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Chapter 3 – Student Solutions Manual
1. A vector
can be represented in the
magnitudeangle
notation (
a
,
θ
), where
a
G
22
x
y
aa
a
=
+
is the magnitude and
1
tan
y
x
a
a
−
⎛⎞
=
⎜⎟
⎝⎠
is the angle
makes with the positive
x
axis.
a
G
(a) Given
A
x
=
−
25.0 m and
A
y
= 40.0 m,
( 25.0 m)
(40.0 m)
47.2 m
A
=−
+
=
(b) Recalling that tan
= tan (
+ 180°), tan
–1
[40/ (– 25)] = – 58° or 122°. Noting that
the vector is in the third quadrant (by the signs of its
x
and
y
components) we see that
122° is the correct answer. The graphical calculator “shortcuts” mentioned above are
designed to correctly choose the right possibility.
3. The
x
and the
y
components of a vector
G
a
lying on the
xy
plane are given by
cos ,
sin
xy
=
=
where
is the magnitude and
is the angle between

=
G
G
a
and the positive
x
axis.
(a) The
x
component of
is given by
a
G
a
x
= 7.3 cos 250° = – 2.5 m.
(b) and the
y
component is given by
a
y
= 7.3 sin 250° = – 6.9 m.
In considering the variety of ways to compute these, we note that the vector is 70° below
the –
x
axis, so the components could also have been found from
a
x
= – 7.3 cos 70° and
a
y
= – 7.3 sin 70°. In a similar vein, we note that the vector is 20° to the left from the –
y
axis, so one could use
a
x
= – 7.3 sin 20° and
a
y
= – 7.3 cos 20° to achieve the same
results.
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 Spring '08
 drty
 mechanics

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