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Forces in 2D - Vector Notation Simplified Revision Notes

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3.1.4 Forces in 2D - Vector Notation

infoNote

When approaching forces in 2D using vector notation, follow these steps:

  1. Express forces as vectors: Represent each force as a vector in component form, e.g., F=(FxFy)\mathbf{F} = \begin{pmatrix} F_x \\ F_y \end{pmatrix} , where FxF_x and Fy F_y are the horizontal and vertical components.
  2. Resolve forces into components: For forces at an angle, resolve them into horizontal and vertical components using:
  • Fx=Fcos(θ)F_x = F \cos(\theta)
  • Fy=Fsin(θ)F_y = F \sin(\theta)
  1. Apply Newton's second law: Use Fnet=ma\mathbf{F}_{\text{net}} = m\mathbf{a} to relate the net force to acceleration, with each force vector summed as:
  • Fx=max\sum F_x = m a_x
  • Fy=may\sum F_y = m a_y
  1. Solve the vector equations: Find unknown forces or accelerations by solving the component equations. This method ensures a structured approach to handling forces using vector notation.

Vector Problems in 3D

infoNote

Example 1: Finding the Position Vector

Problem: A parallelogram ABCDABCD has vertices A(212),B(371),C(642)A\left(\begin{array}{c}2 \\ 1 \\ 2\end{array}\right), B\left(\begin{array}{c}3 \\ 7 \\ 1\end{array}\right), C\left(\begin{array}{c}6 \\ 4 \\ -2\end{array}\right), and DD is unknown. Find the position vector of DD.

Solution:

  1. Use the property of parallelograms: In a parallelogram, opposite sides are equal and parallel. Thus, vector BC=AD\overrightarrow{BC} = \overrightarrow{AD}.
  2. Write the vector equation:
BC=DA\overrightarrow{BC} = \overrightarrow{D} - \overrightarrow{A}CB=DA\overrightarrow{C} - \overrightarrow{B} = \overrightarrow{D} - \overrightarrow{A}(642)(371)=D(212)\left(\begin{array}{c}6 \\ 4 \\ -2\end{array}\right) - \left(\begin{array}{c}3 \\ 7 \\ 1\end{array}\right) = \overrightarrow{D} - \left(\begin{array}{c}2 \\ 1 \\ 2\end{array}\right)

Simplifying:

(395)=D(212)\left(\begin{array}{c}3 \\ -9 \\ 5\end{array}\right) = \overrightarrow{D} - \left(\begin{array}{c}2 \\ 1 \\ 2\end{array}\right)
  1. Solve for D\overrightarrow{D}:
D=(395)+(212)=(587)\overrightarrow{D} = \left(\begin{array}{c}3 \\ -9 \\ 5\end{array}\right) + \left(\begin{array}{c}2 \\ 1 \\ 2\end{array}\right) = \left(\begin{array}{c}5 \\ -8 \\ 7\end{array}\right)

Answer: The position vector of DD is D=(587)\overrightarrow{D} = \left(\begin{array}{c}5 \\ -8 \\ 7\end{array}\right).


infoNote

Example 2: Magnitude of a Vector in 3D

Concept: The magnitude of a vector a\overrightarrow{a} in 3D is given by:

a=x2+y2+z2wherea=(xyz).|\overrightarrow{a}| = \sqrt{x^2 + y^2 + z^2}\quad where \quad \overrightarrow{a} = \left(\begin{array}{c}x \\ y \\ z\end{array}\right).

Problem: If A(279)andB(317)A\left(\begin{array}{c}2 \\7 \\ 9\end{array}\right) and B\left(\begin{array}{c}3 \\ -1 \\ 7\end{array}\right), find AB|\overrightarrow{AB}|.

Solution:

  1. Find the vector AB=BA\overrightarrow{AB} = \overrightarrow{B} - \overrightarrow{A}:
AB=(377)(219)=(182)\overrightarrow{AB} = \left(\begin{array}{c}3 \\ 7 \\ 7\end{array}\right) - \left(\begin{array}{c}2 \\ -1 \\ 9\end{array}\right) = \left(\begin{array}{c}1 \\ -8 \\ -2\end{array}\right)
  1. Calculate the magnitude AB|\overrightarrow{AB}|:
AB=12+82+22=69|\overrightarrow{AB}| = \sqrt{1^2 + 8^2 + 2^2} = \sqrt{69}

Answer: The magnitude AB|\overrightarrow{AB}| is 69\sqrt{69}.

Tips:

infoNote
  1. Resolve angled forces: Always break angled forces into horizontal and vertical components using Fx=Fcos(θ)F_x = F \cos(\theta) and Fy=Fsin(θ)F_y = F \sin(\theta).
  2. Use vector form: Write forces as vectors F=(FxFy)\mathbf{F} = \begin{pmatrix} F_x \\ F_y \end{pmatrix} to keep track of components easily.
  3. Apply equilibrium or motion equations: Use Fx=max\sum F_x = m a_x and Fy=may\sum F_y = m a_y to set up equations for solving unknowns.
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