This preview shows page 1. Sign up to view the full content.
Unformatted text preview: nts of the individual vectors. Once the x and y components of the resultant are
known, we will use the Pythagorean theorem to find the magnitude of the resultant and
trigonometry to find its direction. We will take east as the +x direction and north as the +y
direction.
SOLUTION The x component of the resultant force F is 28 INTRODUCTION AND MATHEMATICAL CONCEPTS b g b g b g Fx = 2240 N cos 34 .0 ° + 3160 N cos 90.0 ° = 2240 N cos 34 .0 °
144 2444 144 2444
4
3
4
3
FAx
FBx
The y component of the resultant force F is b g b g Fy = − 2240 N sin 34 .0 ° + −3160 N
1444 444
2
3 14 4
23
FAy
FBy Using the Pythagorean theorem, we find that the magnitude of the resultant force is
F= Fx2 + Fy2 = 2240
cos
b N g 34.0° 2 b g + − 2240 N sin 34 .0 °−3160 N 2 = 4790 N Using trigonometry, we find that the direction of the resultant force is θ = tan −1 L2240 N g 34.0°+3160 N O 67.2° south of east
b
=
M b sin g 34.0° P
N 2240 N cos
Q ______________________________________________________________________________
47. REASONING AND SOLUTION The horizontal component of the resultant vector is
Rh = Ah + Bh +Ch = (0.00 m) + (15.0 m) + (18.0 m) cos 35.0° = 29.7 m Similarly, the vertical component is
Rv = Av + Bv + Cv = (5.00 m) + (0.00 m) + (– 18.0 m) sin 35.0° = – 5.32 m The magnitude of the resultant vector is R= 2
2
R h + Rv = 29
–5.32
b .7 mg+ b mg =
2 2 30.2 m The angle θ is obtained from θ = tan–1[(5.32 m)/(29.7 m)] = 10.2°
____________________________________________________________________________________________ 48. REASONING The resultant force in part a is FA, because that is the only force applied.
The resultant force R in part b, where the two additional forces with identical magnitudes
(FB = FC = F) are applied, is the sum of all three vectors: R = FA + FB + FC. Because the
magnitude R of the resultant force R is k times larger than the magnitude FA of FA, we have
R = kFA (1) Chapter 1 Problems 29 To solve the problem, we need to find an expression for R in terms of FA and F. Let the +x
direction be in the direction of FA and the +y direction be upward (see the drawing), and
consider the resultant of the two additional force vectors in part b. Because FB and FC are
directed symmetrically about the x axis, and have the same magnitude F, their y components
are equal and opposite. Therefore, they cancel out of the vector sum R, leaving only the x
components of FB and FC. Now, since FA also has only an x component, the resultant R can
be written as the sum of FA and the vector x components of FB and FC:
R = FA + FBx + FCx. Because these vectors all point in the same direction, we can write
down a first expression for the magnitude of the resultant:
R = FA + FBx + FCx (2) +y FB
FA FC 20.0° 20.0° FBx FCx +x R The x components of the vectors FB and FC are adjacent to the 20.0°angles, and so are h related to the common magnitude F of both vectors by Equation 1.2 cos θ = a , with
h ha = FBx and h = F:
FBx = F cos θ and FCx = F cos θ (3) SOLUTION First, we use Equations (3) to replace FBx and FCx in Equation (2):
R = FA + F cos θ + F cos θ = FA + 2 F cos θ (4) Then, we combine Equation (4) with Equation (1) (R = kFA) to eliminate R, and solve for
the desired ratio F/FA:
FA + 2 F cos θ = kFA or 2 F cos θ = kFA − FA or F
k −1
2.00 − 1
=
=
= 0.532
FA 2 cos θ 2 cos 20.0o 2 F cos θ = ( k − 1) FA 30 INTRODUCTION AND MATHEMATICAL CONCEPTS 49. REASONING Using the component method, we find the components of the resultant R
that are due east and due north. The magnitude and direction of the resultant R can be
determined from its components, the Pythagorean theorem, and the tangent function.
SOLUTION The first four rows of the table below give the components of the vectors A,
B, C, and D. Note that east and north have been taken as the positive directions; hence
vectors pointing due west and due south will appear with a negative sign. Vector East/West
Component North/South
Component A
B
C
D + 2.00 km
0
– 2.50 km
0 0
+ 3.75 km
0
–3.00 km R=A+B+C+D – 0.50 km + 0.75 km The fifth row in the table gives the components of R. The magnitude of R is given by the
Pythagorean theorem as
R = (– 0.50 km) 2 + ( + 0.75 km) 2 = 0 . 90 km The angle θ that R makes with the direction due west is
R 0.75 km = 56° north of west
0.50 km θ = tan −1 RN θ RE
_____________________________________________________________________________________________ 50. REASONING According to the component
method for vector addition, the x component of the
resultant vector is the sum of the x component of A
and the x component of B. Similarly, the y
component of the resultant vector is the sum of the
y component of A and the y component of B. The
magnitude R of the resultant can be obtained from
the x and y components of the resultant by using
the Pythagorean theorem. The directional angle θ
can be obtained using trigonometry. North
B
R θ A 30.0° Chapter 1 Problems 31 SOLUTION We find the following results:
Rx = ( 244 km ) cos 30.0o + ( −175 km ) = 3...
View
Full
Document
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

Click to edit the document details