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By the use of these formulas (1 to 4) any two of the quantities u, v, a, t, s may be determined when the other three are given. The student should thoroughly memorize these formulas and exercise himself in applying them to simple problems, such as those on page 69.

97. Force Causing Rectilinear Motion with Constant Acceleration. The kind of motion just discussed is produced whenever a mass is acted on by a constant force in one direction; for in such a case the acceleration is constant and given by the relation

F

a =
m

Thus when a car is drawn along a track by a stretched spring which is kept constantly at the same tension, the motion is with constant acceleration. So also a falling body has this kind of motion, for it is constantly urged downward by its own weight, which is a nearly constant force. When a body slides down an inclined plane, the force urging it down along the plane is the same at one point as at another, and therefore in this case also the acceleration is constant.

98. Falling Bodies.-Freely falling bodies are the most familiar examples of bodies moving with constant acceleration. For a body near the surface of the earth is attracted or urged downward with a certain constant force which we call its weight, and when it is set free so that its weight is the only force acting, it falls with constantly accelerated motion. In ordinary experience, however, where bodies fall through air, the resistance of the air is another force which modifies the motion. If the resistance in a given case were constant, the body would still fall with constant acceleration, but the air resistance increases greatly with the velocity of the falling body, so that, in case of a light body, as the speed increases the air resistance may become equal and opposite to its weight, and when that is the case it falls without acceleration. This is the case with scraps of paper and rain drops.

Strictly speaking, even the weight of a body is not constant as it falls, but increases as it approaches the surface of the earth. The weight of a kilogram one mile above the earth's surface is less by a gram than at sea level, and at the ceiling of a room

3 meters high a kilogram weighs about one milligram less than at the floor. This variation of force with height causes a corresponding increase in the acceleration of a falling body as it ap

FIG. 45.-Fall in vacuo.

proaches the earth's surface; but this is so small, however, that except in case of great heights it may be neglected.

99. Acceleration of Gravity. The early philosophers speculated as to why bodies fell; Galileo was the first to carefully determine how bodies fell. He also showed, contrary to the universally accepted opinion of his day, that except for air resistance. all falling bodies are equally accelerated. A large stone or a small one, an iron cannon ball, a lump of lead, or block of wood when dropped from the top of a tower reach the ground in the same time. If a feather, scraps of paper, and some bits of metal or lead shot are placed in a long tube (Fig. 45) from which the air is exhausted, on quickly inverting the tube all reach the bottom at the same instant. Hence the rate of increase in velocity, or acceleration, is constant at any given place on the earth for all kinds and sizes of bodies.

This constant acceleration is called the acceleration of gravity at the given place, it is usually represented by the symbol g and is measured most accurately by pendulum. experiments.

The value of g at the sea level for the latitude of New York is 980.2 cm. sec., or 32.16 ft. sec. The table on page 108 gives also the values at some other places.

The formulas for falling bodies are therefore obtained from those of $96 by making the acceleration equal to g. Thus

v=u+gt
s = ut+gt2

2gs=v2-u2

When a body is simply dropped, with no initial velocity, u is zero, and we have

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In approximate calculations and in working problems for practice, g may be taken as 980 cm. sec. or 32 ft. sec.

100. Relation between Dyne and Gram.-The force urging downward a freely falling mass m is expressed by the formula

F=mg,

the force being in dynes if C.G.S. units are used. Suppose m=1 gram and g=980, then F-980 dynes; but the force with which a mass of one gram is attracted toward the earth is called the weight of one gram, therefore the weight of 1 gram = 980 dynes, or the force which we have called a dyne is nearly equal to the weight of a milligram at the earth's surface.

The student may show similarly that one poundal is about equal to the weight of a half-ounce; that is, one pound weight at New York 32.16 poundals.

100 a. Gravitation Units of Force.-The weight of a gram or pound is often a convenient unit of force; indeed, engineers in English speaking countries almost always measure forces in pounds; for though the weight of a pound varies from place to place on the earth it is sufficiently constant for ordinary engineering requirements.

If these gravitation units of force are used, formula (1) of $94 becomes

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But most of the formulas in this book are based on the absolute relation F=ma; it will therefore be best for the student in working problems to use consistently either the centimeter-gramsecond system with the force in dynes or the foot-pound-second system with the force in poundals, changing, when required, dynes or poundals into grams or pounds weight by dividing by 980 or 32 as the case may be.

101. Atwood's Machine.-To illustrate experimentally the effect of forces in accelerating masses the form of apparatus known as Atwood's

machine (Fig. 46) may be used. Two equal weights A and B are hung over a very light carefully balanced wheel mounted so that it

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shall run with as little friction as possible. An additional weight or rider w, having two projecting arms, is laid on top of the weight A, which is supported so that it can be liberated at any instant. When the weight A is freed it moves down accelerated by the rider w, until it reaches the ring C which picks off the rider w and allows A to pass freely through. After passing the ring C there is no longer any accelerating force, as the rider is removed, and the weight A continues to move with the velocity which the rider had given to it.

Thus if the ring C is so adjusted that A passes through it exactly two seconds after being liberated, and if D is so placed that A moves from C to D in the next second, then if C and D are found to be 30 cms. apart, we conclude that A acquired a velocity of 30 cms. per second by a force which acted steadily for two seconds. If the same force is now allowed to act for one second, a velocity of only 15 cm. sec. will be acquired. By varying the weight of the rider or using instead of A and B a pair of weights, having double the mass, the following conclusions may be established:

a. The motion is with constant acceleration.

b. The acceleration is proportional to the weight of the rider so long as the moving system A+B+w is constant.

c. If the mass of the moving system is doubled, a given rider will

only cause half as great acceleration as before.

All of these results are expressed, of course, by the formula

F= =ma.

102. Atwood's Machine Problem. Suppose two masses, one 40 and the other 50 grams, are connected by a cord running over a light frictionless pulley as in Atwood's machine, and suppose for simplicity that the mass of the cord and of the pulley may be neglected. It is required to find the acceleration and the tension on the cord.

In this case the whole mass 40+50 moves together and the resultant force which gives it motion is the weight of

50-40=10 grams, or 10g dynes.

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hence the acceleration is one-ninth that of a freely falling body.

40

This result may also be reached by considering that 40 a force of 10 grams acting on a mass of 10 grams gives it an acceleration g, and therefore if that same force act on a mass 9 times as great it will give it an accelerationg.

50

FIG. 47.

50

To find the tension on the cord consider the forces acting on the mass 40. It is urged downward by its own weight 40 grams and upward by the tension on the cord, which we may call T grams. It moves upward with an accelerationg, as has been shown, hence the resultant force must be upward and equal to (T−40) grams or (T-40)g dynes, and we have, since F = ma,

whence

(T-40)g 40 × g

=

T-44 grams' weight.

103. Motion on an Inclined Plane.-When a mass M rests on an inclined plane, the force due to gravity, or its weight, may be resolved into two components, as shown in figure 48, one N perpendicular to the plane and the other F parallel to it. If M is the mass in grams, its weight in dynes is Mg. And from the similarity of the two triangles, we have Mg,N, and F respectively proportional to the sides of the large triangle formed by 1, b, and h. That is, F Mg hl or F = Mg dynes. Thus the force F causing the motion is constant, and the same

h

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