Solução_Calculo_Stewart_6e

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