· Ax= Part II: Vectors À (Ax,y) -(b₁₁by) = A + B = ((x,y) Cx = Ax+Bx = Ay+By 1. A vector A has a magnitude of 10.0 m and points in a direction that makes an angle of 30.0° above the positive x-axis. A second vector B, has a magnitude of 15.0 m and points in a direction 60.0" below the negative x-axis. a) Draw the two vectors and their components. Al cos 30°: 8.66m Ay = 141 sing05m Bx=-181 (05600-7.5m By=-161i0-12m Ax if @ & IAI given A=(Ax, Ag) if given Ax A IATA A Ay tong = 10m A 230° 60° x Ax = Al-case A: AI-sine 15m B b) Find the magnitude of the vector's components. COS300 A 10m (10m) cos 30° 8.66 - A A sin 500= 0 10 m (10m) Sin (30) = 0 S 0 Sin 60° 15m (15m) sin (60°) O -12.99m-0 C0160°= 4. (15m) Cos(60) A -7.5=A Ax = 8.66m Ay = 5.0m By = -7.5m By= -133 c) Using the component method to find the magnitude (II) and direction (Dir()) of the resultant vector C = A + B. C = ((x, y) Cy: Ay+By S+ (-13) -8 Cx = Ax+Bx (x 8.66m+(-7.5m) = 1.16 (1.16)²+(-8)2 65.3456 0 -8 8.08 - 8.08 1.16 tan (11) = = 0 |c|= 8.08m Dir()=87 below tx

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Chapter3: Vectors
Section: Chapter Questions
Problem 3.41P: Express in unit-vector notation the following vectors, each of which has magnitude 17.0 cm. (a)...
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· Ax=
Part II: Vectors
À (Ax,y)
-(b₁₁by)
= A + B = ((x,y)
Cx = Ax+Bx
= Ay+By
1. A vector A has a magnitude of 10.0 m and points in a direction that makes an angle of 30.0° above
the positive x-axis. A second vector B, has a magnitude of 15.0 m and points in a direction 60.0" below
the negative x-axis.
a) Draw the two vectors and their components.
Al cos 30°: 8.66m
Ay = 141 sing05m
Bx=-181 (05600-7.5m
By=-161i0-12m
Ax
if @ & IAI given
A=(Ax, Ag)
if given
Ax A
IATA A
Ay
tong
=
10m
A
230°
60°
x
Ax = Al-case
A: AI-sine
15m
B
b) Find the magnitude of the vector's components.
COS300
A
10m
(10m) cos 30°
8.66 - A
A
sin 500=
0
10 m
(10m) Sin (30) = 0
S
0
Sin 60°
15m
(15m) sin (60°) O
-12.99m-0
C0160°=
4.
(15m) Cos(60) A
-7.5=A
Ax =
8.66m
Ay =
5.0m
By
=
-7.5m
By=
-133
Transcribed Image Text:· Ax= Part II: Vectors À (Ax,y) -(b₁₁by) = A + B = ((x,y) Cx = Ax+Bx = Ay+By 1. A vector A has a magnitude of 10.0 m and points in a direction that makes an angle of 30.0° above the positive x-axis. A second vector B, has a magnitude of 15.0 m and points in a direction 60.0" below the negative x-axis. a) Draw the two vectors and their components. Al cos 30°: 8.66m Ay = 141 sing05m Bx=-181 (05600-7.5m By=-161i0-12m Ax if @ & IAI given A=(Ax, Ag) if given Ax A IATA A Ay tong = 10m A 230° 60° x Ax = Al-case A: AI-sine 15m B b) Find the magnitude of the vector's components. COS300 A 10m (10m) cos 30° 8.66 - A A sin 500= 0 10 m (10m) Sin (30) = 0 S 0 Sin 60° 15m (15m) sin (60°) O -12.99m-0 C0160°= 4. (15m) Cos(60) A -7.5=A Ax = 8.66m Ay = 5.0m By = -7.5m By= -133
c) Using the component method to find the magnitude (II) and direction (Dir()) of the resultant vector
C = A + B.
C = ((x, y)
Cy: Ay+By
S+ (-13)
-8
Cx = Ax+Bx
(x 8.66m+(-7.5m) = 1.16
(1.16)²+(-8)2
65.3456
0
-8
8.08
- 8.08
1.16
tan (11) =
= 0
|c|=
8.08m
Dir()=87 below tx
Transcribed Image Text:c) Using the component method to find the magnitude (II) and direction (Dir()) of the resultant vector C = A + B. C = ((x, y) Cy: Ay+By S+ (-13) -8 Cx = Ax+Bx (x 8.66m+(-7.5m) = 1.16 (1.16)²+(-8)2 65.3456 0 -8 8.08 - 8.08 1.16 tan (11) = = 0 |c|= 8.08m Dir()=87 below tx
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