D_-_3D_Cartesian_Vector

D_-_3D_Cartesian_Vector - FORCE VECTORS Cartesian Vectors...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: FORCE VECTORS Cartesian Vectors (3-Dimensional Vectors) LEARNING OBJECTIVES Be able to resolve each force into its rectangular components Fx, Fy and Fz. Be able to represent the components of force in terms of the Cartesian unit vectors i, j and k. Be able to determine 3-dimensional force resultants using the Cartesian unit vectors. PRE-REQUISITE KNOWLEDGE Units of measurements. Trigonometry concepts. Vector concepts. RIGHT-HANDED COORDINATE SYSTEM A denotes a vector whereas A denotes the magnitude of A RECTANGULAR COMPONENTS OF FORCE VECTOR A=A +A +A Magnitude of Ax : Ax = A cos Magnitude of Ay : Magnitude of Az : Ay = A cos Az = A cos The Directional Cosines are Cos = Ax/A Cos = Ay/A Cos = Az/A CARTESIAN UNIT VECTORS i : the Cartesian unit vectors in x-direction j : the Cartesian unit vectors in y-direction k : the Cartesian unit vectors in z-direction Z 1 unit 1 unit k j Y i X 1 unit 3-D FORCE RESULTANTS FR = F FR = FRx i + FRy j + FRz k FRx = Fx FRy = Fy F = F R 2 FRz = Fz 2 RX +F 2 RY +F 2 RZ + cos2 + cos2 = 1 cos DIRECTIONAL VECTOR The directional vector of the resultant force FR is a unit vector, uF, that indicates the direction of FR FR uF 1 unit FR = F R u F FR FR = ( ) uF = FRX FR i+ FRY FR 2 j+ FRZ FR k ; and , and are called the coordinate direction angles u F = cos i + cos j + cos k FR = FRX + FRY + FRZ and u F = COS + COS + COS = 1 2 2 2 2 2 EXAMPLE 1 Determine the magnitude FR and the coordinate direction angles of the resultant force FR SOLUTION 1 Resolve F1 (the 75 lb force): F1 = 75[0.0i - (24/25)j + (7/25)k F1 = -72.00j + 21.00k Resolve F2 (the 55 lb force): F2 = 55[(cos30)(cos60)i + (cos30)(sin60)j - (sin30)k] F2 = 23.82i + 41.25j 27.50k FR = F1 + F2 = 23.82i + (41.25 72.00)j + (21.00 - 27.5)k = 23.82i - 30.75j - 6.50k SOLUTION 1 - continued We know that FR = F1 + F2 = 23.82i -30.75j -6.50k FR = [23.822 + (-30.75)2 + (6.5)2]0.5 = 39.43 lb. = cos-1(23.82/39.43) = 52.8o; = cos-1(-30.75/39.43) = 141o; and = cos-1(-6.50/39.43) = 99.5o EXAMPLE 2 The pole is subjected to force F whose components are Fx=1.5 kN and Fz=1.25 kN. Assume the direction coordinate angle =75o. Determine the magnitudes (F and Fy) of F and Fy. SOLUTION 2 Use Cos2 + cos2 + cos2 = 1 Given Fx=1.5 kN, Fz=1.25 kN, and =75o 1.5 1.25 2 o + cos 75 + =1 F F F = 2.02kN Fy = F cos 75 = 0.523kN 2 2 ...
View Full Document

This note was uploaded on 07/26/2009 for the course CE 221 taught by Professor Buch during the Spring '08 term at Michigan State University.

Ask a homework question - tutors are online