Er. If a body be accelerated from a state of rest by a uniform force, and describe m feet in the first second of time, it will describe 4m, 9m, 16m.. ..mto feet, in the 2, 3, 4.. ..t first seconds. 33. Cor. i. The space described, reckoning from the beginning of the mo:ien, is half that which would be described in the same time with the last acquired delocity continued uniform. For completing the parallelogram BD, fig. 2, then, from the present article it appears, that the space described in the nine AB, reckoning from the beginning of the motion, may be represented by the triangle ABC. But the space that would be described by a body moving uniformily for the time AB, with the last acquired velocity, may be represented by the rectangle BD, and the area of the triangle ABC, is, by a well-known theorem, equal to one half that of the rectangle BD; hence the proposition is manifest, and in tlie case of wifurin forres we have, generally, 8=fto. 34. Cor. 2. JIS, T, V, represent the space, time, and last acqnired velocity, in any other case of uniform forces, we shall also have, Su{TV. But 8:S :: {tv: {TV :: to: TV; therefore, generally, sxto; or ichen bodies are put in motion by uniform furces, the spucts described in any times, reckoning from the beginning of the motion is euch case, are as the time and last acquired relocity jointly. 35. Cor. 3. The space described in the time GB, fig. 2, is represented by the area GBCN; or, if NM be drawn parallel to GB, the space may be represented by the rectangle GM, together with the triangle NMC. Now GM re presents the space which a budy would describe in the time GB, with the uniform velocity GN; and the triangle NMC, which is similar to the triangle ABC, represents the space through which the body would be moved from a state of rest by the action of the force, in the time GB; thus, the space described in any time, when a body is projected in the direction of the force, is equal to the space which it would have described in that time, with the first velocity continued uniform, together with the space through which it would have beeu moved from a state of rest, in the same time by the action of the force. 36. Cor. 4. If a body be projected in a direction opposite to that in which the uniform force acts, with the velocity BC, and move lill that velocity is destroyed, the whole time of its motion is represented by BA, and the space described by the area ABC. For the time required to destroy any velocity by the action of a uniform force, is equal to the time that would be required to generate the same velocity by the action of the same force (29); and, since the whole times of motion in the two cases are equal, and also if equal times be taken, from the beginning of the motion in oue case, and from the end of the motion in the latter, the velocities at those instants are equal. Since, tben, the whole times of motion are equal, and also the velocities at all corresponding points of times, the whole spaces described are egoal. Also the space described in the time BG is represented, on the same scale, by the area BGNC; that is, by the rectangle BL diminished by the triangle CLN, or CNM. Thus it appears, that the space described in the time BG, is eqnal to that which would have been described with the first velocity continued uniform during that time, diminished by the space through which the body would have been moved from a state of rest in the same time, by the action of an uniform force. 37. The spaces passed over by bodies, urged by any constant and uniform forces, acting during any times, are in the compound ratio of the forces and squares of the times directly, and the body or mass reciprocally, or as the mass and square of the velocity directly, and the force reciprocally. ft . vat , S ft bu Also, since va we have to f' and consequently by sublive stitution, sa j 38. Cor. The same expressions which represent the relations of the forces, spaces, tincs, and velocities, in accelerated motions, represent them, also, when the motions are retarded, and the bodies move till their whole velocities are destroyed. For the time in wbich any velocity is destroyed is equal to the time in which it would be generated by the same force; also, ihe spaces described, on the supposition that the body in the latter case is moved from a state of rest, has been shown to be equal 39. Articles 28, 29, &c. give theorems for resolving all questions relating to motions uniformly accelerated or retarded. Thus, let b = any body or quantity of matter. m = the momentum at the end of that time. Then, from the fundamental relations moc bv, mxft, så tv, vocī we obtain the following Table of the general relations of uniformly accelerated or retarded motions. bsfs m ObvX ft on as creva v (bfs)oc w (uftu) m ft mt pism? mo 630 1 υ ms fs ftv bv m bve bs to x bs btv ft fs mo fsfest uft o ma m? bym?v sectvo 6 f m NON OC &c. faf j fu Tfv' 40. And, from these proportions, those quantities are to be left out which are given, or which are proportional to each other. Thus, if the body or quantity of matter be always the same, then, instead of the relation soc^;, we shall have, more simply, softe, or the space described as the force and square of the time. And if the body be proportional to the force, as all bodies are ia respect of their gravity ; then sX 1°Cvę, or the space described is as the square of the time, or as the square of the velocity. If F be put = {, then will F represent the accelerating force, and substituting this in the foregoing Table, we shall have, V S m ገ MV ms fsa doc m ms m sOC tv XT m m m ON THE LAWS OF GRAVITY, AND OF THE DESCENT OF HEAVY BODIES. 41. The force of gravity, at any given place, is an uniform force, which always acts in a direction perpendicular to the horizon, and accelerates all bodies equally. The same body will, by its gravity, always produce the same effect under the same circumstances : thus it will, at the same place, bend the same spring in the same degree ; it will also fall through the same space in the same time, if the resistance of the air be reinoved ; therefore, the force of gravity is uniform. Also, all bodies which fall freely by this force, descend in lines perpendicular to the horizon; and in an exhausted receiver, they all fall through the same space in the same time; consequently, gravity acts in a direction perpendicular to the horizon, and accelerates all bodies equally. The result of numberless experiments made on the descent of heavy bodies is, that every body which falls freely in vacuo by the force of gravity, descends from rest through 1617 feet in the first second. This fact being established, every thing relating to the descent of bodies when they are accelerated by the force of gravity, and to their ascent when they are retarded by that force, supposing the motions to be in vacuo, may be deduced from the foregoing propositions relating to constant forces. Since in the first second of time a body falls through 16 1 feet, at the end of that time it will have acquired a velocity which, without any further action of gravity would carry it through 326 feet in the next second (33). Therefore, if : denote 161, the space fallen through in one second of time, 2g will denote the velocity generated in that time; then (30 and 31) because the velocities are directly proportional to the times, and the spaces to the squares of the time, we shall have for any other time t. As 1" : 1" :: 29 : 2gt=v, the velocity acquired in t", and 18:1:: g: ge=s, the space passed over in t". So that, for the descents of heavy bodies by the force of gravity, we have these general equations : ว g= 48 Er. 1. How far will a body descend in 3 seconds by the force of gravity. Because s=gt®, and in this case t=3" we have s=16'x3=1443 feet, the distance required. 2. Through what space must a body fall from rest to acquire a velocity of 20 feet per second. 202 Hence we have s= 6.21 feet 4g 4x 165 3. In what time will a body fall 90 feet, and what velocity will be be acquired. 90 First t=N +2.36 seconds nearly. Again v=2wgs=27(164 x 90)=76 feet per second nearly. Again, because the times are as the velocities, and the spaces as the squares of either; therefore If the times be as the nuinbers 1, 2, 3, 4, 5, &c. And the spaces for each time as 1, 3, 5, 7, 9; namely, as the series of the odd numbers, which are the differences of the squares denoting the whole spaces. So that if the first series of natural numbers be seconds of time, Time in Seconds. 1", 2", 3" 4", The velocities in feet will be 326, 641, 96, 1283, The spaces in the whole time 1614, 645, 1445, 257-5, And the space for each second 1679, 481, 807, 112-17. 42. And because the laws for the destruction of motion are the same as those for its generation, by equal forces, but acting in a contrary direction; therefore, 1st. A body thrown directly upwards, with any velocity, will lose equal velocities in equal times. 2d. If a body be projected upwards, with the velocity it acquired in any time by descending freely, it will lose all its velocity in an equal time; and will ascend just to the same height from which it fell, and will describe equal spaces in equal times, in rising and fall. ing, but in an inverse order, and it will have equal velocities at any one and the same point of the line described, both in ascending and descending 43. If a body begin to move in the direction of gravity with any velocity, the whole space described in any time is equal to the space through which the first velocity would carry the body, together with the space through wbich it would fall by the force of gravity in that time. Er. If a body be projected perpendicularly downwards, with a velocity of 20 feet per second, to find the space described in seconds. The space described in 4" with the first velocity is 4 x 20, cr 80 feet ; and the space falleu through in 4" by the action of gravity, is 161 X 16 = 257'5 feet; therefore, the whole space described is 80+2573=337 feet. 44. If a body be projected perpendicularly upwards, the height to which it will ascend in any time is equal to the space through which it would move with the first velocity continued uniform, diniinished by the space through which it would fall by the action of gravity in that time Ex. To what height will a body rise in 3" if projected perpendicularly upwards with a velocity of 100 feet per second ? The space which a body would describe in 3" with the first velocity, is 300 feet; and the space through which the body would fall by the force of gravity in 3", is 161x9=1444 feet; therefore, the height required is 300-1441=155 feet. Eramples on Motions Uniformly varied. 1. What is the difference between the depths of two wells, into each of which, should a stone be dropped at the same instant, the one would strike the bottom at the end of 6, and the other at the end of 10 seconds ? Ans. 1029} feet. 2. Supposing Salisbury steeple to be 400 feet high, in what time would a nusket-ball, let fall from the top of it, reach the ground ? Ans. in 5 seconds nearly. 3. A heavy body was observed to fall 100 feet in the last second of time; from what height was it let fall, and how long was it in motion ? Ans. 335 seconds. 4. A stone being let fall into a well, it was observed that, after being dropped, it was 10 seconds before the sound of the fall at the bottom reached the ear ; what is the depth of the well? Ans. 1270 feet nearly. 5. A drop of rain, in its descent towards the earth, was observed to fall through a space of 400 feet in the last two seconds; from what height did it fall ? Ans. 837,83 feet. 6. If a body be projected perpendicularly upwards with a velocity of 80 fett per second, required its place at the end of 6 seconds ? Ans. 99 feet below the point from which it was projected. |