C_-_2D_Coplanar_Vector

C_-_2D_Coplanar_Vector - FORCE VECTORS FORCE LEARNING...

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Unformatted text preview: FORCE VECTORS FORCE LEARNING OBJECTIVES LEARNING Be able to resolve each force into its Be rectangular components Fx and Fy. rectangular Be able to represent the components of force Be in terms of the Cartesian unit vectors i and j. in Be able to determine coplanar force Be resultants using the Cartesian unit vectors. resultants PRE-REQUISITE KNOWLEDGE PRE-REQUISITE Units of measurements. Trigonometry concepts. Vector concepts. REVIEW (SI UNITS) REVIEW Basic Quantities Length is measured in meter (m) Length Time is measured in second (s) Time Mass is measured in kilogram (kg) Mass Force is measured in Newton (N) REVIEW (FPS) REVIEW Basic Quantities Length is measured in foot (ft) Time is measured in second (s) Force is measured in pound (lb) Mass is measured in slug REVIEW REVIEW Vector Addition Vectors A and B can be added using the parallelogram Vectors law or triangle construction (head to tail) law A A B B R=A+B R=A+B Triangle construction Parallelogram Law REVIEW REVIEW What is the magnitude (FC) of the force FC? FC FA= 35 N Use the cosine law 135 FB=60 N FC2 = FA2 + FB2 - 2(FA)(FB)(cos(135)) A) 43.1 N B) 60.7 N C) 88.3 N D) 95.0 N REVIEW REVIEW Determine the magnitude of the resultant force 4 kN 30o 10 kN Use the cosine law R = 42 + 102 - 2(4)(10)(cos(150)) R R = A) 6.84 kN B) 8.72 kN C) 10.5 kN D) 12.5 kN RECTANGULAR COMPONENTS OF A FORCE VECTOR OF Y F = FX + FY F X = F cos θ FY = F sin θ F θ Fx X Fy CARTESIAN UNIT VECTORS CARTESIAN In the diagram below, i denotes the Cartesian unit vector in the x-direction whereas j denotes the Cartesian unit vector in the y-direction Y 1 unit j i 1 unit X USES OF CARTESIAN UNIT VECTORS VECTORS Y FY F FX F = FX + FY OR F = FXi + FYj F1 = 5i + 2j F F X F2 = 5i − 4j R F3 = −6i The resultant R F R = 4i − 2j COPLANAR FORCE RESULTANTS COPLANAR FR = ∑ F Y FR = FRX i + FRY j F =F 2 R 2 RX FRX = ∑ F X ; FRY = ∑ FY +F 2 RY FRy FR θ FRx X −1 F θ = tan RY F RX COPLANAR FORCE RESULTANTS COPLANAR Y Example F1 = 5 i + 2 j F2 = 5 i - 4 j F3 = -6 i X FR = F1 + F2 + F3 FRx = i(5+5–6)= 4i FRy = j(2 – 4) = -2j F F θ F FR = 4i − 2j FR = 4 + ( − 2 ) 2 2 FR o −2 = 4.47; θ = tan = 333.4 4 −1 REVIEW QUESTION Resolve F along the x and y axes and write it in vector form. F = { ___________ } N y F = 80 N 30° x A) 80 cos (30°) i - 80 sin (30°) j B) 80 sin (30°) i + 80 cos (30°) j D) 80 cos (30°) i + 80 sin (30°) j C) 80 sin (30°) i - 80 cos (30°) j REVIEW QUESTION Determine the magnitude of the resultant vector FR = (F1 + F2) in N when F1 = {10i + 20j } N and F2 = { -20i - 20j } N A) 30 N B) 20 N C) 10 N D) 40 N E) 50 N F1 = {10i + 20j } N F2 = { -20i - 20j } N F1 + F2 = FR = {-10i + 0j}N FR = [(-10)2 + 02]0.5 = 10N EXAMPLE 1 EXAMPLE Determine the magnitude and the direction θ of F1 to lift the 800 N box vertically upward SOLUTION 1 SOLUTION Y 800 N 600 N 3 FW2 5 Draw a free body diagram (FBD) F θ 30 FW1 X 400 N β 4 FW =800 N First, because of symmetry FW1 = FW2; FW = FW1 + FW2 and FW = 800 N SOLUTION 1 SOLUTION Y 800 N 600 N 3 FW2 5 Draw a free body diagram (FBD) F θ 30 FW1 X 400 N β 4 FW =800 N Since the box will be lifted vertically, the components of the resultant force must satisfy F RX = 0.0 N and F RY = 800 N SOLUTION 1 - continued continued Y 800 N 600 N 3 FW2 5 F θ 30 FW1 X 400 N β 4 β = tan-1 (3/4) = 36.9o FW =800 N FRX = 400cos(30) – 600cos(36.9) + F1sin(θ) = 0.0 FRY = 400sin(30) + 600sin(36.9) + F1cos(θ) = 800 SOLUTION 1 Y continued - continued 800 N 600 N 3 5 F θ 30 FW1 X 400 N β 4 F1sin(θ ) = 133.4 N F1cos(θ ) = 239.7 N 2 FW2 FW =800 N F1 (sin 2 θ + cos 2 θ ) = 133.4 2 + 239.7 2 = 75251.65 N 2 F1 = 274.3 N; sin(θ ) = 133.4/274.3 = 0.486; θ = 29o EXAMPLE 2 EXAMPLE Determine the magnitude F of the force F so that the resultant FR of the three forces is as small as possible SOLUTION 2 SOLUTION Solution steps Draw a free body diagram Calculate FRx and FRy Calculate Calculate FR Calculate Differentiate FR with respect to F Differentiate Set dFR/dF equal to zero and solve for FR. Check that the second derivative Check (d2FR/dF2) > 0.0 (d SOLUTION 2 SOLUTION Step 1 – Free Body Diagram Y 20 kN 3 4 45 o X F 12 kN SOLUTION 2 SOLUTION FRx = 20(4/5) - F cos(45) = 16.0 - 0.707F FRy = 20(3/5) - 12 + Fsin(45) = 0.707F FR2 = (FRx )2 + (Fry)2 FR2 = F2 - 22.63F + 256; differentiating 2FR(dFR/dF) = 2F – 22.63 = 0.0; (1) F = 11.31 kN FR2 = (11.31)2 - 22.63 (11.31) + 256 FR = 11.31 kN Substituting in equation 1 yields: 22.62(dFR/dF) = 2F – 22.63; Differentiating 22.62(d2FR/dF2) = 2 (d2F the minimum 0.0 (d Hence, F = 11.31 kN producesR/dF2) = 0.0884 >resultant force. Hence, ...
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This note was uploaded on 10/31/2010 for the course CEE CE 221 taught by Professor Baladi during the Fall '10 term at Michigan State University.

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