SM chapter 3 - Solucionario capitulo 3 Serway 7ma edición PDF

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Solucionario capitulo 3 Serway 7ma edición ...

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3 Vectors ANSWERS TO QUESTIONS

CHAPTER OUTLINE 3.1 3.2 3.3 3.4

Coordinate Systems Vector and Scalar Quantities Some Properties of Vectors Components of a Vector and Unit Vectors

Q3.1

Only force and velocity are vectors. None of the other quantities requires a direction to be described. The answers are (a) yes (b) no (c) no (d) no (e) no (f ) yes (g) no.

Q3.2

The book’s displacement is zero, as it ends up at the point from which it started. The distance traveled is 6.0 meters.   The vector −2D1 will be twice as long as D1 and  in the opposite direction, namely northeast. Adding D 2 , which is about equally long and southwest, we get a sum that is still longer and due east, choice (a).

*Q3.3

*Q3.4

The magnitudes of the vectors being added are constant, and we are considering the magnitude only—not the direction—of the resultant. So we need look only at the angle between the vectors being added in each case. The smaller this angle, the larger the resultant magnitude. Thus the ranking is c = e > a > d > b.

*Q3.5

(a) leftward: negative. (b) upward: positive (c)   rightward: positive (d) downward: negative and A (e) Depending on the signs and angles of B, the sum could be in any quadrant. (f) Now   − A will be in the fourth quadrant, so −A + B will be in the fourth quadrant.

*Q3.6

(i) The magnitude is 10 2 + 10 2 m Ⲑ s, answer (f ). (ii) Having no y component means answer (a).

*Q3.7

The vertical component is opposite the 30° angle, so sin 30° = (vertical component)/50 m and the answer is (h).

*Q3.8

Take the difference of the coordinates of the ends of the vector. Final first means head end first. (i) −4 − 2 = −6 cm, answer ( j) (ii) 1 − (−2) = 3 cm, answer (c)  (i) If the direction-angle of A is between 180 degrees and 270 degrees, its components are both negative: answer (c). If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs: answer (b) or (d).   Vectors A and B are perpendicular to each other.

Q3.9

Q3.10 Q3.11

No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction, not magnitude.

Q3.12

Addition of a vector to a scalar is not defined. Think of numbers of apples and of clouds.

46

Chapter 3

SOLUTIONS TO PROBLEMS Section 3.1 P3.1

Coordinate Systems

x = r cos θ = ( 5.50 m ) cos 240° = ( 5.50 m) (−0.5 ) = −2.75 m y = r sinθ = ( 5.50 m ) sin 240° = ( 5.50 m )( −0.866) = −4.76 m

P3.2

(a)

x = r cosθ and y = r sin θ , therefore x1 = ( 2.50 m ) cos 30.0°, y1 = ( 2.50 m ) sin 30.0°, and

(x 1 , y 1) = ( 2.17, 1.25)

m

x2 = ( 3.80 m) cos 120°, y 2 = ( 3. 80 m ) sin 120°, and

(x 2 , y 2) = (− 1.90, (b) P3.3

3.29) m

d = ( ∆ x )2 + ( ∆ y )2 = 4 .07 2 + 2 .04 2 m = 4 .55 m

The x distance out to the fly is 2.00 m and the y distance up to the fly is 1.00 m. (a)

We can use the Pythagorean theorem to find the distance from the origin to the fly. 2 2 distance = x 2 + y 2 = ( 2.00 m ) + (1.00 m ) = 5.00 m 2 = 2.24 m

(b) P3.4

 ⎛ 1⎞ θ = tan− 1 ⎜ ⎟ = 26 .6 °; r = 2.24 m, 26.6° ⎝ 2⎠

We have 2.00 = r cos 30.0° r=

2.00 = 2 .31 cos 30 .0°

and y = r sin 30.0° = 2.31sin 30.0° = 1.15 . P3.5

⎛ y⎞ We have r = x 2 + y2 and θ = tan− 1 ⎜ ⎟ . ⎝ x⎠ (a)

The radius for this new point is

( −x )2 + y 2

=

x 2 + y2 = r

and its angle is tan −1 (b)

⎛ y⎞ = 180° − θ . ⎝ −x ⎠

( −2x )2 + ( −2y )2 = 2r This point is in the third quadrant if ( x , y) is in the first quadrant or in the fourth quadrant if ( x , y) is in the second quadrant. It is at an angle of 180° +θ .

(c)

( 3x ) 2 + ( −3y ) 2 = 3r This point is in the fourth quadrant if ( x , y) is in the first quadrant or in the third quadrant if ( x , y) is in the second quadrant. It is at an angle of −θ .

Vectors

Section 3.2

Vector and Scalar Quantities

Section 3.3

Some Properties of Vectors

P3.6

 − R = 310 km at 57° S of W (Scale: 1 unit = 20 km)

FIG. P3.6

P3.7

tan 35 .0 ° =

x 100 m

x = (100 m) tan 35. 0° = 70. 0 m FIG. P3.7

P3.8

    Find the resultant F1 + F2 graphically by placing the tail of F2 at the head of F1. The resultant   force vector F1 + F2 is of magnitude 9.5 N and at an angle of 57° above the x axis .

y

F1 + F 2

F2

F1 0 1 2 3 N FIG. P3.8

x

47

48

P3.9

Chapter 3

(a)

 d = − 10.0 iˆ = 10.0 m since the displacement is in a

C

straight line from point A to point B. (b)

The actual distance skated is not equal to the straight-line displacement. The distance follows the curved path of the semi-circle (ACB). s=

P3.10

5.00 m B

A

FIG. P3.9

1 ( 2π r ) = 5π = 15. 7 m 2

(c)

  If the circle is complete, d begins and ends at point A. Hence, d = 0 .

(a)

The large majority of people are standing or sitting at this hour. Their instantaneous footto-head vectors have upward vertical components on the order of 1 m and randomly oriented horizontal components. The citywide sum will be ~ 105 m upward .

(b)

Most people are lying in bed early Saturday morning. We suppose their beds are oriented north, south, east, and west quite at random. Then the horizontal component of their total vector height is very nearly zero. If their compressed pillows give their height vectors vertical components averaging 3 cm, and if one-tenth of one percent of the population are on-duty nurses or police officers, we estimate the total vector height as 5 2 3 ~10 ( 0.03 m ) + 10 ( 1 m) ~ 10 m upward .

P3.11

d

To find these vector expressions graphically, we draw each set of vectors. Measurements of the results are taken using a ruler and protractor. (Scale: 1 unit = 0.5 m ) (a)

  A + B = 5.2 m at 60°

(b)

  A − B = 3.0 m at 330°

(c)

  B − A = 3.0 m at 150°

(d)

 A − 2B = 5.2 m at 300°

FIG. P3.11

Vectors

P3.12

The the solutions for the  three  diagrams    shown  below  represent   graphical    three  vector  sums: R1 = A + B + C, R 2 = B + C + A , and R 3 = C + B + A . We observe that R1 = R2 = R3 , illustrating that the sum of a set of vectors is not affected by the order in which the vectors are added .

100 m B

C

A B

R1

R2

A

C

A R3

C

B

FIG. P3.12

P3.13

The scale drawing for the graphical solution should be similar to the figure to the right. The magnitude and direction of the final displacement from the starting point are obtained by measuring d and θ on the drawing and applying the scale factor used in making the drawing. The results should be FIG. P3.13

d = 420 ft and θ = −3 ° .

Section 3.4 *P3.14

Components of a Vector and Unit Vectors

We assume the floor is level. Take the x axis in the direction of the first displacement. If both of the 90° turns are to the right or both to the left , the displacements add like

(

)

40.0 m iˆ + 15.0 m ˆj − 20.0 m ˆi = 20.0 ˆi + 15. 0 ˆj m to give (a) displacement magnitude (20 + 15 ) 2

2 1Ⲑ 2

m = 25.0 m

at (b) tan (15Ⲑ20) = 36.9° . −1

If one turn is right and the other is left , the displacements add like

(

)

40.0 m iˆ + 15.0 m ˆj + 20.0 m ˆi = 60. 0 ˆi + 15. 0 ˆj m to give (a) displacement magnitude (602 + 152)1Ⲑ2 m = 61.8 m at (b) tan−1(15Ⲑ60) = 14.0˚. Just two answers are possible.

49

50

P3.15

Chapter 3

A x = −25.0 A y = 40.0 A=

2 2 Ax + Ay =

(−25.0 )

2

+ ( 40.0 ) = 47.2 units 2

We observe that tan φ = So

Ay Ax

.

⎛ A ⎞ FIG. P3.15 40 .0 φ = tan −1 ⎜ y ⎟ = tan = tan −1 (1.60 ) = 58.0°. A 25.0 ⎝ x ⎠ The diagram shows that the angle from the +x axis can be found by subtracting from 180°:

θ = 180° − 58° = 122° . P3.16

The person would have to walk 3.10 sin ( 25.0 °) = 1 .31 km north , and 3.10 cos (25.0°) = 2.81 km east .

*P3.17

Let v represent the speed of the camper. The northward component of its velocity is v cos 8.5°. To avoid crowding the minivan we require v cos 8.5° ≥ 28 m Ⲑs. We can satisfy this requirement simply by taking v ≥ (28 m Ⲑs)Ⲑcos 8.5° = 28.3 m Ⲑs.

P3.18

(a)

Her net x (east-west) displacement is −3.00 + 0 + 6.00 = +3.00 blocks, while her net y (north-south) displacement is 0 + 4.00 + 0 = +4.00 blocks. The magnitude of the resultant displacement is R=

( x net) + ( y net) 2

2

=

( 3.00 ) + ( 4.00 ) 2

2

= 5.00 blocks

and the angle the resultant makes with the x axis (eastward direction) is ⎛ 4.00⎞ = tan − 1 (1 .33) = 53 .1 °. θ = tan − 1⎜ ⎝ 3.00 ⎟⎠ The resultant displacement is then 5.00 blocks at 53.1 ° N of E . (b) P3.19

x = r cosθ and y = r sinθ , therefore:

)

(

(a)

x = 12.8 cos 150°, y = 12.8 sin 150°, and ( x , y) = −11 1. iˆ + 6 .40 ˆj m

(b)

x = 3.30 cos 60.0°, y = 3.30 sin 60.0°, and ( x , y) = 1.65 iˆ + 2.86 ˆj cm

(c) P3.20

The total distance traveled is 3 .00 + 4 .00 + 6.00 = 13.0 blocks .

) x = 22.0 cos 215°, y = 22.0 sin 215° , and ( x , y) = ( − 18.0 ˆi − 12.6 ˆj)

x = d cos θ = ( 50.0 m ) cos (120 ) = −25.0 m y = d sin θ = ( 50. 0 m )sin (120 ) = 43. 3 m  d = ( −25.0 m ) ˆi + ( 43. 3 m) ˆj

(

in

Vectors

P3.21

51

Let +x be East and +y be North.

∑ x = 250 + 125 cos 30 ° = 358 m ∑ y = 75 + 125sin 30° − 150 = − 12. 5 m

( ∑ x ) + (∑ y ) = (358 ) + (−12 .5 ) ( ∑ y ) = − 12. 5 = −0. 0349 taan θ = ( ∑ x ) 358 2

d=

2

2

2

= 358 m

θ = − 2. 00°  d = 35 8 m at 2.00° S of E P3.22

The east and north components of the displacement from Dallas (D) to Chicago (C) are the sums of the east and north components of the displacements from Dallas to Atlanta (A) and from Atlanta to Chicago. In equation form: dDC east = d DA east + d AC east = 730 cos 5.00° − 560 sin 21.0° = 527 miles. dDC north = d DA north + d AC north = 730 sin 5.00° + 560 cos 21.0° = 586 miles 2 2 By the Pythagorean theorem, d = (dDC east ) + (dDC north ) = 788 mi .

Then tan θ =

d DC north = 1.11 and θ = 48.0°. dDC east

Thus, Chicago is 788 miles at 48.0° northeast of Dallas . P3.23

   We have B = R − A : Ax = 150 cos120° = − 75. 0 cm Ay = 150 sin 120° = 130 cm Rx = 140 cos 35. 0° = 115 cm

x

Ry = 140 sin 35.0° = 80.3 cm m FIG. P3.23

Therefore,

(

)

 B = [115 −( −75 )] ˆi + [80 .3 − 130 ] ˆj = 190 ˆi − 49 .7 ˆj cm  2 2 B = 190 + 49. 7 = 196 cm

θ = tan − 1 ⎛ − ⎝ P3.24

49.7 ⎞ = −14.7° . 190 ⎠

(a)

See figure to the right.

(b)

   C = A + B = 2.00iˆ + 6.00 jˆ + 3.00iˆ − 2.00 ˆj = 5. 00 ˆi + 4.00jˆ  4⎞ = 6.40 at 38.7° C = 25.0 + 16.0 at tan −1 ⎛ ⎝5⎠    D = A − B = 2.00iˆ + 6.00 jˆ − 3.00 iˆ + 2.00 ˆj = −1.00 ˆi + 8.00 ˆj  D=

( −1 .00 )2 + (8 .00 )2

−1 ⎛ 8 .00 ⎞ at tan ⎜ ⎝ − 1.00⎠⎟

FIG P3 24

52

Chapter 3

(b)

( A + B) = ( 3ˆi − 2 ˆj) + ( −ˆi − 4 ˆj) =   ( A − B) = ( 3ˆi − 2 ˆj) − ( −ˆi − 4 ˆj) =

(c)

  A + B = 2 2 + 6 2 = 6.32

(d)

  A − B = 4 2 + 22 = 4.47

(e)

6 θ A+ B = tan− 1 ⎛ − ⎞ = −71 .6° = 288° ⎝ 2⎠



P3.25

(a)

2ˆi− 6ˆ j



4 ˆi + 2ˆj

2 θ A− B = tan− 1 ⎛ ⎞ = 26.6° ⎝ 4⎠ *P3.26

P3.27

We take the x axis along the slope uphill. Students, get used to this choice! The y axis is perpendicular to the slope, at 35° to the vertical. Then the displacement of the snow makes an angle of 90° − 35° − 20° = 35° with the x axis. (a)

Its component parallel to the surface is 5 m cos 35° = 4.10 m toward the top of the hill .

(b)

Its component perpendicular to the surface is 5 m sin 35° = 2.87 m .

)

(

 d 1 = −3. 50 ˆj m  d 2 = 8.20 cos 45.0° iˆ + 8.20 sin 45.0° ˆj = 5. 80ˆi + 5.80 ˆj m  d 3 = −15 .0iˆ m

(

)

)

(

    R = d 1 + d 2 + d 3 = ( −15.0 + 5.80) iˆ + ( 5 .80 − 3 .50 ) ˆj =

(− 9.20 iˆ + 2.30 ˆj) m

(or 9.20 m west and 2.30 m north) The magnitude of the resultant displacement is  R = R2x + Ry2 = (−9.20 )2 + (2.30 )2 = 9.48 m . ⎛ 2. 30 ⎞ = 166° . The direction is θ = arctan ⎝ −9.20 ⎠ P3.28

Refer to the sketch

A = 10.0

    R = A + B + C = −10.0iˆ − 15.0jˆ + 50.0iˆ = 40. 0ˆi − 15. 0ˆj  2 2 1 /2 R = ⎡⎣( 40. 0) + (−15. 0 ) ⎦⎤ = 42. 7 yards P3.29

East

North

x

y

0m

4.00 m

1.41

1.41

–0.500

–0.866

+0.914

4.55

 R =

R

B = 15.0

x 2 + y 2 at tan−1 ( y Ⲑ x) = 4 .64 m at 78.6° N of E

C = 50.0 FIG. P3.28

Vectors

P3.30

  A = −8.70 iˆ + 15.0 ˆj and B = 13.2 iˆ − 6.60 ˆj       3C = B − A = 21. 9ˆi − 21. 6ˆj A − B + 3C = 0 :  C = 7.30 ˆi − 7.20 ˆj C x = 7.30 cm ; C y = −7.20 cm

or P3.31

(a)

 F  F  F  F

  = F1 + F2 = 120 cos( 60.0°) iˆ + 120 sin( 60. 0° ) jˆ − 80. 0 cos ( 75. 0°) ˆi + 80. 0sin ( 75. 0° )ˆj

(

)

= 60. 0ˆi +104 ˆj − 20 .7 ˆi + 77 .3 ˆj = 39 .3 ˆi +181 ˆj N = 39.3 2 + 181 2 = 185 N

181 ⎞ = 77 .8° θ = tan− 1 ⎛ ⎝ 39 .3 ⎠ (b) P3.32

( −39.3iˆ − 181ˆj ) N

  F3 = −F =

 A = 3.00 m, θ A = 30.0°

 B = 3.00 m, θ B = 90.0°

Ax = A cos θA = 3 .00 cos 30.0 ° = 2.60 m  A = Ax ˆi + Ay ˆj = 2 .60 ˆi +1 .50 ˆj m

(

)

Bx = 0, By = 3.00 m

(

so

)

  A + B = 2.60 iˆ + 1.50 ˆj + 3.00 ˆj = P3.33

A y = A sin θ A = 3 .00 sin 30 .0° = 1.50 m

 B = 3.00 ˆj m

( 2.60 ˆi + 4.50 ˆj) m

 ˆ = 4 .00 ˆi + 6 .00 ˆj + 3 .00 k ˆ B = B x ˆi + B y ˆj + B z k  2 2 2 B = 4 .00 + 6. 00 + 3. 00 = 7.81 4. 00 ⎞ = 59 .2 ° is the angle with the x axis α = cos −1 ⎛ ⎝ 7 .81 ⎠ 6. 0 0⎞ − = 39 .8 ° is the angle with the y axis β = cos 1 ⎛ ⎝ 7 .81 ⎠

P3.34

3. 00 ⎞ γ = cos −1 ⎛ = 67 .4° is the angle with the z axis ⎝ 7 .81 ⎠     (a) D = A + B + C = 2iˆ − 2 ˆj  D = 22 + 22 = 2. 83 m at θ = 315 °

P3.35

(b)

    E = −A − B + C = −6 iˆ + 12 jˆ  E = 62 + 122 = 13.4 m at θ = 117°

(a)

   C =A +B =

(b)

(5.00 ˆi −1.00 ˆj − 3.00 ˆk ) m

 2 2 2 C = ( 5 .00 ) + (1 .00 ) + (3 .00 ) m = 5.92 m    D = 2 A − B = 4.00 ˆi − 11.0 ˆj + 15.0 kˆ m

(

2

)

2

2

53

54

P3.36

Chapter 3

Let the positive x-direction be eastward, the positive y -direction be vertically upward, and the positive z-direction be southward. The total displacement is then  d = 4.80 ˆi + 4.80 ˆj cm + 3. 70ˆj − 3. 70 kˆ cm = 4.80 ˆi + 8.50 ˆj − 3 .70 kˆ cm.

(

(a) (b) P3.37

(a) (b) (c)

P3.38

(

)

(

)

(4.80 )2 + ( 8.50 ) 2 + (−3.70 )2

The magnitude is d =

)

cm = 10 .4 cm .

8 .50 Its angle with the y axis follows from cosθ = , giving θ = 35.5° . 10.4  A = 8.00iˆ + 12.0 ˆj − 4.00kˆ   A B = = 2.00 ˆi + 3. 00 ˆj −1.00 ˆk 4   C = −3A = −24.0 iˆ − 36.0 ˆj + 12.0kˆ

The y coordinate of the airplane is constant and equal to 7. 60 × 103 m whereas the x coordinate is given by x = v i t where vi is the constant speed in the horizontal direction. At t = 30.0 s we have x = 8. 04 × 10 3, so vi = 8 0 40 m 30 s = 268 m s. The position vector as a function of time is  P = ( 268 m s )tˆi + ( 7. 60 × 10 3 m ˆj.

)

 At t = 45.0 s, P = ⎡⎣1 .21 × 10 4 iˆ + 7 .60 × 10 3 jˆ⎦⎤ m. The magnitude is  2 2 P = ( 1. 21 × 104 + ( 7. 60 × 103 m = 1. 43 × 10 4 m

)

)

and the direction is ⎛ 7.60 ×10 3 ⎞ = 32.2 ° above the horizontal . θ = arctan ⎜ ⎝ 1.21 × 10 4 ⎟⎠ P3.39

The position vector from radar station to ship is  S = 17.3 sin 136° iˆ + 17.3 cos 136° jˆ km = 12.0iˆ −12.4 ˆj km.

)

(

)

(

From station to plane, the position vector is  ˆ km, P = 19.6 sin 153 °ˆi +19.6 cos 153 °ˆj + 2.20 k

)

(

or

(

)

 ˆ km. P = 8.90 ˆi − 17.5 ˆj + 2.20 k (a)

To fly to the ship, the plane must undergo displacement

(

)

   ˆ km . D = S − P = 3.12 ˆi + 5.02 ˆj − 2.20 k (b)

The distance the plane must travel is  D= D =

( 3.12 )2 + (5.02 )2 + ( 2.20 )2

km = 6 .31 km .

Vectors

P3.40

(a)

 E = (17.0 cm ) cos 27. 0 °ˆi + (17.0 cm ) sin 27.0 °ˆj

55

y

)

(

 E = 15.1ˆi + 7. 72 ˆj cm (b)

 F = −( 17.0 cm) sin 27 .0 °ˆi + (17 .0 cm ) cos 27.0 °ˆj  F=

( −7.72 ˆi + 15.1ˆj) cm

Note that we do not need to

27.0º 27.0º F

G E

explicitly identify the angle with the positive x axis. (c)

 G = + (17.0 cm ) sin 27 .0 ° ˆi + (17 .0 cm ) cos27 .0 ° ˆj  G=

P3.41

27.0º x FIG. P3.40

( +7.72 ˆi +15.1ˆj) cm

Ax = −3.00 , Ay = 2.00 (a)

 A = Ax ˆi + Ay ˆj = −3. 00ˆi + 2. 00ˆj

(b)

 A =

Ax2 + Ay2 =

tanθ =

Ay Ax

=

( −3.00 )

2

+ (2.00 ) = 3.61 2

2.00 = −0.667, tan − 1( − 0. 667) = − 33. 7° ( −3.00 )

θ is in the 2 nd quadrant, so θ = 180° + (− 33.7°) = 146° . (c)

      Rx = 0, Ry = −4.00 , R = A + B thus B = R − A and Bx = Rx − Ax = 0 − ( −3.00 ) = 3.00 , By = Ry − Ay = −4.00 − 2.00 = −6 .00 .  Therefore, B = 3.00 iˆ − 6.00 ˆj .

P3.42

⎛ 41. 0 km⎞ The hurricane’s first displacement is ⎜ ⎟⎠ ( 3. 00 h ) at 60.0° N of W, and its seco...