Lecturer: Ms. Jwan Khaleel M.
Transcript of Lecturer: Ms. Jwan Khaleel M.
Petroleum and Mining DepartmentSecond Grade- Fall Semester
Statics-General Principles
(Lecture 1)& Vector principle of force
(Lecture 2)
Lecturer: Ms. Jwan Khaleel M. 1
Lecture content:β’ What is Mechanics of Engineering?
β’ Fundamental Concepts
β’ Units of Measurement
β’ Law of Gravitation
β’ Scalars and Vectors
β’ Vector Operations
β’ Vector Addition of Forces
β’ Addition of a System of Coplanar Forces
β’ Solving problems2Lecturer: Jwan Khaleel M.
Learning Outcomes:
At the end of this lecture you will be able to:
β’ Introduce the concept of engineering mechanics and general applications.
β’ Analyze the classification of engineering mechanics.
β’ Evaluate a statement of Newtonβs Laws of Motion and Gravitation.
β’ Use the laws of gravitation.
β’ Analyze the system of forces according to components and resultants.
β’ Express vectors in terms of unit vectors and perpendicular components
β’ Perform vector addition and subtraction.
β’ Compute Parallelogram law and trigonometric rules.
β’ Solve related problems.
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Statics and Dynamics
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Whatis
Engineering Mechanics?
Engineering Mechanics
Statics Dynamics
Is concerned with the equilibrium of
bodies that is either at rest or moves
with constant velocity (no accelerated
motion).
Kinematics Kinetics
Is deal with the accelerated motion of a body
β’ Study of the relations existing between the
forces acting on a body, the mass of the
body, and the motion of the body.
β’ Kinetics is used to predict the motion
caused by given forces or to determine the
forces required to produce a given motion.
Study of motion without paying attention to the forces
causing it.
Or in other words
β’ Study of the geometry of motion.
β’ Relates displacement, velocity, acceleration, and
time without reference to the cause of motion.6
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Mechanics is a branch of the physical sciences that is concerned with the state of rest
or motion of bodies that are subjected to the action of forces.
Applications To Mechanics:
1. Analysis and design of moving structures.
2. Fixed structures subject to shock loads
3. Robotic devices.
4. Machining of turbines and pumps.
5. Automatic control system
6. Rockets.
7. Missiles and spacecraft.
8. Ground and air transportation vehicles. And others.7
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Basic Concept of Mechanics:
β’Time: is a basic quantity in dynamics
β’Mass: is a quantity of matter in a body (statics) measure of the inertia of a body,which its resistance to a change of velocity (dynamics)
β’Force: push or pulls exerted one body to another.
β’Particle: is a body of negligible dimensions.
β’Rigid body:
β’Deformable body:
β’Length: is needed to locate the position of a point in space thereby
describe the size of a physical system.
β’Space: is a geometric region occupied by bodies whose positions
are describes by linear and angular measurement relative to a
coordinate system two or three dimensions
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Newtonβs laws:
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Units of Measurements:
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β’ Units Conversion:
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β’ Prefixes:
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Example 1//
What is the weight in both newton and pounds for 1500 kg beam?
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Example 2// Convert 2 km/h to m/s, How many ft/s is this?
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β’ Law of Gravitation:
The law of gravitation is expressed by the equation.
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Two uniform aluminum spheres are posted as shown. Determine the
gravitational force which sphere A exerts on sphere B. The value of R
is 50 mm. if you know that πΊ = 6.673 Γ 10β11π3
ππ.π 2assume the
sphere is Aluminum so , π = 2690ππ
π3
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Example 3//
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SOLUTION
π = 0.05 π πππ πππ‘β π πβππππ
Radius of Sphere A= 2R= 0.1m
Radius of Sphere A= R= 0.05m
Distance d=8R= 0.4 m
G= 6.673 x10-11m3/kg. s2
π = 2690 ππ/π3
π =4
3π Γ π3
πΉ =πΊππππ‘
π2=πΊππ
π2=
πΊπ4
3πΓ0.13
4
3πΓ0.053
0.42
= 6.61 Γ 10β9π
Because the spheres are located at 30Β° with x-axis
πΉ = πΉπππ ππ + πΉπ ππ ππ
πΉ = 6.61 Γ 10β9πππ 30π + 6.61 Γ 10β9π ππ 30π
πΉ = β 5.72Γ 10β9 π β 3.305 Γ 10β9π N
So the forces are πΉ = β 5.72Γ 10β9 β3.305 Γ 10β9π
Convert the quantities 300 πΌπ. π and 52 slug/ft3 to appropriate SI
units.
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Class activity 1//
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Solution//
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Convert each of the following.
(a) 20 Ib. ft to N . m,
(b) 450 lb/ft3 to kN/m3
(c) 15 ft/h to mm/s.
Class activity 2//
Solution//
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Class activity 3//
The two 100-mm-diameter spheres constructed of different metals
are located in deep space. Determine the gravitational force π which
the copper sphere exerts on the titanium sphere if (a) π = 2π, and
(b) π = 4π. ππ = 8910ππ
π3 and ππ‘ = 3080ππ
π3
SOLUTION
π = 0.05 π πππ πππ‘β π πβππππ
G= 6.673 x10-11m3/kg. s2
ππ = 8910 ππ/π3
ππ‘ = 3080 ππ/π3
ππ‘ = ππ =4
3π Γ π3 =
4
3π Γ 0.053 = 5.236 Γ 10 β4
a) π = 2π
πΉ =πΊππππ‘
π2=πΊππππ‘π
π2=
6.673Γ10β11Γ 8910Γ3080 5.236 Γ10 β4
22= β1.255 Γ 10β10π
b) π = 4π
πΉ =πΊππππ‘
π2=
πΊ 8910 Γ5.236 Γ10 β4 3080 Γ5.236 Γ10 β4
42= β3.14 Γ 10β11π
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β’ What is Force?
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Weight a crane can lift.
β’ External and Internal Effect of Force
β’ External force: Applied or Reactive force
β’ Internal force: deformation through the material
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β’ Force Classifications:
β’ Contact force β’Body force
β’Concentrated force
β’ Distributed force
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β’ Scalars and Vectorsβ’ Scalar: A scalar is any positive or negative physical quantity that can be completely
specified by its magnitude . Examples of scalar quantities include length, mass, and
time.
β’ Vector: A vector is any physical quantity that requires both a magnitude and a direction
for its complete description. Examples of vectors encountered in statics are force,
position, and moment.
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β’ Vector Operation
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Multiplication and Division of a
Vector by a Scalar
Vector Addition
β’ Vector Operation
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β’ Vector Addition of Forces
Concurrent forces: the line of action of two or more forces intersect at that point.
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Parallelogram law
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The law of parallelogram of forces states that if two vectors acting on a particle at the same
time be represented in magnitude and direction by the two adjacent sides of a parallelogram
drawn from a point their resultant vector is represented in magnitude and direction by the
diagonal of the parallelogram drawn from the same point.
Triangle law (Trigonometry)
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Triangle law of vector addition states that when two vectors are represented as two sides
of the triangle with the order of magnitude and direction, then the third side of the triangle
represents the magnitude and direction of the resultant vector.
β’ Finding the Components of a Force
β’Scalar notation:
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β’Cartesian Vector notation:
Where ππ and ππ are vector
components of F in the x- and y-
directions.
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β’Coplanar Force Resultants:
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β’Angles with their axis:
Unit Vector:
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A vector is a quantity that has both magnitude, as well as
direction. A vector that has a magnitude of 1 is a unit vector.
It is also known as Direction Vector.
Unit Vector =Vector
Magnitude of Vector
EXAMPLE// Find the unit vector Τ¦π for the given vector,
β2 ΖΈπ+4 ΖΈπβ4π. Show Unit vector component q.
π =β2 ΖΈπ+4 ΖΈπβ4π
(β2)2+(4)2+(β4)2=β2 ΖΈπ+4 ΖΈπβ4π
36=β2 ΖΈπ+4 ΖΈπβ4π
6=β2 ΖΈπ
6+4 ΖΈπ
6β4π
6=β1
3ΖΈπ +
2
3ΖΈπ β
2
3π
The screw eye in Figure shown a
is subjected to two forces, πΉ1and
πΉ2. Determine the magnitude and
direction of the resultant force.
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Example 1//
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Solution//
Resolve the horizontal 600-lb force in Figure
shown into components acting along the π’
and π£ axes and determine the magnitudes of
these components.
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Example 2//
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Solution//
Example 3//βClass activity ( you have 10 minutes to think/solve it by group)β
Determine the components
of the F force acting along the
u and v axes. If you know:
π1= 70Β°π2= 45Β°π3= 60Β°F = 250 N
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The 1800-N force F is applied to the end
of the I-beam. Express π as a vector
notation π and π.
Example 4//
πΉ = β18003
5π β 1800
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5π = (β1080π β 1440π )N
Solution:
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The end of the boom O in Fig. a is subjected to three concurrent and coplanar forces. Determine the magnitude and direction of the resultant force.
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Example 5//
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Solution:
Example 6//βClass activity ( you have 10 minutes to think/solve it by group)β
The 500-N force π is applied to the vertical pole as shown:
1) Write π in terms of the unit vectors π and π and identify both its vector and scalar components.
2) Determine the scalar components of the force vector π along the xβ²- and yβ²-axes.
3) Determine the scalar components of π along the x-and yβ²-axes.
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Assignment 1// (solve this problems then submit your answer)
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Determine the magnitude of the resultant force acting on the screw eye and its direction
measured clockwise from the x axis.
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Resolve each force acting on
the post into its x and y
components.
Assignment 2// (solve this problems then submit your answer)
Assignment 3 (solve this problems then submit your answer)
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Express each of the three forces
acting on the column in Cartesian
vector form and compute the
magnitude of the resultant force.
Next lecture:
β’ Cartesian Vectors
β’ Addition of Cartesian Vectors
β’ Position Vectors
β’ Force Vector Directed Along a Line
β’ Dot Product
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References:
Engineering Mechanics R.C.
Hibbeler 13th edition (Statics and
Dynamics).
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The end of the lecture
Enjoy your time
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