| 12 | Mechanical Mathematics |
In this chapter, you’ll learn how mathematics is applied to basic problems in the branch of physics known as classical mechanics. This involves the relationship among force, mass, displacement (or distance), velocity (or speed), and acceleration. Ultimately, mechanical phenomena can take the form of work, energy, or power.
WHAT IS FORCE?
Force is a measure of “push.” A cart with a heavy load begins to move steadily when it is pushed (Fig. 12-1A). If the cart has a lighter load, the same “push” will make it go much faster (Fig. 12-1B). Alternatively, greater push can move the heavier load more quickly, too (Fig. 12-1C).
Figure 12-1
At A, a heavy load moves slowly when pushed. At B, a light load is moved more quickly by the same effort. At C, a heavy load moves faster when pushed harder.
Force is needed to start and stop movement. However, in the absence of friction, movement would continue unchanged indefinitely, unless or until some force changed it or stopped it (Fig. 12-2). For simplicity, assume that no friction exists; then force is needed only to start, stop, or change motion. From observation, force is directly proportional to the weight to be moved. Force is also directly proportional to the acceleration (rate at which motion increases) or the deceleration (rate at which motion decreases).
Figure 12-2
Force is needed to slow down or stop movement, as well as to start it or increase it.
UNITS OF FORCE
The unit for measuring force in the foot-pound-second system is the poundal (pdl). It is the force needed to accelerate (or decelerate) a weight (usually called mass) of one pound (1 lb) so that it changes its speed by one foot per second per second (1 ft/s2).
The metric unit of force is the newton (N). A force of 1 N accelerates or decelerates a mass of one kilogram (1 kg) by one meter per second per second (1 m/s2). Another rarely used metric unit is the dyne (dyn), which accelerates a mass of one gram (1 g) by one centimeter per second per second (1 cm/s2).
Whatever units you measure force with, they are equivalent to mass times acceleration. That is proportional to mass multiplied by distance, divided by time squared. That is what “per second per second” means. It isn’t a misprint! It can be translated to “per second, every second” or “per second squared.” You will better understand this concept as you continue through this chapter.
Notation is not always consistent in classical mechanics. For example, some texts will abbreviate second(s) as sec rather than s. Miles per hour is sometimes abbreviated as mph and at other times as mi/h. You’ll see all four of these notational variants in this book. That will help you get used to them all; you’ll know what is meant by the context.
ACCELERATION, SPEED, AND DISTANCE
The “per second per second” part sounds confusing at first, as if I repeated myself by mistake. Look at it in a way that avoids this repetition.
Assume that you travel by car at 40 miles per hour (mph or mi/h). For the next minute, you accelerate steadily. Then one minute (1 min) later, you are moving at 60 mi/h. This acceleration is 20 miles per hour per minute [(mi/h)/min]. Although exact repetition is avoided in this case, “per hour per minute” is still used. This is not a standard unit of acceleration, but it’s a perfectly valid one, and it should help you understand the principle by not using repetitive units.
If acceleration is steady, the average speed during that minute will be midway between the start and finish speeds, that is, 50 mi/h. The distance traveled during that minute will be the same as if this average speed had been used for the whole minute, that is, 5/6 mi.
A standard unit of acceleration is feet per second per second (ft/s2). Suppose that you undergo a steady, straight-line acceleration of 10 ft/s2 from a standstill. At the start, you are not moving. After 1 s, you will have accelerated by 10 ft/s2 to a speed of 10 ft/s. The average speed for the first second will be 5 ft/s, halfway between 0 ft/s and 10 ft/s. So you will travel 5 ft during this first second.
During the next second, the speed will change from 10 ft/s to 20 ft/s, for an average speed of 15 ft/s. Thus, you will travel 15 ft in that second, a total of 20 ft from your starting position. The average speed over 2 s is 10 ft/s.
During the third second, the speed increases from 20 ft/s to 30 ft/s, an average of 25 ft/s, to travel 25 ft, making a total of 45 ft from the starting point. The average speed over 3 s is 15 ft/s.
You can tabulate distances traveled for any number of seconds from the start, as shown in Fig. 12-3A. If you plot the result as a graph, you get a quadratic curve, the familiar parabola (Fig. 12-3B).
Figure 12-3
An example of how acceleration affects distance traveled as time goes by, tabulated (A) and plotted as a graph (B). In this case, the acceleration rate is a constant 10 ft/s2.
If t is the total time from the start in seconds (when the speed is 0, a standstill) and a is the acceleration in feet per second per second, then the speed ν at the end of time t is at (a times t) feet per second. (Speed is symbolized v for “velocity” here, because speed is actually an expression of straight-line velocity.) So the average speed between a starting speed of 0 ft/s and a finishing speed of at ft/s is at/2 ft/s. Because the time is t seconds, the distance d in feet traveled from the start is (at/2) × t, or at2/2. So you can say these things:
Notice that acceleration uses units of length (distance) per time per time, such as feet per second per second (ft/s2) or miles per hour per minute [(mi/h)/min]. Speed results from multiplying acceleration by time, and thus it is length per time, such as feet per second (ft/s) or miles per hour (mph). Distance covered is obtained by multiplying speed by time. For example, feet per second (ft/s) times seconds (s) yields feet because the seconds “cancel out.”
You can also look at it the other way. Speed is a measure of distance covered per unit time. So you would divide feet by seconds to get feet per second (ft/s). Acceleration is a measure of speed change per unit time. You would divide feet per second by seconds to get feet per second per second, or feet per second squared (ft/s2). You can manipulate the above two formulas to get
These formulas apply only when the starting speed is 0 and the starting time is also 0. If either of these things is not the case, or if the acceleration is not constant, things get more complicated.
FORCE AND WORK
Force is a measure of “push” or “pull,” as stated earlier. Work (as a mathematical term) is a measure of what is done by a force. If nothing moves, such as when you lean against something, the force is there, but no work is performed. Work results when an applied force causes movement.
Force is a mass (m) times an acceleration (a). Work (w) is proportional to applied force (F) and distance moved (d). Work is equal to force times distance (Fd), so it must be proportional to mass times acceleration times distance (mad). Distance moved, using constant acceleration from a standstill, is acceleration multiplied by time squared, divided by 2 (at2/2). So work is mass times acceleration times at2/2, or ma2t2/2. Because speed (ν) is acceleration multiplied by time (at), this expression for work can be simplified to
Units of work are the foot-poundal (ft · pdl), the joule (J), and the erg. The foot-poundal represents the work done by a force of 1 pdl moving through 1 ft. The joule, which is the standard unit of work used by most scientists and engineers these days, represents the work done by a force of 1 N moving through 1 m (1 N · m). The erg, occasionally used when the amount of work done is very small, represents 1 dyn moving through 1 cm (1 dyn · cm). The small elevated dot in the composite unit expressions technically represents multiplication.
The formula w = mv2/2 represents the work of bringing a specific mass to a certain speed, regardless of the acceleration. If the acceleration is 10 ft/s2, then 80 ft and 4 s are required to reach a speed of 40 ft/s. If the acceleration is 5 ft/s2, it takes 8 s and 160 ft. Either way, 800,000 ft · pdl will move 1000 lb from standstill to 40 ft/s, although time and distance differ.
WORK AND ENERGY
Work and energy are expressed in the same units. For example, suppose an archer pulls back the string of a bow. The string is pulled back by a force that is equal to (or a little greater than) the tension of the string. The amount of work that is needed to pull the bow string back is stored in the bow as energy. When the archer releases the arrow, the string’s thrust accelerates it to flight speed. This work transfers the energy of the drawn bow to the energy of the arrow in flight.
Energy is a capacity for doing work and, conversely, work is the transfer of energy from one form or place to another. So both use the same units—foot-poundals, joules, or ergs—according to the system of units employed.
When the energy E is in the form of a mass in motion, the appropriate formula is
This formula can be used to determine the work needed to attain this motion, or to determine the energy actually manifest in it.
ENERGY AND POWER
Power is an expression of the rate at which work is done, or the rate at which energy is transferred from one form to another. As work is force applied over a distance, power is force applied over a distance per unit time.
You already know the units used, but an example will illustrate the relations between power and the other quantities. Assume the question relates to power vs. the weight of a car. Suppose that one motor unit develops a power of 290,000 foot-poundals per second (ft · pdl/s), and another has twice the power, or 580,000 ft · pdl/s. Coupled with these power levels, different weights must be moved. One car weighs 1500 lb, and the other weighs 3000 lb.
Energy is expressed in the form mv2/2. So, power must be in the form mv2/2t. Transposing this, using the symbol p for power, we find that the time for a mass to reach a given speed is mv2/2p. From this formula you can find the time taken by the smaller power unit with the smaller weight, the smaller power unit with the greater weight, and the greater power unit with the greater weight. If you want, you can complete the set by taking the greater power with the smaller weight, but we won’t deal with that situation here.
Figure 12-4 tabulates the time needed in each case to reach 44 ft/s, 88 ft/s, and 132 ft/s, which are the speeds that correspond to 30 mi/h, 60 mi/h, and 90 mi/h. Notice that the time needed is related to the square of the speed to be reached. At constant acceleration, the speed is directly proportional to the elapsed time. At constant power, acceleration must diminish as speed increases.
Figure 12-4
An example of how power and weight relate to the time required for two different cars to reach speeds of 30, 60, and 90 mi/h.
GRAVITY AS A SOURCE OF ENERGY
Until now, to keep the units basic (such as the pound, the foot per second, and so on), gravity has been left out of our calculations involving force, work, and power. The constant vertical force of gravity that acts around us, however, provides a convenient means of storing and concentrating energy.
A pile driver illustrates this principle. First, work is done by lifting a weight against the force of gravity (Fig. 12-5A). The weight is not accelerated upward, but it is lifted steadily against a constant force, which is gravity pulling downward. Just as energy or work is force times distance, this energy takes the form of distance lifted times weight (Fig. 12-5B). Moving twice the distance requires twice the work, and therefore stores twice the energy. Speed is not involved—yet. Less power requires more time to do the same amount of work.
Figure 12-5
Principle of the pile driver. At A and B, work lifting a weight stores energy. At C, energy is released when the weight is allowed to fall back down.
When the weight reaches the top, it is released to drop on the pile. Gravity is a mutual pull between earth and any mass. Doubling the mass doubles the pull (weight). So, in free fall, any object will drop at the same acceleration, which is approximately 32 ft/s2 or 9.8 m/s2 at the earth’s surface.
As the object accelerates downward, it stores energy at the rate of mv2/2 as a result of its motion (Fig. 12-5C). When it hits the pile, this stored energy is concentrated for a very short time, thrusting the pile downward.
WEIGHT AS FORCE
Weight provides a steady downward force F on any object with mass. The force is found from the mass m on which gravity acts, multiplied by the acceleration a of gravity, which produces 32 ft/s2 or 9.8 m/s2. That is,
The force needed to prevent an object from falling is equal to the pull of gravity on the weight. As the pull of gravity accelerates the weight downward, the force of gravity on a mass of 1 lb is equal to 32 pdl. On a 2-lb mass, the force is 64 pdl. If the mass is 1 kg, the force is 9.8 N. If the mass is 2 kg, the force is 19.6 N, and so on.
Otherwise stated, when you use gravity on a mass of 1 lb, it becomes a 1-lb weight, exerting a force of 1 lb (32 pdl in absolute nongravitational units). The same thing happens with kilograms and newtons. For this reason, in basic or absolute force units, pounds and kilograms express mass, which is a scientific way of saying how much “stuff” an object has. However, in gravitational units, pounds and kilograms refer to weight. That is the way the layperson usually thinks of mass, anyway.
GRAVITATIONAL MEASURE OF WORK
In gravitational measure of work, force does not have to accelerate a mass. Gravity exerts a force continuously on everything. If something doesn’t fall, it’s so because an equal force supports it, pushing it up.
If a 10-lb object (gravity acting on a 10-lb mass) rests on the floor, the object presses on the floor with a weight of 10 lb, which is a force of 320 pdl. Correspondingly, the floor pushes upward against the object with a force of 320 pdl to prevent it from falling. If a 10-kg object (gravity acting on a 10-kg mass) rests on the floor, the object presses on the floor with a weight of 10 kg, which is a force of 98 N.
You might ask, “Does the floor change its upward force according to what is on it?” The answer, surprisingly enough, is yes. If you hold the 10-lb object, your feet press on the floor, and the floor presses on your feet with a force that is 10 lb more than it would be with only you standing on the floor.
All the time that these forces operate, they balance. They are said to be forces in equilibrium. If the floor cannot provide enough upward push, it collapses, and work—most likely destructive—is done!
ENERGY FOR CONSTANT ACCELERATION
It is simplest to assume that acceleration is constant, which means that speed increases at a uniform rate, such as 1 ft/s2. Then, as well as representing a steady growth in speed, acceleration represents a steady force. However, constant acceleration does not correspond with a constant rate of work.
The faster an object goes, the more power is produced by a given applied force. Remember that work is equal to force times distance. Therefore, exerting or maintaining a constant force at higher and higher speed produces (or requires, depending on the situation) more and more work, energy, or power as time goes by. Figure 12-6 illustrates an example of this.
Figure 12-6
How much energy is needed to maintain constant acceleration?
POTENTIAL AND KINETIC ENERGY
The two forms of mechanical energy are called potential energy and kinetic energy. They can be described as follows.
• Potential energy is a function of the position of an object in the presence of a force. It is proportional to force and distance.
• Kinetic energy exists as a result of an object in motion. It is proportional to mass and speed squared.
Potential energy can be built from movement. But as long as start and finish are both the same or at some constant speed, movement is not involved in the calculation, as it is with kinetic energy. In the pile driver, for instance, a little more force is needed to start its upward movement. While it ascends at a constant rate, force and movement are both constant. A little less force is used to reach the top, if it stops before being released. The overall work needed to lift it is weight times height lifted.
KINETIC ENERGY AND SPEED
At constant acceleration, such as when a weight falls by the pull of gravity, energy builds in proportion to time squared. This happens because energy is proportional to speed squared, as shown in the example of Fig. 12-7.
Figure 12-7
Kinetic energy is proportional to the square of the speed.
Viewed another way, energy is proportional to force times distance. But because distance, at constant acceleration, increases in proportion to time squared, energy also increases in proportion to time squared. Constant acceleration means that the speed grows in direct proportion to the elapsed time.
ACCELERATION AT CONSTANT POWER
The rate of work (power) at constant acceleration increases with speed, requiring progressively greater power during acceleration. By rearranging the formula that relates kinetic energy and power, you get
So if power p and mass m are both constant, the speed v must increase in proportion to the square root (or 1/2 power) of the time t. You can see this by rearranging the above equation as follows:
To illustrate, suppose that constant power enables an accelerated object to reach a speed of 100 ft/s in 20 s from a standstill start. You can calculate the speed at any time during this 20-s period. You can make this calculation without knowing the mass or power involved. After the full 20 s, the speed reaches 100 ft/s. Note that 100 squared is 10,000. Divide this number in proportion to time: 2,000 for 4 s, 4,000 for 8 s, 6,000 for 12 s, and 8,000 for 16 s. Then take the square roots to find the speed at each of these times in feet per second. When you plot the points on a graph and connect them with a smooth curve, you get a graph like Fig. 12-8.
Figure 12-8
Constant power enables an accelerated mass to reach a speed of 100 ft/s in 20 s from a standstill start. Note that the acceleration (slope of the curve) decreases with the passage of time.
In this situation, the acceleration is greatest at the beginning. Then as the speed builds, the acceleration drops. The faster an object goes, the slower is the rate at which its speed increases. Because of this, one-half the final speed is reached in only 1/4 of the time.
HOW A SPRING STORES ENERGY
Another way to store potential energy is with a spring, similar to the principle of the archer’s bow from earlier in this chapter. Assume that a spring supports only the weight that is attached to it, as shown in Fig. 12-9. Now imagine that you progressively apply greater force, in addition to that caused by the 1-lb weight itself. This will compress the spring. Suppose that 3-in compression requires 2 lb of extra force, 6-in compression requires 4 lb of extra force, 9-in compression requires 6 lb of extra force, and 12-in compression requires 8 lb of extra force, as shown.
Figure 12-9
A stressed spring stores energy. In this case the stress takes the form of compression.
The force that is applied to compress the spring by 12 in (or 1 ft) uniformly grows from 0 at the start to 8 lb at the finish. The average force over the 1 ft of compression must therefore be 4 lb. As a result, the energy stored in the spring, when it is compressed by the 8-lb force, is 4 ft · lb. This is equivalent to 4 × 32 = 128 ft · pdl. All this energy will remain stored in the form of potential energy as long as the 8-lb force holds the spring compressed.
HOW A SPRING TRANSFERS ENERGY
Suppose that the 8-lb force holding the spring compressed is suddenly released. When this is done, the spring starts to propel the 1-lb weight upward. The initial force—right at the instant the spring is released—is 8 lb. As the weight moves upward, the accelerating force diminishes. But the speed continues to increase, because as long as stress remains on the spring, it exerts an upward force. The total energy—potential plus kinetic—remains constant in this type of system.
At the instant when the spring is half decompressed (to 6 in), the force has dropped to 4 lb, or 128 pdl. The average force represented in this compression is 2 lb, or 64 pdl, and the distance over which this average force is applied is 6 inches (in), or 0.5 ft. So the remaining potential energy is 64 × 0.5 = 32 ft · pdl. Of the original 128 ft · pdl, 96 ft · pdl has been turned into kinetic energy. This equation for the kinetic energy must be E = mv2/2; m is 1 lb, so v2 must be 2 × 96 = 192. That means that the instantaneous speed v is the square root of 192, or 13.856 ft/s. When the spring is fully decompressed, all the energy is kinetic, so now v2 = 256 and v = 16 ft/s. This is shown in Fig. 12-10.
Figure 12-10
When a compressed spring is released, energy is converted from potential to kinetic.
RESONANCE CYCLE
The transfer of energy from potential to kinetic in the spring-and-weight arrangement, as shown in Fig. 12-10, forms the first part of a resonance cycle. At the instant the spring has become fully decompressed, the weight is moving upward at 16 ft/s. The spring now starts to decelerate the weight, because the spring is going into tension (pulling down, instead of pushing up). For each 3-in movement upward past the neutral position, the spring applies a tension of 2 lb until the upward displacement reaches 1 ft, where the tension becomes 8 lb. This is based on the assumption that the spring constant, or force-vs.-displacement ratio, is the same throughout the entire ±1-ft range of compression and tension discussed here. (This is true of most springs in the real world, unless they are compressed or stretched so far that they are permanently deformed.)
As with the compression, the average force of tension over the range of movement is 4 lb, so the potential energy is once again 4 lb, or 128 ft · pdl. All the energy is again potential, and the weight is momentarily stationary.
Since the weight has reached the upper extreme, an equal acceleration downward starts the second half of the cycle. Another interchange of energy, from potential to kinetic, continues until the neutral position is again reached. At this point, all the energy is kinetic, and the speed is 16 ft/s downward, steady (for the moment), neither accelerating nor decelerating. Then as the weight continues downward, compression starts again, until the weight comes to rest with the spring fully compressed, 1 ft down, with an 8-lb force pushing it back up (Fig. 12-11). And again, as before when the spring was fully compressed, all the energy is potential.
Figure 12-11
As a spring is alternately compressed and decompressed, a resonance cycle occurs.
This up-and-down process, called oscillation, would go on forever in an ideal system. But in practice, the energy gradually dissipates; it transfers to other forms, notably thermal energy (“heat”). Friction absorbs some of the kinetic energy. The nature of the metal from which the spring is made causes some inefficiency in the system. Even air resistance plays a role. The excursion (maximum amount of up-and-down displacement) and speed therefore slowly diminish, and the weight eventually comes to rest.
TRAVEL AND SPEED IN RESONANT SYSTEM
You started with an assumed compression of 1 ft, which led to a maximum speed of 16 ft/s. Suppose the initial compression is only 6 in, or that friction has decreased the excursion to this magnitude. The maximum force is now 4 lb rather than 8 lb (128 pdl instead of 256 pdl). The distance over which the average force was 2 lb (64 pdl) is now compressed to 6 in rather than 12 in. So the maximum potential energy is 32 ft · pdl. When the weight passes through the neutral position, all this energy will become kinetic; v2 will now be 64, so v is 8 ft/s.
Notice that halving the maximum upward or downward extent of travel also halves the maximum speed reached. The object travels one-half the distance at one-half the speed, so it performs the entire cycle in the same time. Interestingly, regardless of the magnitude of the oscillation, resonance still requires the same time to complete a full cycle. This length of time is called the period of oscillation. The number of oscillations per unit time is called the frequency of oscillation. So as a weight-and-spring system “dies down” after having been set in motion, the period and frequency stay the same, even though the excursion gets smaller and smaller.
The principle of resonance is used in the balance wheels of old-fashioned clocks, the pendulums of grandfather clocks, and many similar devices—not just mechanical, but also electrical, electronic, and atomic.
QUESTIONS AND PROBLEMS
This is an open-book quiz. You may refer to the text in this chapter (and earlier ones, too, if you want) when figuring out the answers. Take your time! Consider all values given as exact, so you don’t have to worry about significant figures. The correct answers are in the back of the book.
1. Suppose a car accelerates uniformly from standstill to 40 mi/h in 3 min. How far will it travel in those 3 min?
2. In the next 6 min, the car increases its speed at a steady rate, from 40 mi/h to 60 mi/h. How far will it travel in these 6 min?
3. The same car brakes to a stop in 30 s. If the deceleration is uniform during these 30 s, how far will the car travel before it stops?
4. From the fact that 1 mi = 5280 ft and 1 h = 3600 s, find the speed in miles per hour that corresponds to 88 ft/s.
5. During takeoff, an aircraft builds up a thrust that accelerates it at 16 ft/s2. Its takeoff speed is 240 mi/h. Find the time from releasing the brakes until the plane lifts into the air. How much runway is required?
6. A gun can use cartridges with pellets of two sizes, one that is twice the weight of the other. If the heavier pellet leaves with a muzzle speed of 150 ft/s, find the muzzle speed of the lighter pellet, assuming that the explosive charge develops the same energy in each case.
7. An electric car’s motor and transmission develop constant power during maximum acceleration. This particular car (it’s no drag-racing machine!) can reach 60 mi/h in 20 seconds. In how long will it reach 30 mi/h? 45 mi/h?
8. If the weight of car and driver (in question 7) is 3000 lb, what time is necessary to reach 30 mi/h, 45 mi/h, and 60 mi/h when an additional load of 1000 lb is carried?
9. Find the power developed by the motor and transmission of the same car in foot-poundals per second.
10. A spring-and-weight resonance system can be changed, by altering either the weight or the spring. By figuring the effect of such change on maximum speed reached from a given starting deflection, deduce the effect of (1) doubling the weight and (2) halving it.
11. Using the accompanying graph of distance vs. time (Fig. 12-12), make rough estimates of the speed of the object, in meters per second, for the instants of time corresponding to 1 s, 2 s, 3 s, and 4 s. Explain how you deduced these results.
Figure 12-12
Illustration for problem 11.
12. Using the accompanying graph of speed vs. time (Fig. 12-13), determine the approximate acceleration of the object, in meters per second per second, for the instants of time corresponding to 1 s, 2 s, 3 s, and 4 s. Explain how you deduced these values.
Figure 12-13
Illustration for problems 12 and 13.
13. Assuming the object described by Fig. 12-13 has a constant mass and there is no friction, tension, or gravitational influence, how does the applied force vary qualitatively with time? Explain how you know this. (Think of a mass in outer space, propelled by a small rocket.)
14. Figure 12-14 shows the speed vs. time for an object that moves faster and faster for a while and then slows down. Thus, the curve appears somewhat “bell-shaped.” Qualitatively, how does the acceleration vary with time? When is the acceleration greatest? When is it smallest? What is the approximate acceleration when the elapsed time is 2 s?
Figure 12-14
Illustration for problem 14.
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