*This preview shows
pages
1–3. Sign up
to
view the full content.*

CHAPTER 3 PROBLEM SOLUTIONS
Problem
7. Vectors
A
and
B
in Fig. 3-22 have the same magnitude,
A
. Find the magnitude and direction of (a)
A
!
B
and
(b)
A
+
B
.
Solution
The vectors
A
and
B
in Fig. 3-22 form two sides of a parallelogram, in which
A
!
B
and
A
+
B
are the diagonals, as
shown. Since the magnitudes of
A
and
B
are equal, the parallelogram is a rhombus, and the diagonals are perpendicular
(the converse of this is also true; see Problem 60). Then
A
+
B
is along the perpendicular bisector of the base
A
!
B
of an
isosceles triangle, and vice versa. Using the given angles, we find the magnitudes (a)
A
!
B
=
2
A
sin20
° =
0.684
A
,
and
(b)
A
+
B
=
2
A
cos20
° =
1.88
A
.
In Fig. 3-22, (a)
A
!
B
is up, and (b)
A
+
B
is to the right, but the directions could be
specified relative to
A
,
B
, or some other coordinate system. (This problem can also be readily solved with components and
unit vectors. Figure 3-22 suggests a coordinate system with
x
-axis to the right and
y
-axis up, as shown. Then
A
=
A
(
î
cos
20
° +
ˆ
j
sin
20
°
)
and
B
=
A
(
î
cos
20
°
!
ˆ
j
sin
20
°
)
, from which
A
±
B
are easily obtained.)
Problem
8. Vector
A
has magnitude 1.0 m and points at 35º clock-wise from the
x
-axis. Vector
B
has magnitude 1.8 m. What angle should
B
make with the
x
-axis in order that
A
+
B
be purely vertical?
Solution
The vectors
A
,
B
, and
A
+
B
form a triangle, but given only that
B
=
1.8
A
and
A
+
B
are perpendicular to the
x
-axis, two are possible. In one, the angle opposite side
B is
!
=
125
°
, in the other
=
55
°
, as sketched. For either, the law of sines gives
=
sin
"
1
A
sin
#
B
$
%
&
’
(
)
=
sin
"
1
sin125
°
1.8
$
%
&
’
(
)
=
sin
"
1
sin55
°
1.8
$
%
&
’
(
)
=
27.1
°
. Therefore,
B
makes
an angle of
±
62.9
°
with the negative
x
-axis or
±
117
°
with the
x
-axis (look at suitable
right triangles in the sketch). (This result can also be
obtained from components: since
A
+
B
is vertical,
A
x
+
B
x
=
0,
or
B
x
=
B
cos
x
=
"
A
x
=
"
A
cos35
°
, so
x
=
cos
"
1
(
"
A
cos
35
°
=
B
)
=
cos
"
1
(
"
cos
35
°
=
1
.
8
)
= ±
117
°
,
where
B
y
could be positive
or negative.)

This
** preview**
has intentionally

**sections.**

*blurred***to view the full version.**

*Sign up*