Statics

Statics is the study of bodies and structures that are inequilibrium. For a body to be inequilibrium, there must be no netforce acting on it. In addition, there must be no nettorque acting on it.Figure 17A shows a body in equilibrium under the action of equal and opposite forces.Figure 17B shows a body acted on by equal and opposite forces that produce a net torque, tending to start it rotating. It is therefore not in equilibrium.

When a body has a net force and a net torque acting on it owing to a combination of forces, all the forces acting on the body may be replaced by a single (imaginary) force called theresultant, which acts at a single point on the body, producing the same net force and the same net torque. The body can be brought into equilibrium by applying to it a real force at the same point, equal and opposite to the resultant. This force is called theequilibrant. An example is shown inFigure 18.

The torque on a body due to a given force depends on the reference point chosen, since the torqueτ by definition equalsr ×F, wherer is avector from some chosen reference point to the point of application of the force. Thus, for a body to be at equilibrium, not only must the net force on it be equal to zero but the net torque with respect to any point must also be zero. Fortunately, it is easily shown for a rigid body that, if the net force is zero and the net torque is zero with respect to any one point, then the net torque is also zero with respect to any other point in theframe of reference.

A body is formally regarded as rigid if the distance between any set of two points in it is always constant. In reality no body is perfectly rigid. When equal and opposite forces are applied to a body, it is always deformed slightly. The body’s own tendency to restore thedeformation has the effect of applying counterforces to whatever is applying the forces, thus obeying Newton’s third law. Calling a body rigid means that the changes in thedimensions of the body are small enough to be neglected, even though the force produced by the deformation may not be neglected.

Equal and opposite forces acting on a rigid body may act so as to compress the body (Figure 19A) or to stretch it (Figure 19B). The bodies are then said to be undercompression or undertension, respectively. Strings, chains, and cables are rigid under tension but may collapse under compression. On the other hand, certain building materials, such as brick and mortar, stone, or concrete, tend to be strong under compression but very weak under tension.

The most important application of statics is to study thestability of structures, such asedifices and bridges. In these cases,gravity applies a force to each component of the structure as well as to any bodies the structure may need to support. The force of gravity acts on each bit of mass of which each component is made, but for each rigid component it may be thought of as acting at a single point, thecentre of gravity, which is in these cases the same as the centre of mass.

To give a simple but important example of the application of statics, consider the two situations shown inFigure 20. In each case, a massm is supported by two symmetric members, each making an angleθ with respect to the horizontal. InFigure 20A the members are under tension; inFigure 20B they are under compression. In either case, the force acting along each of the members is shown to be

The force in either case thus becomes intolerably large if the angleθ is allowed to be very small. In other words, the mass cannot be hung from thin horizontal members only capable of carrying either the compression or the tension forces of the mass.

The ancientGreeks built magnificent stonetemples; however, the horizontal stone slabs thatconstituted the roofs of the temples could not support even their own weight over more than a very small span. For this reason, one characteristic that identifies a Greek temple is the many closely spacedpillars needed to hold up the flat roof. The problem posed by equation (71) was solved by the ancientRomans, who incorporated into theirarchitecture the arch, a structure that supports its weight by compression, corresponding toFigure 20B.

Asuspension bridge illustrates the use of tension. The weight of the span and any traffic on it is supported by cables, which are placed under tension by the weight. Corresponding toFigure 20A, the cables are not stretched to be horizontal, but rather they are always hung so as to have substantial curvature.

It should be mentioned in passing thatequilibrium under static forces is not sufficient to guarantee the stability of a structure. It must also be stable against perturbations such as the additional forces that might be imposed, for example, by winds or by earthquakes. Analysis of the stability of structures under such perturbations is an important part of the job of an engineer or architect.

Rotation about a fixed axis

Consider a rigid body that is free to rotate about an axis fixed inspace. Because of the body’sinertia, it resists being set into rotationalmotion, and equally important, once rotating, it resists being brought to rest. Exactly how that inertial resistance depends on the mass and geometry of the body is discussed here.

Take theaxis of rotation to be thez-axis. A vector in thex-y plane from the axis to a bit of mass fixed in the body makes an angleθ with respect to thex-axis. If the body is rotating,θ changes with time, and the body’sangular frequency isω is also known as theangular velocity. Ifω is changing in time, there is also anangular accelerationα, such thatBecause linearmomentump is related to linear speedv byp =mv, wherem is the mass, and because forceF is related to accelerationa byF =ma, it is reasonable to assume that there exists a quantityI that expresses therotational inertia of the rigid body inanalogy to the waym expresses the inertial resistance to changes inlinear motion. One would expect to find that theangular momentum is given byand that thetorque (twisting force) is given by

One can imagine dividing the rigid body into bits of mass labeledm₁,m₂,m₃, and so on. Let the bit of mass at the tip of the vector be calledm_i, as indicated inFigure 21. If the length of thevector from the axis to this bit of mass isR_i, thenm_i’s linearvelocityv_i equalsωR_i (see equation [31]), and its angular momentumL_i equalsm_iv_iR_i (see equation [44]), orm_iR_i²ω. The angular momentum of the rigid body is found by summing all the contributions from all the bits of mass labeledi = 1, 2, 3 . . . :

In a rigid body, the quantity in parentheses in equation (76) is always constant (each bit of massm_i always remains the same distanceR_i from the axis). Thus if the motion is accelerated, thenRecalling thatτ =dL/dt, one may write

(These equations may be written in scalar form, sinceL andτ are always directed along theaxis of rotation in this discussion.) Comparing equations (76) and (78) with (74) and (75), one finds that

The quantityI is called themoment of inertia.

According to equation (79), the effect of a bit of mass on the moment of inertia depends on its distance from the axis. Because of the factorR_i², mass far from the axis makes a bigger contribution than mass close to the axis. It is important to note thatR_i is the distance from the axis, not from a point. Thus, ifx_i andy_i are thex andy coordinates of the massm_i, thenR_i² = x_i² +y_i², regardless of thevalue of thez coordinate. The moments of inertia of some simple uniform bodies are given in theClick Here to see full-size tabletable.

The moment of inertia of any body depends on the axis of rotation. Depending on thesymmetry of the body, there may be as many as three different moments of inertia about mutually perpendicular axes passing through thecentre of mass. If the axis does not pass through the centre of mass, the moment of inertia may be related to that about a parallel axis that does so. LetI_c be the moment of inertia about the parallel axis through the centre of mass,r the distance between the twoaxes, andM the total mass of the body. ThenIn other words, the moment of inertia about an axis that does not pass through the centre of mass is equal to the moment of inertia for rotation about an axis through the centre of mass (I_c) plus a contribution that acts as if the mass were concentrated at the centre of mass, which then rotates about the axis of rotation.

Thedynamics of rigid bodies rotating about fixed axes may be summarized in three equations. The angular momentum isL =Iω, the torque isτ =Iα, and thekinetic energy isK =1/2Iω².