Units, Physical Quantities and Vectors
121
(
)
(0
148.5 m
242.5 m)
94.0 m
y
y
y
y
D
A
B
C
= −
+
+
= −
−
+
= −
2
2
x
y
D
D
D
=
+
2
2
(
108.5 m)
(
94.0 m)
144 m
D
=
−
+ −
=
94.0 m
tan
0.8664
108.5 m
y
x
D
D
θ
−
=
=
=
−
180
40.9
220.9
θ
=
° +
° =
°
(
D
"
is in the third quadrant since both
x
D
and
y
D
are negative.)
Figure 1.73b
The direction of
D
"
can also be specified in terms of
180
40.9 ;
φ
θ
=
−
° =
°
D
"
is 41
°
south of west.
E
VALUATE
:
The vector addition diagram, approximately to scale, is
Vector
D
"
in this diagram
agrees qualitatively with
our calculation using
components.
Figure 1.73c
1.74.
I
DENTIFY
:
Solve for one of the vectors in the vector sum. Use components.
S
ET
U
P
:
Use coordinates for which
x
+
is east and
y
+
is north. The vector displacements are:
2.00 km, 0
of east;
3.50 m, 45
south of east;
=
°
=
°
A
B
$
$
and
5.80 m, 0
east
=
°
R
$
E
XECUTE
:
(
)
(
)(
)
5.80 km
2.00 km
3.50 km
cos45
1.33 km
x
x
x
x
C
R
A
B
=
−
−
=
−
−
° =
;
y
y
y
y
C
R
A
B
=
−
−
(
)(
)
0 km
0 km
3.50 km
sin45
2.47 km
=
−
− −
° =
;
(
)
(
)
2
2
1.33 km
2.47 km
2.81 km
C
=
+
=
;
(
) (
)
1
tan
2.47 km
1.33 km
61.7
north of east.
θ
−
⎡
⎤
=
=
°
⎣
⎦
The vector addition diagram in Figure 1.74 shows good
qualitative agreement with these values.
E
VALUATE
:
The third leg lies in the first quadrant since its
x
and
y
components are both positive.
Figure 1.74
1.75.
I
DENTIFY
:
The sum of the vector forces on the beam sum to zero, so their
x
components and their
y
components
sum to zero. Solve for the components of
F
"
.
S
ET
U
P
:
The forces on the beam are sketched in Figure 1.75a. Choose coordinates as shown in the sketch. The
100N pull makes an angle of 30.0
40.0
70.0
+
=
°
°
°
with the horizontal.
F
"
and the 100N pull have been replaced
by their
x
and
y
components.
E
XECUTE
:
(a)
The sum of the
x
components is equal to zero gives
(100 N)cos70.0
0
x
F
+
=
°
and
34.2 N
x
F
= −
.
The sum of the
y
components is equal to zero gives
(100 N)sin70.0
124 N
0
y
F
+
−
=
°
and
30.0 N
y
F
= +
.
F
"
and
its components are sketched in Figure 1.75b.
2
2
45.5 N
x
y
F
F
F
=
+
=
.
30.0 N
tan
34.2 N
y
x
F
F
φ
=
=
and
41.3
φ
=
°
.
F
"
is
directed at 41.3
°
above the
x
−
axis in Figure 1.75a.
(b)
The vector addition diagram is given in Figure 1.75c.
F
"
determined from the diagram agrees with
F
"
calculated in part (a) using components.
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122
Chapter 1
E
VALUATE
:
The vertical component of the 100 N pull is less than the 124 N weight so
F
"
must have an upward
component if all three forces balance.
Figure 1.75
1.76.
I
DENTIFY
:
The four displacements return her to her starting point, so
(
)
−
D =
A+ B + C
"
"
"
"
, where
A
"
,
B
"
and
C
"
are in the three given displacements and
D
"
is the displacement for her return.
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 Spring '06
 Buchler
 Physics, Addition, Vectors, Vector Space, Dot Product, Force, Cardinal direction

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