STATICS AND EQUILIBRIUM

CONCEPT

Statics, as its name suggests, is the study of bodies at rest. Those bodies may be acted upon by a variety of forces, but as long as the lines of force meet at a common point and their vector sum is equal to zero, the body itself is said to be in a state of equilibrium. Among the topics of significance in the realm of statics is center of gravity, which is relatively easy to calculate for simple bodies, but much more of a challenge where aircraft or ships are concerned. Statics is also applied in analysis of stress on materials—from a picture frame to a skyscraper.

HOW IT WORKS

Equilibrium and Vectors

Essential to calculations in statics is the use of vectors, or quantities that have both magnitude and direction. By contrast, a scalar has only magnitude. If one says that a certain piece of property has an area of one acre, there is no directional component. Nor is there a directional component involved in the act of moving the distance of 1 mi (1.6 km), since no statement has been made as to the direction of that mile. On the other hand, if someone or something experiences a displacement, or change in position, of 1 mi to the northeast, then what was a scalar description has been placed in the language of vectors.

Not only are mass and speed (as opposed to velocity) considered scalars; so too is time. This might seem odd at first glance, but—on Earth at least, and outside any special circumstances posed by quantum mechanics—time can only move forward. Hence, direction is not a factor. By contrast, force, equal to mass multiplied by acceleration, is a vector. So too is weight, a specific type of force equal to mass multiplied by the acceleration due to gravity (32 ft or [9.8 m] / sec²). Force may be in any direction, but the direction of weight is always downward along a vertical plane.

VECTOR SUMS.

Adding scalars is simple, since it involves mere arithmetic. The addition of vectors is more challenging, and usually requires drawing a diagram, for instance, if trying to obtain a vector sum for the velocity of a car that has maintained a uniform speed, but has changed direction several times.

One would begin by representing each vector as an arrow on a graph, with the tail of each vector at the head of the previous one. It would then be possible to draw a vector from the tail of the first to the head of the last. This is the sum of the vectors, known as a resultant, which measures the net change.

Suppose, for instance, that a car travels north 5 mi (8 km), east 2 mi (3.2 km), north 3 mi (4.8 km), east 3 mi, and finally south 3 mi. One must calculate its net displacement—in other words, not the sum of all the miles it has traveled, but the distance and direction between its starting point and its end point. First, one draws the vectors on a piece of graph paper, using a logical system that treats the y axis as the north-south plane, and the x axis as the east-west plane. Each vector should be in the form of an arrow pointing in the appropriate direction.

Having drawn all the vectors, the only remaining one is between the point where the car's journey ends and the starting point—that is, the resultant. The number of sides to the resulting shape is always one more than the number of vectors being added; the final side is the resultant.

In this particular case, the answer is fairly easy. Because the car traveled north 5 mi and ultimately moved east by 5 mi, returning to a position of 5 mi north, the segment from the resultant forms the hypotenuse of an equilateral (that is, all sides equal) right triangle. By applying the Pythagorean theorem, which states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides, one quickly arrives at a figure of 7.07 m (11.4 km) in a northeasterly direction. This is the car's net displacement.

Calculating Force and Tension in Equilibrium

Using vector sums, it is possible to make a number of calculations for objects in equilibrium, but these calculations are somewhat more challenging than those in the car illustration. One form of equilibrium calculation involves finding tension, or the force exerted by a supporting object on an object in equilibrium—a force that is always equal to the amount of weight supported. (Another way of saying this is that if the tension on the supporting object is equal to the weight it supports, then the supported object is in equilibrium.)

In calculations for tension, it is best to treat the supporting object—whether it be a rope, picture hook, horizontal strut or some other item—as though it were weightless. One should begin by drawing a free-body diagram, a sketch showing all the forces acting on the supported object. It is not necessary to show any forces (other than weight) that the object itself exerts, since those do not contribute to its equilibrium.

RESOLVING X AND Y COMPONENTS.

As with the distance vector graph discussed above, next one must equate these forces to the x and y axes. The distance graph example involved only segments already parallel to x and y, but suppose—using the numbers already discussed—the graph had called for the car to move in a perfect 45°-angle to the northeast along a distance of 7.07 mi. It would then have been easy to resolve this distance into an x component (5 mi east) and a y component (5 mi north)—which are equal to the other two sides of the equilateral triangle.

This resolution of x and y components is more challenging for calculations involving equilibrium, but once one understands the principle involved, it is easy to apply. For example, imagine a box suspended by two ropes, neither of which is at a 90°-angle to the box. Instead, each rope is at an acute angle, rather like two segments of a chain holding up a sign.

The x component will always be the product of tension (that is, weight) multiplied by the cosine of the angle. In a right triangle, one angle is always equal to 90°, and thus by definition, the other two angles are acute, or less than 90°. The angle of either rope is acute, and in fact, the rope itself may be considered the hypotenuse of an imaginary triangle. The base of the triangle is the x axis, and the angle between the base and the hypotenuse is the one under consideration.

Hence, we have the use of the cosine, which is the ratio between the adjacent leg (the base) of the triangle and the hypotenuse. Regardless of the size of the triangle, this figure is a constant for any particular angle. Likewise, to calculate the y component of the angle, one uses the sine, or the ratio between the opposite side and the hypotenuse. Keep in mind, once again, that the adjacent leg for the angle is by definition the same as the x axis, just as the opposite leg is the same as the y axis. The cosine (abbreviated cos), then, gives the x component of the angle, as the sine (abbreviated sin) does the y component.

REAL-LIFE APPLICATIONS

Equilibrium and Center of Gravity in Real Objects

Before applying the concept of vector sums to matters involving equilibrium, it is first necessary to clarify the nature of equilibrium itself—what it is and what it is not. Earlier it was stated that an object is in equilibrium if the vector sum of the forces acting on it are equal to zero—as long as those forces meet at a common point.

This is an important stipulation, because it is possible to have lines of force that cancel one another out, but nonetheless cause an object to move. If a force of a certain magnitude is applied to the right side of an object, and a line of force of equal magnitude meets it exactly from the left, then the object is in equilibrium. But if the line of force from the right is applied to the top of the object, and the line of force from the left to the bottom, then they do not meet at a common point, and the object is not in equilibrium. Instead, it is experiencing torque, which will cause it to rotate.

VARIETIES OF EQUILIBRIUM.

There are two basic conditions of equilibrium. The term "translational equilibrium" describes an object that experiences no linear (straight-line) acceleration; on the other hand, an object experiencing no rotational acceleration (a component of torque) is said to be in rotational equilibrium.

Typically, an object at rest in a stable situation experiences both linear and rotational equilibrium. But equilibrium itself is not necessarily stable. An empty glass sitting on a table is in stable equilibrium: if it were tipped over slightly—that is, with a force below a certain threshold—then it would return to its original position. This is true of a glass sitting either upright or upside-down.

Now imagine if the glass were somehow propped along the edge of a book sitting on the table, so that the bottom of the glass formed the hypotenuse of a triangle with the table as its base and the edge of the book as its other side. The glass is in equilibrium now, but unstable equilibrium, meaning that a slight disturbance—a force from which it could recover in a stable situation—would cause it to tip over.

If, on the other hand, the glass were lying on its side, then it would be in a state of neutral equilibrium. In this situation, the application of force alongside the glass will not disturb its equilibrium. The glass will not attempt to seek stable equilibrium, nor will it become more unstable; rather, all other things being equal, it will remain neutral.

CENTER OF GRAVITY.

Center of gravity is the point in an object at which the weight below is equal to the weight above, the weight in front equal to the weight behind, and the weight to the left equal to the weight on the right. Every object has just one center of gravity, and if the object is suspended from that point, it will not rotate.

One interesting aspect of an object's center of gravity is that it does not necessarily have to be within the object itself. When a swimmer is poised in a diving stance, as just before the starting bell in an Olympic competition, the swimmer's center of gravity is to the front—some distance from his or her chest. This is appropriate, since the objective is to get into the water as quickly as possible once the race starts.

By contrast, a sprinter's stance places the center of gravity well within the body, or at least firmly surrounded by the body—specifically, at the place where the sprinter's rib cage touches the forward knee. This, too, fits with the needs of the athlete in the split-second following the starting gun. The sprinter needs to have as much traction as possible to shoot forward, rather than forward and downward, as the swimmer does.

Tension Calculations

In the earlier discussion regarding the method of calculating tension in equilibrium, two of the three steps of this process were given: first, draw a free-body diagram, and second, resolve the forces into x and y components. The third step is to set the force components along each axis equal to zero—since, if the object is truly in equilibrium, the sum of forces will indeed equal zero. This makes it possible, finally, to solve the equations for the net tension.

Imagine a picture that weighs 100 lb (445 N) suspended by a wire, the left side of which may be called segment A, and the right side segment B. The wire itself is not perfectly centered on the picture-hook: A is at a 30° angle, and B on a 45° angle. It is now possible to find the tension on both.

First, one can resolve the horizontal components by the formula F _x = T _Bx + T _Ax = 0, meaning that the x component of force is equal to the product of tension for the x component of B, added to the product of tension for the x component of A, which in turn is equal to zero. Given the 30°-angle of A, _Ax = 0.866, which is the cosine of 30°. _Bx is equal to cos 45°, which equals 0.707. (Recall the earlier discussion of distance, in which a square with sides 5 mi long was described: its hypotenuse was 7.07 mi, and 5/7.07 = 0.707.)

Because A goes off to the left from the point at which the picture is attached to the wire, this places it on the negative portion of the x axis. Therefore, the formula can now be restated as T _B(0.707)−T _A(0.866) = 0. Solving for T _B reveals that it is equal to T _A(0.866/0.707) = (1.22)T _A. This will be substituted for T _B in the formula for the total force along the y component.

However, the y-force formula is somewhat different than for x: since weight is exerted along the y axis, it must be subtracted. Thus, the formula for the y component of force is F _y = T _Ay + T _By−w = 0. (Note that the y components of both A and B are positive: by definition, this must be so for an object suspended from some height.)

Substituting the value for T _B obtained above, (1.22)T _A, makes it possible to complete the equation. Since the sine of 30° is 0.5, and the sine of 45° is 0.707—the same value as its cosine—one can state the equation thus: T _A(0.5) + (1.22)T _A(0.707)−100 lb = 0. This can be restated as T _A(0.5 + (1.22 · 0.707)) = T _A(1.36) = 100 lb. Hence, T _A = (100 lb/1.36) = 73.53 lb. Since T _B = (1.22)T _A, this yields a value of 89.71 lb for T _B.

Note that T _A and T _B actually add up to considerably more than 100 lb. This, however, is known as an algebraic sum—which is very similar to an arithmetic sum, inasmuch as algebra is simply a generalization of arithmetic. What is important here, however, is the vector sum, and the vector sum of T _A and T _B equals 100 lb.

CALCULATING CENTER OF GRAVITY.

Rather than go through another lengthy calculation for center of gravity, we will explain the principles behind such calculations.

It is easy to calculate the center of gravity for a regular shape, such as a cube or sphere—assuming, of course, that the mass and therefore the weight is evenly distributed throughout the object. In such a case, the center of gravity is the geometric center. For an irregular object, however, center of gravity must be calculated.

An analogy regarding United States demographics may help to highlight the difference between geometric center and center of gravity. The geographic center of the U.S., which is analogous to geometric center, is located near the town of Castle Rock in Butte County, South Dakota. (Because Alaska and Hawaii are so far west of the other 48 states—and Alaska, with its great geographic area, is far to the north—the data is skewed in a northwestward direction. The geographic center of the 48 contiguous states is near Lebanon, in Smith County, Kansas.)

The geographic center, like the geometric center of an object, constitutes a sort of balance point in terms of physical area: there is as much U.S. land to the north of Castle Rock, South Dakota, as to the south, and as much to the east as to the west. The population center is more like the center of gravity, because it is a measure, in some sense, of "weight" rather than of volume—though in this case concentration of people is substituted for concentration of weight. Put another way, the population center is the balance point of the population, if it were assumed that every person weighed the same amount.

Naturally, the population center has been shifting westward ever since the first U.S. census in 1790, but it is still skewed toward the east: though there is far more U.S. land west of the Mississippi River, there are still more people east of it. Hence, according to the 1990 U.S. census, the geographic center is some 1,040 mi (1,664 km) in a southeastward direction from the population center: just northwest of Steelville, Missouri, and a few miles west of the Mississippi.

The United States, obviously, is an "irregular object," and calculations for either its geographic or its population center represent the mastery of numerous mathematical principles. How, then, does one find the center of gravity for a much smaller irregular object? There are a number of methods, all rather complex.

To measure center of gravity in purely physical terms, there are a variety of techniques relating to the shape of the object to be measured. There is also a mathematical formula, which involves first treating the object as a conglomeration of several more easily measured objects. Then the x components for the mass of each "sub-object" can be added, and divided by the combined mass of the object as a whole. The same can be done for the y components.

Using Equilibrium Calculations

One reason for making center of gravity calculations is to ensure that the net force on an object passes through that center. If it does not, the object will start to rotate—and for an airplane, for instance, this could be disastrous. Hence, the builders and operators of aircraft make exceedingly detailed, complicated calculations regarding center of gravity. The same is true for shipbuilders and shipping lines: if a ship's center of gravity is not vertically aligned with the focal point of the buoyant force exerted on it by the water, it may well sink.

In the case of ships and airplanes, the shapes are so irregular that center of gravity calculations require intensive analyses of the many components. Hence, a number of companies that supply measurement equipment to the aerospace and maritime industries offer center of gravity measurement instruments that enable engineers to make the necessary calculations.

On dry ground, calculations regarding equilibrium are likewise quite literally a life and death matter. In the earlier illustration, the object in equilibrium was merely a picture hanging from a wire—but what if it were a bridge or a building? The results of inaccurate estimates of net force could affect the lives of many people. Hence, structural engineers make detailed analyses of stress, once again using series of calculations that make the picture-frame illustration above look like the simplest of all arithmetic problems.

WHERE TO LEARN MORE

Beiser, Arthur. Physics, 5th ed. Reading, MA: Addison-Wesley, 1991.

"Determining Center of Gravity" National Aeronautics and Space Administration (Web site). <http://www.grc.nasa.gov/WWW/K-12/airplane/cg.html> (March 19, 2001).

"Equilibrium and Statics" (Web site). <http://www.glenbrook.k12.il.us/gbssci/phys/Class/vectors/u313c.html> (March 19, 2001).

"Exploratorium Snack: Center of Gravity." The Exploratorium (Web site). <http://www.exploratorium.edu/snacks/center_of_gravity.html> (March 19, 2001).

Faivre d'Arcier, Marima. What Is Balance? Illustrated by Volker Theinhardt. New York: Viking Kestrel, 1986.

Taylor, Barbara. Weight and Balance. Photographs by Peter Millard. New York: F. Watts, 1990.

"Where Is Your Center of Gravity?" The K-8 Aeronautics Internet Textbook (Web site). <http://wing.ucdavis.edu/Curriculums/Forces_Motion/center_howto.html> (March 19, 2001).

Wood, Robert W. Mechanics Fundamentals. Illustrated by Bill Wright. Philadelphia, PA: Chelsea House, 1997.

Zubrowski, Bernie. Mobiles: Building and Experimenting with Balancing Toys. Illustrated by Roy Doty. New York: Morrow Junior Books, 1993.

KEY TERMS

ACCELERATION:

A change in velocity.

CENTER OF GRAVITY:

The point on an object at which the total weights on either side of all axes (x, y, and z) are identical. Each object has just one center of gravity, and if it is suspended from that point, it will be in a state of perfect rotational equilibrium.

COSINE:

For an acute (less than 90°) angle in a right triangle, the cosine (abbreviated cos) is the ratio between the adjacent leg and the hypotenuse. Regardless of the size of the triangle, this figure is a constant for any particular angle.

DISPLACEMENT:

Change in position.

EQUILIBRIUM:

A state in which vector sum for all lines of force on an object is equal to zero. An object that experiences no linear acceleration is said to be in translational equilibrium, and one that experiences no rotational acceleration is referred to as being in rotational equilibrium. An object may also be in stable, unstable, or neutral equilibrium.

FORCE:

The product of mass multiplied by acceleration.

FREE-BODY DIAGRAM:

A sketch showing all the outside forces acting on an object in equilibrium.

HYPOTENUSE:

In a right triangle, the side opposite the right angle.

RESULTANT:

The sum of two or more vectors, which measures the net change in distance and direction.

RIGHT TRIANGLE:

A triangle that includes a right (90°) angle. The other two angles are, by definition, acute, or less than 90°.

SCALAR:

A quantity that possesses only magnitude, with no specific direction. Mass, time, and speed are all scalars. The opposite of a scalar is a vector.

SINE:

For an acute (less than 90°) anglein a right triangle, the sine (abbreviated sin) is the ratio between the opposite legand the hypotenuse. Regardless of the size of the triangle, this figure is a constant for any particular angle.

STATICS:

The study of bodies at rest. Those bodies may be acted upon by a variety of forces, but as long as the vector sum for all those lines of force is equal to zero, the body itself is said to be in a state of equilibrium.

TENSION:

The force exerted by a supporting object on an object in equilibrium—a force that is always equal to the amount of weight supported.

VECTOR:

A quantity that possesses both magnitude and direction. Force is a vector; so too is acceleration, a component of force; and likewise weight, a variety of force. The opposite of a vector is a scalar.

VECTOR SUM:

A calculation, made by different methods according to the factor being analyzed—for instance, velocity or force—that yields the net result of all the vectors applied in a particular situation.

VELOCITY:

The speed of an object in a particular direction. Velocity is thus a vector quantity.

WEIGHT:

A measure of the gravitational force on an object; the product of mass multiplied by the acceleration due to gravity.