By equating to zero the coefficients of the unit vectors in each of the two relations (see the above equation), we obtain the desired scalar equations. We observe that as many as three unknown reaction components may be eliminated from these computations through a judicious choice of the point ‘O’. 0 0 0 0 0 0 x y z x y z M M M F F F Equations can be solved for at most 6 unknowns which generally are reactions. From the one vector equation we may obtain the two scalar equations. ( F x) i + ( F y) j + ( F z) k 0 This vector equation will be satisfied only when F x 0 F y 0 F z 0 These equations are the three scalar equations of equilibrium. Johnston, Vector Mechanics for Engineers: Statics (V.1). Normally this will be a system with the origin at the particle and a horizontal x axis and a vertical. The particle will be the object or point where the lines of action of all the forces intersect. Next, we compute all vector products, either by direct calculation or using determinants. For general 3-D body, six scalar equations required to express equilibrium. This equation can be written in terms of its x, y and z components. The general procedure for solving equilibrium of a particle problems in two dimensions is to: Identify the particle. We write and express the forces ‘F’ and position vectors ‘r’ regarding scalar components and unit vectors. These equations can be solved for no more than six unknowns, which generally will represent reactions at supports or connections.ĭuring solving the most of the problems, the above scalar equations will be more conveniently obtained if we first expressed in vector form the conditions for the equilibrium of the rigid body considered. Six scalar equations are required to express the conditions for the equilibrium of a rigid body in the general three-dimensional case, those equations are listed below:
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