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EML 3004C CHAPTER TWO Force Vectors Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-1 EML 3004C 2.1 Scalars and Vectors Force Vectors Scalars : A quantity represented be a number (positive or negative) Ex: Mass, Volume, Length (in the book scalars are represented by italics) Vectors : A quantity which has both A – magnitude (scalar) B – direction (sense) Ex: position, force, moment Line of action Sense Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-2 EML 3004C Forces Classification of Forces Contact 1 – Contacting or surface forces (mechanical) 2 – Non-Contacting or body forces (gravitational, weight) Area 1 – Distributed Force, uniform and non-uniform 2 – Concentrated Force Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-3 EML 3004C Forces Classification of Forces Force System 1 – Concurrent : all forces pass through a point 2 – Coplanar : in the same plane 3 – Parallel : parallel line of action 4 – Collinear : common line of action Three Types 1 – Free (direction, magnitude and sense) 2 – Sliding O 3 – Fixed A Origin Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-4 EML 3004C 2.2 Basic Vector Operations Properties of Vectors 1 – Vector Addition 2 – Vector Subtraction 3 – Vector Multiplication Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-5 EML 3004C Trigonometric Relations of a triangle Phythagorean theorem is valid only for a right angled triangle. For any triangle (not necessarily right angled) A B C sin sin sin Sine Law C 2 A2 B 2 2 AB cos C Cosine Law A B C Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-6 EML 3004C Example Ex: If the angle between F1 and F2 =60o and F1 = 54N and F2 = 60N. Find: the Resultant force and the angle . R F1 120o 60o F2 Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-7 EML 3004C Example… contd R 2 F12 F22 2 F1 F2 cos R 60 54 2 60 54 cos120 2 2 2 R 98.77 98.8 N sin sin 120 F1 R F1 sin sin 60 R 28.26o Namas Chandra Introduction to Mechanical engineering R F1 120o 60o F2 Hibbler Chapter 2-8 EML 3004C Parallelogram Law Properties of Parallelogram A – sum of the three angles in a triangle is 180o B – sum of the interior angles is 360o C- opposite sides are equal 180 Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-9 EML 3004C Vector Addition B B R A A C A B R A B B A R R A B C Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-10 EML 3004C Vector Subtraction Vector Subtraction A B A B A-B A A B Namas Chandra Introduction to Mechanical engineering -B Hibbler Chapter 2-11 EML 3004C 2.3 Vector addition of forces If we consider, only two forces at a time then the result can be obtained using parallelogram law, and by using law of sines and cosines of triangles. Even if we have multiple (say 5) forces, we take two at a time to resolve the resultant one by one. Consider another example in Example 2.4 (a) page 24 Given F1, F2 and To find: Resultant Procedure: Use law of cosines to find the magnitude, law of sines find the angles Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-12 EML 3004C 2.3 Vector addition of forces (example 2.4) Given Resultant R 1kN , 30 To find: F1 and F2 Solution: 1. Draw the vector diagram 2. Use the sine law to find F1 and F2 F1 1000 F1 653N Sin30 Sin130 F2 1000 F2 446 N Sin20 Sin130 Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-13 EML 3004C 2.4 Cartesian Coordinate Systems A much more logical way to add/subtract/manipulate vectors is to represent the vector in Cartesian coordinate system. Here we need to find the components of the vector in x,y and z directions. Simplification of Vector Analysis z z en z k n A y j y x y i x Namas Chandra Introduction to Mechanical engineering x Hibbler Chapter 2-14 EML 3004C Vector Representation in terms unit vector Suppose we know the magnitude of a vector in any arbitrary orientation, how do we represent the vector? Unit Vector : a vector with a unit magnitude A A en A en A Namas Chandra Introduction to Mechanical engineering e^n A Hibbler Chapter 2-15 EML 3004C 2.5 Right-Handed Coordinate System z Right-Handed System If the thumb of the right hand points in the direction of the positive z-axis when the fingers are pointed in the x-direction & curled from the x-axis to the y-axis. Namas Chandra Introduction to Mechanical engineering y x Hibbler Chapter 2-16 EML 3004C Components of a Vector Cartesian (Rectangular) Components of a Vector In 3-D A Ax Ay Az z Az y A Ay A Ay Ax Ax Namas Chandra Introduction to Mechanical engineering x x y In 2-D A Ax Ay Hibbler Chapter 2-17 EML 3004C Cartesian Vectors Cartesian Unit Vectors Ax Ax iˆ A Ax iˆ Ay ˆj Ax kˆ Ay Ay ˆj Az Az kˆ z A z = A zk n k A = Aen en y i Ax = Axi Namas Chandra Introduction to Mechanical engineering x j Ay = Ayj Hibbler Chapter 2-18 EML 3004C Cartesian Vectors Magnitude of a Cartesian Vector A Ax2 Ay2 Az2 z Az A Ay Ax y x Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-19 EML 3004C 2.6. Addition and Subtraction of vectors Fx F cos x Fx cos x F Fy cos y F F cos z z F Fy F cos y Fz F cos z z Fz z F y Fx x Fy y x Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-20 EML 3004C Force Analysis 2 x 2 y z 2 z F = F +F +F Fz F = Fx ˆi+Fy ˆj+Fz kˆ z F = Fcosθ x ˆi+Fcosθ y ˆj+Fcosθ z kˆ ˆ F = F(cosθ x ˆi+cosθ y ˆj+cosθ z k) Û f = cosθ x ˆi+cosθ y ˆj+cosθ z kˆ ˆ F = FU f Namas Chandra Introduction to Mechanical engineering F y Fx Fy y x x Hibbler Chapter 2-21 EML 3004C Force Analysis Ex: d= x 2 +y 2 +z 2 = 62 +102 +82 =14.14ft 6 Fx =Fcosθ x =600 =254lb 14.14 10 Fy =Fcosθ y =600 =424lb 14.14 8 Fz =Fcosθ z =600 =339lb 14.14 ˆ ˆ ˆ F=255i+424j+339klb Namas Chandra Introduction to Mechanical engineering z F = 600 8ft y 6ft x 10ft Hibbler Chapter 2-22 EML 3004C Summary of the Force Analysis F Fx Fy Fz z Fx iˆ Fy ˆj Fz kˆ F cos x iˆ F cos y ˆj F cos z kˆ FZ Fx F cos x F Fy F cos y Fz F cos z x cos 1 Fx F y cos Fy Fx 1 Fy F x y F F Fx Fy Fz 2 2 2 Recall, Cos 2 x Cos 2 y Cos 2 z 1 Namas Chandra Introduction to Mechanical engineering Fz z cos F 1 Hibbler Chapter 2-23 EML 3004C 2.6 Addition/Subtraction of Cartesion vectors Since any vector in 3-D can be expressed as components in x,y,z directions, we just need to add the corresponding components since the components are scalars. A Axiˆ Ay ˆj Az kˆ B Bxiˆ By ˆj Bz kˆ Then the addition R AB R Ax Bx iˆ Ay By ˆj Az Bz kˆ Then the subtraction R A B R Ax Bx iˆ Ay By ˆj Az Bz kˆ Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-24 EML 3004C 2.6 ---Example 2-9, pg 41 Determine the magnitude and the coordinate direction angles of the resultant force on the ring Solution: FR F F1 F2 60 j 80k 50i 100 j 100k 50i 40 j 180k lb FR 50 40 180 2 2 2 191.0 lb Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-25 EML 3004C 2.6 ---Example 2-9…2 Unit vector and direction cosines uFR FR FR 50 ˆ 40 ˆ 180 ˆ 191.0 i 191.0 j 191.0 k 0.2617iˆ 0.2094 ˆj 0.9422kˆ From these components, we can determine the angles cos 0.2617 =74.80 cos 0.2094 =1020 cos 0.9422 =19.60 Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-26 EML 3004C 2.4---Problem 2-27 (pg 35) Four concurrent forces act on the plate. Determine the magnitude of the resultant force and its orientation measured counterclockwise from the positive x axis. Solution: Draw the free body diagram. Then resolve forces in x and y directions. FRx Fx ; FRy Fy ; Namas Chandra Introduction to Mechanical engineering 4 FRx 60 38cos30+ (50) 5 = 67.09 lb 3 FRy = 100 38 cos30 (50) 5 = 51 lb Hibbler Chapter 2-27 EML 3004C 2.4---Problem 2-27…..2 Find the magnitude and direction. Magnitude FR = FRx2 FRy2 = 67.092 512 = 84.3 lb Direction tan 1 FRy FRx = tan Namas Chandra Introduction to Mechanical engineering 1 51 = 37.2 67.09 Hibbler Chapter 2-28 EML 3004C 2.7. Position Vectors z r rB rA (xb, yb, zb) r rB xB iˆ y B ˆj z B kˆ rA x Aiˆ y A ˆj z A kˆ ( xa, ya, za) rB y rA x r xB x A iˆ y B y A ˆj z B z A kˆ •Position vectors can be determined using the coordinates of the end and beginning of the vector Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-29 EML 3004C 2.8---Problem 2-51, page 56 Determine the length of the connecting rod AB by first formulating a Cartesian position vector form A to B and then determining its magnitude. Solution Position vector: r = 16-(-5sin30) i + (0-5cos30)j = 18.5i - 4.330j in Magnitude: r 18.52 4.3302 19.0 in Namas Chandra Introduction to Mechanical engineering Ans Hibbler Chapter 2-30 EML 3004C Force Vector Along a Line A force may be represented by a magnitude & a position Force F is oriented along the vector AB (line AB) B F F u AB u AB z F Unit vector along the line AB A AB AB F F Namas Chandra Introduction to Mechanical engineering AB AB y x Hibbler Chapter 2-31 EML 3004C 2.6---Problem 2-36 (pg 47) The two forces F1 and F2 acting at the end of the pipe have a resultant force FR 120i N. Determine the magnitude and direction angles of F2 . Solution: Force Vector F1 = 85 cos 30sin45i + cos30cos45 j sin 30k = 52.052i + 52.252j - 42.5k N F2 = F2 x i + F2 y j + F2 z k N FR = 120i N FR =F1 +F2 120i = 52.052i + 52.052j - 42.5k F2 xi + F2 y j + F2 z k 120i = 52.052+F2 x i + 52.052+F2 y j+ F2 x 42.5 k Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-32 EML 3004C 2.6---Problem 2-36…..2 Equating i, j, k component yeilds: 52.052 + F2 x 120 Fx 67.948N 52.052+ F2 y 0 F2 y 52.052N F2 x 42.5 0 F2 x 42.5 N F2 67.948i-52.052j+42.5k N Magnitude of F2 F2 6.9487 2 (52.052) 2 42.52 = 95.56N = 95.6N Namas Chandra Introduction to Mechanical engineering Ans Hibbler Chapter 2-33 EML 3004C 2.6---Problem 2-36…..3 Coordinate direction angles: u F2 F2 67.948i-52.052j+42.05k F2 95.56 =0.7110i-0.5447j+0.4447k cos 0.7110 =44.7 Ans cos 0.5447 =123 Ans cos 0.4447 =63.6 Ans Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-34 EML 3004C 2.9. Dot Product Dot Product (Scalar Product) y A B=B A=ABcosθ A Ax A iˆ A(1) cos Ay Ax x A = A x i +A y j +A z k B B = Bx i +B y j +Bz k A B = A x Bx +A y B y +A z Bz A Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-35 EML 3004C 2.9---Problem 2-76, pg 65 A force of F = 80 N is applied to the handle of the wrench. Determine the magnitudes of the components of the force acting along the axis AB of the wrench handle and perpendicular to it. U F cos 30 sin 45i cos 30 cos 45 j+sin30k =-0.6124i + 0.6124j + 0.58k U AB j Hand first rotates 45 in the xy plane and 30 in that plane F = Fuˆ F 80(-0.6124i+0.6124j+0.5k) = (-48.990i+48.99j+40k) N FAB =F u AB ( 48.990i + 48.99j + 40k) (-j) =49.0 N Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-36 EML 3004C 2.9---Problem 2-76…..2 Negative sign indicates that FAB acts in the direction opposite to that of U AB 2 Fper . F 2 FAB = 802 (49.0) 2 63.2N Also, from problem 2-104 =128 FAB F cos 80cos128 49.0N Fper . F sin 80cos128 = 63.2N Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-37 EML 3004C Application of Dot Product Component of a Vector along a line A A cos ˆ AU A ˆ (A U ˆ )U ˆ A U A A A A A A sin A An Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-38 EML 3004C Application of dot product Angle between two vectors B A B AB cos AB cos AB 1 A Namas Chandra Introduction to Mechanical engineering Hibbler Chapter 2-39