EQUILIBRIUM
Consider these objects: (1) a book resting on a table, (2) a hockey puck sliding across a frictionless surface with constant velocity, (3) the rotating blades of a ceiling fan, and (4) the wheel of a bicycle that is traveling along a straight path at constant speed. For each of these four objects:
1. The linear momentum P of its center of mass is constant.
2. Its angular momentum L about its center of mass, or about any other point, is also constant.
We say that such objects are in equilibrium. The two requirements for equilibrium are then
P = a constant and L = a constant. ( Eq. 1 )
We are concerned largely with objects that are not moving in any way – either in translation or in rotation – in the reference frame from which we observe them. Such objects are in static equilibrium. Of the four objects mentioned at the beginning of this section, only one – the book resting on the table – is in static equilibrium.
If a body returns to a state of static equilibrium after having been displaced from it by a force, the body is said to be in stable static equilibrium. A marble placed at the bottom of a hemispherical bowl is an example. If, however, a small force can displace the body and end the equilibrium, the body is in unstable static equilibrium.
For example, suppose we balance a domino as in Figure e-1a, with the domino’s center of mass vertically above a supporting edge. The torque about the supporting edge due to the domino’s weight W is zero, because the line of action W is through that edge: Since the weight then cannot cause the domino to rotate about the edge, the domino is in equilibrium. Of course, even a slight force on it due to some chance disturbance ends the equilibrium: as the weight’s line of action moves to one side of the supporting edge ( as in Figure e-1b, torque due to the weight increases the domino’s rotation. Thus the domino in Figure e-1a is in unstable static equilibrium.
The domino in Figure e-1c is not quite as unstable. To topple this domino, a force would have to rotate it through and beyond the balance position of figure e-1a, in which the center of mass is above a supporting edge. So, a slight force will not topple this domino, but a vigorous flick of the finger against the domino certainly will. (If we arrange a chain of such upright dominos, a finger flick against the first can cause the whole chain to fall.)
Figure e-1 (a) A domino balanced on one edge, with its center of mass vertically above that edge. The domino’s weight W is directed through the supporting edge. (b) If the domino is rotated even slightly from the balanced orientation, its weight W cause a torque that increases the rotation. (c) A domino upright on a narrow side is somewhat more stable than the domino in (a). (d) A square block is even more stable.
The child’s square block in Figure e-1d is even more stable because its center of mass would have to be moved even farther to get it to pass above a supporting edge. A flick of the finger may not topple the block. (This is why you never see a chain of toppling square blocks.) The analysis of static equilibrium is very important in engineering practice. The design engineer must isolate and identify all the external forces and torques that may act on a structure and, by good design and wise choice of materials, make sure that the structure will tolerate these loads. Such analysis is necessary to make sure, for example, that bridges do not collapse under their traffic and wind loads, and that the landing gear of aircraft will survive the shock of rough landings.
THE REQUIREMENTS OF EQUILIBRIUM
The translational motion of a body is governed by Newton’s second law in its linear momentum form.
( Eq. 2 )If the body is in translational equilibrium, that is, if P is a constant, then dP/dt = 0 and we must have
Σ Fext = 0 (balance of force) ( Eq. 3 )
The rotational motion of a body is governed by Newton’s second law in its angular momentum form,
( Eq. 4 )If the body is in rotational equilibrium, that is, if L is a constant, then dL/dt = 0 and we must have
Σ τext = 0 (balance of torque) ( Eq. 5 )
Thus the two requirements for a body to be in equilibrium are as follows:
1. The vector sum of all external forces that act on the body must be zero.
2. The vector sum of all the external torques that act on the body, measured about any possible point, must also be zero.
These requirements obviously hold for static equilibrium as well as the more general equilibrium in which P and L are constant but not zero.
Equation 3 and Equation 5, as vector equations, are each equivalent to three independent scalar equations, one for each direction of the coordinate axes:
Balance of Balance of ( Eq. 6)
Forces Forces
------------- -------------
Σ Fx = 0 Σ τx = 0
Σ Fy = 0 Σ τy = 0
Σ Fz = 0 Σ τz = 0
For convenience, we have dropped the subscript ext. We shall simplify matters by considering only situations in which the forces that act on the body lie in the xy plane. This means that the only torques that can act on the body must tend to cause rotation around an axis parallel to the z axis. With this assumption, we eliminate one force equation and two torque equations from Equation-6, leaving
Σ Fx = 0 (balance of forces) ( Eq.7 )
Σ Fy = 0 (balance of forces) ( Eq.8 )
Σ τz = 0 (balance of forces) ( Eq.9 )
Here, Fx and Fy are, respectively, the x and the y components of the external forces that act on the body, and τz represents the torques that these forces exert either about the z axis or about any axis parallel to it.
A hockey puck sliding at constant velocity over ice satisfies Equations 7,8 and 9 and is thus in equilibrium but not in static equilibrium. For static equilibrium, the linear momentum P of the puck must be not only constant but also zero; the puck must be resting on the ice. Thus there is another requirement for static equilibrium:
3. The linear momentum P of the body must be zero.
THE CENTER OF GRAVITY
The weight W of an extended body is the vector sum of the gravitational forces acting on the individual elements (the atoms) of the body. Instead of considering all those individual elements, we say that a single force W effectively acts at a single point called the center of gravity (cg) of the body. By effectively we mean that if the forces on the individual elements were somehow tuned off and force W at the center of gravity turned on, the net force and the net torque (about any point) acting on the body would not change.
Up until now, we have assumed that the weight W acts at the center of mass (cm) of the body, which is equivalent to assuming that the center of gravity is at the center of mass. Here we show that this assumption is valid provided that acceleration g due to the gravitational force is constant over the body.
Figure e-2a shows an extended body, of mass M, and one of its elements, of mass mi. Each such element has weight mi gi, where gi is the acceleration due to the gravitational force at the location of the element. In Figure e-2a, each weight mi gi produces a torque τi on the element about the origin 0. Using Equation; τ = r ┴F ( Equation 10a) , we can write torque τi as
τi = xi mi gi, ( Eq. 10 )
Figure e-2 (a) An element of mass mi in an extended body has weight mi gi with moment arm xi about the origin 0 of a coordinates system. (b) The weight W of a body is said to act at the center of gravity (cg) of the body. The moment arm of W about 0 is equal to Xcg.
where xi is the moment arm r ┴ of force mi gi. The net torque on all the element of the body is then
τnet = Σ τi = Σ xi mi gi ( Eq. 11 )
Figure 3-2b shows the body’s weight W acting at the body’s center of gravity. From Eq. 10a, the torque about 0 due to W is
τ = xcg W ( Eq. 12 )
where xcg is the moment arm of W. The body’s weight W is equal to the sum of the weight mi gi of its elements. So we can substitute Σ migi for W in Eq. 12 to write
τ = xcg Σ mi gi, ( Eq. 13 )
By definition, the torque due to weight W acting at the center of gravity is equal to the net torque due to the weighs of all the individual elements of the body. So, τ in Eq. 13 is equal to τnet in equation 11, and we can write
xcg mi gi = xi mi gi ( Eq. 14 )
If the acceleration gi are all equal, we can cancel them from the two sides of Eq. 14. Substituting the body’s mass M for Σ mi on the left side of Eq. 14 and then rearranging yield
( Eq. 15 )We see that the right side of Eq. 15 gives the coordinate xcm for the center of mass. We can now write
xcg = xcm ( Eq. 16 )
Thus, the body’s center of gravity and its center of mass have the same x coordinate.
We can generalize this result for three dimensions by using vector notation. The generalized result is: If the gravitational acceleration is constant over a body, the body’s center of gravity is at its center of mass. For the rest of this book, we shall assume that these points coincide.
INDETERMINATE STRUCTURES
We have only three independent equations at our disposal, usually two balance of forces equations and one balance of torques equation about a given axis. Thus if a problem has more than three unknowns, we cannot solve it.
Consider also an unsymmetrical loaded car. What are the forces – all different – on the four tire? Again, we cannot find them because we have only three independent equations with which to work. Similarly, we can solve an equilibrium problem for a table with three legs but not one with four legs. Problems like these, in which there are more unknowns than equations, are called indeterminate.
And yet, solution to indeterminate problems exist in the real world. If you rest the tires of the car on four platform scales, each scale will register a definite reading, the sum of the readings being the weight of the car. What is eluding us in our efforts to solve equations for the individual scale readings?
The problem is that we have assumed – without making a great point of it – that the bodies to which we apply the equation of static equilibrium are perfectly rigid. By this we
Mean that they do not deform when forces are applied to them. Strictly, there are no such bodies. The tires of the car, for example, deform easily under load until the car settles into a position of static equilibrium.
We have all had experience with a wobbly restaurant table, which we usually level by putting folded paper under one of the legs. If a big enough elephant sat on such a table, however, you may be sure that if the table did not collapse, it would deform just like the tires of a car. Its legs would all touch the floor, the forces acting upward upward on the table legs would all assume definite ( and different) values and the table would no longer wobble. But how do we find the values of those forces acting on the legs?
To solve such indeterminate equilibrium problems, we must supplement equilibrium equations with some knowledge elasticity, the branch of physics and engineering that described how real bodies deform when forces are applied to them. The next section provides an introduction to this subject.
ELASTICITY
When a large number of atoms come together to form a metallic solid, such as an iron nail, they settle into equilibrium positions in a three-dimensional lattice, a repetitive arrangement in which each atom has a well-defined equilibrium distance from its nearest neighbors. The atoms are held together by interatomic forces that are represented by springs. The lattice is remarkably rigid, which is another way of saying that the “interatomic springs” are extremely stiff. It is for this reason that we perceive many ordinary objects such as metal ladders, tables, and spoons as perfectly rigid. Of course, some ordinary objects, such as garden hoses or rubber gloves, do not strike us as rigid at all. The atoms that make up these objects do not form a rigid lattice but are aligned in long flexible molecular chains, each chain being only loosely bound to its neighbors.
All real “rigid” bodies are to some extent elastic, which means that we can change their dimensions slightly by pulling, pushing, twisting, or compressing them. To get a feeling for the orders of magnitude involved, consider a vertical steel rod 1 m long and 1 cm in diameter. If you hang a subcompact car from the end of such a rod, the rod will stretch, but only by about 0.5 mm, or 0.05%. Furthermore, the rod will return to its original length when the car is removed.
If you hang two cars from the rod, the rod will be permanently stretched and will not recover its original length when you remove the load. If you hang three cars from the rod, the rod will break. Just before rupture, the elongation of the rod will be less than 0.2%. Although deformations of this size seem small, they are important in engineering practice. (Whether a wing under load will stay on an airplane is obviously important.) The stresses and the strains take different forms, but – over the range of engineering usefulness – stress and strain are proportional to each other. The constant of proportionality is called a modulus of elasticity, so that
Stress = modulus x strain ( Eq. 17 )
In a standard test, the tensile stress on a test cylinder is slowly increased from zero to the point where the cylinder fractures, and the strain is carefully measured and graphed. For a substantial range of applied stresses, the stress – strain relation is linear, and the specimen recovers its original dimensions when the stress is removed; it is here that Equation 17 applies. If the stress is increased beyond the yield strength Sy of the specimen, the specimen becomes permanently deformed. If the stress continues to increase, the specimen eventually rupture, at a stress called the ultimate strength Su.
Tension and Compression
For simple tension or compression, the stress is defined as F/A, the force divided by the area over which it acts. The strain, or unit deformation, is then the dimensionless quantity ∆L/L, the fractional (or sometimes percentage) change in the length of the specimen. If the specimen is long rod and the stress does not exceed the yield strength, then not only the entire rod but also every section of it experiences the same strain when a given stress is applied. Because the strain is dimensionless, the modulus in Equation 17 has the same dimensions as the stress, namely, force per unit area.
The modulus for tensile and compressive stresses is called the Young’s modulus and is represented in engineering practice by the symbol E. Equation 17 becomes
( Eq. 18 )The strain in ∆L/L in a specimen can often be measured conveniently with a strain gauge. These simple and useful devices, which can be attached directly to operating machinery with adhesives, are based on the principle that the electrical properties of the gauge are dependent on the strain it undergoes.
Although the Young’s modulus for an object may be almost the same for tension and compression, the object’s ultimate strength may well be different for the two types of stress. Concrete, for example, is very strong in compression but is so weak in tension that it is almost never used it that manner.
Shearing
In the case of shearing, the stress is also force per unit area, but the force vector lies in the plane of the area rather than at right angles to it. The strain is the dimensionless ratio ∆x/L. The corresponding modulus, which is given the symbol G in engineering practice, is called the shear modulus. For shearing, Equation 17 is written as
( Eq. 19 )Shearing stresses play a critical role in the buckling of shafts that rotate under load and in bone fractures caused by bending.
Hydraulic Stress
The stress is the fluid pressure p on the object, it is a force per unit area. The strain is ∆V/V, where V is the original volume of the specimen and ∆V is the absolute value of the change in volume. The corresponding modulus, with symbol B, is called the bulk modulus of the material. The object is said to be under hydraulic compression, and the pressure can be called the hydraulic stress. For this situation, we write Equation17 as
( Eq. 20 )The bulk modulus of water is 2.2 x 109 N/m2, and that of steel is 16 x 1010 N/m2. The pressure at the bottom of the Pacific Ocean, at its average depth of about 4000m, is 4.0 x 107 N/m2. The fractional compression ∆V/V of a volume of water due to this pressure is 1.8%; that for a steel object is only about 0.025%. In general, solid - -their rigid atomic lattices – are less compressible tightly coupled to their neighbors.
