Statics: Vector Component Addition and Resultant Calculation

Course Announcements and Logistics

• Quiz Schedule and Class Plan:

  • A Module 2 quiz on adding two vectors together occurs in the last $15\,\text{minutes}$ of class.
  • Class exercises focus on adding vectors in two dimensions ($2\text{D}$), mirroring the concepts evaluated on the quiz.
  • The upcoming Tuesday session involves continuing the lecture on statics forces, walking through example problems, and completing a practice quiz.

• SciTech Social Event:

  • An annual barbecue/picnic takes place outside by the Stains Stale statue between the current building and Hong Kong Hall.
  • Food served includes hot dogs provided by the campus food service.
  • Student clubs set up booths to showcase activities and facilitate networking between students and faculty.
  • An icebreaker activity involves cards with faculty facts (e.g., attending a Grateful Dead concert, climbing a mountain, driving an electric car) requiring faculty signatures at $12:30\,\text{PM}$.

• Blackboard Grade Display Mechanics:

  • Quiz grades entered manually appear behind digital quiz settings configured for online course sections.
  • This causes Blackboard to mark quizzes with an expired or missed due date notification for students.
  • Setting the due date as expired is necessary in the current configuration for manually typed grades to remain visible to students.

Fundamental Principles of Vector Components

• Vector Decomposition:

  • A single non-axial vector acting at a point of application can be resolved into two mutually perpendicular component forces: an $x$-component ($F_x$) along the horizontal axis and a $y$-component ($F_y$) along the vertical axis.
  • Recombining these orthogonal component vectors tip-to-tail yields the original vector as the resultant.

• Component Summation Example:

  • Consider a two-vector system:
    • Vector 1 (Quadrant I): $F_{1x} = +75\,\text{units}$, $F_{1y} = +50\,\text{units}$.
    • Vector 2 (Quadrant III): $F_{2x} = -25\,\text{units}$, $F_{2y} = -100\,\text{units}$.
  • Summation of forces in the vertical ($y$) direction:         $\sum F_y = (+50\,\text{units}) - 100\,\text{units} = -50\,\text{units}$         This indicates a net vertical force of $50\,\text{units}$ directed downward.
  • Summation of forces in the horizontal ($x$) direction:         $\sum F_x = (+75\,\text{units}) - 25\,\text{units} = +50\,\text{units}$         This indicates a net horizontal force of $50\,\text{units}$ directed to the right.

Calculating the Resultant Vector

• Magnitude Formulation:

  • Combining the net horizontal component ($\sum F_x$) and net vertical component ($\sum F_y$) forms a right triangle where the hypotenuse represents the total resultant magnitude ($R$).
  • Applying the Pythagorean theorem ($a^2 + b^2 = c^2$) yields the definitive resultant magnitude equation for any multi-vector system:         $R = \sqrt{\left(\sum F_x\right)^2 + \left(\sum F_y\right)^2}$

• Direction Formulation:

  • The directional orientation angle ($\theta$) of the resultant vector is measured from the horizontal axis.
  • Using the inverse tangent function on the right triangle components yields:         $\theta = \arctan\left(\left|\frac{\sum F_y}{\sum F_x}\right|\right)$
  • Direction is specified either using standard quadrant vector notation or as a signed angle relative to the horizontal reference axis.

Trigonometric Methodologies and Diagramming Rules

• Diagramming Standards for Problem Solving:

  • Always draw large, clear vector and component diagrams taking up a full page.
  • Perform mathematical calculations separately on a second page to minimize simple operational errors.
  • Explicitly label every triangle with its opposite side ($O$), adjacent side ($A$), hypotenuse ($H$), and reference angle ($\theta$).
  • The hypotenuse ($H$) is always the longest side, located directly opposite the $90^\circ$ right angle.
  • The reference angle points directly across to the opposite side ($O$) and touches the adjacent side ($A$).

• Component Rule Caveat:

  • $F_x$ is not universally associated with $\cos(\theta)$, nor is $F_y$ universally associated with $\sin(\theta)$.
  • The trigonometric assignment depends strictly on whether the reference angle is measured relative to the horizontal or vertical axis:
    • When angle $\theta$ is measured from the horizontal ($x$-axis): $F_x = H \times \cos(\theta)$ and $F_y = H \times \sin(\theta)$.
    • When angle $\theta$ is measured from the vertical ($y$-axis): $F_x = H \times \sin(\theta)$ and $F_y = H \times \cos(\theta)$.

Multi-Vector System Example

• System Definition:

  • A three-vector concurrent force system is defined as follows:
    • Vector 1: $100\,\text{lbs}$ magnitude at an angle of $30^\circ$ relative to the horizontal (Quadrant I).
    • Vector 2: $50\,\text{lbs}$ magnitude specified by a $3-4-5$ linear slope ratio (Quadrant II).
    • Vector 3: $75\,\text{lbs}$ magnitude at an angle of $10^\circ$ relative to the vertical axis (Quadrant IV).

• Vector 1 Resolution ($100\,\text{lbs}$ at $30^\circ$ from Horizontal):

  • Quadrant I orientation results in positive $x$ and positive $y$ components.
  • Hypotenuse $H = 100\,\text{lbs}$, Angle $\theta = 30^\circ$ from horizontal.
  • Horizontal component: $F_{1x} = +100 \times \cos(30^\circ)$
  • Vertical component: $F_{1y} = +100 \times \sin(30^\circ)$

• Vector 2 Resolution ($50\,\text{lbs}$ with $3-4-5$ Slope):

  • Quadrant II orientation results in a negative $x$ component and a positive $y$ component.
  • Slope triangle calculation via Pythagorean theorem:         $\text{Hypotenuse} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5$
  • Proportional force decomposition using side ratios:
    • Horizontal side ratio ($x$-direction) = $3$ out of $5$.\n * Vertical side ratio (y$-direction) = 4$out of$5$.
  • Horizontal component: $F_{2x} = -\left(\frac{3}{5}\right) \times 50$
  • Vertical component: $F_{2y} = +\left(\frac{4}{5}\right) \times 50$

• Vector 3 Resolution ($75\,\text{lbs}$ at $10^\circ$ from Vertical):

  • Quadrant IV orientation results in a positive $x$ component and a negative $y$ component.
  • Hypotenuse $H = 75\,\text{lbs}$, Angle $\theta = 10^\circ$ measured from the vertical line.
  • Because the angle is referenced to vertical, the opposite side is horizontal ($x$) and the adjacent side is vertical ($y$).
  • Horizontal component: $F_{3x} = +75 \times \sin(10^\circ)$
  • Vertical component: $F_{3y} = -75 \times \cos(10^\circ)$

• Complete Force Summation Equations:

  • Net horizontal force equation:         $\sum F_x = 100 \times \cos(30^\circ) - \left(\frac{3}{5}\right) \times 50 + 75 \times \sin(10^\circ)$
  • Net vertical force equation:         $\sum F_y = 100 \times \sin(30^\circ) + \left(\frac{4}{5}\right) \times 50 - 75 \times \cos(10^\circ)$