UITM LAB PHY406 Experiment 1 Vector PDF

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PHYDEGREE OF FOOD SCIENCE AND TECHNOLOGYLABORATARY 1: VECTORDATE OF SUBMISSION:LECTURE’S NAME: DR. ANISZAWATI AZISPREPARED BY:STUDENT’S NAME STUDENT’S ID CLASSNOR FAIRUZ BINTI ABDUL KAHAR 2021454578 AS2461ANORATHIRAH BINTI MAT JAIS 2021835056 AS2461AEXPERIMENT 1TOPIC: VectorOBJECTIVE:The experiment ...

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PHY406 DEGREE OF FOOD SCIENCE AND TECHNOLOGY LABORATARY 1: VECTOR DATE OF SUBMISSION: LECTURE’S NAME: DR. ANISZAWATI AZIS

PREPARED BY: STUDENT’S NAME NOR FAIRUZ BINTI ABDUL KAHAR NORATHIRAH BINTI MAT JAIS

STUDENT’S ID 2021454578 2021835056

CLASS AS2461A AS2461A

EXPERIMENT 1 TOPIC: Vector OBJECTIVE: The experiment is to illustrate the principles of the methods of vector addition and subtraction by their components. APPARATUS: 1. PhET Simulations https://phet.colorado.edu/sims/html/vector-addition/latest/vector-addition_en.html 2. 3. 4.

Pen Paper Calculator

THEORY:

A vector is a quantity which has both magnitude and direction. In Physics it is usually represented by an arrow the length of which indicates the magnitude while the arrow head indicates the direction. In 2-dimensions, the arrow that represents a vector quantity a can be resolved into two components in the x and y direction labelled as  aₓ and  aᵧ as illustrated in Figure 1.

Figure 1

PROCEDURES: (PART A) 1. Vector PhET simulation was started and Explore 2-D was clicked. 2. A few minutes was spent to understand/ explore the functionalities of the different tabs/ panels. 3. Two vector a and b from the vector panel was dragged to graph paper. Their length and orientations was adjusted. 4. Their magnitudes and the angles was recorded. 5. The x and y-component of the two vectors a and b was calculated using the formula. sₓ was found by using formula. 6. The x-component of the resultant  sᵧ was found by using formula. 7. The y-component of the resultant  8. The observed value obtained from the graph paper was recorded.

(PART B)

Your cat is running around the grass in your back yard. He undergoes successive displacements 7.00 m north, 5.70 m southeast, and 15.0 m east. What is the resultant displacement of the cat? 1) The resultant displacement of the cat was determined by i) The component method using the calculator ii) The 2-D vector simulation 2) The calculation and the PhET simulation image was showed of the resultant displacement

(PART C)

Two displacement vectors are given as:  Y = 11.2 km 26.6 N of W 1)

 X

-

 Y

 X = 18.0 km 56.3 E of N and

was determined by

i) The component method using the calculator

2) The PhET simulation image of the resultant displacement was showed

(PART D)

Figure 3 shows two persons pulling a box. The person on the right pulls with a force  F ₁ of magnitude 11.7 N and direction of θ 1 = 59.0°. The person on the left pulls with a force  F₂ of magnitude 8.2 N and direction of θ 2 = 76.0°. 1) i) The single force that is equivalent to the two forces shown was determined using calculator and PhET simulation ii) The force that a third person would have to exert on the box to make the resultant force equal to zero was determined by using calculator and PhET simulation. The forces are measured in unit N.

2) The calculation and the PhET simulation image of the equivalent force in 1) i) and the force by the third person in 1) ii) was showed.

Figure 3

DATA& RESULT (PART A) 1) PhET SIMULATION

 No First Vector, a of Obs 1. Magnitude | a | = 14.1

Second Vector , b

Magnitude

2.

Angle

θ ° = 45.0

Angle

3.

X-Component

aₓ θ

X-Component

= a cos

= 14.1 cos 45.0 = 10.0

Resultant, s Calculations

| b | = 13.0

X-Component

sₓ = aₓ + bₓ = 10.0+5.0 = 15.0

θ ° = -67.4

Y-Component

sᵧ= aᵧ + bᵧ = 10.0+(-12.0) = -2.0

bₓ = b cos θ = 13.0 cos (-67.4) = 5.0

Magnitude

|s| =

bᵧ = b sin θ = 13.0 sin (-67.4) = -12.0

Angle

√ s 2 ₓ+ s ² ᵧ

−2 ¿ = ¿ 15.02 +¿ √¿ = 15.1

4.

Y-Component

aᵧ = a sin θ = 14.1 sin 45.0 = 10.0

Y-Component

¿ θ=tan ¯ 1 ¿

sᵧ sₓ

) =

¿ tan ¯ 1 ¿

) = -0.13

−2 15.0

2) CALCULATOR No of Obs 1.

2.

3.

First Vector,

Magnitude

Angle

X-Component

Second Vector , b

a

| a | = 14.1

θ ° = 45.0

ax = a cos θ

Magnitude

Angle

X-Component

Resultant, s Calculations

| b | = 13.0

θ ° = -67.4

bx = b cos

θ

XComponent

sx = ax + bx

YComponent

sy = ay + by

Magnitude

|s| =

= 9.97 + (-12.0) = -2.03

√ s 2 ₓ+s ² ᵧ

−2.03 ¿ ¿ = 14.972 +¿ √¿

= 13.0 cos (-67.4°) = 5.0

= 14.1 cos 45.0° = 9.97

= 9.97 + 5.0 = 14.97

=15.11 4.

Y-Component

aᵧ = a sin θ = 14.1 sin 45.0 = 9.97

Y-Component

by = b sin θ = 13.0 sin (-67.4°) = -12.0

¿ θ=tan ¯ 1 ¿

Angle ) =

¿ tan ¯ 1 ¿ −2.03 ) 14.97

= -7.72°

sᵧ sₓ

(PART B) 1) i) CALCULATOR SX = aX + bx + cx = 0.00 + 4.03 + 15.00 = 19.03 Sy = ay + by + cy = 7.00 + (-4.03) + 0.00 = 2.97 |s| =

√ s 2 ₓ+ s ² ᵧ

2.97 ¿ ¿ = 19.032 +¿ √¿ = 19.26 m

ii) PhET SIMULATION SX = aX +bx + cx = 0.00 + 4.03 + 15.00 = 19.0 Sy = ay +by + cy = 7.0 + (-4.0) + 0.0 = 3.0 |s| =

√ s 2 ₓ+s ² ᵧ

3.0 ¿ ¿ = 19.02 +¿ √¿ = 19.2 m

2) i) CALCULATOR Vector | a | = 7.00 m θ = 90.0° | b | = 5.70 m θ = 180° - 135° = 45.0° Because the opposite direction the angle become -45.0 | c | = 15.0 θ = 0.0° Resultant

Magnitude |s| = √ s 2 ₓ+ s ² ᵧ

2.97 ¿ ¿ = 19.032 +¿ √¿ = 19.26 m

x-component(m) ax = a cos θ = 7.00 cos 90.0° = 0.00 bx = b cos θ = 5.70 cos (-45.0) = 4.03

y-component(m) ay = a sin θ = 7.00 sin 90.0° = 7.00 by = b cos θ = 5.70 sin (-45.0) = -4.03

cx = c cos θ = 15.0 cos 0.0° = 15.00 SX = aX + bx + cx = 0.00 + 4.03 + 15.00 = 19.03 Angle

cy = c sin θ = 15.0 cos 0.0° = 0.00 Sy = ay + by + cy = 7.00 + (-4.03) + 0.00 = 2.97

sᵧ ¿ 1 ) θ=tan ¯ ¿ sₓ ¿ −2.97 = ) 1 tan ¯ ¿ 19.03 = 8.87°

ii) PhET SIMULATION

Vector | a | = 7.0 m θ = 90.0° | b | =5.7 m θ = 180° - 135° = 45.0° Because the opposite direction,

x-component(m) ax = a cos θ = 7.0 cos 90.0° = 0.0 bx= b cos θ = 5.7 cos (-45.0°) = 4.0

y-component(m) ay= a sin θ = 7.0 sin 90.0° = 7.0 by= b cos θ = 5.7 sin (-45.0°) = -4.0

the angle become -45.0 | c | = 15.0 θ = 0.0° Resultant

Magnitude |s| = √ s 2 ₓ+s ² ᵧ

cx = c cos θ = 15.0 cos 0.0° = 15.0 SX = aX + bx + cx = 0.00 + 4.03 + 15.00 = 19.0 Angle

cy = c sin θ = 15.0 cos 0.0° = 0.0 Sy = ay + by + cy = 7.0 + (-4.0) + 0.0 = 3.0

sᵧ ¿ 1 θ=tan ¯ ¿ sₓ 3.0 ¿ = tan ¯ 1 ¿ 19.2

3.0 ¿ ¿ = 19.02 +¿ √¿

) )

= 9.0°

= 19.2 m

(PART C) 1) CALCULATOR Vector | a | = 18.0 km θ = 90.0° - 56.3° = 33.7° (E of N) | b | = 11.2 km θ = 180° - 26.6° = 153.4° (N of W) Because of opposite direction, the angle become -26.6°

x-component(km) ax = a cos θ = 18.0 cos 33.7° = 14.98 bx = b cos θ = 11.2 cos (-26.6°) = 10.01

y-component(km) ay = a sin θ = 18.0 sin 33.7° = 9.99 by = b sin θ = 11.2 sin (-26.6°) = -5.01

Resultant

SX = aX + bx = 14.98 + 10.01 = 24.99 Angle

Sy = ay + by = 9.99 + (-5.01) = 4.98

Magnitude |s| = √ s 2 ₓ+s ² ᵧ

4.98 ¿ ¿ = 24.992 +¿ √¿ = 25.48 km

¿ 1 θ=tan ¯ ¿ ¿ = tan ¯ 1 ¿ = 11.06°

sᵧ ) sₓ 4.98 25.48

)

2) PhET SIMULATION

Vector | a | = 18.0 km θ = 90.0° - 56.3°

x-component(km) ax = a cos θ = 18.0 cos 33.7°

y-component(km) ay = a sin θ = 18.0 sin 33.7°

= 33.7° (E of N) | b | = 11.2 km θ = 180° - 26.6° = 153.4° (N of W) Because of opposite direction ,the angle become -26.6°

Resultant

= 15.0 bx = b cos θ = 11.2 cos (-26.6°) = 10.0

= 10.0 by = b sin θ = 11.2 sin (-26.6°) = -5.0

SX = aX + bx = 15.0 + 10.0 = 25.0

Sy = ay + by = 10.0 + (-5.0) = 5.0

Magnitude |s| = √ s 2 ₓ+s ² ᵧ

Angle

sᵧ ¿ θ=tan ¯ 1 ¿ sₓ 5.0 ¿ = 1 tan ¯ ¿ 25.0

5.0 ¿ ¿ = 25.02 +¿ √¿

) )

= 11.3°

= 25.5 km

(PART D) 1) i) CALCULATOR Vector |  F ₁ | = 11.7 N θ = 59.0° |  F ₂ | = 8.2 N θ = 180° - 76.0° = 104.0° Resultant

Magnitude |s| = √ s 2 ₓ+s ² ᵧ

17.99 ¿ ¿ = 4.052 +¿ √¿

x-component (N) Fx = F₁ cos θ = 11.7 cos 59.0° = 6.03 F₂x = F₂ cos θ = 8.2 cos 104.0° = -1.98 SX = FX + F₂x = 6.03 + (-1.98) = 4.05 Angle

¿ θ=tan ¯ 1 ¿ ¿ = tan ¯ 1 ¿ = 77.3°

sᵧ ) sₓ 17.99 4.05

y-component (N) Fy = F₁ sin θ = 11.7 sin 59.0° = 10.03 F₂y = F₂ sin θ = 8.2 sin 104.0° = 7.96 Sy = Fy + F₂y = 10.03 + 7.96 = 17.99

)

= 18.44 N

PhET SIMULATION

ii) CALCULATOR Vector

F ₁∨¿ = 11.7 N ¿ ¿ θ ₁ = 59.0° ¿ F ₂∨¿ = 8.2 N ¿ θ ₂ = 180° - 76.0°

X-component Fₓ = Fₓ cos θ ₁ = 11.7 cos 59.0° = 6.0259 N F₂x = F₂ cos θ₂ = 8.2 cos 104.0° = -1.9838 N

Y-component Fy = F₁ sin θ₁ = 11.7 sin 59.0° = 10.0289 N F₂y = F₂ sin θ₂ = 8.2 sin 104.0° = 7.9564 N

F₃ₓ = F₃ cos θ ° = 18.4 cos 257.5° -3.9825 N

F₃ᵧ = F₃ sin θ ° = 18.4 sin 257.5° = -17.9638 N

= 104.0°

¿ F ₃∨¿ = 18.4 N ¿ θ ₃ = 180° + 77.5° = 257.5°

X-component Sₓ = Fₓ + F₂ₓ + F₃ₓ = 6.0259 + (-1.9838) + (-3.9825) = 0.0596 N

Y-component Sᵧ = Fᵧ + F₂ᵧ + F₃ᵧ = 10.0289 + 7.9564 + (-17.9638) = 0.0215 N

Magnitude |s| =

√s

Angle

sᵧ ¿ 1 ) θ=tan ¯ ¿ sₓ 0.0215 ) = tan¯¹ ( 0.0596

2

ₓ+s ² ᵧ √ ( 0.0596 )2 +(0.0215 ) ²

= = 0.06334 N

= 19.8363°

Vector |  F ₁ | = 11.7 N θ = 59.0° |  F ₂ | =8.2N θ = 180° - 76.0° = 104.0° Resultant

Magnitude |s| = √ s 2 ₓ+ s ² ᵧ

18.0 ¿ ¿ = 4.02 +¿ √¿ = 18.4 N

PhET SIMULATION

x-component (N) Fx = F₁ cos θ = 11.7 cos 59.0° = 6.0 F₂x = F₂ cos θ = 8.2 cos 104.0° = -2.0 SX = FX + F₂x = 6.0 + (-2.0) = 4.0 Angle

y-component (N) Fy = F₁ sin θ = 11.7 sin 59.0° = 10.0 F₂y = F₂ sin θ = 8.2 sin 104.0° = 8.0 Sy = Fy + F₂y =10.0 + 8.0 = 18.0

sᵧ ¿ θ=tan ¯ 1 ¿ sₓ 18.0 ¿ = tan ¯ 1 ¿ 4.0 = 77.5°

) )

Vector

F ₁∨¿ = 11.7 N ¿ ¿ θ ₁ = 59.0° ¿ F ₂∨¿ = 8.2 N ¿ θ ₂ = 180° - 76.0°

X-component Fₓ = Fₓ cos θ ₁ = 11.7 cos 59.0° = 6.0 F₂x = F₂ cos θ₂ = 8.2 cos 104.0° = -2.0

Y-component Fy = F₁ sin θ₁ = 11.7 sin 59.0° = 10.0 F₂y = F₂ sin θ₂ = 8.2 sin 104.0° = 8.0

F₃ₓ = F₃ cos θ ° = 18.4 cos 257.5° = -4.0

F₃ᵧ = F₃ sin θ ° = 18.4 sin 257.5° = -18.0

= 104.0°

¿ F ₃∨¿ = 18.4 N ¿ θ ₃ = 180° + 77.5° = 257.5° X-component Sₓ = Fₓ + F₂ₓ + F₃ₓ = 6.0 + (-2.0) + (-4.0) = 0.0 N

Y-component Sᵧ = Fᵧ + F₂ᵧ + F₃ᵧ = 10.0 + 8.0 + (-18.0) = 0.0 N

Magnitude |s| = √ s 2 ₓ+s ² ᵧ

√

= ( 0 ) +(0 ) = 0.0 N 2

2

Angle

sᵧ ¿ 1 θ=tan ¯ ¿ sₓ 0.0 = tan¯¹ 0.0 ) ¿ =-

)

DISCUSSION: Based on this experiment, the calculation method is more accurate than using PhET simulation 2-D explore. It is because the calculation method gives more specific numbers than the PhET simulation. There is an error in the PhET simulation in which the angle is counted from the x-component that causes a slight miscalculation. By using PhET simulation, the final answer will be rounded to one decimal point automatically which causes it to be not accurate. Meanwhile, by using the calculation method, it can take a more specific number and follow all the numbers given in the calculator to avoid the error. There might be a slight difference between calculation method and PhET simulation 2-D explore. CONCLUSION: The objective is achieved. The principle of vector addition and subtraction by component method can be used to compute and proved. REFERENCES: 1. Nasramudy.Physics Vector.Retrived December 15,2019 from https://vectornotesadress.blogspot.com/?m=1 2. Lew.A.Here's How to Find the Hypotenuse of a Right Triangle.Retrived February 8,2021 from https://tutorme.com/blog/post/how-to-find-hypotenuse/ 3. Jidan.What are vectors in Physics?Retrived November 1,2021 from https://www.physicsread.com/vectors-in-physics/...