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Unformatted text preview: al y components. We will
begin by examining the given information with respect to these criteria, in order to see if
there are obvious reasons why some of the vectors could not be equal. Then we will
compare our choices for the equal vectors by calculating the magnitude and direction and
the scalar components, as needed.
SOLUTION Vectors A and B cannot possibly be equal, because they have different x scalar
components of Ax = 80.0 m and Bx = 60.0 m. Furthermore, vectors B and C cannot possibly
be equal, because they have different magnitudes of B = 75.0 m and C = 100.0 m.
Therefore, we conclude that vectors A and C are the equal vectors. To verify that this is
indeed the case we have two choices. We can either calculate the magnitude and direction
of A (and compare it to the given magnitude and direction of C) or determine the scalar
components of C (and compare them to the given components of A). Either choice will do,
although both are shown below.
+y
The magnitude and direction of A are
A A= 2
Ax + 2
Ay = ( 80.0 m ) 2 + ( 60.0 m ) = 100.0 m θ 2 Ay +x Ax Ay −1 60.0 m = tan = 36.9° 80.0 m Ax θ = tan −1 +y
These results are identical to those given for C.
C The scalar components of C are
36.9° Cx Cy +x Chapter 1 Problems 25 C x = C cos 36.9° = (100.0 m ) cos 36.9° = 80.0 m
C y = C sin 36.9° = (100.0 m ) sin 36.9° = 60.0 m These results are identical to the components given for A. 42. REASONING Because both boats travel at
Bwest
W
101 km per hour, each one ends up
25.0°
(0.500 h)(101 km/h) = 50.5 km from the dock
Bsouth
53.0°
B
after a halfhour. They travel along straight
paths, so this is the magnitude of both
Gsouth
G
displacement vectors: B = G = 50.5 km. Since
the displacement vector G makes an angle of
37° south of due west, its direction can also be
expressed as 90° − 37° = 53° west of south.
Gwest
S
With these angles and the magnitudes of both
vectors in hand, we can consider the westward and southward components of each vector.
SOLUTION
a. First we calculate the magnitudes of the westward component of the displacement of
each boat and then subtract them to find the difference: Magnitude of Bwest = B cos 25.0o = ( 50.5 km ) cos 25.0o = 45.8 km
Magnitude of Gwest = G sin 53.0o = ( 50.5 km ) sin 53.0o = 40.3 km
The blue boat travels farther by the following amount: 45.8 km − 40.3 km = 5.5 km
b. Similarly, we find for the magnitudes of the southward components that Magnitude of Bsouth = B sin 25.0o = ( 50.5 km ) sin 25.0o = 21.3 km
Magnitude of Gsouth = G cos53.0 = ( 50.5 km ) cos53.0 = 30.4 km
o The green boat travels farther by the following amount: 30.4 km − 21.3 km = 9.1 km o 26 INTRODUCTION AND MATHEMATICAL CONCEPTS 43. SSM WWW REASONING AND SOLUTION We take due north to be the direction
of the +y axis. Vectors A and B are the components of the resultant, C. The angle that C
makes with the x axis is then θ = tan −1 (B / A ) . The symbol u denotes the units of the
vectors.
a. Solving for B gives
B = A tan θ = (6.00 u) tan 60.0° = 10.4 u
b. The magnitude of C is C = A 2 + B 2 = ( 6. 00 u) 2 + ( 10 .4 u) 2 = 12 .0 u
______________________________________________________________________________ 44. REASONING AND SOLUTION The force F can be first resolved into two components;
the z component Fz and the projection onto the xy plane, Fp as shown in the figure below on
the left. According to that figure,
Fp = F sin 54.0° = (475 N) sin 54.0°= 384 N
The projection onto the xy plane, Fp, can then be resolved into x and y components.
z
y
F
F 54.0°
54.0° F p F y z 33.0°
F F
x p x a. From the figure on the right,
Fx = Fp cos 33.0° = (384 N) cos 33.0°= 322 N b. Also from the figure on the right,
Fy = Fp sin 33.0° = (384 N) sin 33.0°= 209 N c. From the figure on the left,
Fz = F cos 54.0° = (475 N) cos 54.0°= 279 N Chapter 1 Problems 27 ______________________________________________________________________________
45. SSM REASONING The individual displacements of the golf ball, A, B, and C are
shown in the figure. Their resultant, R, is the displacement that would have been needed to
"hole the ball" on the very first putt. We will use the component method to find R.
N
R C
B θ 20.0 o E A SOLUTION The components of each displacement vector are given in the table below. Vector x Components y Components A
B
C (5.0 m) cos 0° = 5.0 m
(2.1 m) cos 20.0° = 2.0 m
(0.50 m) cos 90.0° = 0 (5.0 m) sin 0° = 0
(2.1 m) sin 20.0° = 0.72 m
(0.50 m) sin 90.0° = 0.50 m R = A+B+C 7.0 m 1.22 m The resultant vector R has magnitude
R = ( 7 .0 m) 2 + ( 1.22 m) 2 = 7 .1 m and the angle θ is 1.22 m = 9.9°
7.0 m θ = tan −1 Thus, the required direction is 9.9° north of east .
______________________________________________________________________________
46. REASONING Using the component method for vector addition, we will find the x
component of the resultant force vector by adding the x components of the individual
vectors. Then we will find the y component of the resultant vector by adding the y
compone...
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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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