Satu tahun cahaya adalah jarak yang ditempuh cahaya dalamruang hampa selama satu tahun. Jika kecepatan cahaya 3 x 108

m/s dan satu tahun sama dengan 365,25 hari. Berapakah Panjang satu tahun cahaya dinyatakan dalam Mm ?

Jawab:

365,25 hari = 365,25 x 86400 s = 31.557.600 s

Jarak yg ditempuh cahaya:

31.557.600 x 3 x108 = 94.672.800 x 108 m

= 9,47 x109 Mm

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▪ Tebal satu lembar kertas kurang dari satu millimeter. Kalian memiliki penggaris dengan skala terkecil satu millimeter. Kalian diminta mengukur tebal selembar kertas denganmenggunakan penggaris tersebut. Dapatkah pengukuran tersebut dilakukan? Jelaskan jawaban kalian.

Jawab:

Tidak bisa. Karena benda yang diukur (kertas) memiliki tebal lebih kecil dari NST alat ukur.

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Jika 1 inci = 2,54 cm tentukan

▪ a) Diameter Velg = 2,54 x 16 = 40,64 cm

▪ b) Diameter luar ban =2(0,4x20,5)+40,64 = 57.04 cm

▪ c) Tebal ban (antara velg sampai, diameter terluar) = 0,4x20,5=8,2 cm

▪ d) Jumlah putaran ban setelahkendaraan berpindah sejauh 500 m. = 50.000/( π x 57.04)= 279 putaran

▪ e) Tentukan tebal ban, diameter luarban dan diameter luar velg jika tertulis180/60/14

Tebal = 0,6x18 = 10,8 cm

D luar ban = 2x10,8 + (2,54x14) =57.16cm

D luar velg = 2,54 x 14 = 35,65 cm

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by

Dr. Eng. Pribadi Mumpuni Adhi

▪ Vector

▪ symbol: A or Ԧ𝐴

▪ Quantity that has magnitude and direction

▪ Required a vector algebra

▪ In diagram: denoted by arrows

▪ Length of arrows: proportional to the magnitude of vector

▪ Arrowhead: indicates direction

▪ Scalar▪ symbol: A

▪ Quantity that has only magnitude.

▪ Required an ordinary algebra

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6

A

B

B’

A’

B”

A”

The arrows from A to B, from A’ to B’, and from A” to B” have the same

magnitude and direction . Thus, they specify identical displacement

vectors and represent the same change of position for the particle. A

vector can be shifted without changing its value if its length and

direction are not changed

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B

A

The displacement vector tells us nothing about the actual path that the

particles takes. All three paths connecting points A and B correspond to

the same displacement vector as shown before. The displacement

vectors represent only the overall effect of the motion, not the motion

itself.

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A particle moves from A to B, then later from B to C

A

B

C

Actual path

Net displacement is the vector sum

Ԧ𝐴 𝐵

𝑅

𝑅 = Ԧ𝐴 + 𝐵

R=A+B

magnitude and

direction of

vector can be

directly

measured

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Tail-to-head

10

A + B = B + A (commutative law)

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(A + B)+ C = A + (B + C) (associative law)

1. a vector if multiplied by -1, magnitude is the same, but the opposite

direction by 180o.

2. Subtraction of vector follow the addition of vector operation.

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Two ways to resolve the A

1. A=Ax + Ay

2.

▪ A vector A with its components :

Ax and Ay is perpendicular with

each other.

▪ Scalar components:

▪ Ax=A cos q

▪ Ay=A sin q

=

+=

−

x

y

yx

A

A

AAA

1

22

tanq

13

Direction of components of

vectors depend on the

coordinate that used

A =Ax + Ay

or

A =A’x + A’y

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15

A car travels 20.0 km due north and then 35.0 km in a direction 60.0° west of north, asshown in the Figure. Find the magnitude and direction of the car’s resultantdisplacement.

Answer:

R = A + B

The magnitude of R can be find by using the law of cosines

θ = 1800 – 600 = 1200

𝑅2 = 𝐴2+ 𝐵2 − 2𝐴𝐵 cos 𝜃

𝑅 = 𝐴2+ 𝐵2 − 2𝐴𝐵 cos 𝜃

= 𝟒𝟖. 𝟐 𝐤𝐦

The direction of R measured from the northerly direction can be obtained from the law of sines

The resultant displacement of the car is 48.2 km in a direction38.9° west of north.

sin 𝛽

𝐵=sin 𝜃

𝑅β = 38.9°

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A small airplane leaves an airport on an overcast day and is later sighted 215 kmaway, in a direction making an angle of 22° east of due north. How far east andnorth is the airplane from the airport when sighted?

Answer:

y

x

22°

θ

𝜃 = 90° − 22° = 68°

𝑑𝑥 = 𝑑 cos 𝜃

𝑑𝑦 = 𝑑 sin 𝜃

= 81 km

= 199 km

Thus the airplane is 81 km east and 199 km north of the airport

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A unit vector is a vector that has a magnitude of exactly 1 and points in a

particular direction.

Unit vector in cartesian coordinate stated by i, j and k which is perpendicular

with each others.

x

y

z

i

j

k

Vector A can be written as:

A

AA

dan

AAA

atau

kAjAiAA

zyx

zyx

=

++=

++=

ˆ

ˆˆˆ

kjiA

A

The magnitude of each unit vector; that is |i| = |j| = |k| = 1

C = A + B

Cx = Ax + Bx

Cy = Ay + By)(tan 1

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x

y

yx

C

C

dan

CCC

−=

+=

q

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and

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There are the following three vectors all components in metersԦ𝑎 = 4.2 Ƹ𝑖 − 1.5 Ƹ𝑗

𝑏 = −1.6 Ƹ𝑖 + 2.9 Ƹ𝑗Ԧ𝑐 = −3.7 Ƹ𝑗

What is their vector sum Ԧ𝑟 , magnitude r, and angle?

Answer:

𝑟𝑥 = 𝑎𝑥 + 𝑏𝑥 + 𝑐𝑥

𝑟𝑥 = 4.2 − 1.6 + 0 = 2.6 m

𝑟𝑦 = 𝑎𝑦 + 𝑏𝑦 + 𝑐𝑦

For x-axis

For y-axis

𝑟𝑥 = −1.5 + 2.9 − 3.7 = −2.3 m

Ԧ𝑟 = 2.6 m Ƹ𝑖 − 2.3 m Ƹ𝑗

𝑟 = 2.6 m 2 + −2.3 m 2 ≈ 3.5m

𝜃 = tan−1−2.3 m

2.6 m= −41°

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A particle undergoes three consecutive displacements: d1 =(15i + 30j + 12k) cm, d2 = (23i -14j - 5.0k) cm, and d3 = (-13i + 15j) cm. Find the components of the resultant displacement and its magnitude.

Answer:

R =d1 + d2 + d3

=(25i + 31j + 7.0k) cm

The resultant displacement has components Rx= 25 cm, Ry = 31 cm, and Rz = 7.0 cm. Its magnitude is

𝑅 = 𝑅𝑥2 + 𝑅𝑦

2 + 𝑅𝑧2

= 40 cm

▪ Multiplication by scalar

Multiplication of vector by a positive scalar a multiplies the magnitude but leaves the direction unchanged

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a (A+B) = aA + aB

A2A

▪Dot product

A.B = AB cos q

A.B = AxBx + AyBy + AzBz

A.B = B.A

A.(B+C) = A.B + A. C

B

Aq

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(Commutative)

(distributive)

Multiplication by vector

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What is the angle φ between a = 3.0i – 4.0j and b = -2.0i + 3.0k

a.b = ab cos φ

Answer:

The magnitude of a and b respectively:a = 5b = 3.61

a.b = (3.0i – 4.0j) . (-2.0i + 3.0k) = -6.0

-6.0 = (5.00)(3.61) cos φ

φ =cos−1(−6.0)

(5.00)(3.61)= 109°

▪Cross product

C = A x B

C = AB sin q

Cx = AyBz – AzBy

Cy = AzBx – AxBz

Cz = AxBy – AyBz

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C

B

Aq

i x i = j x j = k x k =0

i

jki x j = k

j x k = i

k x i =j

▪ Quite complicated:

▪ If you understand about determinant, do as follows:

kBABAjBABAiBABA

iBAjBAiBAkBAjBAkBA

kBjBiBkAjAiABA

xyyxzxxzyzzy

yzxzzyxyzxyx

zyxzyx

ˆ)(ˆ)(ˆ)(

ˆˆˆˆˆˆ

)ˆˆˆ()ˆˆˆ(

−+−+−=

−++−−=

++++=

zyx

zyx

BBB

AAA

kji

BA

ˆˆˆ

=

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▪ c = (3i – 4j) x (-2i + 3k)

▪ = 3i x (-2i) + 3i x 3k + (-4j) x (-2i) + (-4j) x 3k

▪ = -6(0 i) + 9(-j) + 8(-k) – 12(i)

▪ = -12i – 9j – 8k

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If a = 3i – 4j and b = -2i + 3k, what is c = a x b ?

Answer:

Scalar triple product: 𝑨 ∙ 𝑩 × 𝑪

▪ Geometrically, | 𝑨 ∙ 𝑩 × 𝑪 | is the volume of parallelepiped generated by A, B, and C, since |B x C| is the area of the base, and |A cos θ| is the altitude. Evidently

▪ 𝑨 ∙ 𝑩 × 𝑪 = 𝑩 ∙ 𝑪 × 𝑨 = 𝑪 ∙ 𝑨 × 𝑩

▪ 𝑨 ∙ 𝑩 × 𝑪 =

𝐴𝑥 𝐴𝑦 𝐴𝑧𝐵𝑥 𝐵𝑦 𝐵𝑧𝐶𝑥 𝐶𝑦 𝐶𝑧

▪ Note that the dot and cross can be interchanged:𝑨 ∙ 𝑩 × 𝑪 = 𝑨 × 𝑩 ∙ 𝑪

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Vector triple product: 𝑨 × 𝑩 × 𝑪

▪ The vector triple product can be simplified by the so-called BAC-CAB rule:

▪ 𝑨 × 𝑩 × 𝑪 = 𝑩 𝑨 ∙ 𝑪 − 𝑪(𝑨 ∙ 𝑩)

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▪ Answer all the questions in Tutorial I

▪ The answer should be submitted to Ms. Agnes before 18 September 2018, 15:00 WIB

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