Moment of a Force Equations of Sum of Moments Calculating Reaction of a Support

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2. Engineering Science N3

3. Moment is defined as the product of a force and
the perpendicular distance measured from the
point where the force is acting to the turning
point.
Moment, M = F X d
F = Force
d= Perpendicular distance
We shall now define force.
Force
Figure 1 Point A Point B

4. The unit of Force is the Newton.
A Force is applied to a body in order for the body
to move from a point A to a point B.
Mathematically,
Force = mass X acceleration due to gravity.
F = mg
Mass is measured in kilograms.
Mass of a body occupies space and has weight.
The weight of a body acts vertically downwards
towards the ground as a result of acceleration
due to gravity.

5. Acceleration due to gravity, g, is the acceleration
of free fall of an object falling towards the
ground.
In other words, when a body is thrown upwards
into the air, it reaches a maximum height and
then starts to descend towards the ground at an
acceleration equal to g.
g= 9.81m/s2
Acceleration will be explained in the next
chapter.
We shall now take a look at the following
illustrations in order to get a better
understanding of moment.

6. 30N 30N
2m 2m T, Turning Pointa)
T Anti clockwise Moment
Clockwise Moment T
Force
Force
Fig 2 Clockwise and Anticlockwise Moments
b) d
c)

7. The turning point, T, is the support point about
which the beam rotates when a force is applied
to the left hand side or right hand side of the
beam as the case may be.
The law of moments states that the Sum of
Moments acting on a body about a turning point
is zero.
Mathematically,
Clockwise Moments about turning point =
Anticlockwise Moments about the turning point.
We shall now apply this law to Fig 1 a.

8. Clockwise Moments about turning point
= 30N X 2m = 60Nm (60 Newton metre)
Anti Clockwise Moments about turning point =
30N X 2m
30N X 2m = 60Nm (60 Newton meter)
The following Diagrams illustrates the application
of moments in our everyday lives.
Fig 3 a) T
F
b) c) F T F
A B
F
T

9. F
120 o
60 o
30 o
300 mm
8N
300mm
100mm
100mm
O
O
6N
YZ
W X
6N 10N
15N
45 o
45 o
35KN
Fig 4 a R
Fig 4 b
Reaction Support
105 o
600 mm O
Fig 4 c R
300mm O
66N
26N 55N
75N25 o
Fig 4 b
A B C

10. Parallelogram of Forces.
The rectangle WXYZ of Fig 4b illustrates the
Parallelogram of Forces. The force F acting at an
angle 450 C to the horizontal has two components
as shown:
1) Vertical Component XY or WZ
2) Horizontal Component ZY or WX
The rectangle is made up of two similar triangles
WXZ and YZX. Any one of these two triangles may
be chosen in order to evaluate the horizontal or
vertical component of the inclined force F.

11. The Parallelogram of Forces states that the
resultant of two forces acting at a point has a
resultant equivalent to the diagonal of
parallelogram or rectangle formed by the two
forces. Based on this, the force, F, is the resultant
of the two forces:
1) WX Horizontal component
2) XY Vertical component

12. Resolving Forces Acting at an angle.
Considering the right angled triangle ZXY and
applying the trigonometric ratios:
1) Sin 45o = opposite/hypotenuse
0.7071 = FH/F and FH = F0.7071
2) Cos 45o = adjacent/hypotenuse
0.7071 = FV/F and FV = F0.7071
Where FV = Vertical Component of Force
And FH = Horizontal Component of Force
The value of F will now be calculated based on
the table of clockwise and anticlockwise
moments.

13. The equation of Moments:
5Nm = (1.8 + 0.14142 F)Nm
F = 22.63N
The following forces have a moment of 0Nm about
the turning point because the perpendicular
distance from the turning point to the line of action
of the force is zero meters.
These forces are 10N and 8N.
Table 1 Clockwise Moments Anticlockwise Moments
0.1m X 5N = 0.5Nm 0.3m X 6N = 1.8Nm
0.3m X 15N = 4.5Nm 0.2m X F0.7071 N= F X
0.14142Nm
Total 5Nm (1.8 + ( F X 0.14142))Nm

14. Horizontal Forces
Positive x axis
(or Right)
15 N
Sum of Horizontal Forces
Negative x axes
(or Left)
(6 + 8 + Fcos 45)N =
(14 + (FX 0.7071))N=
(14 + (22.63 X 0.7071))N=
30 N
Vertical Forces
Positive x axis
(Upward)
(5 + 10 ) N = 15N
Vertical Forces
Negative y axes
(Downward)
(Fcos 45)N = (FX 0.7071) N
=16N
Table 2
Table 3

15. Calculating the Reaction Support.
We have assumed the reaction support in Fig 4b
to be vertical. It is actually inclined. The angle
which the reaction makes with the horizontal
is shown in the fig 5.
tan ϴ = 1/15
and ϴ = tan-1 (1/15) = 3.81o
The rectangle WXYZ is the vector diagram
representing the parallelogram of forces of
the horizontal and vertical forces. The vector
WY is the resultant of these forces.

16. The reaction support has the same magnitude as
the resultant WY. It acts in the opposite
direction to WY.
For every action there is an equal and opposite
reaction.
The magnitude of the resultant is calculated by
applying Pythagoras Theorem:
R = [(-12 )+ (-152)]0.5
R = 15.03N

17. -30N +15N
+15N
-16N
+y
+x
-y
-x
(-30 + 15)N = -15N
(-16 + 15)N= -1N
+y
-x +x
-y
ϴ
W X
Y
Z
+y
-x +x
-y
ϴ
W X
Y
Z
Fig 5c Reaction Support
Fig 5a Vertical and Horizontal Forces Fig 5b Parallelogram of Forces

18. A skip with a mass of 700kg is hoisted to a height
of 120m by means of a steel rope with a mass
of 3kg/m. Calculate the work done to hoist
the skip through the first 40m of the distance.
120m
40m

19. Work done by raising load through first 40m is
the area under the force distance curve.
Weight : 700kg = 700kg X 9.81m/s2 = 6867N
Weight of Rope :
3kg/m X 120m X 9.81m/s2= 3531.6N
Since the 40m vertical line divides the 120m line
in the ratio 1/3 = 40/120
Force or Weight
(6867 + 3531)N Z
6867N
40m 120m (x metres)
Distance through which load moves
A
D
Y
EX

20. Work done to raise load through the first 40m is
1/3 X Area ZCDY = 1/3 X ½(6867 + 10 388) X
120m
= 344 543.112 Joules
Alternatively, calculate the area by evaluation the
area of the triangle and the rectangle and
then add.