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ECOR 1101 (52)
Lecture 3

# ECOR1101 - Lecture 3.pdf

19 Pages
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Department
Engineering Common Core Courses
Course Code
ECOR 1101
Professor
Jack Vandenberg

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Lecture 3 ECOR1101 Mechanics I ECOR1101 – Mechanics I 1 Cartesian Vectors in 3D • 3D vectors are best represented in Cartesian vector notation • 3D Coordinate System: – Right-handed CS – Thumb  +ve z-axis – Fingers curled around z- axis, sweeps from x-axis to y-axis • Cartesian Unit Vectors – Cartesian unit vectors i, j, k designate vectors in the u  U respectively.tions U – The positive directions of shown in fig.ors are as ECOR1101 – Mechanics I 2 Cartesian Vector Representation z • Resolution of A into the A z Cartesian unit vectors will require two successive A application of parallelogram law A  A  A k k z A  A xi Ay j A i i j x A j A  A x  A y  A z y x • Magnitude of Cartesian vector A A  A2 A y z 2 2 A  Ax y A  A x A yA2 z ECOR1101 – Mechanics I 3 Cartesian Vector Representation z Direction of Cartesian Vector A k • The direction of the z Cartesian vector A is A defined by the angles , , and  it makes with the x, y,  and z axes, respectively. Ax   cos    A y A  y cos   AyDirection cosinesof A A  A x Az cos    A  x ECOR1101 – Mechanics I 4 Cartesian Vector Representation • A vector A can be represented using unit vectors as: A = Au whA,e u is a unAt vector in the direction of A A A A x y A z u A   i  j  k A A A A u A  cos  i  cos  j  cos k • Since the magnitude of a unit vector is 1; 2 2 2 uA  cos   cos   cos   1 2 2 2 cos   cos   cos   1 – Therefore if any two coordinate angles (direction cosines) are known we can find the third. ECOR1101 – Mechanics I 5 Cartesian Vector Representation z • The direction of A can be found from two angles,  and  A (altitude and azimuth) z A zAcos A A =Asin  altitude A xAcos  Asin cos A =Asin  Asin sin y A Asin cos i + Asin sin j + Acos k A y y Ax  A ECOR1101 – Mechanics Izimuth 6 Addition of Cartesian Vectors • Given two vectors A and B; A = A ix+ A jy+ A k z B = B ix+ B jy+ B k z • We can add A and B using Cartesian components as follows: R = A + B = (A +B xi +x(A +B )jy+ (Ay+B )k z z • We can also subtract A and B using Cartesian components as follows: R = A – B = (A – Bx)i +x(A – B )y + (y – B )k z z • General formulation R = F = F i +xF j + yF k z ECOR1101 – Mechanics I 7 Sample Problem F2-18 Determine the resultant force acting on the hook. Solution Procedure 1) Using geometry and trigonometry, write 1 and F2in Cartesian vector form. 2) Then add the two forces (by adding x and y components of the forces). ECOR1101 – Mechanics I 8 Resolve force F . 1 F 1x0 = 0 lb F = 500 (4/5) = 400 lb 1y F 1z500 (3/5) = 300 lb Write F in Cartesian vector form (don’t 1 forget the units!). F = {0i + 400j + 300k} lb 1 ECOR1101 – Mechanics I 9 Resolve force F . 2 We are given only two angles,  and . So we need to resolve F into the z-axis and the xy-plane. 2 o o F2 z F 2in 45  800  sin 45  565 .69 lb. F  F cos 45 o  800  cos 45 o  565 .69 lb. 2 xy 2 F  F cos 30 o  565 .69  cos 30 o  489 .90 lb.
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