VECTORS v Fp Scalar quantities – that can be completely described by a number with the appropriate...
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Transcript of VECTORS v Fp Scalar quantities – that can be completely described by a number with the appropriate...
VECTORVECTORSS v
Fp
Scalar quantities – quantities that can be completely described by a number with the
appropriate units. (They have magnitude only. )Such as length, mass, speed, energy, temperature…
{notation : Δd, m, v, E, to …}
Vector quantities – quantities that require both a magnitude AND a direction for their
complete description. Such as displacement, velocity, force… {Notation : Δd, v, F… }
In a diagram, vectors are represented by arrowswhich point in the direction of the vector and whose length represents the magnitude.
TRIGONOMETRY REFRESHERTRIGONOMETRY REFRESHERANATOMY OF A RIGHT TRIANGLE
Hypotenuse (HYP)
θReference angle
(θ, THETA)
Right angle Adjacent (ADJ)
Opposite(OPP)
TRIGONOMETRY REFRESHERTRIGONOMETRY REFRESHERTRIGONOMETRIC FUNCTIONS
θ
HYPOPP
ADJ
cos ADJ
HYPcos ADJ
HYP tan OPP
ADJtan OPP
ADJ
SOH CAH TOA
sin OPP
HYP
TRIGONOMETRY REFRESHERTRIGONOMETRY REFRESHERIII
IIIIV
All are +Sin is +
Cos is +Tan is +
0o
90o
180o
270o
360o
For a vector, V, when θ is to the positive x-axis :
θ
V
Vx
Vy
sin yVV
cos xVV
tan y
x
VV
Resolving vectors into their Resolving vectors into their componentscomponents
{finding the x- and y components of a vector}{finding the x- and y components of a vector}
To find the x-component
To find the y-component
cos xVV
sin yVV Vy = V sin θ
Vx = V cos θ
Sample problemsSample problems Resolve this velocity into its components.Resolve this velocity into its components.
V = 15 m/s at 60.0o
V = 15 m/s
θ = 60.0o
GIVEN V
θ
Vy
Vx
Vy = V sin θ
Vy = 15 m/s sin 60.0o
Vx = V cos θ
Vx = 15 m/s cos 60.0o
Vx = + 7.5 m/s Vy = + 13 m/s
Sample problemsSample problems Resolve this velocity into its components.Resolve this velocity into its components.
V = 80.0 m/s at 140.0o
θV
Vy
Vx
V = 80.0 m/s
θ = 140.0o
GIVEN
Vy = V sin θ
Vy = 80.0 m/s sin 140.0o
Vy = + 51.4 m/s
Vx = V cos θ
Vx = 80.0 m/s cos 140.0o
Vx = -61.3 m/s
GRAPHIC RESOLUTIONGRAPHIC RESOLUTIONUse a ruler and protractor to draw the two previous vectors to scale. Graph paper is helpful in establishing axes but it is not needed to complete this assignment.
Draw the perpendicular to the x-axis and label the components.
Measure the components in cm and use your scale to convert back to the appropriate units.
GRAPHIC RESOLUTIONGRAPHIC RESOLUTIONFor V = 15 m/s at 60.0o, if I let 1 cm = 0.5 m/s ……
I draw V 30 cm long at 60.0o Vy
Vx
Draw the perpendicular to the x-axis
Label the components
Measure the components and convert
V
θ
xm scmV cm1550 0 5
1. ( ). / = +7.750 m/s
y
m s
cmV cm26 25 0 5
1. ( ). / = +13.12 m/s
15 301
0 5m s cmcm
m s/ ( )
. /
Your turn
Do both vectors and submit
TRIGONOMETRIC VECTOR TRIGONOMETRIC VECTOR ADDITIONADDITION
(FINDING RESULTANTS)(FINDING RESULTANTS)1. Resolve vectors into x- and y-components.
2. Add all of the x-components to find the x-component of the resultant.
3. Add all of the y-components to find the y-component of the resultant.
4. Use tan-1 (arctan) to find the direction, θR.
The signs of the components determine the quadrant.
5. Use cos θR or sin θR to find the magnitude.
θrefθref
θref θref 0o
360o
270o
180o
90o
θR = θrefθR = 180o - θref
θR = 180o + θrefθR = 360o - θref
θR
θRθR
θR
III
IIIIV
ADVANCE TO SAMPLE PROBLEM
SAMPLE PROBLEM (from sheet)SAMPLE PROBLEM (from sheet)
V1 = 320 N at 240.0oV1 = 320 N at 240.0o V2 = 35 N at 30.0oV2 = 35 N at 30.0o
R = V1 + V2Given
V1 = 320 N
V2 = 35 N
θ1 = 240.0o
θ2 = 30.0o
θ2θ1
V1
V2
Rx = V1x + V2x
R = V1 + V2
Rx = V1 cos θ1 + V2 cos θ2
Rx = 320 N cos 240.0o + 35 N cos 30.0o
Rx = (-160 N) + (+30 N)
Rx = -130 N
Find the x-component
From
R = V1 + V2
Ry = V1y + V2y
Ry = V1 sin θ1 + V2 sin θ2
Ry = 320 N sin 240.0o + 35 N sin 30.0o
Ry = (-280 N) + (+18 N)
Ry = -260 N
Find the y-component
From
Rx = -130 N
Ry = -260 N
R = Rx + Ry
Rx
RyRy
R
θRDraw the components and resultant
Find the direction, θR
Tan RRR
y
x
Using
R
y
x
R
R 1
tan ( )
1 260130tan ( )N
N
Rx
RyRy
R
θR
Your calculator only gives youthe reference angle to the x-axis
θref = 63o
θref
Since the x- and y-components are both negative,R is in Quadrant III 63o beyond 180o on the x-axis.
θR = 180o + 63o
θR = 243o
FromR
NN
1 260
130tan ( )
Rx
RyRy
R
θR
θref
Find the magnitude, R
--- using the sin θR
R N
o
260
243sin= 290 N
R
yRR
sin
R Ry
R
sin
Rx
RyRy
R
θR
θref
Find the magnitude, R
--- using the cos θR
R N
o
130
243cos= 290 N
RxRR
cos
R Rx
R
cos
Rx
RyRy
R
θR
θref
Find the magnitude, R
--- using pythagorean theorem
2 2 2
R R Rx y
R R Rx y 2 2
R N N 2 2
130 260( ) ( )
R = 290 N
Rx
RyRy
R
θR
θref
And so, the final answer is
R = 290 N at 243o
Do all of the practice problems on the sheetto prepare yourself for your next quiz.
GRAPHIC VECTOR ADDITIONGRAPHIC VECTOR ADDITIONWhen adding vectors graphically :
1. Draw the vectors to scale.2. Move one of the vectors so that it is “head-to-toe” with the other. (be sure to maintain its magnitude and direction)
3. Draw the resultant from the origin to the head of the last vector you moved.
4. Measure the resultant and use your scale to convert back to the appropriate units.
5. Use a protractor to measure the direction.
GRAPHIC VECTOR ADDITIONGRAPHIC VECTOR ADDITION
Consider a plane flying with a velocity of 220 m/sat 170o in a wind blowing at 45 m/s at 270o.What is the ground velocity of the plane?
The problem is to find the resultant, Vg , of the velocity of the plane, Vp ,in the air and the velocity of the wind, Vw.
Vg =Vp + Vw
Vp = 220 m/s at 170o Vw = 45 m/s at 270o
GRAPHIC VECTOR ADDITIONGRAPHIC VECTOR ADDITION
Vp
VwVwVg
Let 1 cm = 10 m/sVp = 220 m/s (1cm /10 m/s) = 22 cm
Vw = 45 m/s (1cm /10 m/s) = 4.5 cmVg = 21.70 cm (10m/s / 1cm) = 217.0 m/s
The protractor measures 3o beyond 180o so θg =183o
θp
θg
BONUS PROBLEMBONUS PROBLEMA pilot wishes to maintain a ground velocityof 150.0 m/s at 270o and the wind is blowingat 50.0 m/s at 120o. What air speed must hehave and at what bearing (direction)?
Hint : This is actually a vector subtraction.Just as in math, subtraction is the same asadding a negative. A negative vector pointsin the opposite direction. Add or subtract180o to the direction of the vector to be subtracted and then add as usual.
(solve both graphically and trigonometrically to earn maximum credit)
THIS IS THE END OF VECTORS