Introduction to Vectors. Vectors ScalarsVectors Distance Time Mass Work Energy Speed Displacement...
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Transcript of Introduction to Vectors. Vectors ScalarsVectors Distance Time Mass Work Energy Speed Displacement...
Introductionto Vectors
x
x
yy
z
z
Vectors
Scalars Vectors
Distance
Time
Mass
Work
Energy
Speed
Displacement
Velocity
Acceleration
Force
Momentum
Torque
MagnitudeOnly
Magnitudeand
Direction
Ar
A
A
A
A
Vectors have magnitude and direction,but no place
Vectors
Vectors
A
B
Two vectors are equal if they have the same magnitude
and the same direction
BA
Vectors
Scaling Vectors
A
A2
A2
1
Vectors
Scaling a Displacement Vector
d
td v
Vectors
Scaling a Change in Velocity Vector
v
tv a
Vectors
Vector Addition (Graphical)
ABBB
BAC
Vectors
Vector Subtraction (Graphical)
A B B
Vectors
Vector Subtraction (Graphical)
A B
Vectors
Vector Subtraction (Graphical)
A
B
C BAC
BAC
The Pythagorean theorem is a useful method for determining the result of adding two (and only two) vectors that make a right angle to each other.
A
x
y
xA
yA yx AAA
The parallelogram method of vector resolution involves using an accurately drawn, scaled vector diagram to determine the components of the vector.
A
x
y A A
cosAAx
sinAAy
The trigonometric method of vector resolution involves using trigonometric functions to determine the components of the vector.
Vectors
x
A A
A
y
yA
xA
2y
2x
2 AAA
2y
2x AAA
x
y1
A
Atan
Vector Components
A
BR
AxBx
Rx
Ay
By
Ry
x
y
2y
2x RRR
xxx BAR
yyy BAR
Adding Vectors using Vector Components
2y
2x RRR
xxx BAR
yyy BAR
Adding Vectors using Vector Components
A
BR
Ax Bx
Rx
Ay
By
Ry
x
y
α
β
βcosBB
αcosAA
x
x
βsinBB
αsinAA
y
y
Vectors
F2
2
F3
3 F4
4
n
knR FF
1
4321
FFFFF R
F1
1
F1F2
F3
F4
1
2
3 4
F1 = 50 N 1 = 30o
F2 = 100 N 2 = 135o
F3 = 30 N 3 = 250o
F4 = 40 N 4 = 300o
θsinF θcosF kkkk
43.3 25.070.7 70.710.3 28.220.0 34.6
17.7 32.9
22R 9.327.17F
N 4.37FR
Adding Vectors using Vector Components
Vectors
22R 9.327.17F
N 4.37FR
7.17
9.32tan
7.17
9.32tan 1 o7.61
FR = 37.4 N
R
17.7
32.9
180R 7.61180
oR 118
Vectors
F1F2
F3
F4
1
2
3 4
F1 = 90 N 1 = 45o
F2 = 80 N 2 = 150o
F3 =50 N 3 = 220o
F4 = 70 N 4 = 340o
θsinF θcosF kkkk
63.669.338.365.8
63.640.0
23.923.9
21.8 47.6
22R 6.478.21F
N 4.52FR
8.216.47
tan
8.216.47
tan 1R
oR 4.65
FR = 52.4 N
R
47.6
21.8