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FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Engineering Mechanics
Mechanics of Rigid Bodies
Statics Dynamics
KinematicsKinetics
Mechanics of Deformable Bodies
1. Strength of Materials
2. Theory of Elasticity
3. Theory of Plasticity
Mechanics of Fluids
1. Ideal Fluid
2. Viscous Fluid
3. Incompressible Fluid
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Mechanics of Rigid Bodies
Statics
Force Systems
Concurrent
Parallel
Non-Concurrent
Applications
Trusses
Centroids
Friction
Dynamics
Kinematics
Translation
Rotation
Plane Motion
Kinetics
Translation
Rotation
Plane Motion
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
โข Engineering Mechanics โ The science which considers the effects of forces on rigid bodies.
โข Statics โ considers the effects and distribution of forces on rigid bodies which are and remain at rest
โข Dynamics โ considers the motion of rigid bodies caused by the forces acting upon them
โข Kinematics โ deals with pure motion of rigid bodies
โข Kinetics โ relates the motion to applied forces
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
Basic Quantities
Length โ used to locate the position of a point in space and thereby describe the size of a physical system
Time โ is conceived as a succession of events
Mass โ is a measure of the quantity of matter that is used to compare the action of one body with that of another.
Force โ a โpush or pullโ exerted by one body to anotherโข External Force - changes, or tends to change, the state
of motion of a body. (independent on point of application)
โข Internal Force โ produces stress and deformation in the body. (dependent on point of application)
* Principle of Transmissibility โ a force may be moved anywhere along its line of action without changing its external effect on a rigid body.
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
Idealizations
Particle โ has a mass, but a size that can beneglected.
Rigid Body โ can be considered as a largenumber of particles in which all the particlesremain at a fixed distance from one another,both before and after applying a load.
Concentrated Force - represents the effect ofa loading which is assumed to act at a point ona body. We can represent a load by aconcentrated force, provided the area overwhich the load is applied is very smallcompared to the overall size of the body.
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
Newtonโs Three Laws of Motion
First Law (Law of Inertia). A particle originally at rest, or moving in a straight line with constant velocity, tends to remain in this state provided the particle is not subjected to an unbalanced force.
Second Law (Law of Acceleration). A particle acted upon by an unbalanced force experiences an acceleration a that has the same direction as the force and a magnitude that is directly proportional to the force.
Third Law. (Law of Action-Reaction). The mutual forces of action and reaction between two particles are equal, opposite, and collinear
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
Newtonโs Law of Gravitational Attraction
๐น = ๐บ๐1๐2
๐2
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
where
F = force of gravitation between the two particlesG = universal constant of gravitation; according to
experimental evidence, G = 66.73(10-12) m3/(kg ยท s2)m1, m2 = mass of each of the two particlesr = distance between the two particles
Weight โ force on an object due to gravity
W = mg
W= ๐บ๐๐๐ธ
๐2
Fundamental Concepts and Definitions
Units of Measurement
F = ma ; W=mg
CGS: dyne = (g)(cm/s2) MKS: N = (kg)(m/s2) US: lbf = (slug)(ft/s2)kgf = (kgm)(9.8 m/s2) lbf = (lbm)(32.174 ft/s2)kgf = 9.8 N slug = 32.174 lbm
Under Standard Condition: g = 9.8 m/s2
kgf = kgm lbf = lbm
For Non-Standard Condition
kgf = kgm(๐๐๐๐
๐) lbf = lbm (
๐๐๐๐
๐)
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
Force Systems
Force System โ any arrangement when two or more forces act on a body or on a group of related bodies.
โข Coplanar โ the lines of action of all the forces lie in one plane
โข Concurrent โ the lines of action pass through a common point
โข Parallel โ the lines of actions are parallel
โข Non-Concurrent โ the lines of action are neither parallel nor intersect at a common point
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
Axioms of Mechanics
1. The Parallelogram Law: The resultantof two forces is the diagonal of theparallelogram formed on the vectorsthese forces.
2. The forces are in equilibrium only whenequal in magnitude, opposite indirection, and collinear in action.
3. A set of forces in equilibrium may beadded to any system of forces withoutchanging the effect of the originalsystem
4. Action and reaction forces are equalbut oppositely directed.
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Fundamental Concepts and Definitions
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Parallelogram Law Triangle Law
Polygon Law
Fundamental Concepts and Definitions
Scalar and Vector Quantities
Scalars โ quantities which possesmagnitude only and can be addedarithmetically.
Vectors โ quantities which possesmagnitude and direction and can becombined only by geometric (vector)addition.
โข Multiplication or division of a vector bya scalar will change the magnitude ofthe vector. The sense of the vector willchange if the scalar is negative.
โข As a special case, if the vectors arecollinear, the resultant is formed by analgebraic or scalar addition.
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Components of a Force ๐น๐ฅ = ๐น๐๐๐ ๐๐ฅ
๐น๐ฆ = ๐น๐๐๐ ๐๐ฆ
Resultant
๐น = ๐น๐ฅ2 + ๐น๐ฆ
2
Resultant of Three or More Concurrent Forces
๐ = (ฮฃ๐น๐ฅ)2 + (ฮฃ๐น๐ฆ)2
Position of Resultant
๐ก๐๐ฮธ๐ฅ =ฮฃ๐น๐ฆ
ฮฃ๐น๐ฅ
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Resultant of Force Systems
Resultant โ simplest system that can replace the original system without changing the effect on a rigid body
Resultant of Concurrent Forces
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Resultant of Non-Perpendicular Forces
๐ = ๐12 + ๐2
2 + 2๐1๐2๐๐๐ ฮธ
Position of Resultant
๐ก๐๐ฯ =๐1๐ ๐๐ฮธ
๐2+๐1๐๐๐ ฮธ
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Determine the magnitude and direction ofthe resultant of the three forces shown.Determine also the horizontal and verticalcomponent of the resultant.
Solution:๐ ๐ฅ= ฮฃ๐น๐ฅ
= 50๐๐๐ 45 + 75๐๐๐ 75 โ 80๐๐๐ 60๐น๐ = ๐๐. ๐๐ ๐
๐ ๐ฆ = ฮฃ๐น๐ฆ
= 50๐ ๐๐45 + 75๐ ๐๐75 + 80๐ ๐๐60๐น๐ = ๐๐๐. ๐๐ ๐
๐ = (ฮฃ๐น๐ฅ)2 + (ฮฃ๐น๐ฆ)2
= 14.772 + 177.082
๐น = ๐๐๐. ๐๐ ๐
ฮธ = ๐ก๐๐โ1 ฮฃ๐น๐ฆ
ฮฃ๐น๐ฅ= ๐ก๐๐โ1 177.08
14.77
๐ฝ = ๐๐. ๐๐ยฐ
Resultant of Concurrent Forces
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Determine the magnitude and direction of Rif P1 and P2 are 100 lb and 150 lb respectively.P2 lies horizontally while P1 makes 120ยฐ withthe horizontal.
Solution 1:
๐ = (ฮฃ๐น๐ฅ)2 + (ฮฃ๐น๐ฆ)2
= (150 โ 100๐๐๐ 60)2 + 100๐ ๐๐602
๐น = ๐๐๐. ๐๐ ๐ฅ๐
ฯ = ๐ก๐๐โ1 ฮฃ๐น๐ฆ
ฮฃ๐น๐ฅ= ๐ก๐๐โ1 86.60
100
๐ = ๐๐. ๐๐ยฐ
Solution 2:
๐ = ๐12 + ๐2
2 + 2๐1๐2๐๐๐ ฮธ
= 1002 + 1502 + 2(100)(150)๐๐๐ 120๐น = ๐๐๐. ๐๐ ๐ฅ๐
ฯ = ๐ก๐๐โ1(๐1๐ ๐๐ฮธ
๐2+๐1๐๐๐ ฮธ)
๐ = ๐๐. ๐๐ยฐ
Resultant of Concurrent Forces
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Determine the magnitude of R if P1 and P2are 100 lb and 150 lb respectively. ฯ = 41ยฐ
Solution:
Let ฮฑ be the angle opposite R and ฮฒ be the angle opposite P2:
By Sine Law๐
๐ ๐๐ฮฑ=
๐1
๐ ๐๐ฯ=
๐2
๐ ๐๐ฮฒ100
๐ ๐๐41=
150
๐ ๐๐ฮฒฮฒ = 80ยฐฮฑ = 180 โ 80 + 41 = 59100
๐ ๐๐41=
๐
๐ ๐๐59๐น = ๐๐๐. ๐๐ ๐๐
By Cosine Law:
๐ = ๐12 + ๐2
2 โ 2๐1๐2๐๐๐ ฮฑ
= 1002 + 1502 โ 2(100)(150)๐๐๐ 59๐น = ๐๐๐. ๐๐ ๐๐
Resultant of Concurrent Forces
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
A boat moving at 12kph is crossing a river 500 m widein which a current is flowing at 4 kph. In whatdirection should the boat head if it is to reach a pointon the other side of the river directly opposite itsstarting point?
๐ ๐๐ฮธ =4
12
๐ฝ = ๐๐. ๐๐ยฐ, ๐๐๐๐๐๐๐๐
Resultant of Concurrent Forces
Solution:
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Moment of a Force
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Resultant of Non-Concurrent Forces
Moment โ is the measure of the ability of a force to produce turning or twisting about an axis.
๐๐ = ๐น๐where d is the moment arm (perpendicular distance from the axis at point O to the line of action of the force.
The Principle of Moments (Varignonโs Theorem)
The moment of a force is equal to the sum of the moments of its components.๐๐ = ฮฃ(๐น๐)๐๐ = ฮฃ๐ = ๐ ๐ ๐๐ = ๐น1๐1 โ ๐น2๐2 + ๐น3๐3
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Determine the resultant moment ofthe four forces acting on the rodshown below about point O.
Solution:
๐๐ ๐= ฮฃ๐น๐
= โ50๐ 2๐ + 60๐ 0๐+ 20๐ 3๐ ๐๐30๐ โ (40๐)(4๐+ 3๐๐๐ 30๐)๐ด๐น๐ถ
= โ๐๐๐ ๐ ยท ๐ = ๐๐๐ ๐ต โป
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Couple
Couple โ Two parallel, non-collinear forces that are equal in magnitude and opposite in direction๐ = 0 ; ฮฃ๐ โ 0
๐ถ = ๐น๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Equivalent Couples
๐ถ = 100 ๐๐ 2 ๐๐ก = 200 ๐๐ 1 ๐๐ก = 200 ๐๐ ยท ๐๐ก
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Coplanar Force System
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
The force system shown consists ofthe couple C and four forces. If theresultant of this system is a 500-lbยทin.counterclockwise couple, determineP, Q, and C.
Solution:
๐ ๐ฅ= ฮฃ๐น๐ฅ
= โ12
13๐ +
4
5๐ + 80 = 0 (1)
๐ ๐ฆ = ฮฃ๐น๐ฆ
= โ5
13๐ +
3
5๐ โ 20 = 0 (2)
Solving Eqns (1) and (2) simultaneously gives๐ท = ๐๐๐๐๐ ๐๐๐ ๐ธ = ๐๐๐๐๐
๐ถ๐ = ๐๐ = ฮฃ๐๐ด
500 = โ20 3 โ ๐ถ + 80 4 +3
5๐ 6 +
4
5๐(6)
๐ช = ๐๐๐๐ ๐๐ ยท ๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Parallel Force System
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Replace the force and couple momentsystem acting on the beam in the figure byan equivalent resultant force, and findwhere its line of action intersects the beam,measured from point O.
Solution:
๐ ๐ฅ= ฮฃ๐น๐ฅ
= 8 ๐๐3
5= 4.8 ๐๐
๐ ๐ฆ= ฮฃ๐น๐ฆ
= โ4 ๐๐ + 8 ๐๐4
5= 2.4 ๐๐
๐ = (4. 8๐๐)2 + (2.4๐๐)2
๐น = ๐. ๐๐ ๐๐ต
ฮธ = ๐ก๐๐โ1 2.4
4.8= ๐๐. ๐ยฐ
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Resultant of Non-Concurrent Forces
Solution:
๐๐ = ฮฃ๐๐
2.4๐๐ ๐ = โ4๐๐ 1.5๐ โ 15๐๐ ยท ๐ โ
8๐๐3
50.5๐ + 8๐๐
4
5(4.5๐)
๐ = ๐. ๐๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
Equilibrium โ A body is said to be in equilibrium if theresultant of the force system that acts on the bodyvanishes. Equilibrium means that both the resultant forceand the resultant couple are zero.
Free Body Diagram (FBD) - is a sketch of the bodyshowing all forces that act on it. The term free impliesthat all supports have been remove and replaced by theforces (reactions) that they exert on the body.
Forces that Act on a Body1. Reactive Forces (Reactions) - forces that are exerted
on a body by the supports to which it is attached.2. Applied Forces - forces acting on abody that are not
provided by the supports.
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
Conditions of Equilibrium
1. Graphical Condition: Under thiscondition, the forces or vectorsare transformed into a forcepolygon. For equilibrium, theforce polygon must close.
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
2. Directional Condition: If threeor more non-parallel forces orvectors are in equilibrium, thenthey must be concurrent. For atwo-force member, the forcesmust be equal and opposite.
3. Analytical Condition: If forcesor vectors are in equilibrium, thenit must satisfy the three staticequations:
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
Support Reactions
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
Three forces 20 N, 30 N, and 40 N are in equilibrium. Find the largest angle they make with each other.
Solution:
402 = 302 + 202 โ 2 30 20 ๐๐๐ ฮธ๐ฝ = ๐๐๐. ๐๐ยฐ
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
A load of 100 lb is hung from the middleof a rope, which is stretch between tworigid walls 30 ft apart. Due to the load,the rope sags 4 ft in the middle.Determine the tension in the rope.
Solution:
๐ก๐๐ฮธ =15
4ฮธ = 75.068ยฐ
ฮฃ๐น๐ฆ = 0 :
2๐๐๐๐ 75.068 = 100๐ป = ๐๐๐ ๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
A simply supported beam is 5m in length. Itcarries a uniformly distributed load includingits own weight of 300N/m and a concentratedload of 100 N 2m from the left end. Find thereactions if reaction A is at the left end andreaction B is at the right end.
Solution:
ฮฃ๐๐ด = 0 :100๐ 2๐ + 1500๐ 2.5๐ = ๐ ๐ต 5๐
๐น๐ฉ = ๐๐๐ ๐ต
ฮฃ๐น๐ฆ = 0 :
๐ ๐ด + ๐ ๐ต = 100๐ + 1500๐๐น๐จ = ๐๐๐ ๐ต
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
The homogeneous 60-kg disk supported bythe rope AB rests against a rough vertical wall.Using the given FBD, determine the force inthe rope and the reaction at the wall.
Solution:
ฮฃ๐๐ต = 0 :๐ญ๐ช = ๐
ฮฃ๐น๐ฆ = 0 :4
5๐ = 588.6๐
๐ป = ๐๐๐. ๐ ๐ต
ฮฃ๐น๐ฅ = 0 :3
5๐ = ๐๐ถ
๐ต๐ช = ๐๐๐. ๐ ๐ต
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
The homogeneous, 120-kg wooden beam issuspended from ropes at A and B. A powerwrench applies the 500-N ยท m clockwisecouple to tighten a bolt at C. Use the givenFBD to determine the tensions in the ropes.
Solution:
ฮฃ๐๐ด = 0 :๐๐ต(4๐) = 1177.2๐ 3๐ + 500๐๐
๐ป๐ฉ = ๐๐๐๐. ๐ ๐ต
ฮฃ๐น๐ฆ = 0 :
๐๐ด + ๐๐ต = 1177.2๐ป๐จ = ๐๐๐. ๐ ๐ต
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
The structure in Fig. (a) isloaded by the 240-lb ยท in.counterclockwise coupleapplied to member AB.Neglecting the weights of themembers, determine allforces acting on memberBCD.
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Equilibrium of a Rigid Body
Solution:
For Fig. (b) ฮฃ๐๐ด = 0 :๐๐ถ๐๐๐ 30 8 + 240 = ๐๐ท(12)๐๐ท = 0.577๐๐ถ + 20 (๐)
For Fig. (c) ฮฃ๐๐ต = 0 :๐๐ถ๐๐๐ 30 4 + ๐๐ถ๐ ๐๐30(3) = ๐๐ท(8)
๐๐ท = 0.620๐๐ถ ๐
Solving Eqs (1) and (2) simultaneously gives:
๐ต๐ซ = ๐๐๐ ๐๐ ; ๐ป๐ช = ๐๐๐ ๐๐
ฮฃ๐น๐ = 0 :๐๐ท + ๐ต๐ = ๐๐ถ๐๐๐ 30
๐ฉ๐ฟ = ๐๐๐ ๐๐
ฮฃ๐น๐ฆ = 0 :
๐ต๐ฆ = ๐๐ถ๐ ๐๐30
๐ฉ๐ = ๐๐๐ ๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Analysis of Structures
Truss โ is a structure composed of slender membersjoined together at their end joints.
Planar Trusses - lie in a single plane and are often usedto support roofs and bridges
Simple Trusses - constructed by expanding the basictriangular truss
Simple Trusses
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Analysis of Structures
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Analysis of Structures
Truss Wing Spar
Truss-Type Fuselage. A Warren Truss uses mostly diagonal bracing.
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Analysis of Structures
Assumptions for Design
1. The weights of the members are negligible.2. The members are joined together by smooth pins.3. The applied forces act at the joints.
Each member of a truss is a two-force member.
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Analysis of Structures
Method of Joints
When using the method of joints to calculate the forces inthe members of a truss, the equilibrium equations areapplied to individual joints (or pins) of the truss.
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Analysis of Structures
Zero-Force Member โ member that does not carry a loadโ contributes to the stability of the structureโ can carry loads in the event that variations are introduced in
the normal external loading configuration
ฮฃ๐น๐ฆ = 0 :
๐บ๐ถ = 0
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Analysis of Structures
Stable Structuresfor ๐ = 2๐ โ 3 โ stable
๐ < 2๐ โ 3 โ unstable
Determinate Structuresfor r โค 3๐ โ determinate
r > 3๐ โ indeterminate
where:m = no. of membersj = no. of jointsr = no. of reactions
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Analysis of Structures
Using the method of joints, determine the force in each member ofthe truss shown in the figure. Indicate whether the members are intension or compression. (One of the supports is usually designed tobe equivalent to a roller, in order to permit the elongation andcontraction of the truss with temperature changes).
Solution:
ฮฃ๐๐ถ = 0 :๐๐ด 6 + 10 6 = 60(3)
๐๐ด = 20 ๐๐
ฮฃ๐น๐ = 0 :๐ถ๐ = 10 ๐๐
ฮฃ๐น๐ฆ = 0 :
๐ถ๐ฆ + ๐๐ด = 60
๐ถ๐ฆ = 40 ๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Analysis of Structures
at pin A:
ฮฃ๐น๐ฆ = 0 :
๐๐ด + ๐ด๐ต1
2= 0
๐ด๐ต = โ28.28 ๐๐๐จ๐ฉ = ๐๐. ๐๐ ๐๐ต ๐ช
ฮฃ๐น๐ฅ = 0 :
๐ด๐ต1
2+ ๐ด๐ท = 0
๐จ๐ซ = ๐๐ ๐๐ต ๐ป
at pin D:
ฮฃ๐น๐ฆ = 0 :
๐ต๐ท2
5= 60
๐ฉ๐ซ = ๐๐. ๐๐ ๐๐ต ๐ป
ฮฃ๐น๐ฅ = 0 :
๐ต๐ท1
5+ ๐ถ๐ท = ๐ด๐ท
๐ถ๐ท = โ10 ๐๐๐ช๐ซ = ๐๐ ๐๐ต ๐ช
ฮฃ๐น๐ฅ = 0 :๐ต๐ถ + 40 = 0
๐ต๐ถ = โ40 ๐๐๐ฉ๐ช = ๐๐ ๐๐ต ๐ช
at pin C:
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Analysis of Structures
Method of Sections
โข Analyzing the free-body diagram of a part of a trussthat contains two or more joints is called the method ofsections.
โข Principle: If the truss is in equilibrium then anysegment of the truss is also in equilibrium.
โข It permits us to directly determine the force in almostany member instead of proceeding to that member byjoint-to-joint analysis.
โข The cutting plane must not cut more than threemembers whose internal forces are unknown.
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Analysis of Structures
Using the method of sections, determine the forces in members BC, HC, HG, and DF.
Solution:
ฮฃ๐๐ด = 0 :8000 6 + 3000 12 = ๐๐ธ(24)
๐๐ธ = 3500 ๐๐
ฮฃ๐น๐ = 0 :๐ด๐ฅ = 0
ฮฃ๐น๐ฆ = 0 :
๐ด๐ฆ + ๐๐ธ = 8000 + 3000
๐ด๐ฆ = 7500 ๐๐
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Analysis of Structures
Solution:
ฮฃ๐น๐ฆ = 0 :
7500 + ๐ป๐ถ4
5= 8000
๐ฏ๐ช = ๐๐๐ ๐๐ ๐ป
ฮฃ๐๐ป = 0 :7500 6 + ๐ต๐ถ 8 = 0
๐ต๐ถ = โ5625 ๐๐๐ฉ๐ช = ๐๐๐๐ ๐๐ ๐ช
ฮฃ๐๐ถ = 0 :7500 12 = 8000 6 + ๐ป๐บ(8)
๐ฏ๐ฎ = ๐๐๐๐ ๐๐ ๐ป
ฮฃ๐น๐ฆ = 0 :
๐ซ๐ญ = ๐๐๐๐ ๐๐ ๐ป
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Friction
Friction โ force that resists the movement of two contacting surfaces that slide relative toone another.
1. Dry Friction - friction force that exists between two unlubricated solid surfaces.2. Fluid Friction - acts between moving surfaces that are separated by a layer of fluid.
๐ญ = ๐๐ต
where:F = frictional forceฮผ = coefficient of frictionN = normal forceฯ = angle of friction
๐๐๐๐ = ๐
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Friction
๐ญ๐๐๐ = ๐๐๐ต๐น๐๐๐ฅ always opposes impending sliding
๐ญ๐ = ๐๐๐ต๐น๐ always opposes sliding
๐๐ > ๐๐ ; ๐ญ๐ > ๐ญ๐
For very low velocity:๐๐ โ ๐๐ ; ๐ญ๐ โ ๐ญ๐
where:๐น๐๐๐ฅ = maximum static friction๐น๐ = kinetic friction๐น๐ = static frictionฮผ๐ = coefficient of static frictionฮผ๐ = coefficient of kinetic friction
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Friction
The 100-lb block in the figurebelow is at rest on a roughhorizontal plane before the forceP is applied. Determine themagnitude of P that would causeimpending sliding to the right
Solution:
๐ = ๐น = ๐น๐๐๐ฅ = ๐๐ ๐ = 0.5 100๐๐๐ท = ๐๐ ๐๐
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Friction
A 600N block rests in a surface inclined at30ยฐ. Determine the horizontal force Prequired to prevent the block from slidingdown. Angle of friction between the blockand the inclined plane is 15ยฐ.
Solution:
ฮฃ๐น๐ = 0 :๐๐๐๐ ฮธ + ๐น = ๐๐ ๐๐ฮธ๐๐๐๐ ฮธ + ฮผ๐ = ๐๐ ๐๐ฮธ
๐๐๐๐ ฮธ + ๐ก๐๐ฯ๐ = ๐๐ ๐๐ฮธ๐๐๐๐ 30 + ๐ก๐๐15๐ = 600๐ ๐๐30 ๐
ฮฃ๐น๐ฆ = 0 :
โ๐๐ ๐๐ฮธ + ๐ = ๐๐๐๐ ฮธโ๐๐ ๐๐30 + ๐ = 600๐๐๐ 30 ๐
Solving Eqns. (a) and (b) simultaneously gives:
๐ท = ๐๐๐. ๐๐ ๐ต๐ = 600 ๐
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Friction
The uniform 100-lb plank in the figure below isresting on friction surfaces at A and B. Thecoefficients of static friction are shown in the figure.If a 200-lb man starts walking from A toward B,determine the distance x when the plank will start toslide.
Solution:
๐น๐ด = 0.2๐๐ด
๐น๐ต = 0.5๐๐ต
ฮฃ๐น๐ = 0 :๐น๐ด + ๐น๐ต๐๐๐ 40 โ ๐๐ต๐๐๐ 50 = 0 (๐)
ฮฃ๐น๐ฆ = 0 :
๐๐ด + ๐๐ต๐ ๐๐50 + ๐น๐ต๐ ๐๐40 = 300 (๐)ฮฃ๐๐ด = 0 :
๐๐ต๐ ๐๐50 10 + ๐น๐ต๐ ๐๐40 10 โ 200๐ฅ = 100 5 (๐)
Substituting ๐น๐ด and ๐น๐ต to eqns. (a), (b), and (c) and solving simultaneously gives:๐๐ด = 163.3 ๐๐ ; ๐๐ต = 125.7 ๐๐
๐ = ๐. ๐๐ ๐๐
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Force Systems in Space
Six equilibrium equations in three dimensions:
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Force Systems in Space
Assume the three force vectors intersect at a singlepoint.
๐น1 = 4๐ + 2๐ + 5๐๐น2 = โ2๐ + 7๐ โ 3๐๐น3 = 2๐ โ ๐ + 6๐
What is the magnitude of the resultant force vector, R?
Solution:
๐ = ๐น1 + ๐น2 + ๐น3
= 4๐ + 2๐ + 5๐ + โ2๐ + 7๐ โ 3๐ + 2๐ โ ๐ + 6๐๐ = 4๐ + 8๐ + 8๐
๐ = (๐ด๐)2 + (๐ด๐)
2 + (๐ด๐)2
= 42 + 82 + 82
๐น = ๐๐ ๐๐๐๐๐
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Force Systems in Space
Find the reactions for the equipment shelf shown in thesketch. The three applied loads act at the center of thevolume shown. Supports A and B cannot take reactionsin the y direction and support C cannot take a reactionin the x direction.
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Force Systems in Space
Solution:
ฮฃ๐๐ฆ = 0 :
300 5 โ 400 12 + 24๐ 3 = 0๐น๐ = ๐๐๐. ๐ ๐๐
ฮฃ๐น๐ฆ = 0 :
๐น๐ = ๐๐๐ ๐๐
ฮฃ๐๐ฅ = 0 :8๐ 3 + 16๐ 2 + 200 5 โ 400(8) = 0
๐น๐ = ๐๐. ๐๐ ๐๐
ฮฃ๐น๐ง = 0 :๐ 1 + ๐ 2 + ๐ 3 โ 400 = 0
๐น๐ = ๐๐๐. ๐๐ ๐๐
ฮฃ๐๐ง = 0 :24๐ 4 โ 16๐ 6 โ 200 12 + 300 8 = 0
๐น๐ = ๐๐๐ ๐๐
ฮฃ๐น๐ฅ = 0 :๐ 5 + ๐ 6 โ 300 = 0
๐น๐ = ๐
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Centroid and Center of Gravity
CG or Center of Weightโข It is the point at which the resultant of the gravitational forces (weight) act on a
body.โข It is a property of the distribution of weight within the body.
Center of Massโข It is the point through which the resultant inertia force acts on a body.โข It is a property of the distribution of mass within the body.
CG and CM
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Centroid and Center of Gravity
CG by Tabular Summation
CM by Tabular Summation
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Centroid and Center of Gravity
โข It is the point at which area (or volume or line) can be concentratedโข It is the point at which the static moment is zero.โข The centroid represents the geometric center of a body. This point
coincides with the center of mass or the center of gravity only if thematerial composing the body is uniform or homogeneous.
โข Formulas used to locate the center of gravity or the centroid simplyrepresent a balance between the sum of moments of all the parts ofthe system and the moment of the โresultantโ for the system.
โข In some cases the centroid is located at a point that is not on theobject, as in the case of a ring, where the centroid is at its center.Also, this point will lie on any axis of symmetry for the body.
Centroid
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Centroid and Center of Gravity
Centroid by Intergration
Centroid of a Volume
Centroid of an Area Centroid of a Line
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Centroid and Center of Gravity
Find the location of the centroidal axisthat is parallel to the base of the trianglein sketch.
Solution:
๐ด ๐ฆ = ๐ฆ๐๐ = ๐ฆ๐ข๐๐ฆ
By similar triangles:๐ข
๐=
โ โ ๐ฆ
โโ ๐ข =
๐
โโ โ ๐ฆ
Then
๐ด ๐ฆ =๐โ
2 ๐ฆ =
0
โ ๐
โโ โ ๐ฆ ๐ฆ๐๐ฆ
=๐
โ 0
โ(โ๐ฆ๐๐ฆ โ ๐ฆ2 ๐๐ฆ)
=๐
โ
โ๐ฆ2
2โ
๐ฆ3
3
โ
0
=๐
โ
โ3
2โ
โ3
3=
๐โ2
6
๐ฆ =๐โ2
6
2
๐โ
๐ =๐
๐
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Centroid and Center of Gravity
The 16-ft wing of an airplane is subjectedto a lift which varies from zero at the tip to360 lb/ft at the fuselage according to ๐ค =
90๐ฅ1
2 lb/ft where x is measured from thetip. Compute the resultant and its locationfrom the wingtip.
Solution:
๐ = ๐๐๐๐ ๐ข๐๐๐๐ ๐กโ๐ ๐๐ข๐๐ฃ๐
๐ = 0
16
90 ๐ฅ๐๐ฅ
๐น = ๐๐๐๐ ๐๐
๐ฅ = ๐ฅ๐๐
๐ด=
0
16๐ฅ 90 ๐ฅ๐๐ฅ
3480 ๐ = ๐. ๐๐ ๐๐ญ
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Centroid and Center of Gravity
Centroid of Common Geometric Shapes
Area and Centroid
๐ด = ๐๐
๐ฅ =1
2๐
๐ฆ =1
2๐
๐ด =1
2๐โ
๐ฆ =1
3โ
Area and Centroid
๐ด = ฯ๐2
๐ฅ = 0 ๐ฆ = 0
๐ด =1
2ฯ๐2
๐ฅ = 0
๐ฆ =4๐
3ฯ
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Centroid and Center of Gravity
Centroid of Common Geometric Shapes
Area and Centroid
๐ด =1
4ฯ๐2
๐ฅ =4๐
3ฯ
๐ฆ =4๐
3ฯ
๐ด = ฯ๐๐ ๐ฅ = 0
๐ฆ = 0
Area and Centroid
๐ด =1
2ฯ๐๐
๐ฅ = 0
๐ฆ =4๐
3ฯ
๐ด =1
4ฯ๐๐
๐ฅ =4๐
3ฯ
๐ฆ =4๐
3ฯ
Centroid and Center of Gravity
Centroid by Tabular Summation
Centroid of a Composite Area
Centroid of a Composite Curve
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Centroid and Center of Gravity
Find the centroidal axes of the section in the sketch.
Solution:
Element A y x Ay Ax
1 .60 2.85 1.0 1.71 .60
2 .60 .15 1.0 .09 .60
3 .48 1.50 .10 .72 .048
ฮฃA=1.68 ฮฃAy=2.52 ฮฃAx=1.248
๐ฅ =ฮฃ(๐ด๐ฅ)
ฮฃ๐ด=
1.248
1.68=. ๐๐๐ ๐๐
๐ฆ =ฮฃ(๐ด๐ฆ)
ฮฃ๐ด=
2.52
1.68= ๐. ๐๐ ๐๐
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Moment of Inertia
Moment of Inertiaโข Also called the second moment of areaโข For structural cross-sections, the moment of inertia of interest are
those about the centroidal axes.โข Used in determining the stiffness and bending stresses in beams
and the buckling loads of columnsโข For beams, the moment of area of interest is the one about the
bending axis, for columns, it is the minimum moment of inertia
First Moment of Area (Static Moment of Area)
๐ = ๐ฆ๐๐ด = ฮฃ๐๐ฆ
โข Used to find the shear stress distribution over a cross-section of ashear carrying member
Moment of Inertia
Moment of Inertia by Integration Polar Moment of Inertia
Parallel โAxis Theorem (Transfer Formula)
where: ๐ผ๐ = moment of inertia about an arbitrary axis ๐ผ๐ = moment of inertia about the parallel axis that passes through the centroid๐ด = area๐ = distance between the axes (transfer distance)
Moment of Inertia by Tabular Summation
๐ผ๐ฅ = ฮฃ๐ผ ๐ฅ + ฮฃA๐ฆ2
๐ฐ ๐ = ๐ฐ๐ โ ๐ ๐๐
๐ผ๐ฆ = ฮฃ๐ผ ๐ฆ + ฮฃA๐ฅ2
๐ฐ ๐ = ๐ฐ๐ โ ๐ ๐๐
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Moment of Inertia
Centroidal Moment of Inertia (with respect to an axis passing through the centroid):
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Moment of Inertia
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Moment of Inertia
Moment of Inertia with respect to an axis passing through the base:
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Moment of Inertia
A rectangle has a base of 3 cm and a heightof 6 cm. What is its second moment of area(in cm4) about an axis through the center ofgravity and parallel to the base.
Solution:
๐ผ๐ฅ0= ๐ผ ๐ฅ =
๐โ3
12=
3(63)
12๐ฐ๐๐
= ๐๐ ๐๐๐
๐ผ๐ฅ =๐โ3
3=
3(63)
3๐ฐ๐ = ๐๐๐ ๐๐๐
Or by Transfer Formula:๐ผ๐ฅ = ๐ผ ๐ฅ + A๐2 = 54 + 3๐ฅ6 32
๐ฐ๐ = ๐๐๐ ๐๐๐
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Moment of Inertia
Find the moment of inertia of the sectionabout the centroidal axes parallel to axes xand y
Solution:From previous example ๐ฅ = .743 ๐๐๐ ๐ฆ = 1.50
Element A y x Ay2 Ax2 Ix Iy
1 .60 2.85 1.00 4.8735 .60 .0045 .200
2 .60 .15 1.00 .0135 .60 .0045 .200
3 .48 1.50 .10 1.0800 .0048 .2304 .002
1.68 5.967 1.2048 .2394 .402
๐ผ๐ฅ = ฮฃ๐ผ ๐ฅ + ฮฃA๐ฆ2 = .2394 + 5.967 = 6.2064
๐ผ ๐ฅ = ๐ผ๐ฅ โ ๐ด ๐ฆ2 = 6.2064 โ 1.68 1.502
๐ฐ ๐ = ๐. ๐๐๐ ๐๐๐
๐ผ๐ฆ = ฮฃ๐ผ ๐ฆ + ฮฃA๐ฅ2 = .402 + 1.2048 = 1.6068
๐ผ ๐ฆ = ๐ผ๐ฆ โ ๐ด ๐ฅ2 = 1.6068 โ 1.68 .7432
๐ฐ ๐ =. ๐๐๐๐ ๐๐๐
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Fundamental Concepts and Definitions
โข Motion โ Change of position of an object with respect to time and reference point.
โข Translation โ Motion involving change in displacement over a period of time.
โข Rotation โ Motion involving change in angle over a period of time.
โข Rectilinear Motion โ Straight line motion
โข Curvilinear Motion โ Motion along a curved path
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Fundamental Concepts and Definition
โข Displacement, s โ Shortest distance between initial and final position of a particle.
โข Velocity, V โ Instantaneous rate of change of displacement with respect to time.
๐ =๐๐
๐๐ก
โข Speed โ Refers to the magnitude of velocity.
โข Acceleration, a - Instantaneous rate of change of velocity with respect to time.
๐ =๐๐
๐๐ก
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Kinematics of a Particle
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Equations of Motion with Constant Acceleration:
โข ๐ = ๐0๐ก +1
2๐๐ก2
โข ๐ = ๐0 + ๐๐ก
โข ๐2 = ๐02
+ 2๐๐ For a free falling body, ๐0 = 0
โข ๐ฆ =1
2๐๐ก2
Rectilinear Translation
Kinematics of a Particle
A ball is thrown vertically upward at a speed of 20 m/s.
a. How high is it after 3s?
๐ฆ = ๐0๐ก +1
2๐๐ก2
๐ฆ = 20๐
๐ 3๐ + .5 โ9.8
๐
๐ 2 (3๐ )2
๐ = ๐๐. ๐ ๐
b. How high does it rise?
๐2 = ๐02
+ 2๐๐ฆ
0 = 20๐
๐
2
+ 2 โ9.8๐
๐ 2๐ฆ๐๐๐ฅ
๐๐๐๐ = ๐๐. ๐๐ ๐
c. How long does it take to reach the highest point?๐ = ๐0 + ๐๐ก
0 = 20๐
๐ + โ9.8
๐
๐ 2 ๐ก
๐ = ๐. ๐๐ ๐
d. How long does it take for the ball to reach the ground?๐๐๐ = ๐๐ ๐๐๐ = ๐. ๐๐ ๐
e. What is its velocity when it returns to the level from which it started?๐ฝ๐ = ๐ฝ๐ = ๐๐ ๐/๐
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If a particleโs position is given by the expression ๐ฅ ๐ก = 3.4๐ก3 โ 5.4๐ก meters.
a. What is its velocity after t = 3 seconds?
๐ =๐๐ฅ
๐๐ก=
๐
๐๐ก(3.4๐ก3 โ 5.4๐ก)
๐ก = 3๐ฝ = ๐๐. ๐ ๐/๐
b. What is the acceleration of the particle after t = 5 seconds?
๐ =๐๐ฅ
๐๐ก=
๐
๐๐ก3.4๐ก3 โ 5.4๐ก = 10.2๐ก2 โ 5.4
๐ =๐๐
๐๐ก=
๐
๐๐ก(10.2๐ก2 โ 5.4)
๐ก = 5
๐ = ๐๐๐ ๐/๐๐
Kinematics of a Particle
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General Equation of Projectile:
๐ฆ = ๐ฅ๐ก๐๐ฮธ โ๐๐ฅ2
2๐02๐๐๐ 2ฮธ
โข ๐๐ฆ deceases as it goes up, zero at maximum height, and increases as
it goes down.โข ๐๐ฅ is constant.
๐0๐ฅ= ๐๐๐๐๐ ฮธ
๐0๐ฆ= ๐๐๐ ๐๐ฮธ
Curvilinear Translation
A shot is fired at an angle of 45ยฐ with the horizontal and a velocity of 300 fps. Calculate the range of the projectile.
Solution:
๐ =๐0
2๐ ๐๐2ฮธ
๐
=(300 ๐๐ก/๐ )2sin(2๐ฅ45)
32.2 ๐๐ก/๐ 2
๐น = ๐๐๐๐ ๐๐
Kinematics of a Particle
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A projectile leaves with a velocity of 50 m/s at anangle of 30ยฐ with the horizontal. Find the maximumheight that it could reach.
Solution:
๐ป =๐0
2๐ ๐๐2ฮธ
2๐
=(50 ๐/๐ )2(sin30)2
2 9.81๐๐ 2
๐ฏ = ๐๐. ๐๐ ๐
Kinematics of a Particle
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FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
โข ๐ = ๐ฮธโข ๐ = ๐ฯโข ๐ = ๐ฮฑWhere:r = radiusฮธ, ฯ, and ฮฑ are angular displacement, angular velocity, and angular acceleration, respectively.S, V, and a are linear dimensions.
โข Linear velocity acts tangent to the point.โข Linear acceleration has tangential and
normal components.
โข ๐ = ๐๐2 + ๐๐ก
2
โข ๐๐ =๐2
๐
โข ๐๐ก =๐๐
๐๐ก
Also,
โข ฮธ = ฯ0๐ก +1
2ฮฑ๐ก2
โข ฯ = ฯ0 + ฮฑ๐ก
โข ฯ2 = ฯ02
+ 2ฮฑ๐
Rotation
A turbine started from rest to 180 rpm in 6 minutesat a constant acceleration. Find the number ofrevolutions that it makes within the elapsed time.
Solution:
ฯ = ฯ0 + ฮฑ๐ก180 ๐๐๐ = 0 + ฮฑ 6 ๐๐๐๐ข๐ก๐๐
ฮฑ = 30๐๐๐ฃ
๐๐๐2
ฯ2 = ฯ02
+ 2ฮฑ๐
(180 ๐๐๐)2= 0 + 2 30๐๐๐ฃ
๐๐๐2(๐ )
๐ = ๐๐๐ ๐๐๐
Kinematics of a Particle
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A flywheel is 15 cm in diameter acceleratesuniformly from rest to 500 rpm in 20 seconds.What is its angular acceleration?
Solution:
500๐๐๐ฃ
๐๐๐๐ฅ
2ฯ๐๐๐
๐๐๐ฃ๐ฅ
๐๐๐
60 ๐ = 52.36 ๐๐๐/๐
ฯ = ฯ0 + ฮฑ๐ก
52.36๐๐๐
๐ = 0 + ฮฑ 20 ๐
๐ถ = ๐. ๐๐ ๐๐๐ /๐๐
Kinematics of a Particle
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Kinetics of a Particle
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
When a body is subjected to an acceleration, there exists a forceopposite the direction of motion and equal to the product ofmass and acceleration . This force is known as the reverseeffective force, REF or inertial force, ma.
๐ฎ๐ญ = ๐๐ =๐
๐๐
Dโ Alembertโs Principle
Kinetics of a Particle
What force is necessary to accelerate a 30,000 lb railwayelectric car at the rate of 1.25 ft/sec2, if the forcerequired to overcome frictional resistance is 400 lb?
Solution:
ฮฃ๐นโ = ๐๐
๐ โ ๐น =๐
๐๐
๐ โ 400 =30,000 ๐๐
32.2 ๐๐๐ 2 1.25๐๐๐ 2
๐ท = ๐๐๐๐. ๐ ๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Kinetics of a Particle
An elevator weighing 2,000 lb attains an upward velocityof 16 fps in 4 seconds with uniform acceleration. What isthe tension in the supporting cables?
Solution:
๐ = ๐0 + ๐๐ก
16๐๐ก
๐ = 0 + ๐ 4๐
๐ = 4 ๐๐ก/๐ 2
ฮฃ๐น๐ฃ = ๐๐
๐ โ ๐ =๐
๐๐
๐ โ 2000 =2000 ๐๐
32.2 ๐๐๐ 2 4 ๐๐๐ 2
๐ท = ๐๐๐๐. ๐๐ ๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Kinetics of a Particle
Centripetal and Centrifugal Force
๐ถ๐น = ๐๐๐ = ๐๐2
๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
Kinetics of a Particle
A cyclist on a circular track of radius800 ft is traveling at 27 fps. Hisspeed in the tangential directionincreases at the rate of 3 fps2. Whatis the cyclistโs total acceleration?
Solution:
๐๐ =๐2
๐=
(27 ๐๐๐ )2
800 ๐๐ก= 0.91 ๐๐๐/๐ 2
๐ = ๐๐ก2 + ๐๐
2
๐ = 32 + 0.912
๐ = ๐. ๐๐ ๐๐๐๐
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP
FEATI UNIVERSITY โ AERONAUTICAL ENGINEERING REVIEW CENTER NMP