ResultantForceExample 2025-05-30 06:36:32 2025-06-02 05:49:00 Example resultant force % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} %\usepackage{tikz} %\usetikzlibrary{arrows.meta} \section*{Resultant Forces with Vectors} In physics, the \textbf{resultant force} is the single force that represents the combined effect of multiple forces acting on an object. Forces are vector quantities, characterized by both magnitude and direction. To find the resultant force, we add the individual force vectors using vector addition. \subsection*{Example: Two Forces Acting on an Object} Consider an object subjected to two forces: \begin{itemize} \item \(\vec{F_1} = 5 \, \text{N}\) at \(0^\circ\) (along the positive \(x\)-axis). \item \(\vec{F_2} = 3 \, \text{N}\) at \(60^\circ\) from the positive \(x\)-axis. \end{itemize} We aim to find the resultant force \(\vec{F_R} = \vec{F_1} + \vec{F_2}\). \subsubsection*{Step 1: Resolve Forces into Components} Express each force in terms of its \(x\)- and \(y\)-components: \begin{itemize} \item For \(\vec{F_1}\): \[ F_{1x} = 5 \cos(0^\circ) = 5 \, \text{N}, \quad F_{1y} = 5 \sin(0^\circ) = 0 \, \text{N}. \] So, \(\vec{F_1} = (5, 0) \, \text{N}\). \item For \(\vec{F_2}\): \[ F_{2x} = 3 \cos(60^\circ) = 3 \cdot 0.5 = 1.5 \, \text{N}, \quad F_{2y} = 3 \sin(60^\circ) = 3 \cdot \frac{\sqrt{3}}{2} \approx 2.598 \, \text{N}. \] So, \(\vec{F_2} \approx (1.5, 2.598) \, \text{N}\). \end{itemize} \subsubsection*{Step 2: Sum the Components} The components of the resultant force \(\vec{F_R} = (F_{Rx}, F_{Ry})\) are: \[ F_{Rx} = F_{1x} + F_{2x} = 5 + 1.5 = 6.5 \, \text{N}, \] \[ F_{Ry} = F_{1y} + F_{2y} = 0 + 2.598 \approx 2.598 \, \text{N}. \] Thus, \(\vec{F_R} \approx (6.5, 2.598) \, \text{N}\). \subsubsection*{Step 3: Calculate Magnitude and Direction} The magnitude of \(\vec{F_R}\) is: \[ |\vec{F_R}| = \sqrt{F_{Rx}^2 + F_{Ry}^2} = \sqrt{6.5^2 + 2.598^2} = \sqrt{42.25 + 6.75} \approx \sqrt{49} \approx 7 \, \text{N}. \] The direction (angle \(\theta\) relative to the positive \(x\)-axis) is: \[ \theta = \tan^{-1}\left(\frac{F_{Ry}}{F_{Rx}}\right) = \tan^{-1}\left(\frac{2.598}{6.5}\right) \approx \tan^{-1}(0.4) \approx 21.8^\circ. \] Thus, the resultant force is approximately \(7 \, \text{N}\) at \(21.8^\circ\) from the positive \(x\)-axis. \subsection*{Diagram of Forces} The following diagram illustrates the two forces and their resultant: \begin{center} \setlength{\unitlength}{1cm} \begin{picture}(8,5) % Coordinate axes \put(0,1){\vector(1,0){7}} % x-axis \put(7,0.8){$x$} % x label \put(3,0){\vector(0,1){4}} % y-axis \put(2.8,4){$y$} % y label % Force F1 (5 N along x-axis) \thicklines {\color{blue}\put(3,1){\vector(1,0){5}}} % F1 \put(4.5,1.2){$\vec{F_1} = 5 \, \text{N}$} % F1 label % Force F2 (3 N at 60 degrees) {\color{red}\put(3,1){\vector(1,1.732){1.5}}} % F2 (cos 60 = 0.5, sin 60 = 0.866, scaled) \put(4,2.5){$\vec{F_2} = 3 \, \text{N}$} % F2 label % Resultant force F_R (6.5, 2.598) {\color{green}\put(3,1){\vector(1,0.3997){3.25}}} % F_R (scaled by 0.5 for visibility) \put(5,1.8){$\vec{F_R}$} % F_R label % Angle for F2 (60 degrees) \thinlines \put(3,1){\line(1,0){1}} % Base line \put(3,1){\line(1,1.732){0.5}} % 60-degree line \put(3.5,1.2){\arc[0.5]{0}{60}} % Arc for 60 degrees \put(3.7,1.3){$60^\circ$} % Angle label \end{picture} \end{center}