[{"data":1,"prerenderedAt":36},["ShallowReactive",2],{"topic-en-theoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems\u002Fequilibrium-rigid-body-two-dimensions":3},{"topic":4,"trail":17,"children":31,"tasks":32,"alternates":33},{"id":5,"name":6,"locale":7,"path":8,"seo_title":6,"seo_description":9,"seo_text":10,"content_html":11,"content_chunks":12},188,"Equilibrium of a Rigid Body in Two Dimensions","en","theoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems\u002Fequilibrium-rigid-body-two-dimensions","Rigid-body equilibrium in two dimensions: free-body diagrams, support reactions and statics equations for determining unknown forces.","This topic applies statics equations to rigid bodies in two dimensions. It covers free-body diagrams, replacement of constraints by reactions and calculation of unknown forces and support reactions.","\u003Cp>In practical statics problems, \u003Cstrong>equilibrium of a rigid body in two dimensions\u003C\u002Fstrong> is analyzed using a free-body diagram. Real supports and contacts are replaced by their reactions, and the equilibrium equations are then applied to the isolated body.\u003C\u002Fp>\u003Ch2>Free-body diagram\u003C\u002Fh2>\u003Cp>A free-body diagram should contain only the body being analyzed and all external forces and moments acting on it. The physical constraints are removed from the diagram and replaced by their corresponding reactions. Internal forces within the isolated rigid body are not shown.\u003C\u002Fp>\u003Ch2>Solution procedure\u003C\u002Fh2>\u003Col>\u003Cli>\u003Cstrong>Isolate the body.\u003C\u002Fstrong> Identify the object for which the equilibrium equations will be written.\u003C\u002Fli>\u003Cli>\u003Cstrong>Release the constraints.\u003C\u002Fstrong> Replace supports, contacts, cables, and other constraints by their appropriate reactions.\u003C\u002Fli>\u003Cli>\u003Cstrong>Show all loads.\u003C\u002Fstrong> Include concentrated forces, applied couple moments, weight, and equivalent resultants of distributed loads when already determined.\u003C\u002Fli>\u003Cli>\u003Cstrong>Choose axes and sign conventions.\u003C\u002Fstrong> Define positive $x$ and $y$ directions and the positive sense of moments.\u003C\u002Fli>\u003Cli>\u003Cstrong>Write the equilibrium equations.\u003C\u002Fstrong> For a general planar force system use $\\sum F_x=0$, $\\sum F_y=0$, and $\\sum M_O=0$.\u003C\u002Fli>\u003Cli>\u003Cstrong>Solve the equations.\u003C\u002Fstrong> Determine the unknown reactions and other required quantities.\u003C\u002Fli>\u003Cli>\u003Cstrong>Check the result.\u003C\u002Fstrong> Substitute the values into an unused check equation or verify all force and moment sums again.\u003C\u002Fli>\u003C\u002Fol>\u003Ch2>Static determinacy\u003C\u002Fh2>\u003Cp>For one rigid body subjected to a general planar force system, three independent equilibrium equations are available. If the correctly modeled constraints introduce no more than three independent unknown reaction components and those equations determine them uniquely, the problem may be statically determinate. Additional reaction unknowns generally require deformation relations beyond rigid-body statics.\u003C\u002Fp>\u003Ch2>Efficient order of equations\u003C\u002Fh2>\u003Cp>It is often best to begin with a moment equation about a point through which the lines of action of several unknown reactions pass. Those reactions then have zero moment about that point and disappear from the equation. After one unknown is found, force-component equations can be used for the others.\u003C\u002Fp>\u003Ch2>Example: simply supported beam\u003C\u002Fh2>\u003Cp>A 6 m beam is supported by a pin at $A$ and a roller at $B$. A vertical force $P=12$ kN acts at midspan. For this vertical loading, $A_x=0$. Taking moments about $A$ gives $B_y\\cdot6-12\\cdot3=0$, so $B_y=6$ kN. From $\\sum F_y=0$, $A_y+B_y-12=0$, giving $A_y=6$ kN.\u003C\u002Fp>\u003Ch2>Meaning of a negative reaction\u003C\u002Fh2>\u003Cp>The initial direction of an unknown reaction may be assumed. If the calculated value is negative, this is not automatically an error: the actual reaction acts opposite to the assumed direction. The result should still be checked against the physical contact model.\u003C\u002Fp>\u003Ch2>Checking the solution\u003C\u002Fh2>\u003Cp>After calculating the reactions, verify the force and moment sums again, preferably by taking moments about a different point. This helps reveal errors in signs, moment arms, or arithmetic.\u003C\u002Fp>\u003Ch2>Common mistakes\u003C\u002Fh2>\u003Cul>\u003Cli>writing equations before constructing a free-body diagram;\u003C\u002Fli>\u003Cli>modeling a support reaction incorrectly;\u003C\u002Fli>\u003Cli>including forces that act on another body rather than the isolated body;\u003C\u002Fli>\u003Cli>omitting an applied couple moment;\u003C\u002Fli>\u003Cli>using the distance to the force application point as the moment arm without checking perpendicularity.\u003C\u002Fli>\u003C\u002Ful>",[13],{"id":14,"code":15,"type":16,"locale":7},151,"statics-planar-rigid-body-procedure","algorithm",[18,22,26,30],{"id":19,"name":20,"path":21},81,"Theoretical Mechanics","theoretical-mechanics",{"id":23,"name":24,"path":25},136,"Statics","theoretical-mechanics\u002Fstatics",{"id":27,"name":28,"path":29},275,"Equilibrium of Coplanar Systems","theoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems",{"id":5,"name":6,"path":8},[],[],{"en":34,"uk":35},"https:\u002F\u002Fmechclassroom.com\u002Fen\u002Ftopics\u002Ftheoretical-mechanics\u002Fstatics\u002Fequilibrium-of-coplanar-systems\u002Fequilibrium-rigid-body-two-dimensions","https:\u002F\u002Fmechclassroom.com\u002Ftopics\u002Fteoretychna-mekhanika\u002Fstatyka\u002Frivnovaha-ploskykh-system\u002Frivnovaha-tverdoho-tila-na-ploshchyni",1787712535334]