Solução_Calculo_Stewart_6e

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F. 13. By similar triangles, 8 4 √ 3 = w i x ∗ i ⇒ TX.10 SECTION 8.3 APPLICATIONS TO PHYSICS AND ENGINEERING ¤ 365 w i = 2x∗ i √ . The area of the ith 3 rectangular strip is 2x∗ i √ ∆x and the pressure on it is ρg 4 √ 3 − x ∗ i . 3 F = 4 √ 3 0 =4ρg x 2 4 √ 3 0 ρg 4 √ 2x 3 − x √ dx =8ρg 3 4 √ 3 0 xdx− 2ρg √ 4 3 √ x 2 dx 3 − 2ρg 3 √ x 3 4 √ 3 =192ρg − 2ρg 0 3 3 √ 3 64 · 3 √ 3 = 192ρg − 128ρg =64ρg ≈ 64(840)(9.8) ≈ 5.27 × 10 5 N 0 15. (a) The top of the cube has depth d =1m− 20 cm = 80 cm = 0.8m. (b) The area of a strip is 0.2 ∆x and the pressure on it is ρgx ∗ i . F = ρgdA ≈ (1000)(9.8)(0.8)(0.2) 2 = 313.6 ≈ 314 N F = 1 ρgx(0.2) dx =0.2ρg 1 x2 1 =(0.2ρg)(0.18) = 0.036ρg =0.036(1000)(9.8) = 352.8 ≈ 353 N 0.8 2 0.8 17. (a) The area of a strip is 20 ∆x and the pressure on it is δx i . F = 3 0 δx20 dx =20δ 1 2 x2 3 0 =20δ · 9 2 =90δ = 90(62.5) = 5625 lb ≈ 5.63 × 10 3 lb (b) F = 9 0 δx20 dx =20δ 1 2 x2 9 0 =20δ · 81 2 = 810δ = 810(62.5) = 50,625 lb ≈ 5.06 × 104 lb. (c) For the first 3 ft, the length of the side is constant at 40 ft. For 3
F. 366 ¤ CHAPTER 8 FURTHER APPLICATIONS OF INTEGRATION TX.10 21. The moment M of the system about the origin is M = 2 m i x i = m 1 x 1 + m 2 x 2 =40· 2+30· 5 = 230. The mass m of the system is m = 2 m i = m 1 + m 2 =40+30=70. i=1 ThecenterofmassofthesystemisM/m = 230 70 = 23 7 . 23. m = 3 m i =6+5+10=21. i=1 M x = 3 m i y i =6(5)+5(−2) + 10(−1) = 10; M y = 3 m i x i = 6(1) + 5(3) + 10(−2) = 1. i=1 x = My m = 1 Mx and y = 21 m = 10 21 , so the center of mass of the system is 1 , 10 21 21 . 25. Since the region in the figure is symmetric about the y-axis, we know i=1 i=1 that x =0. The region is “bottom-heavy,” so we know that y0.5 and y
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