Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

148 0 CHAPTER 12 PROBLEMS PLUS
(c) The area of a horizontal cross-section of the solid is A ( z) = 1r(z 2 + 1 ), so V = J; A( z )dz = 1r ( ~ z 3 + z] ~ = 4 3 ,..
53 53 54 50 5"- 2 -
jv5 1 = 24
. 32
jv 1j = 23
. 32
. Similarly, jv5j = 24 . 33
, jv.71 :;= 25 . 34
, and in general, jv ,. j = 2 n_ 2 . 3 n_ 3 = 3(ir 2 .
Thus
f Jv,.J = lv1l + lv 2l + f 3(i)"- 2 = 2 + 3 + f 3(~ )"
n = l n = 3 n=l
00 ll
= 5 + E .2.(!~)"- 1 = 5 + ___L_ [sum of a geometric series] = 5 + 15 = 20
n=l
2
G 1 - ~
7. (a) When 8 = 8., the block is not moving, so the sum of the forces on the block
I
must be 0, thus N + F + W = 0. This relationship is illustrated
geometrically in the figure. Since the vectors form a right triangle, we have
- J!l_ 11-.• n ­
tan(O.) - INI - n - fl-s·
(b) We place the block at the origin and sketch the force vectors acting on the block, including the additional horizontal force
H , with initial points at the origin. We then rotate this system so that Flies along the positive x-axis and the inclined plane
is parallel to the x-axis. (See the follo~ing figure.)
N
F
w
IF! is maximal, so jFj = 11-. n for f) > B_,. Then the vectors, in terms of components parallel and perpendicular to the
inclined plane, are
N =n j
W = ( - mg sin 0) i + ( -mg cos fJ) j
F = (11-.n) i
H = (hmin cos B) i + ( - hmin sin B) j
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