at various speeds, because it ended
up at the starting point, the average velocity was zero.
This will always be true when the final displacement is zero.

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Vectors?
M
 On this racetrack a car covers 1.029 miles in distance by
covering a circuit in a lap. After 1 lap the car is back to
the start and has completed the lap at a constant speed
of 30 mph.
 However, when we think of the term “velocity” we
describe not just the speed but the direction. In this case
the “velocity” changes direction all the time. We call it a
vector quantity. This feature is important in helping us
describe the motion of objects clearly!

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2015
Vector equations
An equation is a statement of complete equality. The left hand side must match the right
hand side in both quantity and units. In a vector equation, the vectors on both sides of
the equation must have equal magnitudes and directions.
Take Newton’s second law, for example:
force = mass × acceleration
Force and acceleration are both vectors, so their directions will be equal. Mass is a
scalar: it scales the right-hand side of the equation so that both quantities are equal.
Force is measured in newtons (N), mass in kilograms (kg), and acceleration in ms-2
.
The units on both sides must be equal, so 1 N = 1kgms-2
.

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Displacement vectors
Harry and Sally are exploring the desert. They need to reach an oasis, but choose to
take different routes.
 Harry travels due north, then due
east.
 Sally simply travels in a straight line to
the oasis.
When Harry met Sally at the oasis, they had travelled different distances. However,
because they both reached the same destination from the same starting point, their
overall displacements were the same.
N

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Vector notation
A scalar quantity is often represented by a lower case letter, e.g. speed, v. A vector
quantity can also be represented by a lower case letter, but it is written or printed in one
of the following formats to differentiate it from the scalar equivalent:
The value of a vector can be written in
magnitude and direction form:
e.g. v = (v, θ)
Or as a pair of values called
components:
e.g. a = (8, 6) or
x
y
8
6
8
6
a

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2015
Vector addition
Displacement vectors can always be added ‘nose to tail’ to find a total or resultant
vector.
This can be done approximately by scale
drawing:
It can also be done by calculation, breaking each vector down into perpendicular
components first and then adding these together to find the components of the
resultant:
c + d =
2
3
-2
2
+ =
0
5
a
b
a + b 
10
7 x
y

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R = √ 32
+ 42
Calculating a resultant
When adding two perpendicular vectors, it is often necessary to calculate the exact
magnitude and direction of the resultant vector. This requires the use of Pythagoras’
theorem, and trigonometry.
For example, what is the resultant vector of a vertical displacement of 4 km and a
horizontal displacement of 3km?
direction:
= 5km
tan θ = 4/3
= 53°
θ = tan-1
(4/3)
= √ 25
magnitude:
R2
= 32
+ 42

R
4km
3km

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Just as it is possible to add two vectors together to get a resultant
vector, it is very often
useful to break a ‘diagonal’ vector
into its perpendicular
components.
Vector components
This makes it easier to
describe the motion of an
object, and to do any
relevant calculations.
vertical
component
horizontal
component

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Calculating components
A vector can be separated into perpendicular components given only its magnitude and
its angle from one of the component axes. This requires the use of trigonometry.
For example, what are the horizontal and vertical components of a vector with a magnitude
of 6ms-1
and a direction of 60° from the horizontal?
6ms-1
60°
x
y
cos60° = x / 6
x = 6 × cos60
= 3ms-1
sin60° = y / 6
y = 6 × sin60
= 5.2ms-1

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Pythagoras & Adding Vectors M
TASK: Now try out the same techniques
for two more triangles;
1) opp = 2m, adj = 8m
2) opp = 1.5m, adj = 4m
Answers
1.  = 14 , R =8.25m
2.  = 20.6 , R =4.27m

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Applications... M
TASK: Now try out the same techniques for
two more triangles;
1) What is R &  if opp = 54N, adj = 22N
2) What is adj &  if opp = 54ms-1
, R =
85ms-1
Answers
1.  = 67.83 , R =58.3m
2.  =39.4 , R =65.6ms-1

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Splitting a vector into components...
T
TASK: Put on your maths brain and try and explain using trig why Vx is
related to cos and Vy is related to sin

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Resolving in a context....
TASK:
1) A bird is flying at a speed of 15ms-1
at an angle of
36 to the horizontal. What is its vertical speed.
2) A javelin is thrown at a resultant force of 100N at
a 45  angle. What are the horizontal and
vertical forces on it.
Answers
1. Vy =8.8ms-1
2. Vy =70.7N
Vx = 70.7N

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Questions to try on your own....
P
Answers
.....

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1.How many radians in a circle?
2.How many radians in quarter of a circle?
3.Work out  in degrees for a right-angled triangle if hyp = 5, opp = 4, adj = 3?
4.Work out opp if  = 450
and the adj = 7cm?
5.Work out tan if opp = 53.0m, adj = 42.0m?
6.Work out opp if  = /2 and the adj = 3cm?
7.Work out  in radians for a right-angled triangle if hyp = 5, opp = 4, adj = 3?
8.Rearrange sin=opp/hyp to make  the subject.
9.Work out sin (/2) =, sin (2) = , cos () = , cos(cos-1
(/4)) = , tan(/4) =
10.Use formulae & numbers from your calculator to prove that;
More Trig Practice
adj
opp


tan

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Vector Problems 1 Question 1
Both sides contribute as vectors so double it to If using only pythag
do simply 2 x 15N Sin20 = 10.26 = 10.3N
label upper part of triangle as a.
Hence a2
= (152
+152
-2x152
Cos140)0.5
= 28.19N, then half for
triangle. To 14.1N. Then pythag (152
-14.12
)0.5
= 5.12N
(x2) = 10.23N
a
c
b

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3) A mass of 20.0kg is hung from the midpoint P of a wire. Calculate the
tension in each suspending wire in Newton’s. Assume g = 10ms-2
. (Hint
resolve…..)
Use the idea of point of equilibrium. The forces must balance vertical and
horizontal.
Weight balances the tension so for each wire
Use the 2T Cos70 = 200
T Cos70 = 100
T = 100 / Cos70
= 292.38N

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Vector Problems 1 Question 5/6
5) Magnitude = sqrt(35002
+ 2792
)
= 3511.1m = 3510m
Direction = tan-1
(279/3500)
= 4.557°
= 4.6° south from vertical
6) Vertical motion: 21 sin (43°) = 14.32 ms-1
Horizontal motion: 21 cos (43°) =15.358
ms-1
S = 3 x 15.358 ms-1
= 46.1 ms-1
3500km
0.279km

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2015
Use the parallelogram method to resolve the
forces acting on this object placed. (Hint employ
both cosine & sine rule). (4 marks)

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Use the parallelogram method to find the resultant
acting on this object and angle. (Hint employ both
cosine & sine rule). (4 marks)

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3) Two forces of magnitude 10.0N and F
Newton’s produce a resultant of
magnitude 30.0N in the direction OA.
Find the direction and magnitude of F. (2
marks) (Hint use Pythagoras)

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2015
The graph shows a part completed
vector diagram. You task is to find out
the vector R by mathematical analysis.
The vectors A & B are shown in both
coordinate and bearing formats.
Show working for both a
mathematical method (2 marks) and
drawn out scaled method. (2 marks)
Hint: this is not as hard as it might
initially look!

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2015
Vector Problems 2 Question 4 - stage 1
Finding the components of vectors for vector addition involves forming a right
triangle from each vector and using the standard triangle trigonometry. The
vector sum can be found by combining these components and converting to
polar form.

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Vector Problems 2 Question 4 – stage 2
After finding the components for the vectors A and B, and combining them to
find the components of the resultant vector R, the result can be put in polar form
by

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Vector Problems 2 - Question 5

6) Forces of 60.0N and F
Newton’s act at point O. Find the
magnitude and direction of F if
the resultant force is of
magnitude 30.0N along OX (2
marks)

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Vector Problems 2 - Question 7
Forces of 60.0N and F Newton’s act at
point O.
Find the magnitude and direction of F if
the resultant force is of magnitude
30.0N along OX (2 marks)
SinC
c
SinA
a

c2
= a2
+ b2
- 2abCosC

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2015
Using the Sine or Cosine Rule
30N
In this problem we can solve it with other methods of
trigonometry. Make a triangle of vectors where we are trying
to find length c and angle A;
120
c = F
b = 30N
a = 60N
A
Cosine Rule
c2
= a2
+ b2
- 2abcosC
c2
= 60N2
+ 30N2
– 2x60Nx30N xCos120
c2
= 4500N2
+1800N2
c2
= 6300N
c = 79.4N
c = F = 79.4N
Sine Rule
Use reciprocal version of above
SinC / c = SinA / a
(60N x Sin 120) / 79.4N) = SinA
Sin-1
(60N x Sin 120) / 79.4N) =A
A = 40.876 = 41 
SinC
c
SinA
a


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7) Simultaneous Equations (A*+)
If this problem is to be solved mathematically we must
establish two formulae. One in the vertical and one in the
horizontal; 30N
The key is knowing that the resultant force is 30N
at 0º.
Hence;
Resolving Horizontally;
30N = FCos + 60NCos(360-120)
30N = FCos + 60N x -0.5
30N = FCos - 30N
60N = FCos  Eq 1
Resolving Vertically;
0 = FSin + 60NSin(360-120)
0 = FSin + 60NSin(240)
0 = FSin + 60N x 3/2
0 = FSin -51.96N
51.96N = FSin Eq 2
So by dividing Eq 2 by Eq 1
FSin / FCos  = 51.95N / 60N
tan  = 0.866
 = 40.89
F = 60N / cos 
= 79.4N
Or
F = 51.96N / sin 
= 78.96N

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Science
2018
Mr D Powell
2018
Animated Science
2018
6.2 Balanced Forces
C
CHALLENGE
Why do we have to consider the
direction in which a force acts?
Complete calculations when two (or
more) forces have no overall effect
on a point object?
What is the parallelogram
of forces?
Conduct a precise coplanar forces practical without help
A
B

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Why are the objects not moving... S

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Real Setup…
T F
w=mg

X Y
mg
F 

sin
This simple experiment tests the idea that the vertical component of a
resultant force “F” can be balanced by vertical “mg” force.
So changing “mg” changes F and thus we can use the example as y = mx+c

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Example Practical Results…
3.0 4.0 5.0 6.0 7.0 8.0 9.0 10.0 11.0
0.00
1.00
2.00
3.00
4.00
5.00
6.00
f(x) = 0.4905 x
R² = 1
Weight +/-0.01N
Force on meter (N)
Weight
Added
(N)
Force on Newton
meter +/-0.2N mass +/-1g
Weight +/-
0.01N
Angle +/-1
deg
4.0 200 1.96 24
5.0 250 2.45 24
6.0 300 2.94 24
7.0 350 3.43 24
8.0 400 3.92 24
9.0 450 4.41 26
10.0 500 4.91 27
ave 24.71
Sin (ave) 0.42
In terms of forces Hyp = F
opp = mg
Hence... opp/hyp = adj
mg/F=sin(theta)
Fsin(theta) = mg
F = x
m = sin(theta)
y = mg

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Equilibrium Point of an Object
M

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Example 1
M

W
T
F
Tcos
Tsin

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Example 1 (with numbers)
M
A father pulls a child of weight 60.0N
back with a force of 50.0N.
a) What is the tension in each rope?
b) What is the angle from vertical that
the child is held at?
Answers…
a) (60.02
+50.02
)0.5
= 78.1N
each rope 78.1N /2 = 39.1N
b) tan = 50/60 = 0.833
 = 39.8

W
T
F
Tcos
Tsin

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Inclined Plane Intro….

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Example 2
M
Hint: Draw your triangle 1,2,3….. W, S,
F to make a similar triangle

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Example 2 (with numbers)
M
A block of weight = 30.0N sits on a
slope and is held by frictional
forces of 10.0N
a) What is the support force?
b) What is the angle from vertical
that the slope is at?
Answers…
a) 10.02
+S2
= 30.02
(30.02
– 10.02
)½
= S =28.3N
b) tan = F/S = 10/28.3 = 0.353
= = 19.5
Hint: Draw your triangle 1,2,3….. W, S,
F to make a similar triangle

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Practical Extension...
M
 Assemble the equipment shown below. Use some weights of
different types and measure the angles shown.
 The basic system will adjust in angle to whatever masses you add.
 The tension in each string must equal the force exerted by W1 or W2.
This enables the problem to be solved by resolving
 Try it practically so get a range of angles & weights then prove the
equilibrium.
http://www.walter-fendt.de/html5/phde/equilibriumforces_de.htm

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Example 3
M
1 2
T2sin2
T1sin1
T
2
cos
2
T
1
cos
1
T2
T1
W
P

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2015
Wires Example - Correct
Vertically
1.20N = T1
cos30+ T2
cos60
1.20N = 0.866T1
+ 0.5T2
Eq 1
Horizontally
T1
sin30 = T2
sin60
0.5T1
= 0.866T2
T1
= 1.732T2
Eq 2
Sub 2 into 1;
1.20N = 0.866T1
+ 0.5T2
1.20N = 0.866 x 1.732T2
+0.5T2
1.20N = 1.99T2
T2
= 0.60N
&
T1
= 1.732T2
T1
= 1.732 x 0.6N
T1
= 1.04N
So by turning it on its side we can use the same
formulae (which is confusing but correct)
Left side right side

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WRONG!
Vertically
1.20N = T1
sin30+ T2
sin60
1.20N = 0.5T1
+ 0.866T2
Eq 1
Horizontally
T1
cos30 = T2
cos60
0.866T1
= 0.5T2
T1
= 0.577T2
Eq 2
Sub 2 into 1;
1.20N = 0.866T1
+ 0.5T2
1.20N = 0.866 x 0.577T2
+0.5T2
1.20N = 1T2
T2
= 1.2N
&
T1
= 0.577T2
T1
= 0.577x1.2N
T1
= 0.69N

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2015
Vertically
1.20N = T1
sin60+ T2
sin30
1.20N = 0.866T1
+ 0.5T2
Eq 1
Horizontally
T1
cos60 = T2
cos30
0.5T1
= 0.866T2
T1
= 1.732T2
Eq 2
Sub 2 into 1;
1.20N = 0.866T1
+ 0.5T2
1.20N = 0.866 x 1.732T2
+0.5T2
1.20N = 1.99T2
T2
= 0.60N
&
T1
= 1.732T2
T1
= 1.732 x 0.6N
T1
= 1.04N
If we take the outside angle instead of the inner angle
we can do this and use the triangle;
(90-)
Alternative Can do by using (90- ) as angle;

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2015
Example 1...
M
Using the following
figures the problem
is very simple...
W1 = 5N
W2 = 3N
W3 = 6N
1 = 150
2 = 124
F1 F2
F1sin
F1cos 
F2sin
F2cos 
 
1. If you think that F1 & F2 must resolve and
their components in H & V directions must
balance with the weights to make the point
P be in equilibrium.
2. Define the problem as shown below with
two new angles and triangles. Make sure
you use the correct angles!

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2015
Example 1...
M
 Using the following figures the
problem is very simple...
W1 = 5N W2 = 3N W3 = 6N
1 = 150 2 = 124
=60  = 34 
F1 F2
F1sin
F1cos 
F2sin
F2cos 
 
5
.
2
5
.
2
cos
cos
0
6
67
.
1
33
.
4
sin
sin
0
2
1
3
3
2
1















F
F
H
F
W
F
F
V

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Plenary Question
P
A 2kg chicken rests on point C on
clothes line ACB as shown. What is
the minimum breaking strength of
the line to ensure the bird does
not break the line?
F1
F2
F1sin30
F1cos 30
F2sin45
F2cos45
30 45
N
F
F
N
F
F
N
F
F
V
40
2
20
2
2
2
20
45
sin
30
sin
0
2
1
2
1
2
1








3
2
2
2
2
3
45
cos
30
cos
0
2
1
2
1
2
1
F
F
F
F
F
F
H






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Plenary Question
P
A 2kg chicken rests as point C on clothes line ACB
as shown. What is the minimum breaking
strength of the line to ensure the bird does not
break the line?
N
F
F 40
2
2
1 

3
2
2
1 F
F 
N
F
N
F
F
N
F
F
F
F
N
F
F
68
.
12
732
.
2
64
.
34
2
40
3
3
40
2
3
2
3
2
40
2
2
2
2
2
2
2
1
2
1










N
N
F
F
F
35
.
10
3
2
68
.
12
3
2
1
2
1




F1=10.35N
F2= 12.68N
F1sin30
F1cos 30
F2sin45
F2cos45
30 45
So we have now worked out both the tensions in
the line that are there from the weight of the
chicken. This means that the line must be of a
strength equal or better than F2 = 12.68N

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Resolution of Forces – ICT Link Self Check
http://www.walter-fendt.de/ph14e/forceresol.htm

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2015
Exam Question…
A cyclist rides along a road up an incline at a
steady speed of 9.0 m s–1
. The mass of the rider
and bicycle is 70kg and the bicycle travels 15 m
along the road for every 1.0 m gained in
height.
Neglect energy loss due to frictional forces.
Calculate the component of the weight of the
bicycle and the rider that acts along the incline.
(3 marks)
Downwards…
W = mg = 70kg x 9.81N/kg
= 686.7N (1)
But the component that acts
along the slope is…
F = mg sin 
F = 686.7N x sin 
F = 686.7N x opp/ hyp
F = 686.7N x 1/15 (1)
F = 45.78N
F = 46N (1)
v= 9.0 m s–1
m = 70kg 1m
15m

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2015
9630-AQA-Physics
3.2 Moments, couple & torque
Sir Shakeel Mehmood

Animated Science
2015
3.4.1.2 / 6.3 The principle of moments; 6.4 More on moments; 6.5 Stability
1. Describe the conditions under which a force produces a turning
effect.
2. Explain how the turning effect of a given force can be increased.
3. Explain what is required to balance a force that produces a turning
effect.
4. Explain why the centre of mass is an important idea.
5. Describe the support force on a pivoted body.
6. When a body in equilibrium is supported at two places, state how
much force is exerted on each support.
7. Explain what is meant by a couple.
8. Explain the difference between stable and unstable equilibrium.
9. Assess when a tilted object will topple over.
10. Explain why a vehicle is more stable when its centre of mass is lower.
ICT Links...
Substitution
http://app.mymaths.co.uk/205-resour
ce/substitution-2
Rearranging Equations
ce/rearranging-1
Moments
ce/moments
Real Life Applications
ce/real-life-applications
Mathematical skills:
MS 0.1 Recognise and make use of appropriate units in calculations
MS 0.2 Recognise and use expressions in decimal and standard form
MS 1.1 Use an appropriate number of significant figures
MS 1.5 Identify uncertainties in measurements and use simple techniques to determine uncertainty when data
are combined by addition, subtraction, multiplication, division, and raising to powers
MS 3.2 Plot two variables from experimental or other data
MS 3.3 Understand that y = m x + c represents a linear relationship
MS 3.4 Determine the slope and intercept of a linear graph
MS 4.2 Visualise and represent 2D and 3D forms
MS 4.4 Use Pythagoras’ theorem and the angle sum of a triangle.

Animated
Science
2018
Mr D Powell
2018
Animated Science
2018
6.3 The Principle of
Moments
C
CHALLENGE
I know what conditions a force produce a
turning effect & How can the turning effect of a
given force be increased.
I can explain what is required to
balance a force that produces a
turning effect.
I can explain/ apply the context of
why is the centre of mass an
important idea?
I can weigh a ruler with moments only!
A
B

Animated Science
2015
Centres of mass and gravity
The centre of gravity of an object is a point where the entire weight of the object
seems to act.
An alternative definition is that the centre of mass or centre of gravity of an
object is the point through which a single force has no turning effect on the body.
In a uniform gravitational field the centre of mass is in the same place as the
centre of gravity.
The centre of mass of an object is a point where the entire mass of the object
seems to be concentrated.

Animated Science
2015
Equilibrium
A body persists in equilibrium if no net force or moment acts on it. Forces and
moments are balanced.
Newton’s first law states that a body persists in its state of rest or of uniform motion
unless acted upon by an external unbalanced force.
Bodies in equilibrium are therefore bodies that are at rest or moving at constant velocity
(uniform motion).
F1
F1
F2 F2
equilibrium

Animated Science
2015
The principle of moments
The principle of moments states that (for a body in equilibrium):
total clockwise
moments =
total anticlockwise
moments
4N 6N
5m d
4 × 5 = 6d
This principle can be used in calculations:
What is d?
20 = 6d
d = 20 / 6
d = 3.3m

Animated Science
2015
Human forearm
60N
20N
4cm
16cm
35cm
F
Taking moments about the elbow
joint:
4F = (16 × 20) + (35 × 60)
4F = 2420
F = 605N
The principle of moments can be used to
find out the force, F, that the biceps need to
apply to the forearm in order to carry a
certain weight. When the weight is held
static, the system is in equilibrium.
60N
schematic diagram
weight of arm = 20N

Animated Science
2015
0.2m
3N
The uniform metre rule shown is in equilibrium, with its centre of gravity marked by the
arrow ‘weight’. Find the weight of the metre rule.
0.3m 0.5m
W
total anticlockwise moments = total clockwise moments
3 × 0.2 = weight × 0.3
weight = 0.6 / 0.3
weight = 2N
Finding the weight of a metre rule

Animated Science
2015
Simple Moments….
d1
d2
W0
W2
W1
d3

Animated Science
2015
Work out the mass then density of a ruler...

Animated Science
2015
Example Results + errors!
For a wooden ruler a density is quoted at 750kgm-3
d1 = (0.2  0.001)m  0.5%
d2 = (0.2  0.001)m  0.5%
m = (96  1)g = (0.096  0.001)kg  0.1%
Thickness = (0.0061  0.00001)m  0.16%
Width = (0.00028  0.00001)m  0.04%
Length = (1  0.001)m  0.1%
m1x d1 = m2 x d2
Error in calculated mass = 1.1%
 = m/v
Value for our ruler density...
(562  8) kgm-3
So this must be a different
wood!

Animated
Science
2018
Mr D Powell
2018
Animated Science
2018
6.4 More on Moments –
Bridge Crane
SUCCEED
C
CHALLENGE
I can label/name the support forces on a
pivoted body?
When a body in equilibrium is supported at
two places, I can work out how much force is
exerted on each support?
I understand what is meant by a couple and
can apply it to problems?
I can work out two support and couple problems
without help for a bridge crane!
A
B

Animated Science
2015
Theory of Moments…

Animated Science
2015
Two Support Force Problems
D
d2
W0
Sx
d1
½d
d
Sy

Animated Science
2015
The torque of a couple
torque of a couple = force ×
perpendicular distance between
lines of action of the forces
A point P is chosen arbitrarily. Take moments
about P.
total moment = Fx + F(d – x)
= Fx + Fd – Fx
= Fd
There is a formula specifically for finding the torque of a couple.
d-x
x
d
F
F
p

Animated Science
2015
M
Couples Extras...
If two forces act about a hinge in opposite directions, there is an obvious turning
effect called a couple. The resulting linear force from a couple is zero.
The couple is given by the simple formula:
G = 2 Fs
This strange looking symbol, G, is “gamma”, a Greek capital letter ‘G’. Couples
are measured in Newton metres (Nm)
The turning effect is often called the torque. It is a common measurement made
on motors and engines, alongside the power. Racing engines may be quite
powerful but not have a large amount of torque. This is why it would not make
sense for a racing car to be hitched to a caravan, any more so than trying to win a
Formula 1 race in a 4 x 4.

Animated Science
2015
Practical Investigating the bridge crane
Aims
1. To use simple measurements of distance and to apply the principle of moments to a
model bridge crane.
2. To relate practical measurements to straight line graph theory.
3. To consider measurement errors and how to reduce them.
You require the following equipment:
4. two metre rulers
5. two spring balances calibrated in Newton's
6. a set square
7. slotted masses of weight that can be measured using a balance
8. thread
9. scissors
10. two stands and clamps
11. graph paper
Safety
Ensure the practical arrangement is stable and will not topple over.

Animated Science
2015
Setup Advice....
Ensure that the spring balances are vertical
and that the ruler is horizontal each time
before you make their measurements.
You may need to use a set square and a
second metre ruler to do this.
Use straight line graph theory and the general
equation;
y = mx + c
Theory....
Taking moments about the point where spring
balance S2 supports the bar gives:
S1D = Wd2 + 0.5W0D
S1 = (W/D) d2 + 0.5W0
Y = S1 , x = d2 , m = W/D c = 0.5Wo
D

Animated Science
2015
Setup Advice....
Ensure that the spring balances are
vertical and that the ruler is
horizontal each time before they
make their measurements.
You may need to use a set square
and a second metre ruler to do this.
Use straight line graph theory and
the general equation;
y = mx + c
It is recommended that you carry
out calculations in SI units in order
to avoid confusion.
Results
The weight W should be equal to the gradient of the line  D.
The weight W0
should be equal to 2  the y-intercept.
W = (6N +/- 1.8) N
Wo = (1.4 +/- 0.42)N
S1D = Wd2 + 0.5W0D
S1 = (W/D)d2+ 0.5W0
Y = S1 , x = d2 , m = W/D c = 0.5Wo
D

Animated Science
2015
Answers to Questions
1 (a) Use the second metre ruler to measure the vertical height (above the
bench) of the horizontal ruler at each end. Use the set square against the bench
(assumed horizontal) to check the second metre ruler is vertical. Adjust the clamps
holding the horizontal ruler if necessary and recheck it is horizontal. Use the set
square to check the spring balances are vertical.
(b) The spread of the points about the line of best fit gives an indication of
reliability. If the points are spread too much about the line of best fit, the results
unreliable and the measurements should be repeated.
2) (a) Ensuring the ruler is horizontal and the spring balances are vertical each
time the support forces S1
and S2
are to be measured. Precision of Newton meter.
(b) Repeat each measurement at least twice for each position of the weight
and calculate the mean value of each support force. Use more precise instrument
with better scale.
3) (a) The spring balance does not read zero when it is unloaded.
(b) The gradient would be the same. The y-intercept would be different.

Animated Science
2015
Couples and torques
A couple is a pair of forces acting on a body that are of equal magnitude and opposite
direction, acting parallel to one another, but not along the same line.
The torque of a couple is the rotation force or moment produced.
Forces acting in this way produce a turning force or moment.
F
F
The forces on this beam are a couple,
producing a moment or torque, which will
cause the beam to rotate.
d

Animated Science
2015
Summary Question…
Without a book or any notes in front of you.
1) Express in formula form, in a diagram and
numerically and with text (if possible) the
principles of “equilibrium” that you have
just covered in terms of forces and
moments for
A) One support problem
B) Couple
C) A two support problem
Pass this to another student and ask them
what they think then hand in your work at
the end for checking by your teacher….
NB: You will need to make up your own weights and distances

Animated Science
2015
6.4 Bridge Worked Example Q3….
NB: you can also
take moments from
0.5m in to be more
precise!

Animated
Science
2018
Mr D Powell
2018
Animated Science
2018
6.5 Stability – what is it?
C
CHALLENGE
I can explain what is the difference
between stable and unstable
equilibrium?
I can identify when a tilted object is going
to topple over and quote the
mathematics behind it?
I can mathematically prove why is a
vehicle more stable, the lower its centre
of mass is.
I can resolve for objects on a slope without help!
A
B


Animated Science
2015
What happens here?
W
b
d
F

Animated Science
2015
Equilibrium…
Stable Equilibrium – if an object is displaced it will return to that point
Unstable Equilibrium – if it is displaced slightly it will not return to that point
Toppling…
If centre of mass goes over the pivot point an object will topple…
Centre
of mass
W
b
d
F
W
b
d
F
W
b
d
F

Animated Science
2015
Moments
If we consider moments about the pivot
point….
d = perpendicular distance from line of action
of force
F = force applied
W = weight (acting through middle of base)
b = base of object
Clockwise Moment = Fd
Anticlockwise Moment = W* b/2
Hence we have three scenarios for the relation
of balance or unbalance…
1. Fd < Wb/2
2. Fd = Wb/2
3. Fd > Wb/2
NB: try it out and see
how it feels with a
block
Centre
of mass
W
b
d
F

Animated Science
2015
Moments
Imagine that this box had a base = 1.4m and
the force was applied at distance = 1.8m. The
weight of the box was 800N what is the force
needed to tilt the box to equilibrium?
This scenario means that…
Fd = Wb/2
F = Wb/2d
F = (800N * 1.4m) / (2* 1.8m)
F = 311N
W
b
d
F

Animated Science
2015

esolving Forces Parallel to Slope
F = W x sin 
esolving Perpendicular to Slope
S = Sx + Sy = W x cos 
hyp x sin  = opp
hyp x cos  = adj

Use page 109 Q3 for practice
Inclined Plane - Forces

NB: Sx Not equal to Sy due to position

Animated Science
2015
Example..
The lorry has a wheelbase of 1.8m and
centres of mass unloaded of 0.8m from
the ground. What is the maximum angle
it can drive at before toppling over….
d = 0.8m b/2 = 1.8m/2 = 0.9m
Hence opp/adj = tan 
tan-1
(0.9 / 0.8) = 48.3
Resolving on a Slope... (Harder A-B)
• When trying to work out the angle at
which a lorry will start to topple we
can use the idea of Resolving Forces
Parallel to Slope
• However, it is hard to see, so better to
rotate then turn into a balance
problem
Left = Right
Up = down
Now we can simply say that
Friction = Wsin  (opp) - Eq1
Sx + Sy = Wcos  (adj) - Eq 2
Now divide Eq1 by q2 to F/s = tan 


W
F = b/2 = 0.9m
d = 0.8
Page 103 Worked Example NB: Sy (if higher has lower force as higher)


Animated Science
2015
3.4.1.1/2 / 6.6 Equilibrium rules; 6.7 Statics calculations
1. Addition of vectors by calculation or scale drawing.
2. Resolution of vectors into two components at right angles to each
other.
3. Conditions for equilibrium for two or three coplanar forces acting at a
point. Appreciation of the meaning of equilibrium in the context of an
object at rest.
3.4.1.2
4. Principle of moments.
5. Knowledge that the position of the centre of mass of uniform regular
solid is at its centre.
ICT Links...
Substitution
ce/substitution-2
Rearranging Equations
ce/rearranging-1
Pythagoras Theorem
ce/pythagoras-theorem
Trigonometry – Missing Angles
ce/trig-missing-angles
Trigonometry – Missing Sides
ce/trig-missing-sides
Mathematical skills:
MS 0.6 Use calculators to handle sin x, cos x, and tan x when x is expressed in degrees or radians
MS 4.1 Use angles in regular 2D and 3D structures
MS 4.2 Visualise and represent 2D and 3D forms including 2D representations of 3D objects
MS 4.4 Use Pythagoras’ theorem and the angle sum of a triangle
MS 4.5 Use sin, cos, and tan in physical problems.

Animated
Science
2018
Mr D Powell
2018
Animated Science
2018
6.6 Equilibrium Rules
C
CHALLENGE
I can apply forces on an
object in equilibrium?
I can explain the turning effects of the forces?
I can apply these conditions to predict the
forces acting on a body in equilibrium?
I can apply my ideas to the Physics of Lifting p112
A
B

W
F
S



Animated Science
2015
Visuals…

W
F
S



Animated Science
2015
Scale Diagrams….
Another useful tool in AS is to draw a scale diagram with ruler and compass.
Then we can use the idea of a closed triangle to show equilibrium and resolve
an unknown force.
Just try it out with two vectors and an angle.
Then work out the third force by using simple trigonometry. The actual line
length should match your calculations

Animated Science
2015
Summary Questions….
a) Moments about “X” balance…..
200N x 0.5 = Wo x2m
Wo = 50N
b) Support Forces must match the
total weight down…
Sx = W0 + W = 50N + 200N = 250N
5m
0.5m
0.5m
W= 200N
W0= ?
Basic

Animated Science
2015
a) Moments about “X”
The moment is F x d = 1.2m x 1500N
= 1800Nm
b)
1.2m x 1500N = 1.0m x ?
1.2m x 1500N / 1m = 1800N
Hence, extra force is 1800N
(as applied from 1m away!) W= 1500N
Basic

Animated Science
2015
1. Split the problem into two triangles
working out the angle from vertical
for each
2. Label up sides with adj1 = adj2
Resolve Left = Right
T1sin40 = T2 sin 30
Resolve vertical direction….
2.8N = T1cos40 + T2cos30
Now solve with simultaneous equations
T1 = 1.493N = 1.5N
T 2 = 1.911N = 1.9N
40 30
opp1 opp2
hyp2
hyp1 adj
Medium Physics/ Harder Maths
1
2
T2sin2
T1sin1
T
2
cos
2
T
1
cos
1
T2
T1
W

Animated Science
2015
a) Weight acts down which makes it more
complex!
• But if you turn problem to flat then work out
the component of W acting in line with T to
balance…
• So green triangle you need to work out the
Hyp using  as same angle as slope
Angle of Beam to Horizontal
opp = sin x hyp -> Sin-1
(6m/10m) = 36.9
Up = Down (moments from “X”)
T x 10m = 10m/2 x 15,000N cos (36.9)
= 6000N
T
W = 15kN

W = 15kN
A*+ Harder Physics/
Harder Maths
NB: a simple method using similar
triangles. We know that W acts at 5m, so
6m/2 = 3m so the bottom length is 4m.
Hence moments…
Fd = 15,000 * 4m = 60,000Nm

Animated Science
2015
Angle slope = 36.9
b) Summing forces in the….
x direction on the base of the girder…
Sx (support force) = T*sin(36.9)
Sx = T*sin(36.9) = 6000*sin(36.9)
= 3602N = 3.6kN
Y-direction…
Sy + 6000*cos(36.87) = 15,000
Sy = 15000 - 6000*cos(36.9)
= 10202N = 10.2kN
Resolve Resultant for Support Force…
R = (3.6kN2
+ 10.2kN2
)½
= 10.8kN
T=6000N
3.6kN
10.2kN

Tsin
10.8kN
A*+ Harder Physics/
Harder Maths

Animated Science
2015
Vectors...
A boy runs 400 m due south and then 300 m due east. It takes him 6 minutes.
1. Calculate the total distance travelled by the boy.
2. Calculate the displacement of the boy.
3. Calculate the direction of the displacement.
4. Calculate his average speed.
5. Calculate his average velocity.
1. 400m + 300m = 700m
2. Sqrt((400m)2
+(300m)2
)= 500m (resultant)
3. Opp/adj =tan = tan-1
(300/400) = 37 SE or 143 to NE
4. d/t = s = 700m/(6x60)s = 1.9m/s
5. s/t = v = 500m/(6x60)s = 1.4m/s
400m
300m

Animated Science
2015
Resolving...
A box of mass 18 kg is at rest on a horizontal frictionless surface. A force of 4 N is
applied to the box at an angle of 26o
to the horizontal
1. Show that the horizontal component of this force is 3.6 N.
2. Calculate the acceleration of the box along the horizontal surface.
3. Calculate the horizontal distance travelled by the box in a time of 7 s.
FH = Rcos
FH = 4 x cos 26o
FH = 3.6 N
4 N
FH = 3.6 N
m = 18 kg
a = a
a = FH/m
a = 3.6/18
a = 0.2 ms-2
t = 7 s
a = 0.2 ms-2
u = 0 ms-1
s = s
s = ut + ½at2
s = (0 x 7) + (½ x 0.2 x 72
)
s = 0 + 4.9
s = 4.9 m

Animated Science
2015
Balanced Forces...
A crane on an oil-rig is used to raise a sunken buoy from the seabed. The weight
of the buoy is 4900 N and the buoyancy force (upthrust) acting on it is 1000 N.
When the buoy is being raised vertically at a constant speed, a force of 800 N
acts on it due to water resistance.
What is the size of the force which the vertical cable applies to the buoy?
weight of the buoy = 4900 N
upthrust = 1000 N
water resistance = 800 N
vertical force up = vertical force down
weight of the buoy + water resistance = upthrust + vertical force
vertical force = (weight of the buoy + water resistance) – upthrust
vertical force = (4900 + 800) – 1000
vertical force = 4700 N

Animated
Science
2018
Mr D Powell
2018
Animated Science
2018
6.7 Statics Calculations/ Practice
Questions
C
CHALLENGE
A
B

Animated Science
2015
Conditions for the Equilibrium of Three Non-Parallel Forces
If we say that an object is under the influence of forces which are in equilibrium, we
mean that the object is not accelerating - there is no net force acting. The object may
still be travelling - but at a constant velocity - but in most questions the object will be
stationary.
The following ideas will help you to solve problems that involve a body acted on by
three co-planar forces.
The lines of action of the three forces must all pass through the same point.
The principle of moments: the sum of all the clock-wise moments about any point must
have the same magnitude as the sum of all the anti-clockwise moments about the same
point.
The sum of all the forces acting vertically upwards must have the same magnitude as
the sum of all the forces acting vertically downwards
The sum of all the forces acting horizontally to the right must have the same magnitude
as the sum of all the forces acting horizontally to the left.
If you resolve all of the forces the 'ups' will equal the 'downs' and the 'rights' will equal
the 'lefts

Animated Science
2015
Quick Test what does each one represent...
1. A point on an object where the entire weight of the object seems to act. In a uniform
gravitational field, it is in the same place as the centre of mass.
2. A point on an object where the entire mass of the object seems to be concentrated. In a uniform
gravitational field, it is in the same place as the centre of gravity.
3. A pair of forces acting on a body that are of equal magnitude and opposite direction, acting
parallel to one another, but not along the same line.
4. The state a body is in if no net force or moment acts on the body.
5. The point about which a lever turns. Also called the pivot.
6. The turning effect of a force or forces. Can also be called the torque (symbol τ). It is calculated by
multiplying the force by the perpendicular distance between the pivot and the line of action of
the force. The units of moment are newton metres (Nm).
7. The point about which a lever turns. Also called the fulcrum.
8. The principle stating that the sum of the clockwise moments are equal to the sum of the
anticlockwise moments acting on a body.
9. The turning effect of a force or forces, given the symbol τ. Can also be called the moment. It is
calculated by multiplying the force by the perpendicular distance between the pivot and the line
of action of the force. The units of torque are newton metres (Nm).
10. The rotation force or moment produced by a couple. It can be calculated by multiplying the force
by the perpendicular distance between the lines of action of the forces. Its units are newton
metres (Nm).
11. The force created by the gravitational attraction on a mass. Its units are newtons (N).

Animated Science
2015
Quick Test Answers...
1. centre of gravity – A point on an object where the entire weight of the object
seems to act. In a uniform gravitational field, it is in the same place as the centre
of mass.
2. centre of mass – A point on an object where the entire mass of the object seems
to be concentrated. In a uniform gravitational field, it is in the same place as the
centre of gravity.
3. couple – A pair of forces acting on a body that are of equal magnitude and
opposite direction, acting parallel to one another, but not along the same line.
4. equilibrium – The state a body is in if no net force or moment acts on the body.
5. fulcrum – The point about which a lever turns. Also called the pivot.
6. moment – The turning effect of a force or forces. Can also be called the torque
(symbol τ). It is calculated by multiplying the force by the perpendicular distance
between the pivot and the line of action of the force. The units of moment are
newton metres (Nm).
7. pivot – The point about which a lever turns. Also called the fulcrum.
8. principle of moments – The principle stating that the sum of the clockwise
moments are equal to the sum of the anticlockwise moments acting on a body.
9. torque – The turning effect of a force or forces, given the symbol τ. Can also be
called the moment. It is calculated by multiplying the force by the perpendicular
distance between the pivot and the line of action of the force. The units of
torque are newton metres (Nm).
10. torque of a couple – The rotation force or moment produced by a couple. It can
be calculated by multiplying the force by the perpendicular distance between
the lines of action of the forces. Its units are newton metres (Nm).
11. weight – The force created by the gravitational attraction on a mass. Its units are
newtons (N).

Animated Science
2015
Exam Question ...Jan 07 Spec A Q5
5 (a) Define the moment of a force.
(b) The diagram shows the force, F, acting on a bicycle pedal.
(i) The moment of the force about O is 46Nm in the position shown. Calculate the
value of the force, F. (2 marks)
(ii) Force, F, is constant in magnitude and direction while the pedal is moving
downwards. State and explain how the moment of F changes as the pedal moves
through 80°, from the position shown. (2 marks)

Animated Science
2015
Exam Question ...Jan 07 Spec A Q5
5b)….
(i) The moment of the force about O is 46Nm in the position shown. Calculate the
value of the force, F. (2 marks)
(ii) Force, F, is constant in magnitude and direction while the pedal is moving
downwards. State and explain how the moment of F changes as the pedal moves
through 80°, from the position shown. (2 marks)
46Nm
240N

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