weight of one inch or one foot in length, according as you take the length in inches or in feet. To apply the formulæ we have deduced, we may take Example 1. Now, in this case, the power acts at one end of the lever, and the weight at the other, and the power is required; therefore formula (1) is exactly adapted to this Here W = 100 lbs. a = 60 inches, b 100 × 36+ × 36 ××60o × 3600+27-75 60 60 24 59.2 lbs. the same as before in Ex. 1. But if we take c for the weight of one foot in length, it becomes P = 100 × 3 × 3 × 1 — 1 × 5o × 1 We will now take Example 2d (Fig. 20), where the power is applied at a given point D between A and C, and the weight at a given point E between W and C; and since the power P is required, we must take formula (8). Wr+b2c & aoc = P d Here W = 3 cwt. 336 lbs. a = 8 feet, b=6 feet, c= 123 lbs. weight of one foot of the lever, d=5 feet, and 2 feet; substitute these values in the above, and P × 2 + 1 × 6o × 3. 182 × 3 672 +54-96 r= 336 x 2 + } x 6 x 3 − } 8% x 3 5 = 5 = 126 lbs. which is exactly the same as was found in Example 2d. 50. Let A W C be a lever of the second order, and C its fulcrum; the power multiplied by its distance from the fulcrum, is equal to the weight multiplied by its distance from the fulcrum, together with the whole weight of the lever multiplied by half its length; the lever being considered uniform throughout its length. Example 1. Given the whole weight of the lever 9 lbs. its length A C = 6 feet, a weight of 100 lbs. is put on at 14 feet from the fulcrum; it is required to determine the power acting at A which will keep the lever in equilibrio. (See Fig. 13.) Now 100 lbs. x 14 feet 150 the weight multiplied by its distance. 9 × 3 = 27 half its length. Hence = = the weight of the lever multiplied by 150 + 27 6 =294 lbs. is the weight or power act ing at A which will keep the whole in equilibrio. Or thus, by note to Art. 47: 6: 100 :: 42: 75, the weight upon the fulcrum from the action of the weight. 6: 100 14:25, the power at A which will just support the weight. And the lever being uniform, its whole weight must be considered as acting at the middle of A F; therefore the fulcrum will bear one half of its weight, and the power must support the other. Consequently 75 + 4 = 79 lbs. weight upon the ful crum. And 25+ 4 = 29 lbs. the power necessary to keep the whole in equilibrio, which is exactly the same as before. Example 2. A beam, the weight of which is 12 lbs. and its length 18 feet, is supported at both ends; a weight of 36 lbs. is sus pended at 3 feet from one end, and a weight of 24 lbs. at 8 feet from the other end; required the pressure on each prop or support. (Fig. 21.). For the sake of simplicity, suppose the 36 lbs. weight to be suspended at 3 feet from the end C, and the 24 lbs. at 8 feet from the end A. Then, 18: 36: 15: 30 lbs. the pressure on the support C by the action of the 36 lbs. weight. 18: 24 :: 8 : 103 lbs. the pressure on the support C from the action of the 24 lbs. weight. 30+ 10 = 40 lbs. pressure on the support C from both weights. And 18: 36 :: 3 : 6 lbs. pressure on the support A from the action of the 36 lbs. weight. 18: 24: 10: 13 lbs. pressure on the support A from the 24 lbs. weight. 6+ 13 = 19 lbs. the whole pressure on the support A from both weights. Now half the weight of the lever added to each of the above sums will give the whole pressure on each support. Thus, 403+6=463 lbs. whole pressure on the support C. And 19+6= 25 lbs. the whole pressure on the support A. Example 3. Given the whole length of the lever A C 10 feet (Fig. 22), its weight 15 lbs. a weight of 50 lbs. is suspended at 2 feet from the fulcrum or end C; what power, acting at 3 feet from the other end A, will keep the whole in equilibrio? Now 10-37 feet, the distance of the power from the fulcrum C; therefore 7: 50:: 2 : 143 lbs. the power ne cessary to balance the weight alone; and since the centre of gravity of the lever is 2 feet from the power, and 5 feet from the fulcrum, we have 7:15 5: 10 lbs. the power necessary to sustain the lever. And 14% + 10 = 25 lbs. the power required to sustain both weight and lever. In the same manner we may find the weight sustained by the fulcrum. Thus, 7: 50 :: 5 : 351⁄2 lbs. from the action of the weight. And 7:15:: 2 : 4 lbs. from the action of the lever, Therefore 35+49 40 lbs. the whole pressure on the fulcrum or end C. 51. We may now deduce formulæ for the lever of the second order, retaining the same notation as in Art. 48. We have Pa = W b + 1⁄2 a2c, from which we have When W have P = Pa-ac (4) 2 Pa-2 W b (5) (3) O, or the power just sustains the lever, we a c, and a 2 P C (6) If P = 0, the expression for a is impossible; therefore the equation for P admits of a minimum; and by Dr. Gregory's Mechanics, Art. on the Lever, when P is a mi# 2 W b nimum, a = с = Take Example 1; and in that case W 100, a = 6, b14, and c = 1; consequently we have, by formula We will here give an example when a beam is supported by two posts or props, neither of which are at the ends of the beam. А Example 4. A beam, the weight of which is 84 lbs. and its length A B = 20 feet, is supported by two posts at C and D; the distance of the end A from C tance of the end B from D = 14 feet, and the dis = 24 feet; a weight of 156-lbs. is suspended at E, 2 feet from C; required the pressure on each post. -CD, the distance between the props or posts. And 16 156: 14: 136 lbs. pressure on the prop C from the action of the weight. :: 16: 156 2: 19 lbs. pressure on the prop D from the action of the weight. And A B = 10, and 10-14 = 8 feet, the distance of the centre of gravity from the prop C; also 10-2} = 71⁄2 feet, the distance of the centre of gravity of the beam from the prop D. 16: 84: 81: 445, pressure on the prop D from the action of the lever. 16: 84 :: 71 : 39%, pressure on the prop € from the action of the lever. 136 +39 = 1753 lbs. whole pressure on the prop C. 19 +44643 lbs. whole pressure on the prop D. 52. These examples will sufficiently explain all the different cases of levers of the second order. We will now proceed to give formula for the lever of the third order. Formulæ " The notation remaining the same, we have P a = W b + 1 b2c. |