We stated that one necessary condition for equilibrium is that the net force acting on an object be zero. If the object is treated as a particle, then this is the only condition that must be satisfied for equilibrium. The situation with real (extended) objects is more complex, however, because these objects cannot be treated as particles. For an extended object to be in static equilibrium, a second
condition must be satisfied. This second condition involves the net torque acting on the extended object. Note that equilibrium does not require the absence of motion. For example, a rotating object can have constant angular velocity and still be in equilibrium.
Consider a single force F acting on a rigid object, as shown in Figure 12.1. The effect of the force depends on its point of application P. If r is the position vector of this point relative to O, the torque associated with the force F about O is given by Equation 11.7:
Figure 12.1 A single force F acts on a rigid object at the point P.
Recall from the discussion of the vector product in Section 11.2 that the vector τ is perpendicular to the plane formed by r and F. You can use the right-hand rule to determine the direction of τ : Curl the fingers of your right hand in the direction of rotation that F tends to cause about an axis through O: your thumb then points in the direction of τ. Hence, in Figure 12.1 τ is directed toward you out of the page.
As you can see from Figure 12.1, the tendency of F to rotate the object about an axis through O depends on the moment arm d, as well as on the magnitude of F. Recall that the magnitude of τ is Fd (see Eq. 10.19). Now suppose a rigid object is acted on first by force F1 and later by force F2 . If the two forces have the same magnitude, they will produce the same effect on the object only if they have the same direction and the same line of action. In other words,
two forces F1 and F2 are equivalent if and only if F1 = F2 and if and only if the two produce the same torque about any axis.
The two forces shown in Figure 12.2 are equal in magnitude and opposite in direction. They are not equivalent. The force directed to the right tends to rotate the object clockwise about an axis perpendicular to the diagram through O, whereas the force directed to the left tends to rotate it counterclockwise about that axis.
Figure 12.2 The forces F1 and F2 are not equivalent because they do not produce the same torque about some axis, even though they are equal in magnitude and opposite in direction.
Suppose an object is pivoted about an axis through its center of mass, as shown in Figure 12.3. Two forces of equal magnitude act in opposite directions along parallel lines of action. A pair of forces acting in this manner form what is called a couple. (The two forces shown in Figure 12.2 also form a couple.) Do not make the mistake of thinking that the forces in a couple are a result of Newton’s
third law. They cannot be third-law forces because they act on the same object. Third-law force pairs act on different objects. Because each force produces the same torque Fd, the net torque has a magnitude of 2Fd. Clearly, the object rotates clockwise and undergoes an angular acceleration about the axis. With respect to rotational motion, this is a nonequilibrium situation. The net torque on the object gives rise to an angular acceleration α according to the relationship Στ = 2Fd = Iα (see Eq. 10.21).
Figure 12.3 Two forces of equal magnitude form a couple if their lines of action are different parallel lines. In this case, the object rotates clockwise. The net torque about any axis is 2Fd.
In general, an object is in rotational equilibrium only if its angular acceleration α = 0. Because Στ = Iα for rotation about a fixed axis, our second necessary condition for equilibrium is that the net torque about any axis must be zero. We now have two necessary conditions for equilibrium of an object:
The first condition is a statement of translational equilibrium; it tells us that the linear acceleration of the center of mass of the object must be zero when viewed from an inertial reference frame. The second condition is a statement of rotational equilibrium and tells us that the angular acceleration about any axis must be zero. In the special case of static equilibrium, which is the main subject of this chapter, the object is at rest and so has no linear or angular speed (that is, vCM = 0 and ω = 0).
The two vector expressions given by Equations 12.1 and 12.2 are equivalent, in general, to six scalar equations: three from the first condition for equilibrium, and three from the second (corresponding to x, y, and z components). Hence, in a complex system involving several forces acting in various directions, you could be faced with solving a set of equations with many unknowns. Here, we restrict our discussion to situations in which all the forces lie in the xy plane. (Forces whose vector representations are in the same plane are said to be coplanar.) With this restriction, we must deal with only three scalar equations. Two of these come from balancing the forces in the x and y directions. The third comes from the torque equation—namely, that the net torque about any point in the xy plane must be zero. Hence, the two conditions of equilibrium provide the equations
where the axis of the torque equation is arbitrary, as we now show.
Regardless of the number of forces that are acting, if an object is in translational equilibrium and if the net torque is zero about one axis, then the net torque must also be zero about any other axis. The point can be inside or outside the boundaries of the object. Consider an object being acted on by several forces such that the resultant force ΣF = F1 + F2 + F3 +....... = 0. Figure 12.4 describes this situation (for clarity, only four forces are shown). The point of application of F1 relative to O is specified by the position vector r1 . Similarly, the points of application of F2 , F3, . . . are specified by r2 , r3, . . . (not shown). The net torque about an axis through O is
Figure 12.4 Construction showing that if the net torque is zero about origin O, it is also zero about any other origin, such as O’.
Now consider another arbitrary point O’ having a position vector r’ relative to O. The point of application of F1 relative to O’ is identified by the vector r1 - r’. Likewise, the point of application of F2 relative to O’ is 22 - r’, and so forth. Therefore, the torque about an axis through O’ is
Because the net force is assumed to be zero (given that the object is in translational equilibrium), the last term vanishes, and we see that the torque about O’ is equal to the torque about O. Hence, if an object is in translational equilibrium and the net torque is zero about one point, then the net torque must be zero about any other point.