Analytical Method of Vector Addition Study Guide - General Physics 1

Overview of Analytical Methods in Vector Addition

  • Analytical or mathematical methods provide a precise way to determine the resultant of multiple vectors using geometry and trigonometry.

  • There are two primary categories of analytical methods discussed:

    • Perpendicular Vectors: Solved using the Pythagorean Theorem.

    • Non-Perpendicular Vectors: Solved using the Component Method.

Adding Perpendicular Vectors

  • When vectors are perpendicular to each other, they form the legs of a right-angled triangle.

  • The magnitude of the resultant vector (RR) is calculated using the Pythagorean Theorem:

    • R2=A2+B2R^2 = A^2 + B^2

    • R=A2+B2R = \sqrt{A^2 + B^2}

  • The direction (angle θ\theta) is determined using the tangent function (TOA from SOH CAH TOA):

    • tan⁡(θ)=OppositeAdjacent\tan(\theta) = \frac{\text{Opposite}}{\text{Adjacent}}

    • θ=tan⁡−1(OppositeAdjacent)\theta = \tan^{-1}\left(\frac{\text{Opposite}}{\text{Adjacent}}\right)

Example: Eric's Hike

  • Scenario: Eric leaves a base camp and hikes 11 km11\,\text{km} North and then hikes 11 km11\,\text{km} East.

  • Goal: Determine Eric's resulting displacement.

  • Calculation of Magnitude:

    • d=(11 km)2+(11 km)2d = \sqrt{(11\,\text{km})^2 + (11\,\text{km})^2}

    • d=121+121d = \sqrt{121 + 121}

    • d=242d = \sqrt{242}

    • d=15.56 kmd = 15.56\,\text{km}

  • Calculation of Direction:

    • tan⁡(θ)=11 km11 km=1\tan(\theta) = \frac{11\,\text{km}}{11\,\text{km}} = 1

    • θ=tan⁡−1(1)=45∘\theta = \tan^{-1}(1) = 45^\circ

    • Direction is 45∘45^\circ Northeast (NENE).

  • Final Result: R=15.56 km,45∘NER = 15.56\,\text{km}, 45^\circ NE

Adding Non-Perpendicular Vectors: The Component Method

  • The component method involves breaking down each vector into its horizontal and vertical parts before combining them.

Process Steps

  1. Resolve each vector into its horizontal (xx) and vertical (yy) components using trigonometric functions (sin⁡\sin, cos⁡\cos, tan⁡\tan).

  2. Add all x-components (∑Ax\sum A_x) together, observing proper sign conventions.

  3. Add all y-components (∑Ay\sum A_y) together, observing proper sign conventions.

  4. Find the magnitude of the resultant vector (RR) using the Pythagorean Theorem:

    • R=(∑Ax)2+(∑Ay)2R = \sqrt{(\sum A_x)^2 + (\sum A_y)^2}

  5. Determine the direction (θ\theta) of the resultant using inverse trigonometric functions:

    • θ=tan⁡−1(∑Ay∑Ax)\theta = \tan^{-1}\left(\frac{\sum A_y}{\sum A_x}\right)

Sign Conventions

  • North (N): Positive (++)

  • East (E): Positive (++)

  • South (S): Negative (-$)\n* **West (W):** Negative (-$)

Practice Problems in Computing Components

Problem 1: Force Vector

  • Vector: 150 N150\,\text{N} at an angle of 70∘70^\circ from the horizontal.

  • Horizontal Component (AxA_x):

    • cos⁡(70∘)=Ax150\cos(70^\circ) = \frac{A_x}{150}

    • Ax=150×cos⁡(70∘)=51.30 NA_x = 150 \times \cos(70^\circ) = 51.30\,\text{N}

  • Vertical Component (AyA_y):

    • sin⁡(70∘)=Ay150\sin(70^\circ) = \frac{A_y}{150}

    • Ay=150×sin⁡(70∘)=140.95 NA_y = 150 \times \sin(70^\circ) = 140.95\,\text{N}

Problem 2: Displacement Vector

  • Vector: 50 m50\,\text{m} at an angle of 60∘60^\circ.

  • Horizontal Component (AxA_x):

    • cos⁡(60∘)=Ax50\cos(60^\circ) = \frac{A_x}{50}

    • Ax=50×cos⁡(60∘)=25 mA_x = 50 \times \cos(60^\circ) = 25\,\text{m}

  • Vertical Component (AyA_y):

    • sin⁡(60∘)=Ay50\sin(60^\circ) = \frac{A_y}{50}

    • Ay=50×sin⁡(60∘)=43.30 mA_y = 50 \times \sin(60^\circ) = 43.30\,\text{m}

    • Note: According to sign conventions for specific directions (e.g., South), this may be expressed as −43.30 m-43.30\,\text{m}.

Sample Problem 1: Sum of Two Vectors

Vectors to add:

  • Vector A=5 km,20∘A = 5\,\text{km}, 20^\circ East of North

  • Vector B=6 km,30∘B = 6\,\text{km}, 30^\circ North

Resolution of Component A (20 degrees East of North)

  • Angle is measured from the North (vertical) axis.

  • Ax=5sin⁡(20∘)=1.71 kmA_x = 5 \sin(20^\circ) = 1.71\,\text{km}

  • Ay=5cos⁡(20∘)=4.70 kmA_y = 5 \cos(20^\circ) = 4.70\,\text{km}

Resolution of Component B (30 degrees North of vertical)

  • Bx=6sin⁡(30∘)=3 kmB_x = 6 \sin(30^\circ) = 3\,\text{km}

  • By=6cos⁡(30∘)=5.20 kmB_y = 6 \cos(30^\circ) = 5.20\,\text{km}

Summing Components

  • ∑Ax=Ax+Bx=1.71+5.20=6.91 km\sum A_x = A_x + B_x = 1.71 + 5.20 = 6.91\,\text{km} (Note: and according to slide 13 table summary)

    • Correction per slide table: Vector Bx=5.20 kmB_x = 5.20\,\text{km} and By=3 kmB_y = 3\,\text{km}? (Slide 13 lists: Ax=1.71A_x=1.71, Ay=4.70A_y=4.70, Bx=5.20B_x = 5.20, By=3B_y=3).

    • Total ∑Ax=6.91 km\sum A_x = 6.91\,\text{km}

    • Total ∑Ay=7.7 km\sum A_y = 7.7\,\text{km}

Resultant Calculation

  • Magnitude:

    • R=(6.91 km)2+(7.7 km)2R = \sqrt{(6.91\,\text{km})^2 + (7.7\,\text{km})^2}

    • R=10.35 kmR = 10.35\,\text{km}

  • Direction:

    • θ=tan⁡−1(7.76.91)=48.10∘\theta = \tan^{-1}\left(\frac{7.7}{6.91}\right) = 48.10^\circ

  • Final Result: R=10.35 km,48.10∘R = 10.35\,\text{km}, 48.10^\circ North of East.

Sample Problem 2: Antiparallel Components

Vectors to add:

  • Vector A=5 km,20∘A = 5\,\text{km}, 20^\circ North of East

  • Vector B=6 km,30∘B = 6\,\text{km}, 30^\circ West of North

  • Warning: Horizontal components are antiparallel (pointing in opposite directions).

Resolution of Vector A

  • Ax=5cos⁡(20∘)=4.70 kmA_x = 5 \cos(20^\circ) = 4.70\,\text{km}

  • Ay=5sin⁡(20∘)=1.71 kmA_y = 5 \sin(20^\circ) = 1.71\,\text{km}

Resolution of Vector B

  • Bx=−(6sin⁡(30∘))=−3 kmB_x = -(6 \sin(30^\circ)) = -3\,\text{km} (Negative due to being West)

  • By=6cos⁡(30∘)=5.20 kmB_y = 6 \cos(30^\circ) = 5.20\,\text{km}

Summing Components

  • ∑Ax=4.70+(−3)=1.7 km\sum A_x = 4.70 + (-3) = 1.7\,\text{km}

  • ∑Ay=1.71+5.20=6.91 km\sum A_y = 1.71 + 5.20 = 6.91\,\text{km}

Resultant Calculation

  • Magnitude:

    • R=(1.7)2+(6.91)2R = \sqrt{(1.7)^2 + (6.91)^2}

    • R=7.12 kmR = 7.12\,\text{km}

  • Direction:

    • θ=tan⁡−1(6.911.7)=76.18∘\theta = \tan^{-1}\left(\frac{6.91}{1.7}\right) = 76.18^\circ

  • Final Result: R=7.12 km,76.18∘R = 7.12\,\text{km}, 76.18^\circ North of East.

Verification Methods

  • Results from the analytical method can be verified graphically using:

    • Head-to-Tail Method (Polygon Method): Drawing vectors sequentially and connecting the start and end point.

    • Parallelogram Method: Drawing two vectors from the same origin and forming a parallelogram where the diagonal is the resultant.

  • Scales used in graphical verification: 1 cm:1 km1\,\text{cm} : 1\,\text{km}.

Vincentian Prayer

Dear Lord, teach me the things that are important; to be generous with your gifts; compassionate to those who have less; just in the face of unfair circumstances; true when the world’s values contradict my own; gracious when things don’t go my way; and magnanimous when they do.

May nothing else matter, except faith in your goodness, my neighbors and mine. Hope that things can get better, and charity that always sets things right. May your special love for the poor, the mark of my uniquely Vincentian education, be the work I excel in, the standard I constantly refer to, and my courage when I meet you someday. O Mary conceived without sin. Pray for us who have recourse to thee. St. Vincent de Paul, pray for us. Amen.