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  1. Chapter 2: Concurrent force systems Department of Mechanical Engineering Department of Mechanical Engineering

  2. 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 Department of Mechanical Engineering

  3. 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 Department of Mechanical Engineering

  4. Example direction j i 1000 N α magnitude Point of application Line of action Department of Mechanical Engineering Department of Mechanical Engineering

  5. 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 Department of Mechanical Engineering

  6. 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 Department of Mechanical Engineering

  7. 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 Department of Mechanical Engineering

  8. 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 Department of Mechanical Engineering

  9. 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 Department of Mechanical Engineering

  10. 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 Department of Mechanical Engineering

  11. 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 Department of Mechanical Engineering

  12. 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 Department of Mechanical Engineering

  13. 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 Department of Mechanical Engineering

  14. 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 Department of Mechanical Engineering

  15. Resultant of Forces – Review on vector addition  Vector addition R = B + = + A B B A R A  Triangle method (head-to-tail method) – Note: the tail of the first vector and the head of the last vector 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 A R B Department of Mechanical Engineering Department of Mechanical Engineering

  16. Resultant of Forces – Review on geometric laws Law of Sines A α c Laws of Cosines b γ = + − γ 2 2 2 cos 2 c a b ab β C = + − β 2 2 2 2 cos ac b a c a B = + − α 2 2 2 2 cos a b c ac Department of Mechanical Engineering Department of Mechanical Engineering

  17. Resultant of two concurrent forces Pay attention to the angle and the sign of the last term !!!  The magnitude of the resultant (R) is given by 2 1 F R = = + − γ 2 2 2 cos R F F F F 2 1 2 + + φ 2 2 2 2 cos F F F 1 2 1 2  The direction (relative to the direction of F1) can be given by the law of sines F β sin = φ sin 2 R Department of Mechanical Engineering Department of Mechanical Engineering

  18. 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 Department of Mechanical Engineering

  19. 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 Department of Mechanical Engineering

  20. 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 R 1413 3 . 1413 ≈ = = + + 2 2 2 0 900 600 ( 2 900 )( 600 ) cos 40 R lb – The angle α between the R and the 900-lb force is given by 180 sin( 600 = α α − 0 0 sin 40 ) = 1413 3 . o 15 . 836 – The angle θ therefore is θ = + = 0 0 0 15 . 836 35 50 8 . Department of Mechanical Engineering Department of Mechanical Engineering

  21. Example Problem 2-2 Determine – The resultant R – The angle between the R and the x-axis Department of Mechanical Engineering Department of Mechanical Engineering

  22. 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 Department of Mechanical Engineering

  23. Force components Department of Mechanical Engineering Department of Mechanical Engineering

  24. Resolution of a force into components  The components of a resultant force are not unique !! G B A R + = + = = + = + + I H ( ) C D E F  The direction of the components must be fixed (given) Department of Mechanical Engineering Department of Mechanical Engineering

  25. 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 – Use law of sines to determine the magnitudes of the components Parallel to v 900 F F = = u 45 v o o o sin sin 25 sin 110 o 900 sin 45 0 = = 677 F N u sin 110 0 900 sin 25 = = 405 F N v o sin 110 Department of Mechanical Engineering Department of Mechanical Engineering

  26. 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 Department of Mechanical Engineering

  27. 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 Department of Mechanical Engineering

  28. 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 Department of Mechanical Engineering

  29. The Cartesian coordinate system  The Cartesian coordinate axes 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 z y x Department of Mechanical Engineering Department of Mechanical Engineering

  30. 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 enwill 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 Ae e A A = = direction along n A is in the positive A A = = en n n A A = − = − B B e Be B is in the negative direction along n n n Department of Mechanical Engineering Department of Mechanical Engineering

  31. 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 + F F F i j F x y x y = θ cos F F x y = θ sin F F F y Fy= Fyj 2 2 = + F F F x y F − y θ = 1 tan j F θ Fx= Fxi x x i Department of Mechanical Engineering Department of Mechanical Engineering

  32. The rectangular components in 3D systems z en k = + + Fz= Fzk F = F F F x + y z k + i j F = F F x y z F e F F n + + i j k F F F F θz x y z = = e n F F θ = θy cos θx F F x x Fy= Fyj θ = Fx= Fxi cos F F y y y j θ = cos F F z z i 2 2 = + + 2 = θ + θ + θ F F F F e i j k cos cos cos x x y z n x y z F F F − − − y θ = θ = θ = 1 1 1 cos cos cos x z x y z F F F Department of Mechanical Engineering Department of Mechanical Engineering

  33. Dot Products of two vectors • = • = θ = θ A B B A A B cos cos AB A It’s a scalar !!! Special cosines: θ Cos 0o= 1 Cos 30o= ½ √3 Cos 45o= ½ √2 Cos 60o= 0.5 Cos 90o= 0 B Department of Mechanical Engineering Department of Mechanical Engineering

  34. Dot products and rectangular components  The dot product can be used to obtain the rectangular components of a force (a vector in general) cosθ = • = A e A A A (magnitude) n n n = e A (the vectorial component in the n direction) n n n = • A A e e ( ) n n n The component along en = − A A A t n The component along et Remember, enand etare perpendicular Department of Mechanical Engineering Department of Mechanical Engineering

  35. Cartesian rectangular components  The dot product is particularly useful when the unit vectors are of the Cartesian system (the i, j, k) θ 90 = • = F i cos F F y x F θ = • = − F j cos( ) F F y Fy= Fyj θ = sin F 90-θ j Also, in 3D, θ Fx= Fxi x = • F k F i z = + = + = • + • F F F i j F i i F j j ( ) ( ) F F x y x y Department of Mechanical Engineering Department of Mechanical Engineering

  36. More usage of dot products …  Dot products of two vectors written in Cartesian system B A + = •B A + A B A B x x y y z z  The magnitude of a vector (could be a force vector), here A is the vector magnitude • = = = + + 2 2 A A cos 0 A A A A A A A A x x y y z z  The angle between two vectors (say between vectors A and B)   + +   A B A B A B   − x x y y z z θ = 1 cos AB Department of Mechanical Engineering Department of Mechanical Engineering

  37. The rectangular components of arbitrary direction z k Fz= Fzk en = + + F F F F x + y z k F = F + Ft i j F = F F e θzn x y + z Fn e F F Fy= Fyj Fx= Fxi n n t t θyn θxn = • F e F = n n y j + e + • i j k e ( ) F F F j x i y + z • n F i = • + • e k e F F + x Can you show the following? x n θ y n z n = θ + θ cos cos cos F F F = θ + θ + θ e i j k cos cos cos x xn y yn z zn n xn yn zn Department of Mechanical Engineering Department of Mechanical Engineering

  38. 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 Department of Mechanical Engineering

  39. 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 Department of Mechanical Engineering

  40. Example Problem 2-6 = = β θ − θ cos cos( 90 ) F F F F x y β = = − cos cos( 90 ) F F F F ' ' x y θ = − = o 90 28 62 β β = − = o 62 = 30 32 = 450 cos 62 211 F N x θ = = 450 sin 62 397 = F N y = 450 cos 32 382 F N ' x = = 450 sin 32 238 F N Writing the F in Cartesian vector form: y = + = + F i j e e ( 211 397 ) 382 ( 238 ) N N ' x y ' Department of Mechanical Engineering Department of Mechanical Engineering

  41. Example Problem 2-8 B  Find the angles θx, θy, and θz (θxis the angle between OB and x axis and so on ..)  The x, y, and x scalar components of the force.  The rectangular component Fnof the force along line OA  The rectangular component of the force perpendicular to line OA (say Ft) Department of Mechanical Engineering Department of Mechanical Engineering

  42. Example Problem 2-8 B  To find the angles: – Find the length of the diagonal OB, say d – d = 5.831 m – Use cosines to get the angles 3 − θ = = 1 o cos 59 0 . x . 5 831 4 − θ = = 1 o cos 46 7 . y . 5 831 3 − θ = = 1 o cos 59 0 . z . 5 831 θ = = cos 12 . 862 F F kN  The scalar components in the x, y, and z directions: x x θ = = cos 17 . 150 F F kN y y θ = = cos 12 . 862 F F kN z z = + + F i j k 12 ( . 862 17 . 150 12 . 862 ) kN Department of Mechanical Engineering Department of Mechanical Engineering

  43. 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) = = A OA + + r i j 1 k 3 3 1 + 3 + = = e A + + r i j k 3 1 3 OA r 1 2 2 2 3 A + + i j 36 k 3 3 = = + + i j k . 0 688 . 0 230 . 0 688 – Remember, that any vector can be represented as a multiplication of its magnitude and a unit vector along its line of application . 4 Department of Mechanical Engineering Department of Mechanical Engineering

  44. Example Problem 8-2 = F• e F OA OA  The scalar component of F along OA F OA 688 . 0 862 . 12 + × = = + + • + + i j k i j k 12 ( . 862 17 . 150 12 . 862 ) . 0 ( + 688 . 0 230 . 0 688 ) . 0 × . 0 × = 17 . 150 230 12 . 862 688 21 . 643 F kN OA  The vector component of F along OA F F OA 86 . 14 + = = • = + + e e ( ) . 0 ( 6 . 21 688 . 0 230 . 0 688 ) i j k OA 97 OA + i j k . 4 14 . 86  The vector component of F perpendicular to OA = − = + + − + + F = F F i j k i j k 12 ( j + . 862 k 17 . 150 12 . 862 ) 14 ( . 86 . 4 97 14 . 86 ) OA t − + i ( 2 12 . 18 2 )  The scalar component of F perpendicular to OA = ( | = − + − = ) 2 − + + ) 2 − = 2 2 2 F i j | ) k | | 2 12 . 18 2 ( 12 . 18 ( 12 . 50 F kN t t Check: Department of Mechanical Engineering Department of Mechanical Engineering = + = + ≈ 2 2 2 2 21 . 643 12 . 50 25 F F F kN OA t

  45. Resultants by rectangular components  The Cartesian rectangular components of forces can be utilized to obtain the resultant of the forces •Adding the x vector components, we obtain the x vector component of the resultant ∑ x x y = = + R F F F 1 2 x x F1 •Adding the y vector components, we obtain the y vector component of the resultant ∑ y y F1y F2x = = + R F F F 1 2 y y x F1x •The resultant can be obtained by performing the vector addition of these two vector components R R R y x = + = F2 F2y R + i j R x y Department of Mechanical Engineering Department of Mechanical Engineering

  46. Resultants by rectangular components  The scalar components of the resultant F F R 2 1 x x x + = F F R 2 1 y y y + = = = + + = = i i ( ) F F ( F F R R 1 2 x x x j j ) 1 2 y y y  The magnitude of the resultant = + 2 x 2 y R R R  The angles formed by the resultant and the Cartesian axes R x x cos = θ θ R − − y = 1 1 cos y R R  All of the above results can be easily extended for 3D system Department of Mechanical Engineering Department of Mechanical Engineering

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

  48. HW Problem 2-20  Determine the non-rectangular components of R Department of Mechanical Engineering Department of Mechanical Engineering

  49. HW Problem 2-37 Determine the components of F1and F2 in x-y and x’-y’ systems Department of Mechanical Engineering Department of Mechanical Engineering

  50. 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 Department of Mechanical Engineering

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