Chapter 2: Concurrent force systems
Department of Mechanical Engineering Objectives
To understand the basic characteristics of forces 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 Mechanical Engineering 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 space 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 system?
A number of forces (in 2D or 3D system) that is treated as a group: 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 motion » 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 times, the rigid body 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
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 position 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