Download Notes 2

Dr. Mark Sullivan
Analytical methods
Part II
BASIC TRIGONOMETRY
Imagine two sticks, the bottom one being 1
meter long. As you open them up, two
things happen;
1) The angle they make at the join will
increase.
Height
2) The height measured from the top
stick to the end of the meter long base
one increases.
Angle
There is a direct correlation between the
angle and the height. A table of the
heights is below.
1m
Question:
A)
If the bottom stick was 2 meters
long, how would the table opposite change?
Answer;
Angle
Height
0
10
20
30
40
50
60
70
80
0.00 m
0.18 m
0.36 m
0.58 m
0.84 m
1.19 m
1.73 m
2.75 m
5.67 m
From your answer to the last question you can see that we only need to multiply the height that
we get from the table above by the length of the base (or distance to the object) to find the
height of something. This saves a lot of shinning up towers!
Height  Base  Height derived from angle 
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Dr. Mark Sullivan
Analytical methods
The Height of the tree
is found by;
h = d tan 
(This last symbol is a
Greek letter
pronounced Theeta.)
SAQ
Set A
Set B
A tower block is 125 m distant. The angle of
elevation of the top of the tower is 38. How
high is the tower?
A tower block is 143 m distant. The angle of
elevation of the top of the tower is 32. How
high is the tower?
A flag pole is 35 m distant. The angle of
elevation of the top of the tower is 48. How
high is the tower?
A flag pole is 43 m distant. The angle of
elevation of the top of the tower is 42. How
high is the tower?
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Dr. Mark Sullivan
Analytical methods
A Statue is 224 m distant. The angle of
elevation of the top of the tower is 18. How
high is the tower?
A Statue is 231 m distant. The angle of
elevation of the top of the tower is 20. How
high is the tower?
A Hilltop is 1.45 km distant. The angle of
elevation of the top of the tower is 8. How
high is the tower?
A Hilltop is 16.8 km distant. The angle of
elevation of the top of the tower is 3. How
high is the tower?
A Monument is 35 m distant. The angle of
elevation of the top of the tower is 58. How
high is the tower?
A Monument is 83 m distant. The angle of
elevation of the top of the tower is 42. How
high is the tower?
Formulae like
Height  Base  Height derived from angle  can be swapped around as long as
you remember one golden rule; if you move something to the other side of the = sign, it must do
the opposite job. For example, take the quantity ‘Base’ in the above formula. Its job, so to speak,
is to multiply the ‘Height derived from angle’ quantity. When we move it to the other side it must
do the opposite of multiply which is divide.
So re-arranging gives;
Height

Base
Height derived
from angle 
This method of dividing the height (or opposite) by the base (or adjacent) is called ‘taking the
tangent’ and is very useful in surveying techniques.
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Dr. Mark Sullivan
Analytical methods
For any right-angled triangle, i.e. not one necessarily sitting on the ground, we can it
label like this;
In this case the tangent (tan) can
be written as;
Hypotenuse
Opposite
the angle
tan  
Opp
Adj
Angle
Adjacent to
the angle
This shouldn’t be too much of a surprise, after all we derived the tangent from the vertical
measurement of height above a horizontal 1 m stick. In other words; the Opposite and the
Adjacent.
With similar logic, two other arrangements can be written. These are Sine and Cosine or sin and
cos. Pronounced ‘Sign’ and ‘Koz’
sin  
Opp
Hyp
cos 
Adj
Hyp
Imagine our two sticks again. As you open
them up, two things happen;
Slope
1) The angle they make at the join will
increase.
2) The distance measured from the top of
the bob back along the top stick to the
angle increases.
There is a direct correlation between the
angle and the slope. A table of the slope
lengths is below.
Table of the Sine of angles
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Angle
1m
Angle
0
10
20
30
40
50
60
70
80
Height
0.00 m
0.17 m
0.34 m
0.50 m
0.64 m
0.77 m
0.87 m
0.94 m
0.98 m
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Dr. Mark Sullivan
Analytical methods
Write down the following (first bit is already
done)
.3
5

4
Sin  =
=
3
5
0.6
Find the angle which gives this value
_____________________________
Cos  =
____
=
Find the angle which gives this value
Tan  =
=
____
Find the angle which gives this value
_____________________________
_____________________________
SAQ
Find the angles marked  below, the triangles being right-angled.
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Dr. Mark Sullivan
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5

4
3.5

1.8
51.58

42.36
In ABC, C = 90, B =
23.3 and AC = 11.2 cm.
Find AB.
In ABC, B = 90, A =
67.5 and AC = 0.86 m.
Find BC.
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Dr. Mark Sullivan
Analytical methods
SINE AND COSINE RULES
The Sin, Cos and Tan functions only work for right-angled triangles. This isn’t too
helpful for surveying as land seldom has nice tidy right-angles in it.
Much more useful for us are two rules which apply to any triangle weather it has
a right-angle or not. They’re very imaginatively called the Sine rule and the
Cosine rule.
a
b
c


sin A sin B sin C
Sine rule
1) Where to use it
Use when you have
a) A side and its opposite angle
b) Any other side or angle
2) How to use it
Write down a fraction with a known side on the top and the sin of the known angle
opposite that side on the bottom.
Write down an ‘=’ sign and then another fraction with a side and opposite angle as you
did in the first step except this time one must be known and the other must be what you’re
trying to find.
Rearrange the equation so that the quantity you want to find is isolated and work out
its value.
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Dr. Mark Sullivan
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Worked example;
b
Next step; Write down a fraction with a
known side on the top and the sin of the known
angle opposite that side on the bottom.
a
30
40
10
sin 110
10m
The span of a roof
is 10 m with
c
pitches of 30 and
40. Find the
length of the slope a.
Next, write down an ‘=’ sign and then
First step; Do we have two angles? Yes,
then find the third. No, then skip the
step.
must be what you’re trying to find;
To find the missing angle use 'sum of
angles'.
another fraction with a side and opposite
angle as you did in the first step except
this time one must be known and the other
10
a

sin 110 sin 30
Finally, rearrange the equation so that the
quantity you want to find is isolated
a
30 + 40 + C = 180
C = 180 - (30 + 40)
10 sin 30
sin 110
a
C = 110
10  0.5
0.9397
= 5.231 m
= 5.230 m (to the nearest mm) .
Try these (SAQ)
Set A
Set B
The span of a roof is 12 m with pitches
of 35 and 45. Find the length of the
slope a.
The span of a roof is 14 m with pitches
of 33 and 47. Find the length of the
slope a.
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Dr. Mark Sullivan
Analytical methods
The span of a roof is 8 m with pitches
of 20 and 50. Find the length of the
slope a.
The span of a roof is 9 m with pitches
of 25 and 40. Find the length of the
slope a.
In triangle ABC, a = 15cm, b = 25cm,
and angle A = 47. What is angle B
(angle opposite side b).
In triangle ABC, a = 12cm, b = 20cm,
and angle A = 35. What is angle B
(angle opposite side b).
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Analytical methods
c 2  a 2  b 2  2ab cos c
Cosine rule
3) Where to use it
Use when you have
a) Two sides and the angle in-between
4) How to use it
This can be used in two forms; if you want to find the length of a side then this
formula is just a matter of plugging numbers into the calculator. Here is a re-wording of the
Cosine rule that might be more useful.
Side 
First side 2  Second side 2  2 First side  Second side  Cos of angle between
if, however, you want to find an angle, then this version of the Cosine rule is the one
to use;
 a2  b2  c2 

Angle  Cos 1 
2
ab


Worked example
C
25
100m
A
Using the cosine rule:
130m
B
Two points a and b lie on opposite sides
of a gully. A point c is 100 m from a and
130 m from b and angle C = 25. How far
horizontally is a from b?
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c2 = a2 + b2 - 2 a x b x cos C
c2 = 1002 + 1302 - 2 x 100 x 130 x cos
25
c = 57,76 m
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Dr. Mark Sullivan
Analytical methods
Try these (SAQ)
Set A
Set B
Two points a and b lie on opposite sides of a
gully. A point c is 80 m from a and 90 m from b
and angle C = 35. How far is a from b?
Two points a and b lie on opposite sides of a
gully. A point c is 70 m from a and 100 m from
b and angle C = 45. How far is a from b?
Two points a and b lie on opposite sides of a
road. A point c is 180 m from a and 130 m from
b and angle C = 46. How far is a from b?
Two points a and b lie on opposite sides of a
road. A point c is 140 m from a and 110 m from
b and angle C = 54. How far is a from b?
Two points a and b lie on opposite sides of a
chip shop. A point c is 18.345 m from a and
13.042 m from b and angle C = 46 34’ 40”. How
far is a from b?
Two points a and b lie on opposite sides of a
chip shop. A point c is 12.382 m from a and
10.012 m from b and angle C = 36 44’ 50”. How
far is a from b?
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Dr. Mark Sullivan
Analytical methods
Sine and Cosine rules
In triangle ABC, angle A = 55, a = 7.2, and c =
6. Find the angles B and C and the length of
the side b.
In triangle ABC, angle A = 40, a = 6.2, and c =
5. Find the angles B and C and the length of
the side b.
In triangle ABC, c = 8, angle A = 30, and angle
B = 42. Find the length of the sides b and a.
In triangle ABC, c = 12, angle A = 40, and angle
B = 52. Find the length of the sides b and a.
Two points of interest; c & d are on opposite sides of a wood with no line of sight between them.
A surveyor takes measurements at a & b and draws the following sketch (not to scale). Find the
distance between c & d.
a
15 m
57
58
b
48
61
c
d
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Dr. Mark Sullivan
Analytical methods
ERRORS.
In all applications, numbers come in two parts. One is the value, the other is its accuracy or more
correctly its probable error.
With a standard ruler,
Could you measure something accurately to the nearest centimetre?
What about to the nearest millimetre?
To the nearest tenth of a millimetre?
Almost certainly.
Very probably.
No chance!
A ruler is only accurate to about 1 mm, possibly 0.5 mm if you use it carefully but beyond that,
confidence is lost. Correctly then, all measurements should be recorded with an appropriate
error estimate, for example;
105 mm
 0.5 mm
With a tape measure on-site, do you think you would be able to measure a distance of about 100
m to an accuracy of 1 mm ? 1 cm? 1.5 cm? 1 m?
SAQ
Try this;
A tower block is 625 m distant. The angle of elevation of the top of the tower is 18 34’ 10”. How
high is the tower?
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Dr. Mark Sullivan
Analytical methods
Your collegue does the same exercise at the same distace and finds an angle of elevation greater
than yours. What is the difference in your calculated heights if his measurments are
a) 1 larger
b) 1’ larger
c) 10” larger
d) 1” larger
What would be an acceptable level of error and what accurassy will you have to achieve with your
measurments to be of professional standard?
(One or two sentences justifying your
value(s))
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Dr. Mark Sullivan
Analytical methods
SIX - FIGURE GRID REFERENCES
A six figure grid reference is
normally the one you would
choose to use to locate a point
(say a building).
To quote a six figure reference, first locate the square that
you are interested in. Then imagine dividing that square into 10
equally spaced intervals along and up the square. This would
form an imaginary grid of 100 squares within the grid square.
Then quote a further reference using these numbers. For
example, an object in the exact centre of a square would have a
reference along the lines of ‘something something 5 something
something 5’.
The first two numbers would be the eastings of the grid square, the second two 'somethings'
would be the northings of the grid square. The 5 and 5 are there because it would be 5 in and 5 up
within the blue grid square (i.e. in the centre). For ease of use, when we write this, we write all
the eastings together, then all the northings together.
Have a look at the example below and see if it helps. Here we
are trying to find the six figure grid reference of the church
with a tower in Bishop's Tachbrook. The large cross represent
a distance of 5 tenths of the way into the square (in other
words halfway) horizontally and vertically.
Eastings First (Vertical Lines) –
between 31 & 32 so 31something. Estimate (or
measure for accuracy) the
number of tenths of a square it
is along. Think of it as 31.4
across
Northings Second (Horizontal
Lines) – between 61 & 61 so 62something. Estimate (or
measure for accuracy) the
number of tenths of a square it
is up. Think of it as 61.4 up
The reference is 314614.
Further examples are: 
Mill Mound
314617
New House Fm
304621
Make sure you can locate these on the map.
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Dr. Mark Sullivan
Analytical methods
NAVIGATING FROM ONE POINT TO ANOTHER
To find the distance between two points for which you have map references we need to reacquaint
ourselves with good old Pythagoras. You remember him, the ancient Greek, about 500BC who did
stuff with triangles.
Pythagoras said that if you square the small sides of a right-angled triangle, their sum total will be
eqal to the square of the long side;
c
a
c2 = a2 + b2
b
Using the basic rule of algebra, you can swap things about as long as they do the opposite job on
the other side of the equals, I can move the squaring 2 from the a to the other side where it does
the opposite job and becomes a root.
c  a 2  b2
Now consider our triangle on a
grid map with its points at
positions A & B.
58
If we want to know the distance
between point A & B its like
finding the hypotenuse of a
right-angled triangle.
57
All we need are the distances
East-West between them and
North-South.
B
N


56
55 21
A
22
23
24
25
Consider points A (220560) and B (250570)
The first 3 digits refer to the didtance East-West and the last 3 to the distance North-South of
each point from a datum point. The difference between them is the length of the base and height
of our triangle.
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Dr. Mark Sullivan
Analytical methods
E-W
N-S
B=
250
570
A=
220
560
30
10
Difference =
Now using Pythagoras’ formula in the form c  a 2  b 2 we can calculate the length of c, the
distance between A & B.
Using the calculator;
 ( 30 x2 + 10 x2
)
=.
.
31.6227766
Remember that the map references were given to a tenth of a 1 km square accuracy, which is an
accuracy of a tenth of a km.
If that’s what the references were measured to, then that is the units of the answer. To swap
from tenths of a km to a more recognisable unit say km, then divide by ten (move the decimal one
place to the left).
.
3.16227766 km.
Presented here in km, the first number after the decimal is 1 and represents tenths of a km. The
next number after the decimal, 6, represents hundredths of a km but the references were given
only to a tenth of a km. Any numbers after the first, tenths, place are therefore irrelevant
The distance from A to B is =
3.1 km  100 m
SAQ
Find the distance between the points used on the last map;
Mill Mound
314617
New House Fm
304621
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Dr. Mark Sullivan
Analytical methods
Baring of B from A
Consider again our triangle on a
grid map with its points at
positions A & B (reproduced
here).
If we want to know the baring
of point B from A we can use
the tangent function.
We know the distances EastWest between them and NorthSouth.
tan  
58
B
57
N

56

A
55
North  South
East  West
21
22
23
24
25
Again using the rule of algebra, I will move the tan function to the other side where it swaps to
anti-tan or inverse tan;
 North  South 

 East  West 
  tan 1 
in the case of the example in the diagram, we already know the E-W & N-S distances so we can
use the calculator again;
SHIFT tan ( 10  30 ) =.
.
18.43494882
this is the value of the angle  in decimal degrees. To swap this to degrees, minutes and seconds;
press;
SHIFT  ‘ “
.
18 26’ 5.82”
The final step is to convert this angle to one reading from North. This is angle  in the diagram
(pronounced Theeta).
We know that it’s a right-angle between West and North i.e. 90 so we can just subtract  from
90 to find the whole circle baring.
 = 90 - 
 = 90 - 18 26’ 5.82”
 = 71 33’ 54.18”
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Dr. Mark Sullivan
Analytical methods
SAQ
Set A
Set B
Find the distance and baring of point B from
point A for the following;
Find the distance and baring of point Z from
point Y for the following;
A = 258369
B = 266381
Y = 258369
Z = 366381
A = 369478
B = 464481
Y = 269255
Z = 465281
A = 470585
B = 567681
Y = 270364
Z = 364387
A = 581692
B = 661381
Y = 281471
Z = 299381
A = 681392
B = 661381
Y = 381471
Z = 299381
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Dr. Mark Sullivan
Analytical methods
Finding a position from a baring
Point B lies 2.8 km on a baring of 030 from an observer located at 220560. what is the position
of point B?
Now consider our triangle on the
grid map again with its points at
positions A & B.
58
we know the distance between
point A & B is 2.8 km in this
example.
57
We also know that angle  is
030 (whole circle barings are
always quoted with three digits
even if the first is a zero)
56
B
N


A
55 21
If  = 030 or 30 then angle  =
90 - 30 = 60
22
23
24
25
Northings;
Eastings;
The Northing or height of the triangle is
opposite the angle that is known.
The Easting or base of the triangle is adjacent
to the angle that is known.
The distance between the points A & B is the
length of the hypotenuse.
The distance between the points A & B is the
length of the hypotenuse.
The sine function is defined as being equal to
the opposite side divided by the hypotenuse;
The cosine function is defined as being equal to
the adjacent side divided by the hypotenuse;
sin  
opposite
Hypotenuse
sin  
Rearranging this we get;
adjacent
Hypotenuse
Rearranging this we get;
Opposite = Hypotenuse  sin 
Adjacent = Hypotenuse  cos 
= 2.8 km  sin 60
= 2.8 km  cos 60
= 2.42 km
= 1.4 km
= 2.4 km  0.1 km
= 1.4 km  0.1 km
Now add on Northings and Eastings;
Position of A =
220560
This is;
220
plus
+2.4 +1.4
Position of B;
244
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560
574

244574
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Dr. Mark Sullivan
Analytical methods
SAQ
Set A
Set B
Find the position of point B given its baring and
distance from A;
Find the position of point Z given its baring and
distance from Y;
2.3 km baring 030
2.1 km baring 020
12.4 km baring 040
2.9 km baring 030
24.7 km baring 056 24’
7.5 km baring 075 43’
3.1 km baring 189
4.7 km baring 230
5.6 km baring 280
62.4 km baring 330
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Dr. Mark Sullivan
Analytical methods
VECTORS
Breaking a direction down into North and East directions is called ‘resolving into vectors’ and can
be used for a whole load more than just working out where you are.
Before we can proceed further we need to explore the idea of a vector and for that matter
scalars also. All quantities, it turns out, fall into one of two categories; scalars or vectors.
A scalar has only its numeric value for example, 1kg of sugar, 3 mins 20 secs delay or 34 C.
A vector, on the other hand, is a quantity which expresses both magnitude and direction for
example, a velocity of 45 km/s in a Northerly direction, a force of 300 kN at 45 to the vertical
or a fresh Easterly breeze.
Graphically we represent a vector as an arrow. In typeset
notation a vector is represented by a boldface character,
while in handwriting an arrow is drawn over the character
representing the vector.
Ax = A sin 
Ay = A cos 
A

The angle θ represents the whole circle
orientation of a two dimensional vector
which determines the quantities Ax and
Ay through the cos and sin functions.
The quantities Ax, & Ay, represent the
Cartesian components of the vectors in the
figure opposite. A vector can be represented
either by its Cartesian components, which are
just the projections of the vector onto the map
grid, or by its direction and magnitude (polar
co-ordinates). The direction of a vector in two
dimensions is best represented by the
clockwise angle of the vector relative to North
or the y axis.
Use this space to derive expressions for the Cartesian quantities Ax and Ay from the polar
quantities A and .
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Dr. Mark Sullivan
Analytical methods
Method of employing trigonometric functions to determine components of a vector:
1.
Draw a sketch (no scale needed) of the vector in the indicated direction; label its
magnitude and the angle which it makes with the horizontal.
2. draw a rectangle about the vector so that the vector is the diagonal of the rectangle;
beginning at the tail of the vector, sketch vertical and horizontal lines; then sketch
horizontal and vertical lines at the head of the vector; the sketched lines will meet to
form a parallelogram.
3. Draw the components of the vector; the components are the sides of the rectangle; be
sure to place arrowheads on these components to indicate their direction (up, down, left,
right).
4. Meaningfully label the components of the vectors with symbols to indicate which
component is being represented by which side.
5. Determine the length of the side opposite the indicated angle using trig functions;
substitute the magnitude of the vector for the length of the hypotenuse; use some
algebra to solve the equation for the length of the side opposite the indicated angle.
6. Repeat the above step using the cosine function to determine the length of the side
adjacent to the indicated angle.
The above method is illustrated below for determining the components of the force acting upon
Fido. As the 60-Newton tensional force acts upward and rightward on Fido at an angle of 40
degrees, the components of this force can be determined using trigonometric functions.
If we use vectors to represent force instead of movement across the ground, we can use them to
work out stresses on a structure.
Consider the following example;
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Dr. Mark Sullivan
Analytical methods
1.
A 3m uniform beam of mass 5 kg/m is supported horizontally above the ground by a hanger
at one end attached to a wall and by a rope at the other. The other end of the rope is fixed to the
wall vertically above the beam.
a) Derive an expression for the tension in the rope.
b) Calculate the tension for a rope fixed 60 cm, 1m, 2m and 3m above the beam.
c) Comment on how the tension varies with height
____________________________________________
First let’s draw a sketch of the structure.
T cos 

T
T sin 
½Mg
½ Mg
The tension in the rope, T, is shown in its component form as a vertical and a horizontal force T
sin  and T cos .
The load is the weight of the beam whose mass, M, is acted upon by gravity resulting in a
downward force of Mg (mass times the force due to gravity, g). the beam is, however, supported
at both ends so this force is supported equally at each end; ½Mg.
It is clear from the diagram that the upward force due to the tension in the rope must be exactly
equal to the downward force produced by gravity acting on the beam;
½Mg = T sin 
The question asks for an expression for tension, T, so we must re-arrange the equation for T;
1
Mg
2
T
sin 
By convention we write it the other way round;
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Mg
T
2 sin 

T
Mg
2 sin 
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Dr. Mark Sullivan
Analytical methods
b) Calculate the tension for a rope fixed 60 cm, 1m, 2m and 3m above the beam.
Look first at the expression we just worked out; T 
Mg
and identify which bits the question
2 sin 
is asking us to do repeatedly. In this case the only bit that will change as we put the different
lengths in will be , the angle of the rope. All other parts we only have to work out once.
T
Mg
2 sin 
where M = 5 kg/m  3 m = 5
kg m
kg
 3 m  15
 15 kg .
m
m
G = 9.81 m/s2
So
Mg = 15 kg  9.81 m/s2 = 15 kg  9.81
kg m
m
 147.15 2
2
s
s
kg m/s2 is the fundamental units of force. We call these units a Newton, N.
Now do the working out for the different heights given. This is best done in a table;
Height
Length
Height = tan 
Length
60 cm
3m
0.6/3 = 0.2
1m
3m
1/3
2m
3m
3m
3m
sin 
Mg/(2 sin )
11 18’ 35.76”
0.1961
375.19 N
= 0.333’
18 26’ 5.82”
0.3162
232.69 N
2/3
= 0.666’
33 41’ 24.24”
0.5547
132.64 N
3/3
= 1.0
45
0.7071
104.05 N

The tensions for the different height fixings are;
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60 cm
375.19 N
1m
232.69 N
2m
132.64 N
3m
104.05 N
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Dr. Mark Sullivan
Analytical methods
c) Comment on how the tension varies with height
In order to say something about the results obtained in part b, it would be useful to ‘see’ them
The clue is in the question, we were asked to find tensions for different heights so plot height
versus tension. The heights do not all go in regular steps so a scatter plot is the right one here.
On an x-y axis we plot the values of height along the x-axis and tension up the y-axis
Plot of rope tension versus height of fixing.
400
Tension in rope /N
375.160361
300
232.664579
200
132.639218
104.050763
100
0
0
1
2
3
4
Height of fixing /m
notice that the line connecting the points is curved. This indicates that at small values a small
change in fixing height leads to a rapid change of tension. At large heights, the tension is low and
not very sensitive to change.
SAQ
Try some for yourselves;
1.
A 5 m uniform beam of mass 12 kg/m is supported horizontally above the ground by a hanger
at one end attached to a wall and by a rope at the other. The other end of the rope is fixed to the
wall vertically above the beam.
a) Derive an expression for the tension in the rope.
b) Calculate the tension for the rope fixed 60 cm, 1m, 2m and 3m above the beam.
c) Comment on how the tension varies with height
Set A
Set B
Your beam is 8 m long & has a mass of 12 kg/m.
Your beam is 6 m long & has a mass of 18 kg/m.
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Dr. Mark Sullivan
Analytical methods
Answer;-
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Dr. Mark Sullivan
Analytical methods
2. A rope is laid across two parallel rails. Three equal masses are suspended, one at either
end and one in the middle between the two rails. Calculate the angle that the string makes
with the vertical between the central mass and one of the rails.
Answer;-
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Dr. Mark Sullivan
Analytical methods
3. A telegraph pole has three wires attached to it. Their bearings from the pole are 315,
270 and 210 and the forces in the wires are 1000N, 1100N and 1050N respectfully.
Calculate the bearing a support stay should be fixed at in order that there are no
unbalanced lateral forces acting on the pole.
Answer;-
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