A consequence of Newton’s third law:
“For every force which exerts on , exerts an equal and oppositely directed force on .”
is that the total external force acting on a system is equal to the total mass times the acceleration of the center of mass. In particular, if the system is in equilibrium (i.e., every particle is at rest or moving with constant velocity) the total external force must be zero. However, as we can see from the example illustrated in Figure 1, the vanishing of the total external force is not sufficient to ensure that the system will be in equilibrium. Under the influence of the pair of equal and opposite forces applied to its ends, the rod in Figure 1 will start to spin. In this chapter we shall be concerned with what conditions must be satisfied in order that a body may remain in equilibrium. In this exposition we shall discuss how the body moves and spins when subjected to arbitrary forces.
The discussion of equilibrium relies heavily on the concept of torque, and the discussion of nonequilibrium situations also requires the concept of angular momentum which will be subsequently defined.
Our experience tells us that when a force acts on an extended body (e.g., a seesaw), the effect of the force depends not only on the magnitude and direction of the force, but also on the location of the point where the force acts. If we introduce an origin and draw a vector from to the point where the force acts, then the vector cross-product is called “the torque produced by the force about the origin ”. The torque depends on the location of the origin since the vector changes if we change the origin.
We denote torque by the vector (Greek letter “tau”); that is, The direction of is perpendicular to the plane containing and and (according to the right-hand rule) would point out of the page if and are the vectors shown in Figure 3. The magnitude of is , which can also be written or , where is the magnitude of the component of which is perpendicular to , and is the magnitude of the component of which is perpendicular to (see Figure 3).
Let us consider a system (any collection of particles) which is in equilibrium (i.e., each particle is in equilibrium). We enumerate the particles by an index . The total force on each particle is zero.
Furthermore, if we choose an origin and take the cross-product of both sides of (1) with (the vector from to the position of the th particle), we find
We recall that if we add up (1) for all , the internal forces cancel out as a result of Newton’s third law and we obtain where is the total external force on the system. Can we also say that the internal forces produce no net torque on the system?
To examine this question we decompose into an external and internal part where is the force exerted on by . If we now add up the torque equations (2) for all the particles we obtain
The terms in the double sum do not obviously cancel in pairs. For example, the terms () and () give which can be combined (using Newton’s third law ) to give . This cross-product will vanish if is parallel or antiparallel to . But is the vector from particle #2 to #1. Thus if the force between particles is parallel or antiparallel to the line between the particles, the cross-product will vanish and the internal forces will not contribute to the total torque. Forces which act along the line between the particles are called central forces; familiar examples are the gravitational force and the electrostatic force (which are mathematically of the same form). There are forces in nature which are not central forces, the most familiar Example being magnetic forces. Even in this case, however, it can be shown by a more elaborate argument that the internal forces make no contribution to the total torque. Were this not true then an isolated system would, under some circumstances, begin to rotate faster and faster and could continuously perform work without any energy input.
Accordingly, we assert (though a completely general proof is beyond the scope of this discussion) that, if a system is in equilibrium, the internal forces make no contribution to the total torque. Thus, where is the torque produced by the external forces which act on the system.
In most of the examples which we shall consider here, one of the external forces acting on a system is the force of gravity. This force acts on every particle of the system, and different particles are at different distances from the origin . However, there is an important theorem which makes it easy to calculate the torque produced by gravity.1
Theorem. For purposes of computing torques, the entire gravitational force on a system may be considered to act at the center of mass. (As illustrated in Figure 4, the gravitational torque on the system is where is the vector from the origin to the of the system, is the total mass, and is a unit vector pointing vertically up.)
The proof follows immediately from the definition of the center of mass,
The gravitational torque is
Inserting (3) into (4) we obtain which is the desired result. With the aid of this theorem we can now work out some examples.
A uniform beam of weight and length rests on two supports, one at the left end and the other at distance from the left end. Find the force on the beam exerted by each support.
We introduce unit vectors: pointing vertically up, pointing to the right, and into the paper as shown in Figure 5. The forces acting on the beam are acting on the left end, acting at distance from the left end, and the force of gravity which may be considered to act at the midpoint. Since the total force on the beam is zero, we have
The total torque around any origin must be zero. If we take the origin at the left end of the beam, the force contributes no torque and we obtain
Thus and therefore . Using (5) we find . We could equally well have taken torques about some other origin, for example, the other support. In this case we find which yields . Thus we see that the problem can be solved by writing one force equation and one torque equation, or by writing torque equations around two origins.
If the total force on a system is zero, and if the torque around some particular origin is zero, then it follows that the torque around any other origin is zero. To see this, let and be two origins, and let be the vector from to . Suppose the total external force is zero: and the external torque around is zero: where is the vector from to the th particle. The torque around is where is the vector from to the th particle. From Figure 6 we see that and thus . Therefore, if and vanish, then also vanishes.
This means that all the information about a given system is contained in the force equation plus the torque equation about one origin. Additional equations obtained by taking torques about other origins will be algebraic consequences of the force equation and the first torque equation.
We have shown that the vanishing of the total external force and torque are necessary conditions for the equilibrium of a system. Are these conditions also sufficient for equilibrium? If the system is not a rigid body the answer is evidently “no,” as can be seen by consideration of many simple examples, such as that illustrated in Figure 7. The total force and torque are zero, and the two particles will accelerate toward each other. If the system is a rigid body, and if at one instant all the particles of the system are at rest, it can be shown that all particles of the system will remain at rest if the total external force and torque vanish. The proof, which is given in the next chapter, involves an analysis of the possible motions of a rigid body (which are greatly restricted compared with the motions of an arbitrary collection of particles).
All the statics examples which we shall discuss are two dimensional; that is, all the particles and all the forces are in a plane (which we take as the plane of the paper). Accordingly, all torques are perpendicular to this plane and (in the notation of Example 1) are proportional to or . A torque proportional to is directed into the page and is frequently called a clockwise torque; a torque proportional to is directed out of the page and is frequently called a counterclockwise torque. Thus we can omit all vectors from the torque equation, provided that we remember to put opposite signs in front of clockwise and counterclockwise torques.
In Figure 5 the force produces a counterclockwise torque around the left end of the beam. The force produces a clockwise torque. Note that in Figure 5 the force is “trying” to turn the beam in a counterclockwise sense around the left end, whereas the force is “trying” to turn the beam in a clockwise sense around the left end. If we take the right end of the beam as our origin, then and produce clockwise torques and produces a counterclockwise torque. Students who have trouble with the signs should simply calculate the cross-products.
A uniform rod , of weight , is attached to a wall by a smooth hinge at and is kept horizontal by a weightless wire which makes angle with the horizontal. Calculate the tension in the wire and the horizontal and vertical force components exerted by the wall on the rod.
In Figure 9 we exhibit all the forces acting on the rod. and are the magnitudes of the vertical and horizontal forces exerted by the wall at ; these forces are assumed to have the directions indicated by the arrows. If turns out to be negative, the equations would be telling us that the vertical force exerted by the wall is directed down rather than up; similarly, a negative value for would mean that the wall exerts a horizontal force directed to the left. Note that there are three unknowns (, , ) and three equations which state the equilibrium conditions for the rod (two components of the force equation and one torque equation). Thus the problem is mathematically determinate.
The horizontal and vertical force equations are
If we take torques around , we find where is the length of the rod, and thus . Substituting into (6) we find and . The value of could have been found directly by taking torques around .
Most introductory courses devote very little time to statics and consequently few students acquire adequate technique to solve problems like this one, and still fewer can solve it efficiently. Nevertheless it is recommended that the student study this Example which illustrates a number of important points.
The geometry of this Example is similar to that of (Example 2); is a rod of weight , is a rod of weight , and the joints at , , and are smooth hinges. Calculate the horizontal and vertical forces exerted by the wall at and at and the horizontal and vertical forces exerted by each rod on the other at .
In this problem there are many unknown forces. The algebra will be greatly simplified if, from the outset, we eliminate some of the unknowns by making use of the force equations and Newton’s third law. It is also important to realize that there are three different systems (the rod , the rod , and the system consisting of both rods) for which one can write force and torque equations. (However, not all equations are algebraically independent. If the force and torque on any two of these systems vanish, the force and torque on the third system will also vanish.)
In Figure 10 we have exhibited all the external forces which act on (the forces at the joint are internal forces in the system ). We denote the horizontal and vertical forces at by and , with assumed directions as indicated by the arrows. The horizontal and vertical forces at are then determined by the requirement that the total horizontal and total vertical forces on must vanish. Similarly, Figure 11 shows the forces acting on (note that the forces at the right end are the forces exerted by rod on rod ). Figure 12 shows the forces acting on rod . Thus, by using the force equations we have reduced the number of unknowns to two.
We can determine most easily by taking torques on around the origin , obtaining ( length of ) and thus . An easy way to calculate is to take torques on around the origin (Figure 10), obtaining and thus . One can, of course, write other torque equations which lead to the same values of and . Inserting the values of and into Figures 10–12 we obtain all the forces.
A common error is to assume that the force which the rod exerts on the wall is parallel to the rod . If this were so then the force which the wall exerts on would also be parallel (or antiparallel) to and would produce no torque about ; thus the only torque on around the origin would be the torque produced by and the rod could not be in equilibrium (unless in which case the force exerted by on the wall is parallel to ).
It is important also to understand what is meant by a “smooth hinge,” and why we usually assume that a joint is smoothly hinged. We assume that the rods are connected to each other, and to brackets on the wall, by means of pins which are perpendicular to the plane of the paper and pass through circular holes in the rods (see Figures 13–15). It is assumed that the surface of contact between the pin and the hole is well lubricated so that the only forces which act at that surface are perpendicular to the surface. Thus, if we take the center of the hole as origin, we see that the pin exerts no net torque on the rod (but usually does exert a net force). If a rod is attached to a wall bracket by means of a sufficiently rusty hinge (Figure 15), the rod can remain in a horizontal position without additional support. If we take our origin at the center of the hole, the weight produces a clockwise torque on the rod; however, the tangential component of the force which the pin exerts on the surface of the hole produces a counterclockwise torque which (if the hinge is sufficiently rusty) has the same magnitude as the gravitational torque, and thus the rod is in equilibrium.