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144 2444
4
3
14 3
24
B
Ax
x
R y = ( 244 km ) sin 30.0o + ( 0 km ) = 122 km
144 2444
4
3
123
B
Ay
y
2
R = Rx + R 2 =
y ( 36 km )2 + (122 km )2 = 127 km Ry o
−1 122 km = tan = 74
R 36 km x
______________________________________________________________________________ θ = tan −1 51. REASONING If we let the directions due east and due north be the positive directions,
then the desired displacement A has components A E = (4.8 km ) cos 42° = 3.57 km
A N = ( 4.8 km ) sin 42° = 3.21 km
while the actual displacement B has components
B E = (2.4 km ) cos 22° = 2.23 km
B N = ( 2.4 km ) sin 22° = 0.90 km
Therefore, to reach the research station, the research team must go
3.57 km – 2.23 km = 1.34 km, eastward and
3.21 km – 0.90 km = 2.31 km, northward SOLUTION
a. From the Pythagorean theorem, we find that the
magnitude of the displacement vector required to
bring the team to the research station is
R = (1.34 km) 2 + (2.31 km) 2 = 2.7 km b. The angle θ is given by N R
θ
1.34 km 2.31 km 32 INTRODUCTION AND MATHEMATICAL CONCEPTS θ = tan –1 2.31 km = 6.0 × 10 1 degrees, north of east 1.34 km _____________________________________________________________________________________________ 52. REASONING Let A be the vector from
base camp to the first team, B the vector
Second team
from base camp to the second team, and
N (y )
C the vector from the first team’s position
First team
C
to the second team’s position. C is the
vector whose magnitude and direction are
θ
given by the first team’s GPS unit. Since
B=A+C
35°
A
you can get from the base camp to the
19°
second team’s position either by traveling W
E (x )
along vector B alone, or by traveling first
Base camp
along A and then along C, we know that
B is the vector sum of the other two:
B = A + C. The reading on the first
team’s GPS unit is then C = B − A. The components of C are found from the components of
A and B: Cx = Bx − Ax, Cy = By − Ay. Once we have these components, we can calculate the
magnitude and direction of C, as shown on the GPS readout. Because the first team is
northwest of camp, and the second team is northeast, we expect the vector C to be directed
east and either north or south. SOLUTION Let east serve as the positive x direction and north as the positive y direction.
We then calculate the components of C, noting that B’s components are both positive, and
A’s xcomponent is negative:
C x = ( 29 km) sin 35 − ( −38 km) cos 19 = 53 km
144
244
3 144 2444
4
3
B
A
o x o x C y = (29 km) cos 35o − (38 km) sin 19o = 11 km
144
244
3 144
244
3
By
A
y The second team is therefore 53 km east of the first team (since Cx is positive), and 11 km
north (since Cy is positive). The straightline distance between the teams can be calculated
with the Pythagorean theorem (Equation 1.7): C = Cx2 + C y 2 = ( 53 km )2 + (11 km )2 = 54 km The direction of the vector C is to be measured relative to due east, so we apply the inverse
tangent function (Equation 1.6) to get the angle θ: Chapter 1 Problems 33 Cy −1 11 km o = tan = 12
Cx 53 km θ = tan −1 53. SSM REASONING Since the finish line is coincident with the starting line, the net
displacement of the sailboat is zero. Hence the sum of the components of the displacement
vectors of the individual legs must be zero. In the drawing in the text, the directions to the
right and upward are taken as positive.
SOLUTION In the horizontal direction Rh = Ah + Bh + Ch + Dh = 0
Rh = (3.20 km) cos 40.0° – (5.10 km) cos 35.0° – (4.80 km) cos 23.0° + D cos θ = 0
D cos θ = 6.14 km (1) In the vertical direction Rv = Av + Bv + Cv + Dv = 0.
Rv = (3.20 km) sin 40.0° + (5.10 km) sin 35.0° – (4.80 km) sin 23.0° – D sin θ = 0.
D sin θ = 3.11 km (2) Dividing (2) by (1) gives
tan θ = (3.11 km)/(6.14 km) or θ = 26.9 ° Solving (1) gives
D = (6.14 km)/cos 26.9° = 6.88 km 54. REASONING We know that the three displacement vectors have a resultant of zero, so
that A + B + C = 0. This means that the sum of the x components of the vectors and the sum
of the y components of the vectors are separately equal to zero. From these two equations
we will be able to determine the magnitudes of vectors B and C. The directions east and
north are, respectively, the +x and +y directions.
SOLUTION Setting the sum of the x components of the vectors and the sum of the y
components of the vectors separately equal to zero, we have 34 INTRODUCTION AND MATHEMATICAL CONCEPTS 1
mg
0B
41.0
− C cos
°=
b550 42cos 25.3° + 1sin243° + b44235.03g 0
4
144 444
1
44
Ax Bx Cx 1550 g
0
− B cos 41.0 +
35.0
b44m2sin 25.3° + b44244° g C sin243° = 0
1 4 444 1
3 14
Ay Cy By These two equations contain two unknown variables, B and C. They can be solved
simultaneously to show that
a. B = 5550 m
and b. C = 6160 m
______________________________________________________________________________
55. REASONING The drawing shows the vectors A,
B, and C. Since these vectors add to give a resultant
that is zero, we can write that A + B + C = 0. This
addition will be carried out by the component
method. This means t...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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