Notes on Dimensional Analysis, Vector Components, and Graphical Addition

Dimensional Analysis, Vector Components, and Graphical Addition

  • Start with dimensional consistency

    • If a variable has units of meters per second (m/s), then any term that represents a quantity with the same dimension must also have m/s
    • Example setup: if t is cubed in an expression, t^3 has units of seconds^3 (s^3)
    • If you divide by a term with units s^3, the remaining units change accordingly (e.g., if you divide by s^3, the time unit contributes s^−3 to the denominator; be explicit about units to see the resulting unit of the quantity)
    • Concrete outcome shown in the transcript: dividing by seconds^3 yielded a unit of meters per second to the fourth, i.e.
      a ext{ has units } rac{m}{s^4}

  • General approach to unit analysis you can rely on

    • When unsure, write out the algebraic equation fully and solve for the unknown, grouping all known terms on one side
    • Use unit cancellation to verify each step and track what cancels and what remains
    • This practice is common in exam-style questions (often multiple choice) where unit consistency is key

  • Unit conversions and pitfalls to avoid

    • Base conversion example: 12 inches = 1 foot; therefore 12 in per 1 ft
    • Important caution: when units are squared or raised to any power, you must square (or raise) the entire conversion factor
    • If you have inches^2, convert using
      (1extft/12extin)2(1 ext{ ft} / 12 ext{ in})^2
    • Equivalently, 1 in^2 = (1/144) ft^2
    • Always ensure that the squared (or higher-power) units cancel properly in your calculation to avoid leftover units

  • Vectors: why direction matters (polar vs rectangular form)

    • Many quantities in motion are vectors; they require both magnitude and direction
    • Vector direction is essential to fully describe the quantity; a size alone is not enough
    • Two common representations:
    • Polar form: magnitude R with an angle θ from a reference axis
    • Rectangular (I-hat, J-hat) form: components along x and y, i.e., vectors expressed as R<em>x i^+R</em>y j^R<em>x \,\hat{i} + R</em>y \,\hat{j}
    • To add vectors quantitatively, decompose each into x- and y-components and then sum the components separately

  • Using SOHCAHTOA for components

    • When resolving a vector into components, you typically use a right triangle formed by the vector and axes
    • If θ is the angle from the reference axis:
    • x-component: Rx=Rcos⁡θR_x = R \cos\theta
    • y-component: Ry=Rsin⁡θR_y = R \sin\theta
    • Remember sign conventions:
    • Positive x to the right, positive y up
    • Depending on quadrant, components may be negative
    • Distinguish which side is adjacent vs opposite for a given angle; this determines whether you use cosine or sine for a component
    • Quick reminder: if the x-component lies along the adjacent side to θ, use cos; if it lies along the opposite side, use sin

  • Worked example: polar to rectangular components for a vector (example values from the transcript)

    • A vector (call it c) has components reported as:
    • cx=4.43(positive x)c_x = 4.43\quad (\text{positive x})
    • cy=−6.57(negative y)c_y = -6.57\quad (\text{negative y})
    • Magnitude and angle of c (polar form):
    • Magnitude: ∣c∣=7.85 m|c| = 7.85\text{ m}
    • Angle: θc=−56.8∘\theta_c = -56.8^{\circ} (56.8 degrees below the +x axis)
    • The polar form can be recovered from rectangular components via:
    • Magnitude: ∣c∣=c<em>x2+c</em>y2=7.85 m|c| = \sqrt{c<em>x^2 + c</em>y^2} = 7.85\,\text{m}
    • Angle: θ<em>c=tan⁡−1(c</em>ycx)=tan⁡−1(−6.574.43)=−56.8∘\theta<em>c = \tan^{-1}\left(\frac{c</em>y}{c_x}\right) = \tan^{-1}\left(\frac{-6.57}{4.43}\right) = -56.8^{\circ}
    • Once you have the angle in the standard form (angle from +x axis, CCW positive), you can present the direction as e.g. "56.8° below the +x axis" or as a negative angle

  • Vector addition: a + b + c = d

    • You can solve for d in two ways: graphical (tip-to-tail) or algebraic (component-wise)

  • Graphical (tip-to-tail) method (step-by-step as described in the transcript)

    • Use graph paper and an origin as the reference point
    • Draw vector a from the origin according to its magnitude and direction
    • Example note: vector a has magnitude 10 and some direction (the transcript uses an angle like 30°, depending on the problem setup)
    • From the tip of a, draw vector b according to its magnitude and direction
    • Example: vector b is due west with a magnitude of 6 cm (horizontal to the left, no vertical component)
    • Hence, components: b<em>x=−6 cm,b</em>y=0b<em>x = -6\text{ cm}, \quad b</em>y = 0
    • From the tip of b, draw vector c according to its magnitude and direction
    • For the example, c has components approximately: c<em>x≈4.43 cm,c</em>y≈−6.57 cmc<em>x \approx 4.43\text{ cm}, \quad c</em>y \approx -6.57\text{ cm}
    • The end point after placing a, then b, then c gives the resultant d (tip of c)
    • Measure the resultant vector d (its magnitude and direction) to obtain the polar form of d
    • Important practical notes:
    • When building the graph, you do not draw all three vectors from the origin; you place them tip-to-tail in sequence
    • Use a protractor and ruler to ensure angles and lengths are scaled appropriately
    • For a vector at a given angle, you may need to determine the components using the appropriate trigonometric function depending on which side is adjacent vs opposite

  • Algebraic (component) method for d = a + b + c

    • Resolve each vector into x- and y-components first
    • For a vector with magnitude R and angle θ from +x axis,
      • a<em>x=Rcos⁡θ,a</em>y=Rsin⁡θa<em>x = R\cos\theta, \quad a</em>y = R\sin\theta
    • Repeat for b and c with their respective magnitudes and angles
    • Sum components separately:
    • d<em>x=a</em>x+b<em>x+c</em>xd<em>x = a</em>x + b<em>x + c</em>x
    • d<em>y=a</em>y+b<em>y+c</em>yd<em>y = a</em>y + b<em>y + c</em>y
    • Then convert the resultant components back to polar form
    • Magnitude: ∣d∣=d<em>x2+d</em>y2|d| = \sqrt{d<em>x^2 + d</em>y^2}
    • Angle from +x axis: θ<em>d=tan⁡−1(d</em>ydx)\theta<em>d = \tan^{-1}\left(\frac{d</em>y}{d_x}\right)
      • If desired, express the angle as negative if it lies below the +x axis, e.g. "56.8° below the +x axis" or keep as a standard negative angle
    • Worked example notes from transcript:
    • Signs observed: $ax$ positive, $bx$ negative, $cx$ positive; $cy$ negative; $ay$ and $by$ signs depend on their directions
    • For the given numbers: $ax = 10\cos(30^{\circ})$, $ay = 10\sin(30^{\circ})$, $bx = -6$, $by = 0$, $cx = 4.43$, $cy = -6.57$
    • Then $dx = ax + bx + cx$ and $dy = ay + by + cy$ with the correct signs
    • The resulting vector d is often found to have a magnitude around a value like ∣d∣≈7.85 m|d| \approx 7.85\text{ m} with an angle about θd≈−56.8∘\theta_d \approx -56.8^{\circ} (i.e., 56.8° below the +x axis) in the example
    • Important sign-check tips during computation:
    • If using a triangle to find a component, ensure which side is opposite vs adjacent for your chosen angle
    • Double-check that the x-components sum to a positive or negative value according to the direction of the resultant along the x-axis
    • Double-check that the y-components sum to the intended vertical direction (positive up, negative down)

  • Practical tips and problem setup notes from the transcript

    • Before diving into heavy algebra, try to anticipate the direction and quadrant of the resultant vector
    • When using a graphical method, start from a specific vector and add sequentially (tip-to-tail) rather than drawing all vectors from a single origin
    • When moving between representations (polar ⇄ rectangular), remember:
    • Polar to rectangular: x=Rcos⁡θ,y=Rsin⁡θx = R\cos\theta, \quad y = R\sin\theta
    • Rectangular to polar: ∣R∣=x2+y2,θ=tan⁡−1(yx)|R| = \sqrt{x^2 + y^2}, \quad \theta = \tan^{-1}\left(\frac{y}{x}\right) and interpret the angle with respect to the +x axis
    • If an angle is given as, for example, below the +x axis, you can express it as a negative angle (e.g., −56.8∘-56.8^{\circ}) or describe it verbally as “56.8° below the +x axis.” Either is acceptable as long as unambiguous
    • Use degrees on calculators when working with typical classroom problems (and switch to radians only if explicitly required by the problem)

  • Summary takeaway

    • Dimensional analysis and unit consistency are foundational; always check units as you simplify
    • Vectors require both magnitude and direction; convert between polar and rectangular forms to add them correctly
    • The component method is robust for adding multiple vectors (a, b, c) by summing x- and y-components, then converting back to polar if needed
    • The graphical tip-to-tail method provides a visual check and intuition for the resultant vector
    • Always be mindful of signs and the orientation of angles; SOHCAHTOA remains a guiding principle for resolving components

  • Quick reference formulas

    • Component forms from polar:
    • a<em>x=Rcos⁡θ,a</em>y=Rsin⁡θa<em>x = R\cos\theta,\quad a</em>y = R\sin\theta
    • Resultant components:
    • d<em>x=a</em>x+b<em>x+c</em>x,d<em>y=a</em>y+b<em>y+c</em>yd<em>x = a</em>x + b<em>x + c</em>x, \quad d<em>y = a</em>y + b<em>y + c</em>y
    • Polar from rectangular:
    • ∣d∣=d<em>x2+d</em>y2,θ<em>d=tan⁡−1(d</em>ydx)|d| = \sqrt{d<em>x^2 + d</em>y^2}, \quad \theta<em>d = \tan^{-1}\left(\frac{d</em>y}{d_x}\right)
    • Direction convention: angle measured from the +x axis, counterclockwise; negative angle indicates below the +x axis

  • Note on exam-style practice hinted in the transcript

    • Problems frequently ask you to:
    • Convert between representations, compute components, and obtain a final magnitude and angle
    • Use both graphical (tip-to-tail) and algebraic (component) methods to verify results
    • Be comfortable with common vectors like a = 10 at some angle, b = 6 cm due west, etc., and interpret their components accordingly