Statics: Equilibrium of Rigid Bodies



“Equations are more important to me, because politics is for the present, but an equation is something for eternity”
Albert Einstein
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[Inertial reference frames are frames moving at constant velocity with respect to one another]

 Linear acceleration,  acm= 0, therefore  Fext = Macm = 0, where

eqn1
or

eqn2

In this course we will limit ourselves to problems in which all forces are acting in a plane, meaning that there will only be x and y equations.

  • ROTATIONAL EQUILIBRIUM

Angular acceleration = 0, therefore eqn5, where

eqn3

or
eqn4

In this course we will limit ourselves to problems in which all forces act in a plane, which means there can only be torques about an axis perpendicular to this plane (typically the z axis).  Therefore, only the z torque equation above will be necessary.

    • exclThe choice of axis about which to calculate torques is arbitrary.  If the sum of the torques about one axis is zero the body is not rotating; using a different axis to calculate torques will not cause the body to rotate, thus the total torque must still be zero.

    • hotAlthough all axes are arbitrary in any specific situation there are typically certain choices of axes about which to calculate torques which will make your life easier.

    1. Choose an axis through which as many forces as possible act, the torque of these forces about such an axis are zero.

    1. Choose an axis such that one or more of the unknown forces in the problem act through this axis.

  • excl General Comments
    • If all forces acting on a body are concurrent then rotational equilibrium is assured, we need only apply the conditions for translational equilbrium.
    • If there is more than one body in the problem consider the equilibrium of each body separately.



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extreme see saw


"In the long run we are all dead"
John Maynard Keynes – A tract on Monetary Reform (1923)


 

Dr. C. L. Davis

Physics Department

University of Louisville

email: c.l.davis@louisville.edu