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Cal Poly State University, SLO
D. Niebuhr
Physics Department
Winter 2008
Phys 141
Homework Solutions
Chapter 3, 4
Assigned Problems:
Chapter 3
: 3, 7, 10, 18, 22, 26, 31, 45, 47
Chapter 4
: 4, 5, 8, 16, 25, 28, 32, 34, 41, 46
3.3.
Visualize:
Solve:
(a)
C
=
A

B
if
B
r
is directed in the same direction as
A
r
. Size does not matter, except that
A
B
because
C
as a magnitude cannot be negative.
(b)
C
=
A
+
B
if
B
r
is directed opposite to the direction of
A
r
. Size does not matter.
3.7.
Visualize:
The figure shows the components
v
x
and
v
y
, and the angle
θ
.
Solve:
We have,
sin40 ,
y
v
v
= 
°
or
10 m/s
sin40 ,
v

= 
°
or
15.56 m/s.
v
=
Thus the
x
component is
cos40
(15.56 m/s ) cos40
11.9 m/s.
x
v
v
=
° =
° =
Assess:
Note that we had to insert the minus sign manually with
y
v
since the vector is in the fourth quadrant.
3.10.
Visualize:
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View Full DocumentCal Poly State University, SLO
D. Niebuhr
Physics Department
Winter 2008
Phys 141
We will follow the rules given in the Tactics Box 3.1.
Solve:
(a)
(2 km)sin30
1 km
x
r
= 
° = 
(2 km)cos30
1.73 km
y
r
=
° =
(b)
(5 cm/s)sin90
5 cm/s
x
v
= 
° = 
(5 cm/s)cos90
0 cm/s
y
v
=
° =
(c)
2
2
(10 m/s )sin40
6.43 m/s
x
a
= 
° = 
2
2
(10 m/s )cos40
7.66 m/s
y
a
= 
° = 
(d)
(50 N)sin36.9
30.0 N
x
F
=
° =
(50 N)cos36.9
40.0 N
y
F
=
° =
Assess:
The components have the same units as the vectors. Note the minus signs we have manually inserted
according to the Tactics Box 3.1.
3.18.
Visualize:
Solve:
(
a)
We have
5
2
A
i
j
=
+
r
and
3
5 .
B
i
j
= 

r
This means 2
10
4
A
i
j
=
+
r
and 3
9
15 .
B
i
j
= 

r
Hence,
2
3
E
A
B
=
+
r
r
r
±
1
11 .
i
j
=

(b)
Vectors
A
r
,
,
B
r
and
E
r
are shown in the above figure.
(c)
From the
E
r
vector,
1
x
E
=
and
11
y
E
= 
. Therefore, the magnitude and direction of
E
v
are
2
2
(1)
( 11)
11.05
E
=
+ 
=
1
1
1
tan
tan
5.19


11
x
y
E
E
φ


=
=
=
°
Assess:
Note that
is the angle made with the
y
axis, and that is why
1
tan (
/
)
x
y
E
E

=
rather than
1
tan (
/
)
y
x
E
E

which would be the case if
were the angle made with the
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 Winter '06
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