A secn2 x dx tan2 x dx xn ex dx xn ex n 1 tan3 x

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Unformatted text preview: 0 = 6000π ft·lb 0 -2 9 (10 − x)62.4(300)dx 13. (a) W = 10 9 0 9 (10 − x)dx = 18, 720 x 0 = 926,640 ft·lb (b) to empty the pool in one hour would require 926,640/3600 = 257.4 ft·lb of work per second so hp of motor = 257.4/550 = 0.468 0 20 14. When the rocket is x ft above the ground 15 3000 total weight = weight of rocket + weight of fuel = 3 + [40 − 2(x/1000)] = 43 − x/500 tons, x Rocket 3000 (43 − x/500)dx = 120, 000 ft·tons W= 0 0 10 - x Exercise Set 8.7 280 100 15(100 − x)dx 15. W = 16. Let F (x) be the force needed to hold charge A at position x, then c c F (x) = , F (−a) = 2 = k , 2 (a − x) 4a 0 = 75, 000 ft·lb Pulley so c = 4a2 k . 100 0 W= −a 100 - x 4a2 k (a − x)−2 dx = 2ak J A −a x B 0 a x Chain 0 17. (a) 150 = k/(4000)2 , k = 2.4 × 109 , w(x) = k/x2 = 2,400,000,000/x2 lb (b) 6000 = k/(4000)2 , k = 9.6 × 1010 , w(x) = 9.6 × 1010 /(x + 4000)2 lb 5000 (c) W = 9.6(1010 )x−2 dx = 4,800,000 mi·lb = 2.5344 × 1010 ft·lb 4000 18. (a) 20 = k/(1080)2 , k = 2.3328 × 107 , weight = w(x + 1080) = 2.3328 · 107 /(x + 1080)2 lb 10.8 [2.3328 · 107 /(x + 1080)2 ] dx = 213.86 mi·lb = 1,129,188 ft·lb (b) W = 0 19. W = F · d = (6.40 × 105 )(3.00 × 103 ) = 1.92 × 109 J; from Theorem 8.6.4, 2 2 vf = 2W/m + vi = 2(1.92 · 109 )/(4 · 105 ) + 202 = 10,000, vf = 100 m/s 20. W = F · d = (2.00 × 105 )(2.00 × 105 ) = 4 × 1010 J; from Theorem 8.6.4, 2 2 vf = 2W/m + vi = 8 · 1010 /(2 · 103 ) + 108 ≈ 11.832 m/s. 21. (a) The kinetic energy would have decreased by (b) (4.5 × 1014 )/(4.2 × 1015 ) ≈ 0.107 1 1 mv 2 = 4 · 106 (15000)2 = 4.5 × 1014 J 2 2 1000 (0.107) ≈ 8.24 bombs (c) 13 EXERCISE SET 8.7 1. (a) F = ρhA = 62.4(5)(100) = 31,200 lb 2 P = ρh = 62.4(5) = 312 lb/ft 2. (a) F = P A = 6 · 105 (160) = 9.6 × 107 N (b) F = ρhA = 9810(10)(25) = 2,452,500 N P = ρh = 9810(10) = 98.1 kPa (b) F = P A = 100(60) = 6000 lb 2 62.4x(4)dx 3. F = 0 0 x 2 x dx = 499.2 lb = 249.6 0 2 4 281 Chapter 8 3 4. F = 9810x(4)dx 0 1 x dx = 39240 4 1 x 3 1 3 = 156, 960 N 5 0 5 y x y= √25 – x 2 5 9810x(2 25 − x2 )dx 5. F = 2 √25 – x 2 0 5 x(25 − x ) 2 1/2 = 19, 620 dx 0 = 8.175 × 105 N √ 23 F= 0 √ 23 w(x) x 4 2√ 62.4x √ (2 3 − x) dx 3 124.8 =√ 3 4 0 6. By similar triangles √ 2 3−x 2√ w(x) √ = , w(x) = √ (2 3 − x), 4 23 3 4 2√ 3 √ (2 3x − x2 )dx = 499.2 lb 0 7. By similar triangles 0 10 − x w(x) = 6 8 2 x 3 w(x) = (10 − x), 4 10 9810x F= 2 6 w(x) 8 3 (10 − x) dx 4 10 10 (10x − x2 )dx = 1,098,720 N = 7357.5 2 0 8. w(x) = 16 + 2u(x), but 12 − x 1 u(x) = so u(x) = (12 − x), 4 8 2 4 u (x) 4 4 x w(x) w(x) = 16 + (12 − x) = 28 − x, 12 16 12 62.4x(28 − x)dx F= 4 12 (28x − x2 )dx = 77, 209.6 lb. = 62.4 4 b 9. Yes: if ρ2 = 2ρ1 then F2 = b ρ2 h(x)w(x) dx = a b 2ρ1 h(x)w(x) dx = 2 a ρ1 h(x)w(x) dx = 2F1 . a Exercise Set 8.7 282 2 50x(2 4 − x2 )dx 10. F = 0 0 y 2 2 x(4 − x2 )1/2 dx = 100 y = √ 4 – x2 0 = 800/3 lb 2√ 4 – x 2 x 11. Find the forces on the upper and lower halves and add them: 0 w1 (x) x √ =√ , w1 (x) = 2x 2a 2a/2 √ 2a/2 F1 = √ 2a/2 ρx(2x)dx = 2ρ 0 √2a / 2 √ x2 dx = 2ρa3 /6, x F2 = √ √ ρx[2( 2a − x)]dx = 2ρ 2a/2 √ 2a √ a √2a 0 √ √ w2 (x) 2a − x √ =√ , w2 (x) = 2( 2a − x) 2a 2a/2 √ 2a a x √ √ ( 2ax − x2 )dx = 2ρa3 /3, 2a/2 √ √ √ F = F1 + F2 = 2ρa3 /6 + 2ρa3 /3 = ρa3 / 2 lb 12. h(x) = x sin 60◦ = 100 F= √ 3x/2, 200 √ 9810( 3x/2)(200)dx 0 0 √ = 981000 3 100 60° 100 0 √ = 4,905,000 3 N 13. h (x) x x dx √ √ √ 162 + 42 = 272 = 4 17 is the other dimension of the bottom. √ (h(x) − 4)/4 = x/(4 17) √ h(x) = x/ 17 + 4, √ 4 17 F= 0 √ 62.4(x/ 17 + 4)10dx √ 4 17 = 624 16 4 0 h (x) 4 x √ (x/ 17 + 4)dx 4 4√17 0 √ = 14, 976 17 lb h+2 14. F = ρ0 x(2)dx 0 h h h+2 x dx = 2ρ0 h = 4ρ0 (h + 1) h x h+2 2 2 10 a w1(x) a√ w2(x) 2a 283 Chapter 8 15. (a) From Exercise 14, F = 4ρ0 (h + 1) so (assuming that ρ0 is constant) dF/dt = 4ρ0 (dh/dt) which is a positive constant if dh/dt is a positive constant. (b) If dh/dt = 20 then dF/dt = 80ρ0 lb/min from part (a). EXERCISE SET 8.8 1. (a) (b) (c) (d) (e) (f ) 2. (a) csch(−1) ≈ −0.8509 (b) sech(ln 2) = 0.8 sinh 3 ≈ 10.0179 cosh(−2) ≈ 3.7622 tanh(ln 4) = 15/17 ≈ 0.8824 sinh−1 (−2) ≈ −1.4436 cosh−1 3 ≈ 1.7627 3 tanh−1 ≈ 0.9730 4 3. (a) sinh(ln 3) = (c) coth 1 ≈ 1.3130 1 (d) sech−1 ≈ 1.3170 2 (e) coth−1 3 ≈ 0.3466 √ (f ) csch−1 (− 3) ≈ −0.5493 1 1 ln 3 (e − e− ln 3 ) = 2 2 3− 1 3 4 3 = 5 4 (b) cosh(− ln 2) = 1 − ln 2 1 (e + eln 2 ) = 2 2 (c) tanh(2 ln 5) = 312 25 − 1/25 e2 ln 5 − e−2 ln 5 = = 2 ln 5 + e−2 ln 5 e 25 + 1/25 313 (d) sinh(−3 ln 2) = 1 +2 2 1 1 −3 ln 2 (e − e3 ln 2 ) = 2 2 = 1 −8 8 4. (a) 1 1 ln x (e + e− ln x ) = 2 2 x+ 1 x = 1 1 ln x (e − e− ln x ) = 2 2 x− 1 x = x2 − 1 ,x>0 2x (c) x2 − 1/x2 x4 − 1 e2 ln x − e−2 ln x ,x>0 =2 =4 e2 ln x + e−2 ln x x + 1/x2 x +1 (d) 1 − ln x 1 (e + eln x ) = 2 2 63 16 x2 + 1 ,x>0 2x (b) =− 5. sinh x0 1 +x x = 1 + x2 ,x>0 2x cosh x0 tanh x0 coth x0 sech x0 csch x0 (a) 2 √5 2/√ 5 √ 5/2 1/ √ 5 1/ 2 (b) 3/ 4 5/ 4 3/ 5 5/3 4/5 4/3 (c) 4/3 5/ 3 4/5 5/4 3/ 5 3/ 4...
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This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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