WORLD FOR CIVIL

Topic 1.3 - Rotational Equilibrium

The second condition for equilibrium is rotational equilibrium. We can see the need for this second condition if we look at the diagram 1.3a. In this diagram, if we apply the 1^st condition of equilibrium and sum the forces in the y-direction, we obtain zero. (+100 lb. - 100 lb. = 0). This would indicate that the object is in translational equilibrium. However, we almost instinctively recognize that the object certain will not remain at rest, and will experience rotational motion (and rotational acceleration).
Please notice that the object actually is in translational equilibrium. That is, even though it rotates, it rotates about the center of mass of the bar, and the center of mass of the bar will not move.

The second condition for equilibrium states that if we are to have rotational equilibrium, the sum of the Torque acting on the structure must be zero. Torque (or Moments) is normally covered in the first semester of a General College Physics course. (For an overview and review of Torque, please select: Subtopic 1.31 Torque)

The 2nd condition for equilibrium may be written:
or in three dimensions: , ,

Since, most of our problems will be dealing with structures in two dimensions, our normal rotational equilibrium condition will be: . That is, the sum of the torque acting on a structure with respect to any point selected must equal zero. Where, in two dimensions, one can only have counterclockwise(+) or clockwise(-) torque.

Now let's take a moment to go very slowly through an introductory problem involving both conditions of equilibrium, and applying our statics problem solving techniques. Select Torque Example 1.
Select Torque Example 2; Select Torque Example 3