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By the end of this section, you will be able to:
  • Identify the physical conditions of static equilibrium.
  • Draw a free-body diagram for a rigid body acted on by forces.
  • Explain how the conditions for equilibrium allow us to solve statics problems.

We say that a rigid body is in equilibrium    when both its linear and angular acceleration are zero relative to an inertial frame of reference. This means that a body in equilibrium can be moving, but if so, its linear and angular velocities must be constant. We say that a rigid body is in static equilibrium    when it is at rest in our selected frame of reference . Notice that the distinction between the state of rest and a state of uniform motion is artificial—that is, an object may be at rest in our selected frame of reference, yet to an observer moving at constant velocity relative to our frame, the same object appears to be in uniform motion with constant velocity. Because the motion is relative , what is in static equilibrium to us is in dynamic equilibrium to the moving observer, and vice versa. Since the laws of physics are identical for all inertial reference frames, in an inertial frame of reference, there is no distinction between static equilibrium and equilibrium.

According to Newton’s second law of motion, the linear acceleration of a rigid body is caused by a net force acting on it, or

k F k = m a CM .

Here, the sum is of all external forces acting on the body, where m is its mass and a CM is the linear acceleration of its center of mass (a concept we discussed in Linear Momentum and Collisions on linear momentum and collisions). In equilibrium, the linear acceleration is zero. If we set the acceleration to zero in [link] , we obtain the following equation:

First equilibrium condition

The first equilibrium condition for the static equilibrium of a rigid body expresses translational equilibrium:

k F k = 0 .

The first equilibrium condition, [link] , is the equilibrium condition for forces, which we encountered when studying applications of Newton’s laws.

This vector equation is equivalent to the following three scalar equations for the components of the net force:

k F k x = 0 , k F k y = 0 , k F k z = 0 .

Analogously to [link] , we can state that the rotational acceleration α of a rigid body about a fixed axis of rotation is caused by the net torque acting on the body, or

k τ k = I α .

Here I is the rotational inertia of the body in rotation about this axis and the summation is over all torques τ k of external forces in [link] . In equilibrium, the rotational acceleration is zero. By setting to zero the right-hand side of [link] , we obtain the second equilibrium condition:

Second equilibrium condition

The second equilibrium condition for the static equilibrium of a rigid body expresses rotational equilibrium:

k τ k = 0 .

The second equilibrium condition, [link] , is the equilibrium condition for torques that we encountered when we studied rotational dynamics. It is worth noting that this equation for equilibrium is generally valid for rotational equilibrium about any axis of rotation (fixed or otherwise). Again, this vector equation is equivalent to three scalar equations for the vector components of the net torque:

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Source:  OpenStax, University physics volume 1. OpenStax CNX. Sep 19, 2016 Download for free at http://cnx.org/content/col12031/1.5
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