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Ecor 1101 lec3

# Ecor 1101 lec3 - Lecture 02 Vectors Resultant&...

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Unformatted text preview: Lecture 02 Vectors Resultant & Components Sec\$ons 2.1 to 2.4 Pages 17 to 42 Prac\$ce Problems 2.3, 2.15, 2.30, F2.7, 2.34, 2.58 MES Assignment 01 Due 11 Jan 20101 Scalars & Vectors Scalar: Amount, quan\$ty, size. Single number Examples: mass, volume, length Vector: More than one characteris\$c or aKribute. Magnitude (Size) Direc\$on (Angle) Point of Applica\$on (Loca\$on) Sense (Tension/compression) Examples: Force, posi\$on, velocity, accelera\$on 2 Force as Vector Magnitude = Size Direc\$on = Angle P of A = Loca\$on Sense = T or C Line of Ac\$on Head = Sense Sense: Push or Pull ? Tail = Point of Applica\$on Length = Magnitude Direc\$on Angle Tension or Compression ? Referenc e 3 VECTOR ADDITION Parallelogram Law Triangle method (\$p to tail) 4 Sine/Cosine Law c A b C B Sin a Sin b Sin c = = A B C ! 2 2 a A = B + C " 2BC Cos a 5 2 Determine the magnitude of the resultant and its direc\$on CCW from the +x axis y Example 4 kN x 30o 10 kN 6 Solu=on y 10 60o 4 10 kN 4 kN R 30o x 30o R 120o Parallelogram Law Triangle of forces Sine Law Cosine Law 7 Triangle of Forces 10 120o 4 Cosine Law R 2 = 10 2 + 42 - 2(10)(4)Cos 120o R R = 156 = 12.5 kN Sine Law Sin 120o Sin = R 10 10 Sin = Sin 120o = 0.693 R = Sin-1 (0.693) = 43.9o 8 Components of a Vector 9 Resolu=on of a force into components v Fv F Fv Fu u 10 Example Find the components of the 500 lb force in the direc\$on of AB and AC for = 30o 11 Vector Force Characteris\$cs Magnitude Direc\$on P of A Sense T c S A a b Size Angle Loca\$on Ten/Comp Size Newtons or lbs Direc\$on Angle (degrees) Direc\$on Ra\$o a/b Loca\$on Point of contact Sense Push or Pull ? 2 2 L c = a +b b Sin A = c a Cos A = c 12 Special Case (Rectangular Components) Y Fx = F Cos Fy = F Sin Fy Fx F Tan = Fy Fx X 13 CARTESIAN VECTOR NOTATION `resolve' vectors into components using the x and y axes system. Each component of the vector is shown as a magnitude and a direc\$on. The direc\$ons are based on the x and y axes. We use the "unit vectors" i and j to designate the x and y axes. 14 The x and y axes are always perpendicular to each other. Together, they can be directed at any inclina\$on. ! ! ! F = FX i + FY j ! "FX % F =# & \$ FY ' F= 2 FX + 2 FY * FY (X = tan , / + FX . )1 15 Resultant of multiple vectors Step 1 resolve each force into its x and y components F1x = F1 Cos "1x F2x = F2 Cos "2x F3x = F3 Cos "3x F1y = F1 Sin "1x F2y = F2 Sin "2x F3y = F3 Sin "3x 16 Step 2 add all the x components together and add all the y components together. These two totals become the resultant vector. R x = " Fix R y = " Fiy i=1 i=1 3 3 ! 17 Step 3 find the magnitude and angle of the resultant vector. y R= Rx R R Ry x 2 Rx + #1 2 Ry \$ Ry ' "R = tan & ) % Rx ( ! 18 EXAMPLE Find the resultant Plan: a) Resolve the forces in their x-y components. b) Add the respec\$ve components to get the resultant vector. c) Find magnitude and angle from the resultant components. 19 EXAMPLE F1 = { 15 sin 40 i + 15 cos 40 j } kN = { 9.642 i + 11.49 j } kN F2 = { -(12/13)26 i + (5/13)26 j } kN = { -24 i + 10 j } kN F3 = { 36 cos 30 i 36 sin 30 j } kN = { 31.18 i 18 j } kN 20 Summing up all the i and j components respec\$vely, we get, FR = { (9.642 24 + 31.18) i + (11.49 + 10 18) j } kN = { 16.82 i + 3.49 j } kN y FR EXAMPLE FR = ((16.82)2 + (3.49)2)1/2 = 17.2 kN = tan-1(3.49/16.82) = 11.7 x 21 ...
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