Chapter 14: Static Equilibrium 14.2 Lever Law Figures 14.3 Force diagrams for each body. Example 14.4.5 The KneeChapter 14: Static Equilibrium The proof of the correctness of a new rule can be attained by the repeated application of it, the frequent comparison with experience, the putting of it to the test under the most diverse circumstances. This process, would in the natural course of events, be carried out in time. The discoverer, however hastens to reach his goal more quickly. He compares the results that flow from his rule with all the experiences with which he is familiar, with all older rules, repeatedly tested in times gone by, and watches to see if he does not light on contradictions. In this procedure, the greatest credit is, as it should be, conceded to the oldest and most familiar experiences, the most thoroughly tested rules. Our instinctive experiences, those generalizations that are made involuntarily, by the irresistible force of the innumerable facts that press upon us, enjoy a peculiar authority; and this is perfectly warranted by the consideration that it is precisely the elimination of subjective caprice and of individual error that is the object aimed at. Ernst Mach, The Science of Mechanics1 14.1 Introduction: Static Equilibrium We have already seen in Section 4.4 that if the vector sum of the forces acting on a point-like object is zero then the object will continue in its state of rest, or of uniform motion in a straight line. If the object is in uniform motion we can always change reference frames so that the object will be at rest. In Section 9.2, we showed that for a collection of point-like objects the sum of the external forces may be regarded as acting at the center of mass (Equation 9.2.20). If that sum is zero, the center of mass will continue in its state of rest, or of uniform motion in a straight line. In Section 13.1 we introduced the idea of a rigid body, and again showed that in addition to the fact that the sum of the external forces may be regarded as acting at the center of mass, forces like the gravitational force that acts at every point in the body may be treated as acting at the center of mass. However for an extended rigid body it matters where the force is applied because even though the sum of the forces on the body may be zero, a non-zero sum of torques on the body may still produce angular acceleration. In particular for fixed axis rotation, the torque on the object is proportional to the angular acceleration (Equation 13.3.15). It is possible for a body that is not constrained to rotate about a fixed axis that the sum of the torques may be zero and the body still undergoes rotation. Thus, we would like to restrict ourselves to the special case in which in an inertial reference frame both the center of mass of the body is at rest and the body does not undergo any rotation, a condition that is called static equilibrium of an extended object. The two sufficient and necessary conditions for a rigid body to be in static equilibrium are: 1 Ernst Mach, The Science of Mechanics: A Critical and Historical Account of Its Development, translated by Thomas J. McCormack, Sixth Edition with Revisions through the Ninth German Edition, Open Court Publishing, Illinois. 8/25/2008 1(1) The sum of the forces acting on the rigid body is zero, total 1 2=+ +⋅⋅⋅=FFF 0GGGG. (14.1.1) (2) The vector sum of the torques about any point in a rigid body is zero, S ,total ,1 ,2SSS=+ +⋅⋅⋅= 0GGGGτττ . (14.1.2) When a body is in static equilibrium, the torques about any two points are equal (see the Appendix to this chapter). As a result, when solving static equilibrium problems, wisely choosing the point about which to compute torque can greatly simplify a given problem. 14.2 Lever Law Let’s consider a uniform rigid beam of mass balanced on a pivot at the center of mass of the beam. We place two point-like bodies 1 and 2 of masses and on the beam, at distances and respectively from the pivot, so that the beam is static (that is, the beam is not rotating. See Figure 14.1. The finite extent of the bodies, as represented in the figure, is not part of this derivation). Bm1m2m1d2d Figure 14.4 Pivoted Lever. Let’s consider the forces acting on the beam. The earth attracts the beam downward. This gravitational force acts on every atom in the beam, but we can summarize its action by stating that the gravitational force near the surface of the earth gravity Bm=FgGG is concentrated at a point in the beam called the center of gravity of the beam, which is identical to the center of mass of the uniform beam. There is also a contact force pivotFG between the pivot and the beam, acting upwards on the beam at the pivot point. The bodies 1 and 2 exert normal forces downwards on the beam, B,1 1≡NNGG, and , with magnitudes , and , respectively. Note that the normal forces are not the gravitational forces acting on the bodies, but contact forces between the beam and the body. (In this case, they are mathematically the same, due to the horizontal configuration of the beam and the fact that all objects are in static equilibrium.) The distances and are called the moment arms with respect to the pivot point for the B,2 2≡NNGG1N2N1d2d8/25/2008 2forces and , respectively. The force diagram on the beam is shown in Figure 14.2. Note that the pivot force 1NG2NGpivotFG and the force of gravity BmgG each has a zero moment arm about the pivot point. Figure 14.2 The Lever Law in action. Because we assume the beam is not moving, the sum of the forces in the vertical direction acting on the beam is therefore zero: pivot B 1 20FmgNN−−− =. (14.2.1) The force diagrams on the bodies are shown in Figure 14.3. Note the magnitude of the normal forces on the bodies are also and since these are each part of an action-reaction pair, , and 1N2NB,1 1,B=−NNGGB,2 2,B=−NNGG. Figures 14.3 Force diagrams for each body. The condition that the forces sum to zero is not sufficient to completely predict the motion of the beam. All we can deduce is that the center of mass of the system is at rest (or moving with a uniform velocity). In order for the beam not to rotate the sum of the torques about any point must be zero. In particular the sum of the torques about the pivot point must be zero. Since the moment arm of the gravitational force and the pivot force is zero, only the two normal forces
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