rigid body

INTRODUCTION :
Conditions for Rigid-Body Equilibrium

The rigid body in figure (a), which is fixed in the x, y, z, reference at constant velocity. A free-body diagram of the arbitrary ίth particle of the body is shown in figure (b). There are two types of forces which act on it. The resultant internal forces, f , is caused by interaction with adjacent particles. The resultant external forces F represents, for example, the effects of gravitational, electrical, magnetic, or contact forces between the ίth particles is in equilibrium, then appliying not included within the body. If the particle is in equilibrium, then applying Newton’s first law we have When the equation of equilibrium is applied to each of the other particles of the body, similar equations will result. If all these equations are added together vectorially, we obtain

The summation of the internal forces will equal zero since the internal forces between particles within the body will occur in equal but opposite collinear pairs, Newton’s third law. Consequently, only the sum of the external forces will remain; and therefore, letting ∑F = ∑f, the above equation can be written as

Let consider the moments of the forces acting on the ίth particles about the arbitrary point O, figure (b). using the above particles equilibrium equation and the distributive law of the vector cross product we have

Similar equations can be written for the other particles of the body, and adding them together vectorially , we obtain The second term is zero since , as started above, the internal forces occur in equal but opposite collinear pairs, and therefore the resultant moment of each pairs of forces about point 0 is zero. Hence , using the notation ∑Mo = ∑r × ∑F , we have

Hence the two equations of equilibrium for a rigid body can be summarized as follows:

(1)
These equations require that a rigid body will remain in equilibrium provided the sum of all external external forces acting on the body is equal to zero and the sum moments of external forces about a point is equal to zero. The fact that these conditions are necessary for equilibrium has now been proven. They are also sufficient for maintaining equilibrium. To show this, lets assume that the body is in equilibrium and the force system acting on the body satisfies Eqs (1). Suppose that an additional force F’ is applied to the body. As a result, the equilibrium equations become

Where M’o is the moment of F’ about 0. Since ∑F = 0 and ∑Mo = 0,
Than we require F’ = 0 (also ∑M’o = 0 ). Consequently , the additional force F’ is not require, and indeed Eqs (1) are also sufficient condition maintaining equilibrium.