AS Physics Unit 7
Forces in Equilibrium
Mr D Powell
7
Forces in Equilibrium
The addition of vectors by calculation or scale drawing. Calculations will be limited to two
perpendicular vectors. The resolution of vectors into two components at right angles to each
other; examples should include the components of forces along and perpendicular to an
inclined plane.
Conditions for equilibrium for two or three coplanar forces acting at a point; problems may be
solved either by using resolved forces or by using a closed triangle.
Moment of a force about a point defined as force × perpendicular distance from the point to
the line of action of the force; torque.
Couple of a pair of equal and opposite forces defined as force × perpendicular distance
between the lines of action of the forces.
The principle of moments and its applications in simple balanced situations.
Centre of mass; calculations of the position of the centre of mass of a regular lamina are not
expected.
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Chapter Map
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7.1 Vectors and Scalars
1.
What is a vector quantity?
2.
How do we represent vectors?
3.
How do we add and resolve vectors?
Representing a vector
Addition of vectors using a scale diagram
Addition of two perpendicular vectors using a
calculator
Resolving a vector into two perpendicular
components
Spec 3.2.1
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A Definition....
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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Displacement and velocity
A runner completes one lap of an athletics
track.
What distance has she run?
400 m
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?
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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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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 = 1 kgms-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.
N

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.
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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)
y
6
a
Or as a pair of values called
components:
e.g. a = (8, 6) or
8
8
x
6
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Vector addition
Displacement vectors can always be added ‘nose to tail’ to find a total or resultant
vector.
y
b
This can be done approximately by scale
drawing:
a
a+b
10
x
7
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
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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 3 km and a
horizontal displacement of 4 km?
magnitude:
R2 = 32 + 42
R
4 km
R = √ 32 + 42
= √ 25
θ
direction:
tan θ = 4/3
θ = tan-1(4/3)
= 53°
= 5 km
3 km
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Vector components
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.
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 6 ms-1 and a direction of 60° from the horizontal?
cos60° = x / 6
x = 6 × cos60
sin60° = y / 6
y = 6 × sin60
6 ms-1
y
= 3 ms-1
= 5.2 ms-1
60°
x
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Pythagoras & Adding Vectors
TASK: Now try out the same techniques
for two more triangles;
1) opp = 2m, adj = 8m
Answers
1.
2.
 = 14 , R =8.25m
 = 20.6 , R =4.27m
2) opp = 1.5m, adj = 4m
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Applications...
TASK: Now try out the same techniques for
two more triangles;
1) What is R &  if opp = 54N, adj = 22N
Answers
1.
2.
 = 67.83 , R =58.3m
 =39.4 , R =65.6ms-1
2) What is adj &  if opp = 54ms-1, R =
85ms-1
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Splitting a vector into components...
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....
Answers
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.
1. Vy =8.8ms-1
2. Vy =70.7N
Vx = 70.7N
2) A javelin is thrown at a resultant force of 100N at
a 45  angle. What are the horizontal and vertical
forces on it.
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Questions to try on your own....
Answers
.....
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More Trig Practice
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;
opp
tan 
adj
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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+152-2x152Cos140)0.5 = 28.19N, then half for
triangle. To 14.1N. Then pythag (152-14.12)0.5 = 5.12N
(x2) = 10.23N
a
b
c
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Vector Problems 1 Question 2
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Vector Problems 1 Question 3
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
3500km
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
0.279km
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Vector Problems 2 Question 2
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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Vector Problems 2 Question 2
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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Vector Problems 2 Question 3
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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Vector Problems 2 Question 4
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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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
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5a
N
5b
N
N
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)
Vector Problems 2 - Question 7
Forces of 60.0N and F Newton’s act at
point O.
a
c

SinA SinC
Find the magnitude and direction of F if
the resultant force is of magnitude
30.0N along OX (2 marks)
c2 = a2 + b2 - 2abCosC
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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;
b = 30N
120
A
a
c

SinA SinC
a = 60N
c=F
Sine Rule
Cosine Rule
Use reciprocal version of above
c2 = a2 + b2 - 2abCosC
c2 = 60N2 + 30N2 – 2x60Nx30N xCos120
c2 = 4500N2 +1800N2
c2 = 6300N
c = 79.4N
c = F = 79.4N
SinC / c = SinA / a
(60N x Sin 120) / 79.4N) = SinA
Sin-1 (60N x Sin 120) / 79.4N) =A
A = 40.876 = 41 
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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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Wires Example - Correct
So by turning it on its side we can use the same
formulae (which is confusing but correct)
Left side
Vertically
1.20N = T1cos30+ T2cos60
1.20N = 0.866T1+ 0.5T2
Horizontally
T1sin30 = T2sin60
0.5T1= 0.866T2
T1= 1.732T2
Eq 2
Eq 1
right side
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
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WRONG!
Vertically
1.20N = T1sin30+ T2sin60
1.20N = 0.5T1+ 0.866T2
Horizontally
T1cos30 = T2cos60
0.866T1= 0.5T2
T1= 0.577T2
Eq 2
Eq 1
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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Alternative Can do by using (90- ) as angle;
If we take the outside angle instead of the inner
angle we can do this and use the triangle;
(90-)
Vertically
1.20N = T1sin60+ T2sin30
1.20N = 0.866T1+ 0.5T2
Horizontally
T1cos60 = T2cos30
0.5T1= 0.866T2
T1= 1.732T2
Eq 2
Eq 1
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
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7.2 Balanced Forces
1.
Why do we have to consider the direction
in which a force acts?
2.
When do two (or more) forces have no
overall effect on a point object?
3.
What is the parallelogram of forces?
Equilibrium of a point object
Testing three forces in equilibrium
Practical - Coplanar forces
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Why are the objects not moving...
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Equilibrium Point of an Object
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Example 1
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Example 1
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Example 2
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Practical...
 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.
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Example 3
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Example 1...
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!
Using the following
figures the problem
is very simple...
W1 = 5N
W2 = 3N
W3 = 6N
1 = 150
2 = 124
F1sin
F1
F2

F2sin

F2cos 
F1cos 
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Example 1...
V  0
 Using the following figures the
problem is very simple...
F1 sin   F2 sin   W3  F3
4.33  1.67  6
W1 = 5N W2 = 3N
W3 = 6N
1 = 150
2 = 124
=60  = 34 
H  0
F1 cos  F2 cos 
2.5  2.5
F1sin
F1
F2

F1cos 

F2sin
F2cos 
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Plenary Question
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?
H  0
V  0
F1 cos30  F2 cos 45
F1 sin 30  F2 sin 45  20N
3
2
F1
 F2
2
2
2
F1  F2
3
F1
2
 F2
 20N
2
2
F1  F2 2  40N
F1sin30
F2
F1
30
F1cos 30
F2sin45
45
F2cos45
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Plenary Question
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?
F1  F2 2  40N
F1  F2
F1  F2 2  40N
F1  F2
F2
2
3
2
 F2 2  40N
3
F2  3F2 
3  40N
2
34.64
F2 
 12.68N
2.732
2
3
F1  F2
2
3
2
F1  12.68N 
 10.35N
3
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
F2= 12.68N
F1=10.35N
F2sin45
F1sin30
30
F1cos 30
45
F2cos45
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Index
Practical Results…
Weight +/0.01N
1.96
2.45
2.94
3.43
3.92
4.41
4.91
ave
Sin (ave)
In terms of forces Hyp = F
opp = mg
Hence... opp/hyp = adj
mg/F=sin(theta)
Angle +/-1
deg
24
24
24
24
24
26
27
24.71
0.42
Weight +/-0.01N
6.00
y = 0.4905x
R² = 1
5.00
Weight Added (N)
Force on Newton
mass +/-1g
meter +/-0.2N
4.0
200
5.0
250
6.0
300
7.0
350
8.0
400
9.0
450
10.0
500
4.00
3.00
2.00
1.00
Fsin(theta) = mg
F=x
m = sin(theta)
y = mg
0.00
0.0
2.0
4.0
6.0
8.0
10.0
Force on meter (N)
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12.0
Resolution of Forces – ICT Link Self Check
http://www.walter-fendt.de/ph14e/forceresol.htm
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7.3 Principle of Moments
1.
Under what conditions does a force produce a turning effect?
2.
How can the turning effect of a given force be increased?
3.
What is required to balance a force that produces a turning effect?
4.
Why is the centre of mass an important idea?
Turning effects, The principle of moments, Centre of mass, Centre of mass
tests, Calculating the weight of a metre rule
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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.
The centre of mass of an object is a point where the entire mass of the object
seems to be concentrated.
In a uniform gravitational field the centre of mass is in the same place as the
centre of gravity.
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.
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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
F2
F2
F1
equilibrium
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The principle of moments
The principle of moments states that (for a body in equilibrium):
total clockwise
moments
total anticlockwise
moments
=
This principle can be used in calculations:
5m
What is d?
d
4 × 5 = 6d
4N
6N
20 = 6d
d = 20 / 6
d = 3.3 m
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Human forearm
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.
weight of arm = 20 N
60 N
Taking moments about the elbow
joint:
4 cm
F
schematic diagram
4F = (16 × 20) + (35 × 60)
16 cm
20 N
4F = 2420
60 N
35 cm
F = 605 N
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Finding the weight of a metre rule
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.
0.2 m
0.3 m
0.5 m
W
3N
total anticlockwise moments = total clockwise moments
3 × 0.2 = weight × 0.3
weight = 0.6 / 0.3
weight = 2 N
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Work out the mass then Density of a ruler...
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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 = (0.028 +/- 0.00001)m +/- 0.04%
Value for our ruler density...
Length = (1 +/- 0.001)m +/- 0.1%
m1xd1 = m2xd2
Error in mass = 1.1%
(562.1 +/- 8) kgm-3
So this must be a different
wood!
Density = m/v
Error in density = 1.1%+ 0.16% + 0.04% + 0.1% = 1.4%
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7.4 More on Moments
1.
What can we say about the support force
on a pivoted body?
2.
When a body in equilibrium is supported
at two places, how much force is exerted
on each support?
3.
What is meant by a couple?
Support forces
- Single support problems
- Two support problems
Couples
ISA preparation:
practical task – practice in recording
measurements and plotting a graph
written task Section A and B –
opportunities to discuss reliability
and to analyse data and errors
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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:
1.
2.
3.
4.
5.
6.
7.
8.
two metre rulers
two spring balances calibrated in newton
a set square
slotted masses of weight that can be measured using either balance
thread
scissors
two stands and clamps
graph paper
Safety
Ensure the practical arrangement is stable and will not topple over.
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Index
Setup Advice....
Ensure that the spring balances are vertical
and that the ruler is horizontal each time
before you make their measurements.
D
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 = Wd2/D + 0.5W0
Y = S1 , x = d2 , m = W/D c = 0.5Wo
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Setup Advice....
S1D = Wd2 + 0.5W0D
S1 = Wd2/D + 0.5W0
Ensure that the spring balances are
vertical and that the ruler is
Y = S1 , x = d2 , m = W/D c = 0.5Wo
horizontal each time before they
make their measurements.
You may need to use a set square
and a second metre ruler to do this.
D
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
W = (6N +/- 1.8) N
Wo = (1.4 +/- 0.42)N
The weight W should be equal to the gradient of the line  D.
The weight W0 should be equal to 2  the y-intercept.
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Index
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.
Mr Powell 2009
Index
Couples...
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.
Mr Powell 2009
Index
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.
Forces acting in this way produce a turning force or moment.
The torque of a couple is the rotation force or moment produced.
The forces on this beam are a couple,
producing a moment or torque, which will
cause the beam to rotate.
F
d
F
Mr Powell 2009
Index
The torque of a couple
There is a formula specifically for finding the torque of a couple.
A point P is chosen arbitrarily. Take moments
about P.
F
P
x
total moment = Fx + F(d – x)
d–x
F
d
= Fx + Fd – Fx
= Fd
torque of a couple = force ×
perpendicular distance between
lines of action of the forces
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Index
7.5 Stability
1.
What is the difference between stable
and unstable equilibrium?
2.
When is a tilted object going to topple
over?
3.
Why is a vehicle more stable, the lower
its centre of mass?
Mr Powell 2009
Index
Mr Powell 2009
Index
Forces Triangles...
Resolving Forces Parallel to Slope
F = W x sin 
hyp x sin  = opp
hyp x cos  = adj
Resolving Perpendicular to Slope
Sx + Sy = W x cos 


Mr Powell 2009
Index
7.6 Equilibrium Rules
1.
What condition must apply to the forces
on an object in equilibrium?
2.
What condition must apply to the turning
effects of the forces?
3.
How can we apply these conditions to
predict the forces acting on a body in
equilibrium?
Free body force diagrams
The triangle of forces
The conditions for equilibrium of a body
Mr Powell 2009
Index
Visuals…
Mr Powell 2009
Index
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).
Mr Powell 2009
Index
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).
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Index
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)
Mr Powell 2009
Index
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)
Mr Powell 2009
Index
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
Mr Powell 2009
Index