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Example 2. A: B::3;2, 2A=3B, A=3B÷÷2, and B= 2A÷3. Again, ab:;5; 4, 4a=5b, and b-4a÷5. By

Bxa-b 2A

Χα

4a

Art. 122, velocity lost by A=: = (24 xa-1a)

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A+B

3

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the veloci.

ty gained by A, and the velocity gained by B=3a÷25.

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Example 3. The velocity lost by A=(Bxa+b)÷A+B =12B÷A+B; therefore a=7, and b=5, which momenta lost 12 AB÷(A+B)=7A÷4. Hence 48AB=7A2+ 7AB, and 48B=7A+7B, 41B=7A: therefore, A: B::41 :7=the answer.

Example 4. The velocity gained by B= (A× A—b) ÷ (A+B)=(6Xa-7)÷11. By the question, A=6, B=5, and b=7 (6+a-7)÷11. 77-8a-42, 119=6a, and a= 119÷6-19=the answer.

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Example 5. The velocity which A loses by impact = (A−B×a+2Bb) — (A+B) = (5+24)÷7-29÷7. velocity of B after impact=(B-Axb+2Aa)÷(A+B) = (B4+40) 7=36÷7. Therefore, velocity of A volocity of A: 29-736÷7::29:36.

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Example 6. By Art. 130, the velocity of A after impact (A−B×a—2Bb) ÷ (A+B) = (as A=3B, and b=2a), (3B-BXa-4Ba)÷(3B+B)=(2Ba—4Ba) ÷ 4B= — 2 Ba÷4B-2a÷4=-a÷4. Again, the velocity of B af. ter impact (A-B×b+2Aa) ÷ (A+B)=(3B−B × b + 3Bb)÷(3B+B)=5Bb÷4B=5b÷4.

Example 7. By Art. 135, the velocity of the last body= 2n-1 a = 25 ÷ ÷ (4) 5 = 7 11.

1+r

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Example 8. ab::32, 2a=36, and b=2a-÷÷3; also, A=2B and, by the question, perfect elasticity: imperfect elasticity:4 3, 3Xperfect elasticity=4Ximperfect elastici

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ty, and imperfect elasticity=3×perfect elast.÷4. By Art. 137, the velocity of Aafter impact-a-1+m ×B×a-b =(a-7

A+B

÷ 4×B×a÷3)÷3B=(3Ba−7 Ba÷12)÷3B=29 Ba÷ 36B=29a÷36=the velocity of A after impact. Again, by Art. 137, the velocity of B after impact=b+1+m×A+a−b

A+B

B-mAxb+1+mXAa = 76Ab-48A=76b: 48=196

A+B

÷12=the velocity of B after impact.

Example 9. By Art. 137, the velocity of A after impact= (A-mBXa)(A+B) = (as A=3B)(3-mxa)+4=3a 5; hence 15a-5am-12a, 3-5m, and m 3-5: therefore, perfect elasticity imperfect elasticity::1:3÷5::5; 3.

QUESTIONS FOR PRACTICE ON THE LEVER.

Example 1. Let x=the required weight, then (Art. 168.) 10:x::5:2, 5x=20, and x=4 pounds=the answer.

Example 2. (Art. 170.) 112 500::20.x, 112x=1000 and x 8 feet 11 inches the answer.

Example 3. By Art. 171, P=(W×BC)÷AC+({w× BC2—AC )÷(AB÷AC); hence, P=(21×3)÷7+(3× 9-49)÷70 (630÷70)-(120÷70)=510÷70=51÷7= 72=the answer.

Example 4. 10 pounds and 9 ounces=169 ounces; and 12 pounds and 4 ounces 196 ounces. 169 × 196=33124, whose square root=182 ounces=11lbs. 6 oz.=the answer.

Example 5. See note on page 120. Let x=the pressure on A; then 531::x: 150-x, 750—5x=34x, 83x=750, and x=88 the weight that A sustains: And 150—88,4,= 6113 the weight that B sustains.

Example 6. Let x=the distance of the moveable weight from the point of graduation, hence; 5:12::6;x, 5x= 72, and x=143, therefore the graduation begins 23 inches from the fulcrum on the shorter arm. To find the graduation, 12:14: graduation=3 inches. Again, let the greatest weigh then, 69835 x, and x=82 pounds=the greatest weight. Again, let x=the least weight, then, 6: 83::5: x, 6x=42, and, x=7=the least weight.

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Example 7. See Art. 174. P:p::2xsin 60° 3X sin 45°, and Pp::2√3÷2:: 3√(as sin 60°=cos 30°) -3, and sin 45° 12(Trig. 96.): therefore, P:p:: √3(3x1)=√2:: √6:3.

Example 8. See Fig. 93, and Art. 178, Cor. 2. cos BCD =P÷Ŵ. But, by the question, P: W::1: √2; hence, P÷W=1÷√2 = sin 45° = cos 45° (Trig. 96.) Then, if BCD=45°. BCA=180°-45°=135° the answer.

Example 9. See Fig. 90. Let x=the weight at E; then, 3:x:: 1×2×4:9 × 11 × 12, 8x=3564, and x=4451 pounds.

QUESTIONS FOR PRACTICE ON THE WHEEL, AND AXLE,

AND PULLEY.

Example 1. P: W::r: R; or 3:x::21:36, 2}x=108, and x=431=the answer.

Example 2. P:W::r: R::5:150::x:60, 150x=300, and x=2. 2×2=4 inches the diameter of the axle.

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Example 3. P: W::r+t;R+t, and x: 320::4+1:36 +1:537, 37x=1600, and x=43, 24 pounds.

Example 4. P: W::axrxr'r': PQXRXR'×R". (Art. 186.) 10æ::4×4×4×3:30×24×18×12, 10: a:192 155520, 192x=1555200, and x=8100 pounds.

Example 5. P: W::1:n, x:112::1;12, 12x=112, and a 9 pounds.

Example 6. P: W::1 cos 45°::1:2× √ √ therefore, P: W::1: √2=the answer.

Example 8. See example 3d. P=W÷(221)=1020÷ 28-1-1020 (256-1)=1020-255-4 pounds.

Example 7. See example 2d. n=(log. W-log. D)÷ log. 2 (log. 640-log. 5)÷log. 2-(2×80618-0.6989700) 0.30103007 pulleys.

QUESTIONS FOR PRACTICE ON THE INCLINED PLANE AND

SCREW.

Example 1. By Art. 198, P: W::hight of the plane : length of the plane, that is, 150 W::20: 30, 20 W=4500, and W 225 pounds.

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Example 2. By Art. 200, P: W::sin 50°: R=sin 90°; that is, 500 W::0871557:1, W=500÷.0871557=573 6.7 pounds.

Example 3. By Art. 209, P: W::distance of the threads circumference: Or, by Cor. under the same article, W= 2paP÷d=2×72×3.14159×50=22619.448 pounds.

Example 4. See example 3d, and Fig. 166. Let BC=a 18 inches, CD=d=1=distance of the threads, R=radius of the wheel=24, r=radius of the axle=6, S=the number of strings=4, Q=the force exerted by the screw on the wheel, and D=the force exerted by the wheel on the pulley; then, P: W::d: 2pa, Q: Dr R, D: W::1 S, and by compounding, P: W::dr: 2paRS, W=2pa RSP÷dr=(2×3.14159×18×24×4×100)÷6=180957. 25=the weight raised in a perpendicular direction. Again, by Arts. 196, 198, P. W::sin 30° R::180957.25 :: 1, and W=361912.5 pounds=161 tons3+=the force that the man would exert on the ship.

QUESTIONS ON THE MOTION OF BODIES ON INCLINED PLANES.

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Example 1. H=250, L=400, T=4", m=16, and S= 200. S=H÷LXmT2 (250÷400) × 16×16 160.82 feet the required space. T= √L×S÷√mH=√400 × 200 ✓ 16×250=√80000÷√4020.833

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4.4 seconds.

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V=(HX4mXS)÷L=√250×64×360

√400=√5790000÷√400=√14475=120 feet 1 inch in

one second.

Example 2. V=(H÷L) x 2mT= × 32X5=531 feet in 1".

Example 3. H-V2÷4m=250000641-3886 = L : therefore L=3886×2=7772 feet.

Example 4. R: 9: :sin 5° : H; or 10: 0.95424 :: 8.940 30: H=0.7844; therefore, √mH√7844 × 16 × 5280 =258.08 feet. T=9 miles 258,08=(9×5280)÷258.08 184.12=3′ 4.12′′. V 258.08 × 2=516.16 feet per se

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Example 5. R: 42240: :sin 3° 14' : H-2383 feet. T L÷√mH=4224÷÷√ 16×2383=215.8-3′ 35.8′′.

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Example 6. V=2√mH, and 3 H=2 of 2383=1787.25. V=2✓16, 1787.25=339.0392. Momentum = V×Q= 339.0392 × 1000=339039.2=the force.

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Example 7. V=2√mH=2√]6,1×50=56.6. sin 60° : 50: R L 57.72; and the other plane-100. T=L÷ √mH=57.72÷28.3=time up the plane. T=L÷√mH =100÷28.3=3.52=time down the plane; therefore, 203

+2.052=5.55 seconds.

T=L÷√mH

T=

Example 8. Sin 45° R::30: L=42.43. =42.43÷21.96=1.931=time down the whole plane. L÷√mH=21.215÷15.53=1.376 = time down the first

half; therefore, 1.931-1.376=.555 seconds.

Example 9. See example 11. Fig. 134. AE (AC— AB)2÷4AC=752÷600=5625÷600-9.37 feet from the

top of the plane.

QUESTIONS FOR PRACTICE ON THE PEndulum.

Example 1. By Art. 255, x=(2 × L)÷2=(9.8695 × 39. 02074)÷2=192.56=the velocity per second at St. Thomas. Making use of the same formula, and substituting numbers, (9.8695×391386) ÷ 2=193.1402=the velocity per second at London. 9.8695 X 39.21469-193.5187-the ve. locity per second at Spitzbergen. (9.8695 × 39.10168)÷2 192.9612=the velocity per second at New York.

Example 2. By Art. 253, L=2m÷=32.1666÷9.86 95-39.108 or 391, very nearly.

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Example 3. By the rule contained in the note on page 188, length of pendulum vibrating seconds=39.1x=9, 775 inches. Length of pendulum vibrating seconds=39. 1x,', 2.444 inches. Length of pendulum vibrating 2 seconds=39.1X4 = 156.4 inches. Length of pendulum vibrating once an hour 39.1×12960000=5667360000 in. ches 7997.7 miles.

Example 4. By example 4th page 187, the hight of the mountain=rt (T-t) (3956x3)+(3600-3)=11868÷ 3597 3.3 miles.

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