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

SECTION 12.2 VECTORS 0 117
33. The given force vectors can be expressed in terms of their horizontal and vertical components as -300 i and
~00 cos 6~ 0 i + 200 sin 60°j = 200 ( ~) i + 200 ( ~) j = 100 i + 100 v'3 j . The resultant force F is the sum of
these two vectors: F = ( -300 + 100) i + (0 + 100 v'3) j = - 200 i + 100J3j. Then we have
IFI ~ V(-200) 2 + (100v'3) 2 = .j70,000 = 100 V7 ~ 264.6 N. Let() be the angle F makes with the
positive · · x-ax1s. · Th en tan () = lOO _ J3 = - . J3 an d t h e termma . I pomt . o f F 1· 1es m . t h e secon d qua d rant, so
200 2
() = tan- 1 ( - ~) + 180° ~ -40.9° f. 180° = 139.1°.
35. With respect to the water's surface, the woman's velocity is the vector sum of the velocity of the ship with respect
to the water, and the woman's velocity with respect to the ship. If we let north be the positive y-direction, then
v = (0, 22) + (-3, 0) = (-3, 22). The woman's speed is lvl = .jg + 484 ~ 22.2 mi/ h. The vector v makes an angle()
with the east, where()= tan- 1 ( ~;) ~ 98°. Therefore, the woman's direction is about N(98- 90) 0 W = N8°W.
37. Let T 1 and T 2 represent the tension vectors in each side of the
clothesline as shown in the figure. T 1 and T 2 have equal vertical
components and opposite horizontal components, so we can write
T 1 = - a i + b j and T 2 = a i + b j [a, b > 0]. By similar triangles, !!_ = 0·08 => a = 50b. The force due to gravity
a 4
acting on .Ule shirt has magnitude 0.8g ~ (0.8)(9.8) = 7.84 N, hence we have w = - 7.84j. The resultant T 1 + T 2
of the tensile forces counterbalances w , so T 1 + T 2 = -w => (-a i + bj ) + (a i + bj ) = 7.84j =>
( -50b i + bj ) + (50b i + bj ) = 2bj = 7.84j => b = 7·: 4 = 3.92 and a= 50b = 196. Thus the tensions are
T 1 = - a i + bj = - 196i + 3.92j and T 2 = ai + bj = 196i + 3.92j .
Alternatively, we can find the value of() and proceed as in Example 7.
39 .. (a) Set up coordinate axes so that the boatman is ·at the origin,- the canal is
y
bordered by the y-axis and the line x = 3, and the current flows in the
negative y-direction. The boatman wants to reach the point (3, 2). Let() be
the angle, measured from the positive y-axis, in the direction he should
steer. (See the figure.)
X
In still water, the boat has velocity v b = (13 sjn B, 13 cos()) and the velocity of the current is V c (0, -3.5), so the true path
of the boat is determined by the velocity vector v = v b + V c = (13 sin B, 13 cos() -
3.5). Let t be the time (in hours)
after the boat departs; then the position of the boat at time t is given by tv and the boat crosses the canal when
tv = (13sin0, 13cos0- 3.5) t = (3, 2). Thus 13(sinO)t = 3 =>
3
t = - 13
. ()and (13cosB- 3.5) t = 2.
sm ·