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This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1854 Excerpt: ... in t seconds, neglecting the inertia of the pulley. The pressure causing motion is 9--7 = 2 ounces, and the weight moved is 9 + 7 = 16 ounces, therefore (Art. 71), 2 the accelerating force = g.--= 4 nearly. 16 Substituting this in the general formula = f&, we have = 2fi. Hence, in one second the space described is 2 feet, in two seconds 8 feet, in three seconds 18 feet, and so on. Ex. 6. A weight of 9 lbs. is drawn along a smooth horizontal table by a weight of 1 lb. hanging vertically by a string passing over a pulley at the edge of the table; find the space described from rest in 3 seconds, the velocity acquired in 4 seconds, and the space described in the 5th second. The pressure causing motion is 1 lb., the weight moved is 9 + 1 = 10 lbs. Therefore accelerating force, = g. = 3-2, space described in 3 seconds = l-6 X 9 = 14-4 ft. velocity acquired in 4 seconds = 3-2 X 4 = 12-8. The space described in the 5th second is f multiplied by the 5th odd number, or 9;.'. space described in 5th second = 1-6 x 9 = 14-4 feet. EXAMPLES. (In the following examples take g = 32.) If a body fall from rest under the action of gravity, Find 1. The space described in the 21st second. Ans. 656 ft. 2. The velocity acquired in 8 seconds. Ans. 256. 3. The space described in 6 seconds. Ans. 576 ft. 4. The time of acquiring a velocity 160. Ans. 5 sec. 5. The space described in acquiring a velocity 224. Ans. 784 ft. Find 6. The time of describing 400 ft. Ans. 5 sec., 7. The velocity acquired in describing 576 ft. Ans. 192. 8. A weight of 8J lbs. descends drawing up another of 7J lbs. over a fixed pulley, find the space described in 5 seconds. Ans. 25 ft. 9. A weight of 13 lbs. is drawn along a smooth horizontal table by a weight of 3 lbs. find...
