Canadian University of Dubai Engineering Mechanics Worksheet

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Homework 1
SCI 220 Engineering Mechanics
Homework 1
1.
2.
A and B
1
Homework 1
3.
If the particle is subjected to the three forces and is in equilibrium,
Deterimine F3
4.
Determine the three forces acting on the lamp in figure below and draw the forces acting
on the 8kg suspended lamp.
2
Homework 1
5.
Hint: determine the unit vector of each tension.
6.
3
SCI-220 Engineering Mechanics
Spring 2020
Free Body Diagram
Moment of a Force
Principle of Moments
SCI-220 Eng Mechanics
Announcements

Textbook
– “Engineering Mechanics,” Statics / Dynamics, R. C. Hibbeler, Prentice Hall
Spring 2020
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SCI-220 Eng Mechanics






The Free-Body Diagram
Particle Equilibrium
Three-Dimensional Force Systems
Moment of a Force – Scalar Formation
Moment of Force – Vector Formulation
Principle of Moments
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SCI-220 Eng Mechanics
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SCI-220 Eng Mechanics
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SCI-220 Eng Mechanics
WHY TREAT A BODY AS A PARTICLE
In a real world there is no such things as
particles, which occupy zero volume in
space. However, when the body doesn’t
rotate, we can treat the body as a particle.
Even, when the body rotates but its angular
velocity remain constant, we can still treat
the body as particle, since it rotational
quantities remain unchanged during the
motion.
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Spring 2020
SCI-220 Eng Mechanics
Condition for the Equilibrium of a Particle
• Particle is at equilibrium if
– at rest
– moving at a constant velocity
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SCI-220 Eng Mechanics
Condition for the Equilibrium of a Particle
• Newton’s first law of motion
Equilibrium
∑F = 0
where ∑F is the vector sum of all the
forces acting on the particle
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Spring 2020
SCI-220 Eng Mechanics
Condition for the Equilibrium of a Particle
• Newton’s second law of motion
∑F = ma
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SCI-220 Eng Mechanics
The Free-Body Diagram FBD
Best representation of all the unknown forces (∑F)
which acts on a body
A sketch showing the particle “free” from the
surroundings with all the forces acting on it
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SCI-220 Eng Mechanics
The Free-Body Diagram FBD
Two common connections in particle
equilibrium
– Spring
– Cables and Pulleys
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SCI-220 Eng Mechanics
The Free-Body Diagram
Spring
• Spring constant or stiffness k,
defines the elasticity of the spring
• Magnitude of force F when spring
is elongated or compressed
F = ks
stretched
Spring 2020
spring’s deformed length l
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SCI-220 Eng Mechanics
3.2 The Free-Body Diagram
• Spring deformed distance, s
where s is determined from the difference in
spring’s deformed length l and its undeformed
length lo
s = l – lo
– If s is positive, F “pull”
onto the spring
– If s is negative, F “push”
onto the spring
stretched
Spring 2020
spring’s deformed length l
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SCI-220 Eng Mechanics
The Free-Body Diagram
Example
Given lo = 0.4m and k = 500N/m
Magnitude of force F
deformed
To stretch it until l = 0.6m, A force, F = ks
deformed
F=(500N/m)(0.6m – 0.4m) = 100N is needed
To compress it until l = 0.2m,
A force, F = ks
deformed
=(500N/m)(0.2m – 0.4m)
= -100N is needed
stretched
Spring 2020
spring’s deformed length l
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SCI-220 Eng Mechanics
The Free-Body Diagram
Cables and Pulley (Assumptions)
• Negligible weight and cannot stretch.
• Only support tension or pulling force.
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SCI-220 Eng Mechanics
The Free-Body Diagram
Cables and Pulley (Assumptions)
• Tension always acts in the direction of
the cable
• Tension force in a continuous cable
must have a constant magnitude for
equilibrium
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SCI-220 Eng Mechanics
The Free-Body Diagram
• Cables and Pulley
For any angle θ, the cable is subjected to
a constant tension T, throughout its length
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SCI-220 Eng Mechanics
The Free-Body Diagram
Procedure for Drawing a FBD
1. Draw outlined shape
– Isolate particle from its surroundings
2. Show all the forces
– Indicate all the forces
– Active forces: set the particle in motion
– Reactive forces: result of constraints and
supports that tend to prevent motion
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SCI-220 Eng Mechanics
The Free-Body Diagram
Procedure for Drawing a FBD
3. Identify each forces
– Known forces should be labeled with
proper magnitude and direction
– Letters are used to represent magnitude
and directions of unknown forces
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SCI-220 Eng Mechanics
The Free-Body Diagram
Example 3.1
The sphere has a mass of 6kg and is
supported.
Draw a free-body diagram of
i)the sphere,
ii)the cord CE
iii)and the knot at C.
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SCI-220 Eng Mechanics
The Free-Body Diagram
Solution
FBD at Sphere
• Two forces acting,
weight and the force
on cord CE.
• Weight of 6kg
x(9.81m/s2) = 58.9N
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SCI-220 Eng Mechanics
The Free-Body Diagram
Solution
Cord CE
• Two forces acting, force
of the sphere and force
of the knot
• Newton’s Third Law: FCE
is equal but opposite
• FCE and FEC pull the cord
in tension
• For equilibrium, FCE =
FEC
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The Free-Body Diagram
Solution
FBD at Knot
• Three forces acting, force by cord CBA, cord CE
and spring CD
• Important to know that
the weight of the sphere
does not act directly on
the knot but subjected to
by the cord CE
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SCI-220 Eng Mechanics
FBD & Coordinate System
In solving equilibrium equation, it is
good to resolve forces into a
component of their coordinate system.
The problems could be solved in either
1D, 2D or 3D depends on the
movement, required accuracies and
etc.
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SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar (2D) Systems
A particle is subjected to coplanar forces in the x-y plane
Equilibrium
resolve forces into a component
equilibrium equations
Spring 2020
of their coordinate system
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SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar (2D) Systems
A particle is subjected to coplanar forces in the x-y plane
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SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar (2D) Systems
• Procedure for Analysis
1. Free-Body Diagram
– Establish the x, y axes in any suitable
orientation
– Label all the unknown and known forces
magnitudes and directions
– Sense of the unknown force can be
assumed
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SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar (2D) Systems
• Procedure for Analysis
2) Equations of Equilibrium
– Apply the equations of equilibrium
∑Fx = 0
∑Fy = 0
– Components are positive if they are
directed along the positive axis and
negative axis are negative, if directed
along the negative axis
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SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar (2D) Systems
Example 3.2
Determine the tension in
cables AB and AD for
equilibrium of the 250kg
engine.
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SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar
(2D) Systems
To solve for tension in AB and AD, you need to
establish equilibrium equation. You need to
establish the Cartesian coordinate system.
Where to locate the origin of the coordinate
system?
Locating the origin of coordinate system at point
D, make it difficult for determining the point A
in later stage (distance of A from D were not 30
Spring 2020
given).
SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar (2D) Systems
The best is to locate origin of the coordinate
system at point A. By doing so, it is easy for
you to resolve the component of tension force
in AD and AB in x and y direction of 2D
Cartesian coordinate system.
Always try to solve the equilibrium AT A POINT
where all the forces of the system acts.
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3.3 Equilibrium in Coplanar
(2D) Systems
Solution
FBD at Point A
– Initially, two forces acting, forces
of cables AB and AD
– Engine Weight
= (250kg)(9.81m/s2)
= 2.452kN supported by cable CA
– Finally, three forces acting, forces
TB and TD and engine weight
on cable CA
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SCI-220 Eng Mechanics
3.3 Equilibrium in Coplanar
(2D) Systems
resolve forces into a component
Solution equilibrium equations
+→ ∑Fx = 0;
TBcos30° – TD = 0
+↑ ∑Fy = 0;
TBsin30° – 2.452kN = 0
Solving,
TB = 4.90kN
TD = 4.25kN
*Note: Neglect the weights of the cables since they
are small compared to the weight of the engine 33
Spring 2020
SCI-220 Eng Mechanics
3.4 Equilibrium in 3D Force Systems
• Make use of the three scalar equations to
solve for unknowns such as angles or
magnitudes of forces
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SCI-220 Eng Mechanics
3.4 Equilibrium in 3D Force Systems
• Make use of the three scalar equations to
solve for unknowns such as angles or
magnitudes of forces
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SCI-220 Eng Mechanics
3.4 Three-Dimensional Force Systems
• For particle equilibrium
∑F = 0
• Resolving into i, j, k components
∑Fxi + ∑Fyj + ∑Fzk = 0
• Three scalar equations representing algebraic sums of
the x, y, z forces
∑Fxi = 0
∑Fyj = 0
∑Fzk = 0
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SCI-220 Eng Mechanics
3.4 Equilibrium in 3D Force Systems
Example 3.4
A 90N load is suspended from the hook. The
load is supported by two cables and a spring
having a stiffness k = 500N/m.
Determine the force in the
cables and the stretch of the
spring for equilibrium. Cable
AD lies in the x-y plane and
cable AC lies in the x-z plane.
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SCI-220 Eng Mechanics
3.4 Equilibrium in 3D Force Systems
Establish the coordinate system where the origin is
located at point A.
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SCI-220 Eng Mechanics
3.4 Equilibrium in 3D Force Systems
Solution
FBD at Point A
– Point A chosen as the forces are
concurrent at this point
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SCI-220 Eng Mechanics
3.4 Equilibrium in 3D Force Systems
Solution
Equations of Equilibrium,
∑Fx = 0;
FDsin30° – (4/5)FC = 0
∑Fy = 0;
-FDcos30° + FB = 0
∑Fz = 0;
(3/5)FC – 90N = 0
Solving,
FC = 150N
FD = 240N
FB = 208N
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SCI-220 Eng Mechanics
3.4 Equilibrium in 3D Force Systems
Solution
For the stretch of the spring,
FB = ksAB
208N = 500N/m(sAB)
sAB = 0.416m
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SCI-220 Eng Mechanics
Example
1. When a particle is in equilibrium, the sum of forces
acting on it equals ___ . (Choose the most appropriate
answer)
A) A constant
B) A positive number
C) Zero
D) A negative number E) An integer
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SCI-220 Eng Mechanics
Example
5. Select the correct FBD of particle A.
30
A 40
100 kg
F1
A
A)
F2
B)
30
40°
100 kg
A
F
C)
30
°
D)
A
100 kg
F1
30
°
F2
40°
A
100 kg
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SCI-220 Eng Mechanics
Example
7. Particle P is in equilibrium with five (5) forces acting on
it in 3-D space. How many scalar equations of
equilibrium can be written for point P?
A)2 B) 3 C) 4 D) 5 E) 6
8. In 3-D, when a particle is in equilibrium, which of the
following equations apply?
A) ( Fx) i + ( Fy) j + ( Fz) k = 0
B)  F = 0
C)  Fx =  Fy =  Fz = 0
D) All of the above.
E) None of the above.
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SCI-220 Eng Mechanics
Moment of a Force
tendency of rotation
caused by Fx
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SCI-220 Eng Mechanics
4.1 Moment of a Force – Scalar Formation
• Moment of a force (Torque) about a point
or axis – a measure of the tendency of
the force to cause a body to rotate
about an axis passing through a
point.

Torque – tendency of rotation caused by Fx
or simple moment (Mo) z
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SCI-220 Eng Mechanics
4.1 Moment of a Force – Scalar Formation
Magnitude

tendency of rotation
caused by Fx
For magnitude of MO, Force vector F
MO = Fd (Nm)
Counter clock wise
where d = perpendicular distance
from O to its line of action of force
Direction

Direction using “right hand rule”
Sense of rotation
Clock wise or Counter clock wise.
Spring 2020
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SCI-220 Eng Mechanics
4.1 Moment of a Force •
For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
– Scalar Formation
tendency of rotation
caused by Fx
– Vector Formation
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4.1 Moment of a Force – Scalar Formation
Resultant Moment
• Resultant moment, MRo = moments of
all the forces
MRo = ∑Fd

For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
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SCI-220 Eng Mechanics
Example 4.1
For each case, determine the moment of the
force about an axis that pass thorough
point O.
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SCI-220 Eng Mechanics
Solution

For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
Line of action is extended as a dashed line to establish
moment arm d.
Tendency to rotate is indicated and the orbit is shown as
a colored curl.
(a)Mo = (100 N )( 2m) = 200 N.m(CW )
clock wise
tendency of rotation
caused by Fx
Spring 2020
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SCI-220 Eng Mechanics
Solution

For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
(b)Mo = (50N )(0.75m) = 37.5N.m(CW )
clock wise
Extend force
line of action
(c )Mo = (40N )(4m + 2 cos30 m) = 229N.m(CW )
clock wise
tendency of rotation
caused by Fx
Spring 2020
Extend force
line of action
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SCI-220 Eng Mechanics
Solution

For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
(d )Mo = (60N )(1sin 45 m) = 42.4N.m(CCW )
tendency of rotation
caused by Fx
Extend force
line of action
(e)Mo = (7kN)( 4m − 1m) = 21.0kN.m(CCW )
Extend force
line of action

For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
Spring 2020
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SCI-220 Eng Mechanics

Example 4.5
For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
Determine the moment of the force about point O.
tendency of rotation
caused by Fx
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SCI-220 Eng Mechanics
Solution

For magnitude of MO, Force vector F
MO = Fd (Nm)
where d = perpendicular distance
The moment arm d can be found from trigonometry,
d = (3)sin 75 = 2.898 m
Thus,
M O = Fd = (5)(2.898) = 14.5 kN  m
Since the force tends to rotate or orbit clockwise about
point O, the moment is directed into the page.
Spring 2020
tendency of rotation
caused by Fx
55
4.4 Principles of Moments (This is good for
2D vector analysis)
• Also known as Varignon’s Theorem
“Moment of a force about a point is equal
to the sum of the moments of the
forces’ components about the point”
• Since F = F1 + F2,
MO = r X F
= r X (F1 + F2)
= r X F1 + r X F2
Principles of Moments
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SCI-220 Eng Mechanics
4.2 Cross Product
Direction
• Vector C has a direction that is
perpendicular to the plane containing A and
B such that C is specified by the right hand
rule
• Expressing vector C when
magnitude and direction are known
C = A X B = (AB sinθ)uC
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4.2 Cross Product
Laws of Operations
1. Commutative law is not valid
AXB≠BXA
Rather,
AXB=-BXA
• Cross product A X B yields a
vector opposite in direction to C
B X A = -C
Spring 2020
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SCI-220 Eng Mechanics
4.4 Principles of Moments (This is good for
2D vector analysis)
Resolve the Force components
Force F2 line of action passing through the
moment is equal to zero
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SCI-220 Eng Mechanics
4.4 Principles of Moments (This is good for
2D vector analysis)
Principles of Moments

MO = r X F
= r X (F1 + F2) Resolve the Force components
= r X F1 + r X F2
Add the moments
Spring 2020
SCI-220 Eng Mechanics
4.4 Principles of
Moments (This is
good for 2D vector
analysis)
Resolve the Force components
Add the moments
Force F line of action passing
through the moment is equal
to zero
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SCI-220 Eng Mechanics
Chapter Review
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Chapter Review
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Chapter Review
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SCI-220 Eng Mechanics
Next lecture
• More Examples
Spring 2020
65

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