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.
    $a ext{ has units } \frac{m}{s^4}$

• 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
    $(1 ext{ ft} / 12 ext{ in})^2$
  • 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 $R<em>x \,\hat{i} + R</em>y \,\hat{j}$
  • 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: $R_x = R \cos\theta$
  • y-component: $R_y = R \sin\theta$
  • 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:
  • $c_x = 4.43\quad (\text{positive x})$
  • $c_y = -6.57\quad (\text{negative y})$
  • Magnitude and angle of c (polar form):
  • Magnitude: $|c| = 7.85\text{ m}$
  • Angle: $\theta_c = -56.8^{\circ}$ (56.8 degrees below the +x axis)
  • The polar form can be recovered from rectangular components via:
  • Magnitude: $|c| = \sqrt{c<em>x^2 + c</em>y^2} = 7.85\,\text{m}$
  • Angle: $\theta<em>c = \tan^{-1}\left(\frac{c</em>y}{c_x}\right) = \tan^{-1}\left(\frac{-6.57}{4.43}\right) = -56.8^{\circ}$
  • 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: $b<em>x = -6\text{ cm}, \quad b</em>y = 0$
  • From the tip of b, draw vector c according to its magnitude and direction
  • For the example, c has components approximately: $c<em>x \approx 4.43\text{ cm}, \quad c</em>y \approx -6.57\text{ cm}$
  • 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,
    • $a<em>x = R\cos\theta, \quad a</em>y = R\sin\theta$
  • Repeat for b and c with their respective magnitudes and angles
  • Sum components separately:
  • $d<em>x = a</em>x + b<em>x + c</em>x$
  • $d<em>y = a</em>y + b<em>y + c</em>y$
  • Then convert the resultant components back to polar form
  • Magnitude: $|d| = \sqrt{d<em>x^2 + d</em>y^2}$
  • Angle from +x axis: $\theta<em>d = \tan^{-1}\left(\frac{d</em>y}{d_x}\right)$
    • 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: $a<em>x$ positive, $b</em>x$ negative, $c<em>x$ positive; $c</em>y$ negative; $a<em>y$ and $b</em>y$ signs depend on their directions
  • For the given numbers: $a<em>x = 10\cos(30^{\circ})$, $a</em>y = 10\sin(30^{\circ})$, $b<em>x = -6$, $b</em>y = 0$, $c<em>x = 4.43$, $c</em>y = -6.57$
  • Then $d<em>x = a</em>x + b<em>x + c</em>x$ and $d<em>y = a</em>y + b<em>y + c</em>y$ with the correct signs
  • The resulting vector d is often found to have a magnitude around a value like $|d| \approx 7.85\text{ m}$ with an angle about $\theta_d \approx -56.8^{\circ}$ (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: $x = R\cos\theta, \quad y = R\sin\theta$
  • Rectangular to polar: $|R| = \sqrt{x^2 + y^2}, \quad \theta = \tan^{-1}\left(\frac{y}{x}\right)$ 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., $-56.8^{\circ}$) 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:
  • $a<em>x = R\cos\theta,\quad a</em>y = R\sin\theta$
  • Resultant components:
  • $d<em>x = a</em>x + b<em>x + c</em>x, \quad d<em>y = a</em>y + b<em>y + c</em>y$
  • Polar from rectangular:
  • $|d| = \sqrt{d<em>x^2 + d</em>y^2}, \quad \theta<em>d = \tan^{-1}\left(\frac{d</em>y}{d_x}\right)$
  • 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