1.30:
7.8 km, 38 north of east
1.32:
a) b) c) d)
11.1 m @ 28.5 m @ 11.1 m @ 28.5 m @
77.6 202 258 22
1.33:
144 m, 41 south of west.
1.34:
1.35:
A; A x 12.0 msin 37.0 7.2 m, A y 12.0 m cos 37.0 9.6 m.
B ; B x 15.0 m cos 40.0 11 .5 m, B y 15 .0 m sin 40.0
C ; C x 6.0 m cos 60.0 1.36:
(a)
tan θ
A y A X
9.6 m.
3.0 m, C y 6.0 m sin 60.0 5.2 m.
1.00 m
2.00 m
0.500
θ
tan 1 0.500 360
(b)
tan θ θ
(c )
tan θ θ
(d)
A y
1.00 m
26.6
2.00 m
A x
A x
1.00 m
2.00 m
tan 1 0.500 180 A y
A x
1.00
333
0.500
tan 1 0.500 26.6
A y
tan θ
0.500 26.6
m
2.00 m
153
0.500
tan 1 0.500 180 26.6 207 1.37: Take the + x-direction to be forward and the + y-direction to be upward. Then the second force has components F 2 x F 2 cos 32.4 433 N and F 2 y F 2 sin32.4 275 N. θ
The first force has components F 1 x 725 N and F 1 y 0. F x F 1 x F 2 x 1158 N and F y F 1 y F 2 y 275 N
The resultant force is 1190 N in the direction 13.4 above the forward direction.
1.38:
(The figure is given with the solution to Exercise 1.31). o
The net northward displacement is (2.6 km) + (3.1 km) sin 45 = 4.8 km, and the net o eastward displacement is (4.0 km) + (3.1 km) cos 45 = 6.2 km. The magnitude of the resultant displacement is arctan
(4.8 km)2 (6.2 km)2 = 7.8 km, and the direction is
64..28 = 38o north of east.
Using components as a check for any graphical method, the components of B x 14.4 m and B y 10.8 m, A has one component, A x 12 m .
1.39:
B
are
a) The x - and y - components of the sum are 2.4 m and 10.8 m, for a magnitude of
2.4 m2 10.8 m2 11.1 m, , and an angle of 10.8 77.6 .
2.4
b) The magnitude and direction of A + B are the same as B + A. c) The x- and y-components of the vector difference are – 26.4 m and 10.8 m, for a magnitude of 28.5 m and a direction arctan
1026..84 202 . Note that 180
must be added to
.8 arctan1026..84 22 in order to give an angle in the third quadrant. arctan 10 26.4
d) B A 14.4 mi 10.8 m j 12.0 mi 26.4 mi 10.8 mj . ˆ
Magnitude
1.40:
ˆ
ˆ
ˆ
ˆ
26.4 m2 10 .8 m2 28 .5 m at and angle of arctan
10.8
22 .2 . 26.4
Using Equations (1.8) and (1.9), the magnitude and direction of each of the given vectors is:
(8.6 cm) 2 (5.20 cm) 2 = 10.0 cm, arctan
a)
58.20.60 = 148.8o (which is 180o
– 31.2o).
o
b)
(9.7 m) 2 (2.45 m) 2 = 10.0 m, arctan
c)
(7.75 km) 2 (2.70 km) 2
29.45.7 = 14o + 180o = 194o.
= 8.21 km, arctan
7.275.7 = 340.8o (which is
o
360 – 19.2 ). 1.41:
The total northward displacement is 3.25 km 1.50 km 1.75 km, , and the total westward displacement is 4.75 km . The magnitude of the net displacement is
1.75 km2 4.75 km2 5.06 km. The south and west displacements are the same, so The direction of the net displacement is 69.80 West of North.
1.42:
a) The x- and y-components of the sum are 1.30 cm + 4.10 cm = 5.40 cm, 2.25 cm + ( – 3.75 cm) = – 1.50 cm.
b) Using Equations (1-8) and (1-9),
(5.4 0 cm) 2 (1.50 cm) 2
=
5.60 cm, arctan
15..5040 = 344.5o ccw.
c) Similarly, 4.10 cm – (1.30 cm) = 2.80 cm, – 3.75 cm – (2.25 cm) = – 6.00 cm. d)
(2.80 cm) 2 (6.0 cm) 2
1.43:
a) The magnitude of A B
6.62 cm, arctan
=
26.80.00 = 295o (which is 360o – 65o).
is
2.80 cm cos 60.0 1.90 cm cos 60.0 2 2.48 cm 2 2.80 cm sin 60.0 1.90 cm sin 60.0
and the angle is
2.80 cm sin 60.0 1.90 cmsin 60.0 18 arctan 2.80 cm cos 60.0 1 . 90 cm cos 60 . 0
b) The magnitude of A B is
2.80 cm cos 60.0 1.90 cm cos 60.0 2 4.10 cm 2 2.80 cm sin 60.0 1.90 cm sin 60.0
and the angle is
2.80 cm sin 60.0 1.90 cmsin 60.0 84 arctan 2.80 cm cos 60.0 1 . 90 cm cos 60 . 0 c) B A A B ; the magnitude is 4.10 cm and the angle is 84 180 264 .
1.44:
A =
( – 12.0 m)
More precisely,
i . ˆ
A 12.0 m cos 180 i 12.0 m sin 180
j .
B 18.0 m cos 37 i 18.0 m sin 37 j 14.4 m i
ˆ
ˆ
ˆ
10.8 m j ˆ
1.45:
A
12.0 m sin 37.0 i 12.0 m cos 37.0 j 7.2 m i 9.6 m j
ˆ
ˆ
ˆ
ˆ
B 15.0 m cos 40.0 i 15.0 m sin 40.0 j 11.5 m i 9.6 m j
ˆ
ˆ
ˆ
ˆ
C 6.0 m cos 60.0 i 6.0 m sin 60.0 j 3.0 m i 5.2 m j
ˆ
1.46:
a)
A
ˆ
ˆ
ˆ
3.60 m cos 70.0 i 3.60 msin 70.0 j 1.23 mi 3.38 m j
ˆ
ˆ
ˆ
ˆ
B 2.40 m cos 30.0 i 2.40 m sin 30.0 j 2.08 mi 1.20 m j
b)
ˆ
ˆ
ˆ
ˆ
C 3.00 A 4.00B
3.001.23 m i 3.003.38 m j 4.00 2.08 m i 4.00 1.20 m j ˆ
ˆ
ˆ
ˆ
12.01 m i 14.94 j ˆ
c)
ˆ
(Note that in adding components, the fourth figure becomes significant.) From Equations (1.8) and (1.9), 14.94 m 2 2 C 12.01 m 14.94 m 19.17 m, arctan 51.2 12.01 m
1.47:
b) c)
4.002 3.002 5.00,
A
a)
B
5.002 2.002 5.39
A B 4.00 3.00i 5.00 2.00 j 1.00i 5.00 j ˆ
ˆ
1.00 2 5.002 5.10,
5.00 101 .3 - 1.00
arctan
ˆ
ˆ
d)
1.48:
a) i j k 12 12 12 3 1 so it is not a unit vector ˆ
ˆ
ˆ
b)
A
A x A y A z 2
2
2
If any component is greater than + 1 or less than – 1,
A 1,
so it cannot be a unit
vector. A can have negative components since the minus sign goes away when the component is squared.
c)
A
1
a 3.0 a 4.0 1 2
2
2
2
a a
1.49:
A B A x
B A
1
0.20
5.0
a) Let A A x i A y j , ˆ
25 1
2
B B x i B y j .
ˆ
ˆ
ˆ
B x i A y B y j ˆ
ˆ
B x A x i B y A y j ˆ
ˆ
Scalar addition is commutative, so A B B A
A B A x B x B A
A y B y
B x A x B y A y
Scalar multiplication is commutative, so A B B A
B A B A
B A i B A
B A j B A
B A k
A B A y B z A z B y i A z B x A x B z j A x B y A y B x k
b)
y
z
ˆ
ˆ
ˆ
z
y
ˆ
z x
x
z
x
y
ˆ
ˆ
y
x
Comparison of each component in each vector product shows that one is the negative of the other.
1.50:
Method 1: Pr oduct of magnitudes cos θ AB cos θ 12 m 15 m cos 93 9.4 m 2
BC cos θ 15 m 6 m cos 80 15.6 m 2
AC cos θ 12 m 6 m cos 187 71.5 m 2
Method 2: (Sum of products of components) 2 A B 7.22 (11.49) (9.58)( 9.64) 9.4 m BC
(11.49)( 3.0) (9.64)( 5.20) 15.6 m 2
AC
(7.22)(3.0) (9.58)( 5.20) 71.5 m 2
1.51:
a) From Eq.(1.21),
A B 4.005.00 3.00 2.00 14.00.
b)
AB
1.52:
AB cos θ , so θ arccos 14 .00 5.00 5.39 arccos.5195 58 .7 .
For all of these pairs of vectors, the angle is found from combining Equations (1.18) and (1.21), to give the angle as
A B A B A y B y arccos x x . arccos AB AB In the intermediate calculations given here, the significant figures in the dot products and in the magnitudes of the vectors are suppressed.
a)
A B 22, A
40, B 13,
and so
22 165 13
arccos 40
b)
c)
A B 60, A
.
60 34, B 136, arccos 28 34 136
A B 0, 90.
.
