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EGR 240 Chapter 2 part 1

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Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
1
Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
2
Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
3
Scalar Vs Vector
Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
4
Scalar Vs Vector
Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
5
Addition of planar forces
Line of action
10lb
A
Fixed axis
1. Point of application
2. Magnitude
3. Direction :
• Line of action
• sense
30°
Line of action
10lb
A
30°
Fixed axis
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Natural & Applied Science Division ‐ Engineering
6
Line of action
F
Angle
A
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Line of reference
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P
A
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R=P+Q
P
Q
A
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Q
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Joining tip to tail
Q
R=Q+P
P
P
A
Q
Dr. Hamadi M. Belghith
P
R=P+Q
Q
A
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A
9
P
‐P
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Natural & Applied Science Division ‐ Engineering
10
Production of a Scalar and a Vector
P
P
P
P
P
2P
P P 3P
……….. P n P
We define the production kP of a
scalar k and a vector P as a vector
having the same direction as P if k is
positive or a direction opposite of P
if k is negative and the magnitude
equal to the product of P and the
absolute value of k.
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P
Natural & Applied Science Division ‐ Engineering
2P
‐2P
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Associative addition
S
P Q
S P
P
Q
Q
S
P
Commutative addition
Q S
P Q
S S
P
Q
This expression shows that the
order of addition is immaterial
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Natural & Applied Science Division ‐ Engineering
12
Since a particle A acted by several coplanar
forces, since all forces pass through A, we said
the forces are CONCURRENT.
P
S
WE CAN ADD ALL THE VECTORS USING THE
POLYGON RULE joining tip to tail of all
vectors
R has the same effect as the given forces.
As we saw before the order of adding all
forces is immaterial
Dr. Hamadi M. Belghith
Q
A
S
P
Q
Q
P
S
A
Natural & Applied Science Division ‐ Engineering
R
A
R
13
Two forces are applied as shown to a bracket support.
Determine the magnitude and direction of their resultant using:
1. The parallelogram Law
2. The triangle Law.
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Natural & Applied Science Division ‐ Engineering
14
PROBLEM 2.6
A telephone cable is clamped at A to the pole AB. Knowing that the
tension in the left-hand portion of the cable is T1 = 800 N,
determine by trigonometry
(a) the required tension T2 in the right-hand portion if
the resultant R of the forces exerted by the cable at A is to be
vertical,
(b) the corresponding magnitude of R.
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We have seen that we can add 2 or more forces acting on a particles and
obtain the RESULTANT that has the same effect on this particle
#
The single forces acting on a particle may be replaced by two or more
forces that, together, have the same effect on this particle, This forces are
called
COMPONENTS
F
P
Of the original forces
A
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Natural & Applied Science Division ‐ Engineering
Q
16
y
F
𝐹
𝐹 and 𝐹 called rectangular components
Rectangular components of F, axes rotated
from horizontal and vertical.

O
x
𝐹
y
F
x

𝐹
𝐹
O
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Natural & Applied Science Division ‐ Engineering
17
𝐹
𝐹 cos 𝛼
y
𝐹
𝐹 sin 𝛼
tan 𝛼
𝐹
Dr. Hamadi M. Belghith
𝐹
F
𝐹
𝐹𝑦
𝐹

O
𝐹
x
𝐹
Natural & Applied Science Division ‐ Engineering
18
Y
Doing projection of each force to both
axis X and Y, we obtain
𝑅
Py
P
Sy
𝐹
S
Px
Sx
𝑅
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Qy
𝐹
Natural & Applied Science Division ‐ Engineering
X
Qx
Q
19
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Natural & Applied Science Division ‐ Engineering
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PROBLEM 2.25
Member BC exerts on member AC a force P directed along line BC.
Knowing that P must have a 325-N horizontal component, determine
(a) the magnitude of the force P,
(b) its vertical component.
Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
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Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
22
Problems 2.1 and 2.36
Please see problem statement in next slide
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Natural & Applied Science Division ‐ Engineering
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PROBLEM 2.1
Two forces are applied as shown to a hook.
Determine graphically the magnitude and
direction of their resultant using
(a) The parallelogram law, (b) the triangle rule.
Dr. Hamadi M. Belghith
Natural & Applied Science Division ‐ Engineering
24
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