[{"data":1,"prerenderedAt":37},["ShallowReactive",2],{"topic-en-theoretical-mechanics\u002Fstatics\u002Fthree-dimensional-statics\u002Fthree-dimensional-force-systems":3},{"topic":4,"trail":18,"children":32,"tasks":33,"alternates":34},{"id":5,"name":6,"locale":7,"path":8,"seo_title":9,"seo_description":10,"seo_text":11,"content_html":12,"content_chunks":13},191,"Three-Dimensional Force Systems","en","theoretical-mechanics\u002Fstatics\u002Fthree-dimensional-statics\u002Fthree-dimensional-force-systems","Three-Dimensional Force Systems — Statics","Three-dimensional force systems in statics: force components, moments about axes, resultant force vector and resultant moment in 3D problems.","This topic extends the main concepts of statics to three-dimensional force systems. It covers force components in space, moments and system characteristics required for spatial equilibrium analysis.","\u003Cp>A \u003Cstrong>three-dimensional force system\u003C\u002Fstrong> is one whose force lines of action are not confined to a single plane. Its analytical description requires three coordinate axes, vector moments, and the rules of the vector cross product.\u003C\u002Fp>\u003Ch2>Force in three dimensions\u003C\u002Fh2>\u003Cp>In Cartesian coordinates, a force is written as $\\vec F=F_x\\vec i+F_y\\vec j+F_z\\vec k$. The three components uniquely define the force vector, and their signs determine the component directions along the coordinate axes.\u003C\u002Fp>\u003Cp>If a three-dimensional force is specified by its Cartesian components $F_x$, $F_y$, and $F_z$, its magnitude is:\u003C\u002Fp>\u003Cp>$$F=\\sqrt{F_x^2+F_y^2+F_z^2}.$$\u003C\u002Fp>\u003Cp>The components are algebraic quantities whose signs determine the vector direction, while the force magnitude is always nonnegative.\u003C\u002Fp>\u003Cdiv data-formula-calculator-config=\"eyJ0aXRsZSI6IkZvcmNlIE1hZ25pdHVkZSBmcm9tIFRocmVlIENvbXBvbmVudHMiLCJmb3JtdWxhIjoiRiA9IOKImihG4oKTwrIgKyBG4bWnwrIgKyBGX3rCsikiLCJ2YXJpYWJsZXMiOlt7ImtleSI6ImZvcmNlIiwic3ltYm9sIjoiRiIsImxhYmVsIjoiRm9yY2UgbWFnbml0dWRlIiwicXVhbnRpdHkiOiJmb3JjZSIsImRlZmF1bHRVbml0IjoiTiJ9LHsia2V5IjoiZngiLCJzeW1ib2wiOiJG4oKTIiwibGFiZWwiOiJ4IGNvbXBvbmVudCIsInF1YW50aXR5IjoiZm9yY2UiLCJkZWZhdWx0VW5pdCI6Ik4ifSx7ImtleSI6ImZ5Iiwic3ltYm9sIjoiRuG1pyIsImxhYmVsIjoieSBjb21wb25lbnQiLCJxdWFudGl0eSI6ImZvcmNlIiwiZGVmYXVsdFVuaXQiOiJOIn0seyJrZXkiOiJmeiIsInN5bWJvbCI6IkZfeiIsImxhYmVsIjoieiBjb21wb25lbnQiLCJxdWFudGl0eSI6ImZvcmNlIiwiZGVmYXVsdFVuaXQiOiJOIn1dLCJzb2x2ZSI6eyJmb3JjZSI6Ik1hdGguc3FydChmeCAqIGZ4ICsgZnkgKiBmeSArIGZ6ICogZnopIiwiZngiOiJNYXRoLnNxcnQoZm9yY2UgKiBmb3JjZSAtIGZ5ICogZnkgLSBmeiAqIGZ6KSIsImZ5IjoiTWF0aC5zcXJ0KGZvcmNlICogZm9yY2UgLSBmeCAqIGZ4IC0gZnogKiBmeikiLCJmeiI6Ik1hdGguc3FydChmb3JjZSAqIGZvcmNlIC0gZnggKiBmeCAtIGZ5ICogZnkpIn19\">\u003C\u002Fdiv>\u003Ch2>Direction cosines\u003C\u002Fh2>\u003Cp>If $\\alpha$, $\\beta$, and $\\gamma$ are the angles between the force vector and the positive $x$, $y$, and $z$ axes, then $F_x=F\\cos\\alpha$, $F_y=F\\cos\\beta$, and $F_z=F\\cos\\gamma$. The direction cosines satisfy $\\cos^2\\alpha+\\cos^2\\beta+\\cos^2\\gamma=1$.\u003C\u002Fp>\u003Ch2>Force along a specified line\u003C\u002Fh2>\u003Cp>If a force is directed from point $A(x_A,y_A,z_A)$ toward point $B(x_B,y_B,z_B)$, first form the vector $\\overrightarrow{AB}$. Normalize it to obtain $\\vec e_{AB}=\\overrightarrow{AB}\u002F|\\overrightarrow{AB}|$, then write the force as $\\vec F=F\\vec e_{AB}$.\u003C\u002Fp>\u003Ch2>Moment of a force about a point\u003C\u002Fh2>\u003Cp>The moment of a force $\\vec F$ applied at point $A$ about point $O$ is $\\vec M_O=\\vec r\\times\\vec F$, where $\\vec r=\\overrightarrow{OA}$. The moment vector is perpendicular to the plane formed by $\\vec r$ and $\\vec F$, with its direction given by the right-hand rule.\u003C\u002Fp>\u003Ch2>Moment of a force about an axis\u003C\u002Fh2>\u003Cp>The moment about a coordinate or arbitrary axis equals the projection of the force-moment vector onto that axis. If $\\vec e$ is a unit vector along the axis, then $M_{axis}=\\vec e\\cdot(\\vec r\\times\\vec F)$.\u003C\u002Fp>\u003Ch2>Resultant force vector and resultant moment\u003C\u002Fh2>\u003Cp>For a three-dimensional force system, the resultant force vector is $\\vec R=\\sum\\vec F_i$, and the resultant moment about point $O$ is $\\vec M_O=\\sum(\\vec r_i\\times\\vec F_i)+\\sum\\vec M_j$. In a 3D problem, each of these vectors has three Cartesian components.\u003C\u002Fp>\u003Ch2>Example\u003C\u002Fh2>\u003Cp>Suppose a force has components $F_x=3$ kN, $F_y=4$ kN, and $F_z=12$ kN. Its magnitude is $F=\\sqrt{3^2+4^2+12^2}=13$ kN. The force vector can be written as $\\vec F=(3\\vec i+4\\vec j+12\\vec k)$ kN.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>using a planar moment sign convention for a three-dimensional moment vector;\u003C\u002Fli>\u003Cli>failing to normalize the vector between two points before multiplying it by a specified force magnitude;\u003C\u002Fli>\u003Cli>confusing the moment about a point with its projection onto an axis;\u003C\u002Fli>\u003Cli>reversing the order in $\\vec r\\times\\vec F$, which reverses the moment direction;\u003C\u002Fli>\u003Cli>omitting one of the three force or moment components.\u003C\u002Fli>\u003C\u002Ful>",[14],{"id":15,"code":16,"type":17,"locale":7},157,"statics-force-magnitude-3d","formula",[19,23,27,31],{"id":20,"name":21,"path":22},81,"Theoretical Mechanics","theoretical-mechanics",{"id":24,"name":25,"path":26},136,"Statics","theoretical-mechanics\u002Fstatics",{"id":28,"name":29,"path":30},276,"Three-Dimensional Statics","theoretical-mechanics\u002Fstatics\u002Fthree-dimensional-statics",{"id":5,"name":6,"path":8},[],[],{"en":35,"uk":36},"https:\u002F\u002Fmechclassroom.com\u002Fen\u002Ftopics\u002Ftheoretical-mechanics\u002Fstatics\u002Fthree-dimensional-statics\u002Fthree-dimensional-force-systems","https:\u002F\u002Fmechclassroom.com\u002Ftopics\u002Fteoretychna-mekhanika\u002Fstatyka\u002Fprostorova-statyka\u002Fprostorova-systema-syl",1787712535636]