70. To find the velocity acquired by a falling body, per second, (or by a stream of water, having the perpendicular descent given) at the end of any giv en period of time.

RULE.

1. The velocity acquired at the end of any period is equal to twice the mean velocity, with which it passed during that period.

Or, 2. Multiply the perpendicular space fallen through by 64, and the square root of the product is the velocity required.

If a ball fall through a space of 484 feet in 5 seconds, with what velocity will it strike?

By the former part of the rule.

484 5.5=88, and

88x2=176, Ans.

By the latter part, with

out regarding the time.

✔484x64 176, Ans.

71. There is a sluice, or flume, one end of which is 24 feet lower than the other: What is the velocity of the stream per second?

2.5x64-160, and 160-12-649 feet, Ans.

72. The velocity, with which a falling body strikes, given, to find the space fallen through. RULE.

Divide the square of the velocity by 64, and the quotient will be the height required.

If a ball strike the ground with a velocity of 56 feet per second, from what height did it fall?

56x56-64-49 feet, Ans. 73. The mean velocity of a fluid, or stream, is 12.649 feet per second: What is the perpendicular fall of the stream?

12-649x12-649÷64-24 feet, Ans.

74. The weight of a body, and the space fallen through, given, to find the force with which it will strike.

RULE.

The momentum, or force, with which a falling body strikes, is equal to its weight multiplied by its velocity; therefore, find the velocity, by Problem 70, and multiply it by the weight, which will produce the force required.

If the rammer, used for driving the piles of Charlestown bridge, weighed 24 tons, or 4500lb. and fell through a space of 10 feet, with what force did it strike the pile?

✔ 10x64 = 25.3 = velocity, and 25.3 × 4500 = 113850lb. momentum, Answer.

75. The weight and momentum, or striking force, given, to find the space fallen

through.
RULE.

Divide the momentum by the weight, and the quotient will be the velocity; then divide the square of the velocity by 64, and the quotient will be the space fallen through.

If

If the aforementioned rammer weighed 4500lb. and struck with a force of 113850lb.: From what height did it fall?

113850-4500=25.3, and 25-3x25-3-64-10 feet, Ans. 76. If it were required to know with what quantity of motion, momentum or force, a fluid, moving with a given velocity, strikes upon a fixed obstacle.

RULE.

By Problem 72 find the fall, which will produce the given velocity; multiply that height by 62.5lb. Avoird. for clean river water, by 63 lb. for dirty water, and by 64 for sea water.

Suppose a stream of clear water to move at the rate of 5 feet per second, and to meet with a fixed obstacle (or bulk head) 15 feet wide and 4 feet high: What is the momentary, instantaneous pressure of

the stream?

5x5÷64-2 and 25÷64-39 of a foot, for the perpendicular fall of the water. Now 62·5x⋅39=24-375lb. the pressure upon each square foot, which, multiplied by 60, (the number of square feet in the obstacle) gives 1462-5lb. going with the given velocity of 5 feet per second; therefore, 1462-5x5=7312.5lb. Ans.*

77. The velocity of water, spouting through a sluice, or aperture in a reservoir, or bulk head, is the same that a body would acquire by falling through a perpendicular space equal to that between the top of the water in the reservoir and the aperture.

What is the velocity of water, issuing from a head of 5 feet deep? By Problem 70th 64x5= 320, and 320 18 feet, nearly, Ans. 78. If the velocity of a stream issuing through the bulk head of a mill, be 16 feet per second, what head of water is there.

16x16÷644 feet, Ans.

79. The quantity of water discharged from a hole in a vessel, is as the square root of the height of water above the aperture.

A miller has a head of water 4 feet above the sluice: How high must the water be raised above the opening, so that half as much again water may be discharged from the sluice in the same time?

42, and half as much again as 2, is 2+1=3, for the square root of the required depth; therefore, 3X3-9 feet high, Ans.

OF PENDULUMS.

80. The time of a vibration, in a cycloid, is to the time of a heavy body's descent through half its length, as the circumference of a circle to its diameter, that is, as 3-1416 to 1: therefore, (as a body descends freely, by gravity, through about 193-5 inches in the first second) to find the length of a pendulum vibrating seconds.

RULE. AS 3.1416×3-1416: 1x1 :: 1935: 19-6 inches, the half length, and 19-6×2=39.2 inches, the length.

18. To

• Water, being a yielding fubftance, lofes two thirds of its power in producing effects.

81. To find the length of a pendulum, that will swing any given time. RULE.-Multiply the square of the seconds in any given time by 39.2 and the product will be the length required, in inches. Required the lengths of several pendulums, which will respectively swing seconds, seconds, seconds, minutes and hours?

•25x25x39*2=2.45 inches for seconds. 5x 5×39.2=9·8 inches for seconds. 1x1x39-2=39-2 inches for seconds, as above; 60x 60x39.2 the inches in 2 miles and 1200 feet, for minutes; and 1 hour 3600 seconds, therefore 3300×3600×39-2 = the inches in 8018 miles and 96 feet, for hours, Ans.

82. What is the difference between the length of a pendulum, which vibrates half seconds and one which swings three seconds?

3X3X39-2-5X 5x39.2=343 feet, Ans.

83. To find the time which a pendulum of any given length will swing. RULE. Divide the given length by 39-2, and the quotient will be the square of the time in seconds.

Or, as 6.261 (the square root of 39-2) is to the square root of the given length, so is 1 second to the time of 1 oscillation: that is, divide the square root of the given length by 6-261, and the quotient will be the time of one vibration of that pendulum.

How often will a pendulum of 9-8 inches vibrate in a second? By the former part of the rule, 9.8÷392-25 of a second, and ✔255 of a second, the time of one vibration, that is, it vibrates half seconds, or 60+5=120 times in a minute.

By the latter part. √9.8=3·13, and 39.26.261, therefore, 3.136.2615 of a second.

84. I observed, that while a stone was falling from a precipice, a string, (with a bullet at the end) which measured 25 inches, (to the middle of the ball,) had made 5 vibrations: What was the height of the precipice?

2539-26377+, and 6377-7986 of a second, the time of one vibration, and •7986x5=4 seconds, nearly, the time of the stone's descent; then 4x4≈16, and 16×16-256 feet, Ans.

85. To find the true depth of a well, by dropping a stone into it, also the time of the stone's descent and of the sound's ascent.

RULE.-1. Take a line of any length, and by the last Problem find the time from the dropping of the stone till you hear it strike the bottom.

2. Multiply 73088 (=16x4x1142; 1142 feet being the distance, which sound moves in a second) by the number of seconds till you hear the stone strike the bottom.

3. To this product add 1304164 (= the square of 1142) and from the square root of the sum take 1142.

4. Divide the square of the remainder by 64 (=16x4) and the quotient will be the depth of the well in feet.

5. Divide

5. Divide the depth by 1142, and the quotient will be the time of the sound's ascent, which, being taken from the whole time, will leave the time of the stone's descent in seconds.

Suppose I drop a stone into a well, and a string with a plummet, which measured to the middle of the ball, 25 inches, made 5 vibrations before I heard the stone strike the bottom: Required the depth, time of the stone's descent, and of the sound's ascent :

25÷39•2=6377, and ✔·6377=7986, and 7986×5=4 seconds to the hearing of it strike; then, 75088×4+1304164-1142=121·53; and 121.53x121∙53÷64–23077 feet, the depth, and 23077÷1142 2 of a second, the time of the sound's ascent, and 4-2=3.8 seconds, 'the time of the stone's descent.

OF THE LEVER OR STEELYARD.

86. It is a principle in mechanicks, that the power is to the weight, as the velocity of the weight, to the velocity of the power. Therefore, to find what weight may be raised or balanced by any given

power, say;

As the distance between the body to be raised or balanced, and the fulcrum, or prop, is to the distance between the prop and the point where the power is applied; so is the power to the weight which it will balance.

If a man, weighing 160lb. rest on the end of a lever 10 feet long, what weight will he balance on the other end, supposing the prop one foot from the weight?

The distance between the weight and prop being 1 foot, the distance from the prop to the power is 10-19 feet; therefore, as 1 foot: 9 feet :: 160lb.: 1440lb. Ans.

87. If a weight of 1440lb. were to be raised with a lever 10 feet long, and the prop fixed one foot from the weight, what power or weight, applied to the other end of the lever would balance it?

As 91: 1440 160 lb. Ans. 88. If a weight of 1440lb. be placed 1 foot from the prop, at what distance from the prop must a power of 160lb. be applied, to balance it? As 160 1440 1 9 feet, Ans. 89. At what distance from a weight of 1440lb. must a prop be placed, so as that a power of 160 lb. applied 9 feet from the prop may balance it? As 1440 160 :: 9: 1 foot, Ans.

90. In giving directions for making a chaise, the length of the shafts between the axletree and backband, being settled at 9 feet, a dispute arose whereabout on the shafts the centre of the body should be fixed. The chaise maker advised to place it 30 inches before the axletree; others supposed 20 inches would be a sufficient incumbrance for the horse: Now, supposing two passengers to weigh 3 cwt. and the body of the chaise cwt. more: What will the beast in both these cases bear, more than his harness ?

Weight of the chaise and passengers 34 cwt. 180 inches.

In. lb.

Then, as 108: 420 ::

= 420ib. and 9 feet = In. lb.

30: 116 Ans. 120: 773S Ог

OF THE WHEEL AND AXLE,

91. The proportion for the wheel and axle (in which the power is applied to the circumference of the wheel, and the weight is raised by a rope, which coils about the axle as the wheel turns round) is, as the diameter of the axle is to the diameter of the wheel, so is the power applied to the wheel, to the weight suspended by the axle.

A mechanick would make a windlass in such a manner, as that Ilb. applied to the wheel, should be equal to 10lb. suspended from the axle; now, supposing the axle to be 6 inches diameter, required the diameter of the wheel?

lb. in. lb. in.

::

As 10 6 1 60 inversely, the diameter required. 92. Suppose the diameter of the wheel to be 60 inches: Required the diameter of the axle, so as that 1lb. on the wheel may balance 10lb. on the axle ?

Inversely, as 1 : 60 :: 10: 6 diameter required. 93. Suppose the diameter of the axle 6 inches, and that of the wheel 60 inches, what power at the wheel will balance 10lb. at the axle ? in. lb. in. lb. Inversely, 6 10 60 1 Ans.

94. Suppose the diameter of the wheel 60 inches, and that of the axle 6 inches: what weight at the axle will balance 1lb. at the wheel? in. lb. in. lb.

Inversely, as 60: 1; 6: 10 Ans.

OF THE SCREW.

95. The power is to the weight, which is to be raised, as the distance between two threads of the screw, is to the circumference of a circle described by the power applied at the end of the lever.

RULE. Find the circumference of the circle described by the end of the lever; then, as that circumference is to the distance between the spiral threads of the screw: so is the weight to be raised, to the power which will raise it, abating the friction, which is not proportional to the quantity of surface; but to the weight of the incumbent part; and, at a medium, part of the effect of the machine is destroyed by it, sometimes more and sometimes less.

There is a screw, whose threads are an inch asunder; the lever by which it is turned 30 inches long, and the weight to be raised a ton, or 2240lb.: What power or force must be applied to the end of the lever, sufficient to turn the screw-that is, to raise the weight. The lever being the semidiameter of the circle, the diameter is 60 inches; then, 3·1416x60=188-496 inches, the circumference: in. lb. lb.

in.

Therefore, as 188-496: 1 :: 2240: 11·88, Ans.

96. Let the lever be 30 inches, (the circumference of which is found to be 188-496) the threads 1 inch asunder, and the power 11·88 lb. Required the weight to be raised?

As 1: 188-496 :: 11:58: 2240 nearly, Ans.

2...U

97. Let