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Class Notes
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Canada
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Carleton University
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ECOR 1101
(52)

Jack Vandenberg
(13)

Lecture 3

School

Carleton University
Department

Engineering Common Core Courses

Course Code

ECOR 1101

Professor

Jack Vandenberg

Description

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 A2 A y
z
2 2
A Ax 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 AyDirection 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 xAcos Asin cos
A =Asin 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 +xF 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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