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CIE A-Level Maths Study Notes

3.1.2 Vector Analysis of Forces

In the field of Mathematics, a comprehensive understanding of vector analysis in relation to forces is imperative. This subject area encompasses an in-depth examination of forces represented as vectors, including their component representation in two dimensions, and the computation of resultant forces through vector addition. Mastery of these concepts is crucial for a thorough understanding of the broader principles of mechanics and physics.

Concept of Force as a Vector

  • Vector Nature: Force is a vector, meaning it has magnitude and direction. Unlike scalars, which only have magnitude.
  • Physics Representation: Drawn as arrows. Arrow length = magnitude. Arrow direction = force direction.
Force Vector

Image courtesy of Calc Workshop

Components of Forces in Two Dimensions

  • Resolving Forces: Break down force into horizontal and vertical parts for 2D analysis.
  • Mathematical Approach:
    • Horizontal Component: Fx=Fcos(θ)F_x = F \cos(\theta)
    • Vertical Component: Fy=Fsin(θ)F_y = F \sin(\theta)
  • Mechanics Example: Essential for calculating things like cable tension or ramp force.

Calculating Resultants Using Vector Addition

  • Resultant Force Concept: Single force summing up multiple forces' effects on a body.
  • Algebraic Vector Addition:
    • Add horizontal and vertical components separately.
    • For forces F1=(F1x,F1y)\vec{F}1 = (F{1x}, F{1y}) and F2=(F2x,F2y)\vec{F}2 = (F{2x}, F{2y}), resultant R=(F1x+F2x,F1y+F2y)\vec{R} = (F{1x} + F{2x}, F{1y} + F{2y}).
  • Avoid Scale Drawing: Use algebra for accuracy, especially in complex or exam situations.

Application and Example Problems

Example 1: Resolving Forces

Question: Force of 50 N at 30°. Find horizontal and vertical components.

Solution:

  • Convert Angle: 30° to radians = 30×π18030 \times \frac{\pi}{180}.
  • Horizontal Component (Fx):Fx=50cos(30°in radians)( F_x ): F_x = 50 \cos(30° \, \text{in radians}).
  • Vertical Component (Fy):Fy=50sin(30°in radians)( F_y ): F_y = 50 \sin(30° \, \text{in radians}).
  • Results: Fx43.3N,Fy=25N F_x \approx 43.3 N, F_y = 25 N.
Resolving Forces graph

Example 2: Finding the Resultant Force

Question: Forces of 40 N at 0° and 30 N at 90°. Calculate the resultant force.

Solution:

  • Force 1 (40 N, 0°):
    • Horizontal Component (F1x):F1x=40N( F{1x} ): F{1x} = 40 N.
    • Vertical Component (F1y):F1y=0N( F{1y} ): F{1y} = 0 N.
  • Force 2 (30 N, 90°):
    • Horizontal Component (F2x):F2x=0N( F{2x} ): F{2x} = 0 N .
    • Vertical Component (F2y):F2y=30N( F{2y} ): F{2y} = 30 N.
  • Resultant Force (R):R=(40+0,0+30)=(40,30)N( \vec{R} ): \vec{R} = (40 + 0, 0 + 30) = (40, 30) N.
  • Magnitude of Resultant Force: R=402+30250N |\vec{R}| = \sqrt{40^2 + 30^2} \approx 50 N.
Resultant Force graph
Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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