D_-_3D_Cartesian_Vector

# D_-_3D_Cartesian_Vector - FORCE VECTORS Cartesian...

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

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

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern