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Chapter 2: Concurrent

Department of Mechanical Objectives

 To understand the basic characteristics of  To understand the classification of force systems  To understand some force principles

 To know how to obtain the resultant of forces in 2D and 3D systems  To know how to obtain the components of forces in 2D and 3D systems

Department of Characteristics of forces  Force: Vector with magnitude and direction  Magnitude – a positive numerical value representing the size or amount of the force

 Directions – the slope and the sense of a line segment used to represent the force – Described by angles or dimensions – A negative sign usually represents opposite direction  Point of application – A point where the force is applied – A line of action = a straight line extending through the point of application in the direction of the force  The force is a physical quantity that needs to be represented using a mathematical quantity

Department of Mechanical Engineering Example

direction j

i

1000 N

α magnitude

Point of application Line of action

Department of Mechanical Engineering Vector to represent Force

 A vector is the mathematical representation that best describes a force

 A vector is characterized by its magnitude and direction/sense

 Math operations and manipulations of vectors can be used in the force analysis

Department of Mechanical Engineering Free, sliding, and fixed vectors

 Vectors have magnitudes, slopes, and senses, and lines of applications

 A free vector – The application line does not pass a certain point in  A sliding vector – The application line passes a certain point in space  A fixed vector – The application line passes a certain point in space – The application point of the vector is fixed

Department of Mechanical Engineering Vector/force notation

 The symbol representing the force  bold face or underlined letters

 The magnitude of the force  lightface (in the text book, + italic) A = A or A = A

Department of Mechanical Engineering Classification of forces

 Based on the characteristic of the interacting bodies: – Contacting vs. Non-contacting forces  Surface force (contacting force) – Examples: » Pushing/pulling force » Frictions  Body force (non-contacting force) – Examples: » Gravitational force » Electromagnetic force

Department of Mechanical Engineering Classification of forces

 Based on the area (or volume) over which the force is acting – Distributed vs. Concentrated forces  Distributed force – The application area is relatively large compare to the whole loaded body – Uniform vs. Non-uniform  Concentrated force – The application area is relatively small compare to the whole loaded body

Department of Mechanical Engineering What is a force ?

 A number of forces (in 2D or 3D system) that is treated as a :  A concurrent force system – All of the action lines intersect at a common point  A coplanar force system – All of the forces lie in the same plane  A parallel force system – All of the action lines are parallel  A collinear force system – All of the forces share a common line of action

Department of Mechanical Engineering The external and internal effects  A force exerted on the body has two effects: – External effects » Change of » Resisting forces (reactions) – Internal effects » The tendency of the body to deform  develop strain, stresses – If the force system does not produce change of motion » The forces are said to be in balance » The body is said to be in (mechanical) equilibrium

Department of Mechanical Engineering External and internal effects  Example 1: The body changes in motion a F

Not fixed, no (horizontal) support

 Example 2: The body deforms and produces (support) reactions  The forces must be in balance F Fixed support Support Reactions

Department of Mechanical Engineering Principle for force systems

 Two or more force systems are equivalent when their applications to a body produce the same external effect  Transmissibility  Reduction = – A process to create a simpler equivalent system – to reduce the number of forces by obtaining the “resultant” of the forces  Resolution = – The opposite of reduction – to find “the components” of a force vector  “breaking up” the resultant forces

Department of Mechanical Engineering Principle of Transmissibility

 Many , the assumption is taken  only the external effects are the interest  The external effect of a force on a rigid body is the same for all points of application of the force along its line of action

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Department of Mechanical Engineering Resultant of Forces – Review on vector addition

 Vector addition B R = A + B = B + A A R  Triangle method (head-to-tail A method) R – Note: the tail of the first vector and the head of the last vector B become the tail and head of the resultant  principle of the force polygon/triangle  Parallelogram method – Note: the resultant is the diagonal of the parallelogram formed by the vectors being summed

Department of Mechanical Engineering Resultant of Forces – Review on geometric laws

 Law of Sines A

α

 Laws of Cosines c b

c2 = a2 + b2 − 2abcosγ γ 2 2 2 β C b = a + c − 2ac cos β a B a2 = b2 + c2 − 2ac cosα

Department of Mechanical Engineering Resultant of two concurrent forces

Pay attention to the angle  The magnitude of the resultant (R) is given by and the sign of the last 2 = 2 + 2 − γ R F1 F2 2F1F2 cos term !!! 2 2 2 R = F1 + F2 + 2F1F2 cosφ

 The direction (relative to the direction of F1) can be given by the law of sines F sinφ sin β = 2 R

Department of Mechanical Engineering Resultant of three concurrent forces and more

 Basically it is a repetition of finding resultant of two forces  The sequence of the addition process is arbitrary  The “force polygons” may be different  The final resultant has to be the same

Department of Mechanical Engineering Resultant of more than two forces

 The polygon method becomes tedious when dealing with three and more forces  It’s getting worse when we deal with 3D cases  It is preferable to use “rectangular-component” method

Department of Mechanical Engineering Example Problem 2-1

 Determine: – The resultant force (R) – The angle θ between the R and the x-axis  Answer: – The magnitude of R is given by R2 = 9002 + 6002 + 2(900)(600)cos 400 R =1413.3 ≈1413lb – The angle α between the R and the 900-lb force is given by sinα sin(1800 − 400 ) = 600 1413.3 α =15.836o – The angle θ therefore is

0 0 0 θ =15.836 + 35 = 50.8 Department of Mechanical Engineering Example Problem 2-2

 Determine – The resultant R – The angle between the R and the x-axis

Department of Mechanical Engineering Another example

 If the resultant of the force system is zero, determine

– The force FB

– The angle between the FB and the x-axis

Department of Mechanical Engineering Force components

Department of Mechanical Engineering Resolution of a force into components

 The components of a resultant force are not unique !! R = A + B = (G + I) + H = C + D = E + F

 The direction of the components must be fixed (given)

Department of Mechanical Engineering How to obtain the components of a force (arbitrary component directions)? Parallel to u  Steps: – Draw lines parallel to u and v crossing the tip of the R – Together with the original u and v lines, these two lines produce the parallelogram – The sides of the parallelogram represent the components of R Parallel to v – Use law of sines to determine the magnitudes of the components F F 900 u = v = sin 45o sin 25o sin110o

900sin 45o F = = 677N u sin1100 900sin 250 F = = 405N v sin110o Department of Mechanical Engineering Example Problem 2-5

 Determine the components of F = 100 kN along the bars AB and AC

 Hints: – Construct the force triangle/parallelogram – Determine the angles α, β, γ – Utilize the law of sines

Department of Mechanical Engineering Another example

 Determine the magnitude of the components of R in the directions along u and v, when R = 1500 N

Department of Mechanical Engineering Rectangular components of a force

 What and Why rectangular components? – Rectangular components  all of the components are perpendicular to each other (mutually perpendicular) – Why? One of the angle is 90o ==> simple  Utilization of unit vectors  Rectangular components in 2D and 3D  Utilization of the Cartesian c.s.

Arbitrary rectangular

Department of Mechanical Engineering The Cartesian coordinate system

 The Cartesian coordinate axes z are arranged following the right-hand system (shown on the right)  The setting of the system is arbitrary, but the results of the analysis must be independent of the chosen system x y

Department of Mechanical Engineering Unit vectors

 A dimensionless vector of unit magnitude  The very basic coordinate system used to specify coordinates in the space is the Cartesian c.s.  The unit vectors along the Cartesian coordinate axis x, y and z are i, j, k, respectively

 The symbol en will be used to indicate a unit vector in some n- direction (not x, y, nor z)  Any vector can be represented as a multiplication of a magnitude and a unit vector A is in the positive direction along n A = A e = Ae A A n n e = = n A A B is in the negative direction along n B = − B en = −Ben Department of Mechanical Engineering The rectangular components of a force in 2D system

 While the components must be perpendicular to each other, the directions do not have to be parallel or perpendicular to the horizontal or vertical directions

F = Fx + Fy = Fxi + Fy j = θ y Fx F cos F Fy = F sinθ

Fy = Fy j 2 2 F = Fx + Fy

−1 Fy j θ = tan θ F = F i x x x Fx i Department of Mechanical Engineering The rectangular components in 3D systems z k en

Fz = Fz k F = Fx + Fy + Fz

= Fxi + Fy j+ Fzk

F = Fen F F F i + F j+ F k θ e = = x y z z n F F

θ θy Fx = F cosθ x x F = F j Fx = Fx i y y Fy = F cosθ y j y Fz = F cosθ z

2 2 2 i F = F + F + F x x y z en = cosθ xi + cosθ y j+ cosθ zk

F Fy F θ = cos−1 x θ = cos−1 θ = cos−1 z x F y F z F Department of Mechanical Engineering Dot Products of two vectors

A • B = B • A = A B cosθ = AB cosθ

A It’s a scalar !!! Special cosines:

θ Cos 0o = 1 Cos 30o = ½ √3 Cos 45o = ½ √2 B Cos 60o = 0.5 Cos 90o = 0

Department of Mechanical Engineering Dot products and rectangular components  The dot product can be used to obtain the rectangular components of a force (a vector in general)

An = A •en = Acosθn (magnitude)

An = Anen (the vectorial component in the n direction)

An = (A •en )en The component along en

At = A − An The component along et

Remember, en and et are perpendicular

Department of Mechanical Engineering Cartesian rectangular components

 The dot product is particularly useful when the unit vectors are of the Cartesian system (the i, j, k)

y Fx = F • i = F cosθ F Fy = F • j = F cos(90 −θ ) Fy = Fy j = F sinθ

90-θ j Also, in 3D, θ F = F i x x x i Fz = F •k

F = Fx + Fy = Fxi + Fy j = (F • i)i + (F • j)j Department of Mechanical Engineering More usage of dot products …

 Dot products of two vectors written in Cartesian system

A • B = Ax Bx + Ay By + Az Bz  The magnitude of a vector (could be a force vector), here A is the vector magnitude

2 2 A • A = A cos0 = A = Ax Ax + Ay Ay + Az Az  The angle between two vectors (say between vectors A and B) −1 Ax Bx + Ay By + Az Bz  θ = cos    AB  Department of Mechanical Engineering The rectangular components of arbitrary direction z

k Fz = Fz k en F = F + F + F x y z F F = Fxi + Fy j+ Fzk t θzn Fn F = F j F = Fnen + Ftet Fx = Fx i y y θ θxn yn

Fn = F •en j y = (Fxi + Fy j+ Fzk) •en i = Fxi •en + Fy j•en + Fzk •en x Can you show the following?

= Fx cosθ xn + Fy cosθ yn + Fz cosθ zn en = cosθ xni + cosθ yn j+ cosθ znk

Department of Mechanical Engineering Summarizing ….

 The components of a force resultant are not unique  Graphical methods (triangular or parallelogram methods) combined with law of sinus and law of cosines can be used to obtain components in arbitrary direction  Rectangular components are components of a force (vector) that perpendicular to each other  The dot product can be used to – obtain rectangular components of a force vector – obtain the magnitude of a force vector (by performing self- dot-product) – Obtain the angle between two (force) vectors

Department of Mechanical Engineering Example Problem 2-6

 Find the x and y scalar components of the force  Find the x’ and y’ scalar components of the force  Express the force F in Cartesian vector form for the xy- and x’y’- axes

Department of Mechanical Engineering Example Problem 2-6

Fx = F cosθ Fy = F cos(90 −θ )

Fx' = F cos β Fy' = F cos(90 − β ) θ = 90 − 28 = 62o β β = 62 − 30 = 32o F = 450cos62 = 211N x θ

Fy = 450sin 62 = 397N

Fx' = 450cos32 = 382N Writing the F in Cartesian vector form: Fy = 450sin 32 = 238N

F = (211i + 397j)N = (382e + 238e )N x' y' Department of Mechanical Engineering Example Problem 2-8 B

 Find the angles θx, θy, and θz (θx is the angle between OB and x axis and so on ..)  The x, y, and x scalar components of the force.

 The rectangular component Fn of the force along line OA  The rectangular component of the force perpendicular to line OA (say Ft)

Department of Mechanical Engineering Example Problem 2-8 B

 To find the angles:

– Find the length of the 3 θ = cos−1 = 59.0o diagonal OB, say d x 5.831

– d = 5.831 m −1 4 o θ y = cos = 46.7 – Use cosines to get the 5.831 3 angles θ = cos−1 = 59.0o z 5.831

 The scalar components in the Fx = F cosθ x =12.862kN

x, y, and z directions: Fy = F cosθ y =17.150kN

Fz = F cosθ z =12.862kN F = (12.862i +17.150j+12.862k)kN Department of Mechanical Engineering Example Problem 2-8

 To find the rectangular component Fn of the force along line OA: – Needs the unit vector along OA – Method 1 : Follow the method described in the book – Method 2: utilize the vector of A (basically vector OA) r 3i +1j+ 3k e = A = OA r 2 2 2 OA = rA = 3i +1j+ 3k A 3 +1 + 3 3i +1j+ 3k – Remember, that any vector can = = 0.688i + 0.230j+ 0.688k be represented as a multiplication 4.36 of its magnitude and a unit vector along its line of application

Department of Mechanical Engineering Example Problem 8-2

FOA = F •eOA  The scalar component of F along OA

FOA = (12.862i +17.150j+12.862k) • (0.688i + 0.230j+ 0.688k)

FOA =12.862×0.688 +17.150×0.230 +12.862×0.688 = 21.643kN

 The vector component of F along OA

FOA = (F •eOA )eOA = 21.6(0.688i + 0.230 j + 0.688k) =14.86i + 4.97j+14.86k

 The vector component of F perpendicular to OA

Ft = F − FOA = (12.862i +17.150j+12.862k) − (14.86i + 4.97j+14.86k) = (−2i +12.18j+ 2k)

 The scalar component of F perpendicular to OA

2 2 2 Ft =| Ft |=| (−2i +12.18j− 2k) |= (−2) +12.18 + (−2) =12.50kN

2 2 2 2 Department of Mechanical Engineering Check: F = FOA + Ft = 21.643 +12.50 ≈ 25kN Resultants by rectangular components

 The Cartesian rectangular components of forces can be utilized to obtain the resultant of the forces y •Adding the x vector components, we obtain the x vector component of the resultant

R x = ∑Fx = F1x + F2x F1 •Adding the y vector components, we obtain the y F 1y vector component of the resultant

F2x R y = ∑Fy = F1y + F2 y x F1x •The resultant can be obtained by performing the F2 vector addition of these two vector components

F2y R = R x + R y = Rxi + Ry j Department of Mechanical Engineering Resultants by rectangular components

 The scalar components of the resultant

R x = F1x + F2x = (F1x + F2x )i = Rxi

R y = F1 y + F2 y = (F1y + F2 y )j = Ry j

 The magnitude of the resultant 2 2 R = Rx + Ry

 The angles formed by the resultant and the Cartesian axes R R θ = cos−1 x θ = cos−1 y x R y R  All of the above results can be easily extended for 3D system

Department of Mechanical Engineering Please do example problems 2-9, 2-10, and 2-11

Department of Mechanical Engineering HW Problem 2-20

 Determine the non-rectangular components of R

Department of Mechanical Engineering HW Problem 2-37

 Determine the

components of F1 and F2 in x-y and x’-y’ systems

Department of Mechanical Engineering HW Problem 2-44

 Express the cable tension in Cartesian form  Determine the magnitude of the rectangular component of the cable force  Determine the angle α between cables AD and BD

Typo in the problem!!!

B(4.9,-7.6,0) C(-7.6,-4.6,0)

Don’t worry if you don’t get the solution in the back of the book Department of Mechanical Engineering HW Problem 2-46

 Determine the scalar components  Express the force in Cartesian vector form  Determine the angle α between the force and line AB

Department of Mechanical Engineering HW problems 2-55

 Given: F1 = 500 lb, F2 = 300 lb, F3 = 200 lb  Determine the resultant  Express the resultant in the Cartesian format  Find the angles formed by the resultant and the coordinate axes

Department of Mechanical Engineering HW Problem 2-49

 Given T1 and T2 are 650 lb,

 Determine P so that the resultant of T1, T2 and P is zero

Department of Mechanical Engineering