This document provides information about units of measurement, forces, moments, and calculating moments using two different methods. It discusses the conversion of various units like kgf, hp, bar, ton, and hectare. Forces are defined as vectors with magnitude, point of application, line of action, and direction. An example problem is given to calculate the moment at point O of a cantilever beam with a 10N force applied at angle Ø using the scalar product and vector product formulas. Both methods yield a net moment of 0 Nm, verifying the equivalence of the two approaches.

1. 1
Chapter I : MOMENTS
I-1Conversion of units
1kgf=1daN 1hp = 0.745699872 kW
1bar=100,000Pa (N/m2
) 1 MPa = 1,000,000 Pa =10bars
1ton=1000N 1daN/mm2
=10MPa or N/mm2
1hectar=10,000m2
1 dm3
= 0.001 m3
= 1 L
I-2 Forces
Forces (P, A0/1 or A1/0) are vectors so they have:
 Points of application(P.A.)( G and A )
 line of action or support or direction
 Sense of direction (→)
 Magnitude or intensity(I)
I-3 Bill of external forces
Upon isolating 1 from o we have the following bill of external forces:
Force P.A Direction sense Intensity(N)
A0/1 A AG or ?
P G AG or 20
I-4 Representing forces on Cartesian coordinates
G
A
1
0
α
y
x
R
R =Xi+Yj = ǁRǁ(Cosαi+sinαj)
ǁRǁ= √ (X2
+Y2
)

2. 2
I-5 Moments (M)
A moment is the near turning effect of an object. It is usually calculated in two
different ways:
a) scalar product formula
b) vector product formula
Consider the cantilever beam below, with a force of magnitude 10N applied
to its free end at an angle Ø, at point A.
By Pythagoras theorem on OIA, IA=4m
Y
X
Ø
3m
10N
A
1
O
I
2
5m
G
Assumption
Mass of piece is 1.2kg, g=10N/kg
Point O: pivot or turning point.
REVISION
 Pythagoras theorem: R2
=X2
+ Y2
 Sine rule:
𝑄
𝑆𝑖𝑛𝑞
=
𝑃
𝑆𝑖𝑛𝑝
=
𝑅
𝑆𝑖𝑛𝑟
 Cosine rule: R2
=P2
+Q2
-2PQCosr
 Sine curve
 Cosine curve
 Sine, cosine and tangent of angles (SOHCAHTOA)
 Isosceles and equilateral triangles.
P Q
R
r
q p

3. 3
Point A: point of action of the force
Point I: point of intersection of the perpendicular distance (OI) from pivot
to line of action (AI) of force.
Upon isolating piece 1
The force at A can be denoted as follows:
ǁA2/1 ǁ= 10N is known as its magnitude, or scalar form of the force.
A2/1 = Ax2/1i+ Ay2/1j
= ǁA2/1ǁ (CosØi +SinØj) is the denotation for its vector form.
Its weight can be denoted as: P=mg
P=-12Nj
ǁ P ǁ=12N
ǁOG ǁ=2.5, OG=2.5i
The link or joint force can be denoted as:
O0/1 =Ox0/1+Oy0/1
Similarly the distance OI can be express in both the scalar and vector form
as follows:
ǁ OI ǁ=3m, OI = OIX +OIY =OGi+ GIj
Method 1: scalar product formula
a) Moment=force x perpendicular distance from pivot to line of action of the force
b) Moment=perpendicular component of the force to the distance from pivot x
the distance.
Conventionally anticlockwise moment is positive and vice versa, denoted
5m
3m α
90-α
XWe can find α by trigonometry
Cosα =
x
3
Cos 90 − α = Sinα =
x
4
tanα =
3
4
α=37˚
I
O
OI=3Sin37⁰i-3cos37⁰j
4m
y
x
+
+
+

4. 4
We can determine the moments at O, denoted (M0)
a) M0 =ǁ A2/1ǁ x ǁ OI ǁ+ ǁPǁ x ǁOGǁ
M0 =10x3-12x 2.5=0N
b) M0 =10sinØ x5-12x 2.5=0
→50sinØ=30
SinØ=3/5 → Ø=37˚ (Hence)
Method 2: vector product method
MO (A2/1)= A2/1 ˄ OI= (Ax2/1 + Ay2/1) ˄ (OIX +OIY )= OIY X Ax2/1- OIX X Ay2/1
→ǁA2/1ǁ(cosØi +sinØj)˄ (3Sin37⁰i-3cos37⁰j)
→ (10cosØi + 10sinØj) ˄ Sin i − os )
→-30cosØcos37⁰-30sinØsin37⁰ with Ø=37⁰
→-30(cos2
Ø+sin2
Ø) =-30Nm hence
Likewise
MO(P)= P˄OG= 12j˄2.5i=30 hence
→MO =MO (P+A2/1)=-30+30=0Nm( hence verified that the two methods are the
same)
+
+