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Unformatted text preview: Q3.3 Q3.4 Q3.5 Q3.6 Q3.7 Q3.8
Q3.9 Vectors ANSWERS TO QUESTIONS Q3.1 No. The sum of two vectors can only be zero if they are in
opposite directions and have the same magnitude. If you walk
10 meters north and then 6 meters south, you won't end up
where you started. Q3.2 No, the magnitude of the displacement is always less than or
equal to the distance traveled. If two displacements in the same
direction are added, then the magnitude of their sum will be
equal to the distance traveled. Two vectors in any other
orientation will give a displacement less than the distance
traveled. If you first walk 3 meters east, and then 4 meters
south, you will have walked a total distance of 7 meters, but
you will only be 5 meters from your starting point. The largest possible magnitude of R = A + B is 7 units, found when A and B point in the same
direction. The smallest magnitude of R = A + B is 3 units, found when A and B have opposite
directions. Only force and velocity are vectors. None of the other quantities requires a direction to be described. If the directionangle of A is between 180 degrees and 2.70 degrees, its components are both
negative. If a vector is in the second quadrant or the fourth quadrant, its components have opposite signs. The book's displacement is zero, as it ends up at the point from which it started. The distance
traveled is 6.0 meters. 85 miles. The magnitude of the displacement is the distance from the starting point, the 2.60mile
mark, to the ending point, the 175mile marl<. Vectors A and B are perpendicular to each other. No, the magnitude of a vector is always positive. A minus sign in a vector only indicates direction,
not magnitude. 55 56
Q3.10 Q3.11 Q3.12 Q3.13 Q3.14 Vectors Any vector that points along a line at 45° to the x and y axes has components equal in magnitude. Ax =Bx and Ay =By. Addition of a vector to a scalar is not defined. Think of apples and oranges. One difficulty arises in determining the individual components. The relationships between a vector
and its components such as Ax = A cos 6, are based on righttriangle trigonometry. Another problem would be in determining the magnitude or the direction of a vector from its components. Again, A = 112492, + A; only holds true if the two component vectors, Ax and Ay, are perpendicular. If the direction of a vector is specified by giving the angle of the vector measured clockwise from the
positive yaxis, then the xcomponent of the vector is equal to the sine of the angle multiplied by the
magnitude of the vector. SOLUTIONS TO PROBLEMS Section 3.1 P3.1 P3.2 P3.3 Coordinate Systems x = 7 cos 6 = (5.50 m) cos 240° 2 (5.50 m)(—0.5) = —2.75 m
y = r sin6 = (5.50 m)sin 2400 = (5.50 m)(—0.866) = —4.76 m (a) x = rcos 6 and y = r sin6, therefore
x1 = (2.50 m)cos 30.00, y1 = (2.50 m)sin30.0°, and (x1, = (2.17, m x2 = (3.80 m)cos 120°, y2 = (3.80 m)sin 120°, and (x2, h): (—1.90, 3.29) m . (b) d = ./(Ax)2 +(Ay)2 = J16.6 +4.16 = 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. distance = .le +y2 = (2.00 m)2 +(1.00 m)2 = 5.00 m2 = 2.24 m (b) 6 = tan’1 = 26.60; r = 2.24 m, 26.60 Chapter 3 57 P3.4 (a) d = \/(x2 — x1)2 + (yZ — y1)2 = \/(2.00 — [—3.00])2 + (400 — 3.00)Z d = J250 +490 = 8.60 m (b) 71 = (2.00)2 +(400)Z = J200 = 4.47 m 61 =tan1[—ﬂ)= —63.4°
200 1'2 = ,/(—3.00)2 +(300)Z = J180 = 4.24 m 62 2 135° measured from the +x axis. P3.5 We have 2.00 = 7 cos 30.0o r——= 2.31 and y = rsin 30.0°= 2.31 sin30.0°= 1.15 . P3.6 We have 7 = ﬁxz +y2 and 6 : Jamil]. X (a) The radius for this new point is ‘l(—x)2 +yZ =1lx2 +yZ = r and its angle is tan1[i)= 1800—6 . (b) 1/(—2x)2 +(—2.y)Z = Zr . 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°+6 . (c) (3x)Z +(—3y)Z = 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 —6 . 58 Vectors
Section 3.2 Vector and Scalar Quantities
Section 3.3 Some Properties of Vectors
P3.7 tan 35.00: x {fr/H
100 In x" x
x = (100 In) tan35.0°= 70.0 In ,ﬁﬁﬁﬂ"
I
FIG. P3.7
P3.8 R = 14 km
6 = 65° N of E
R 13 km
_
1 km 6 km
FIG. P3.8
1’39 —R = 310 km at 57° 5 of W
(Scale: 1 unit 2 2.0 km)
P3.10 (a) Using graphical methods, place the tail of vector B at the head of vector A. The new
vector A + B has a magnitude of 6.1 at 112° from the xaxis. (b) The vector difference A — B is found by
placing the negative of vector B at the
head of vector A. The resultant vector A — B has magnitude 14.8 units at an FIG. P3.10 angle of 22° from the + xaxis. P3.11 P3.12 P3.13 (a) ‘dH—iooi straight line from point A to point B. (b) The actual distance skated is not equal to the straightline = 10.0 m since the displacement is in a displacement. The distance follows the curved path of the semicircle (ACB).
1
s = 5(2711'): 57! = 15.7 In
(c) If the circle is complete, d begins and ends at point A. Hence, d  = 0 . 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 xaxis . Chapter 3 FIG. P3.11 59 x
0 1 2. 3 N
FIG. P3.12
(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 (b) Most people are lying in bed early Saturday morning. We suppose their beds are oriented ~105 m upward . north, south, east, 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 onetenth of one percent of the population are onduty nurses or police officers, we estimate the total vector height as ~ 105(0.03 m)+ 102 (1 m) ~ 103 m upward . 60
P3.14 P3.15 >*P3.16 Vectors Your sketch should be drawn to scale, and
should look somewhat like that pictured to
the right. The angle from the westward
direction, 6, can be measured to be 4° N of W , and the distance R from the sketch can be converted according to the
7.9 In . scale to be 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 2 0.5 m) (a) A + B = 5.2mat 60° (b) A — B = 3.0 m at 330° (c) B — A = 3.0 m at 150° (d) A _ 213 = 5.2 m at 300°. FIG. P3.14 TTT—T
. I II (a) +++++»
. I II + I 7+7+7+7
. I A+B I I I I I
.l._l_4.__l._4._.l I I I
Lima; I I I I I I I I I I I
mcmcmcmcmcmcmclclcmclcmca (C) r—T T—TTT
I .
III B‘A‘T‘
FA+7 e+e+e
I ~++ +
kB+. I.
. ,
I.A ..
+r+i 7+7+r
I —+—+ + 4
L,L,;,l,i ;, ,J
FIG.P3.15 The three diagrams shown below represent the graphical solutions for the three vector sums:
R1 = A+B+C, R2 = B+C+A, and R3 = C+B+A . You should 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. FIG. P3.16 Chapter 3 61 P3.17 The scale drawing for the graphical solution _ _ _
should be similar to the figure to the right. The _ _ _
magnitude and direction of the final displacement I ‘ 7 e — ' ‘
from the starting point are obtained by measuring ‘ . . . ,
d and 60n the drawing and applying the scale factor used in making the drawing. The results ,14.141,1,1.1_;_L_L_L_L_L_L_1_L_;_L_1_J
should be (Scale: 1 unit 2 20 ft)
d = 420 ft and 6 2 —3° FIG. P3.17
Section 3.4 Components of a Vector and Unit Vectors
P3.18 Coordinates of the superhero are:
x = (100 m) cos(—30.0°) = 86.6 m
y = (100 m)sin(—30.0°) = —50.0 m
P3.19 Ax = —25.0 _ y
Ay = 40.0 I
_ l 2 2 _ I _ 2 2 _ .
A — Ax + Ay — ( 25.0) + (40.0) — 47.2 un1ts 40.0 A
We observe that l ¢ 9
M l
y
tan 2 .
¢ M. l
FIG. P3.19
50
A
¢ = tan’1 y = tan 400 = tan’1 (1.60) = 58.00.
le 1 25.0
The diagram shows that the angle from the +x axis can be found by subtracting from 180°:
6 = 180°—58°= 122° .
P3.20 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 . 62 Vectors P3.21 x = 7 cos 6 and y = 7 sin 6, therefore: (a) x = 12.8 cos 150°, y = 12.8sin1500, and (x, y) = (—11.13 6.403) m (b) x = 3.30 cos 6000, y = 3.30 sin 60.00, and (x, y) = (1.653 —— 2.863) cm (c) x = 22.0 cos 215°, y = 220501 2150, and (x, y) = (—18.0i — 12.63) in P3.22 x = d cos 6 = (50.0 m) cos(120) = —25.0 m
y = dsinQ = (50.0 m)sin(120) = 43.3 m d = (—25.0 m)3+(433 m)3 >*P3.23 (a) Her net x (eastwest) 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 = \/(xnet)2 + (ym)Z = \/(3.00)2 + (4.00)2 = 5.00 blocks and the angle the resultant makes with the xaxis (eastward direction) is 6 = tan1[3‘—) = tan’1 (1.33) = 53.10. The resultant displacement is then 5.00 blocks at 53.10 N of E . (b) The total distance traveled is 3.00 + 4.00 + 6.00 = 13.0 blocks . >*P3.2.4 Let 2 east and 2 north. The unicyclist's displacement is, in meters N 2803:2203:3603 3003 1203:603 403 903:703. R: —1103+5503 = 110 m 2 + 550 m 2 at tan’1 110 m west of north
E
550 m = 561 m at 11.30 west of north. FIG. P3.24
The crow’ s velocity is Ax _ 561 mat 11.30 W ofN
At 40s 2 14.0 m/s at 11.30 west of north . V_ Chapter 3 63
P3.25 +x East, +y North 2x = 2.50 + 12.5 cos 30°: 358 m
2y 2 75 + 125 sin30°—150 = —12.5 m d = (2202 +(Zy)2 = 1R3525)Z +(—12.5)Z = 358 m tan6 = y) = — = —0.0349
(2 x) 358
6 = —2.00° d 2358 m at 2.000 S ofE P3.26 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 2 dDA east + dAC east 2 730 cos 5.000—560 sin 2.1.00 2 52.7 miles.
dDC north 2 dDA north + dAC north 2 730 sin 5.00°+ 560 cos 2.1.00 2 586 miles. By the Pythagorean theorem, d = 4(ch east)Z +(dDC north)2 2 788 mi. d
Then tan6 = DC—“Orth = 1.11 and 6 = 4800.
DC east Thus, Chicago is 788 miles at 48.00 northeast of Dallas . P3.27 (a) See figure to the right. (b) c — A : B — 2.003 : 6.00} : 3.003 2.00} — 5.003 : 4.00} c = 425.0 + 16.0 at 1.1111(3) = 6.40 at 38.70 5 D = A — B = 2.00% +6.00} — 3.003 + 2.00} = —1.00§ + 8.00} D = (—1.00)2 +(800)Z at mfg)
—1.00 FIG. P3.27 D = 8.06 at (1800—8290) = 8.06 at 97.2.0 P3 .28 d 2
\/(x1 +962 +x3)2 +(y1 +y2 +y3) = 3.00 5.00 :600 2 : 2.00 : 3.00 : 1.00 2 =J52.0 = 7.21m
J< 6 =tan1[@)= 56.30 64 Vectors
P3.29 We have BzR—A: y Ax = 150 c03120°= —75.0 cm B
Ay = 150 sin120°= 130 cm
Rx = 140 cos 35.00 = 115 cm 3500
Ry = 140 sin35.0° = 80.3 cm FIG. P3.29 Therefore, B = [115 — (—75)]i + [80.3 — 130]} = (1905 — 49.73) cm
‘3‘: \/1902 +4972 = 196 cm 6=tan1[—ﬂ)= —14.7° .
190 P330 A = —8.703 + 15.0} and B = 13.23 — 6.60} A—B+3C=0:
3C=B—A=21.9i—21.6§ C = 7.303 — 7.20} 01‘ Cx = 7.30 cm ;Cy = —7.20 cm P331 (a) (A +13) = (33— 2§)+(—i—4§) = 23—6} (b) (A B)—(3i 2}) (i 4})— 43:2}
(c) ]A+B‘=1/22+6Z = 6.32
(d) A—B=\/4Z+22 = 4.47 (e) alMBI = tan1[—%)= —71.6° = 288° 2
9 =1 ’1 — = 26.60
AiB an [4) P332 (a) D=A+B+C=23+43 [D] = Wl2Z +4Z = 4.47 m at a = 634° (b) E=—A—B+C=—6i+6} E=x/6Z +6Z = 8.49 mat 621350 Chapter 3 65 P333 d1 = (—3.503) m
d2 = 8.20 c0s45.0°i+8.20 sin 45.00} = (5.80i +5803) m d3 = (—15.0i) m de1 :d2 :d3 =( 15.0 : 5.80)i : (5.80 3.50)}: (—9.203+2.30§)m (or 9.20 In west and 2.30 In north) The magnitude of the resultant displacement is [R] = JR; +11; = ‘/(—9.20)2 +(2.30)Z = 9.48 m . The direction is 6 = arctan[ 2'30 )2 166° . —9.20
P334 Refer to the sketch ‘Al 2 10,0
R=A+B+C = 71003715039003 R
= 40.03 — 15.0}
[R] = [(40.0)2 +(—15.0)2]1/Z = 42.7 yards \CI 2500
FIG. P334
P335 (a) F = P1 + F2 F = 120 c0s(60.0°)i + 120 sin(60.00)} — 80.0 c0s(75.0°)i + 80.0 sin(75.00)}
F = 60.0i + 104} — 20.73 + 77.3} = (39.33 + 1813) N
m =x/393Z +181Z = 185 \I 6 = tan’1 = 77.80
39.3 (b) P3 = —F = (—39.33— 1813) \I P336 = 4.64 In at 78.60 N of E 66 Vectors
P3.37 A = 3.00 In, (9A = 30.00 B = 3.00 In, (93 = 90.00
Ax = A cos (9A 2 3.00 cos 30.00 = 2.60 In Ay = A sin19A = 3.00 sin30.0O = 1.50 In A = Axi+Ay3=(2.601+ 1.503) m Bx = 0, By =3.00 m s0 B = 3.003 m A + B = (2.603 + 1.503) + 3.003 = (2.603 + 4.503) m P338 Let the positive xdirection be eastward, the positive ydirection be vertically upward, and the
positive zdirection be southward. The total displacement is then d = (4.803 + 4.803) cm + (3.703 — 3.7012) cm = (4803 +8503 — 3.7012) cm. (a) The magnitude is d = (4.80)2 —i—(8.50)Z —i—(—3.70)Z cm: 10.4 cm . (b) Its angle with the yaxis follows from cos (9 = %, giving (9 = 35.50 . P339 B = Bxi+By3+B212 = 4003 +6003 + 30012 )B) =3/400Z +6.002 +3.002 = 7.81 or = c0s1[ﬂj = 59.2.0
7.81 5 = cosiﬂj = 39.80 7 = cosiwj = 67.40 P3.40 The y coordinate of the airplane is constant and equal to 7.60 x 103 In whereas the x coordinate is
given by x = vit where vi is the constant speed in the horizontal direction. At t: 30.0 s we have x = 8.04>< 103, so vi = 2.68 m/s. The position vector as a function of time
is A P = (268 m/s)t3+(7.60 X 103 111)]. At t: 45.0 s, P = [1.21 X 1043+7.60 X 103 3] m. The magnitude is IJ=,/(1.21><104)2+(7.60><103)Z m: 1.43x104 m and the direction is 32.2.O above the horizontal . [7.60x103]
(9 = arctan — = 1.21 x104 P3 .42 P3 .43 P3 .44 (a) A = 8.00i + 12.0} — 4.0012 (b) B = — = 2.003 + 3.00} — 1.0012 (c) C = —3A = —24.03 —36.03 + 12012 R = 75.0 cos 240°i+ 75.0sin 240°} + 125 cos 135oi+ 125 sin135° } + 100 cos 1600i + 100 sin160°] R: 37.5} 65.0} 8843:8843 94.0}:342} R = —2203 + 57.6} R = 1[(—220)Z +576Z at arctan[%) above the —xaxis R = 227 paces at 1650 (a) C = A + B = (5.003 — 1.00} — 30012) no 1c121/(500)Z +0.00)Z +(3.00)Z m: 5.92m (b) D = 2A — B = (4.003 — 11.0} + 15012) no \D=,/(4.00)Z +(110)Z +(150)Z m: 19.0 m The position vector from radar station to ship is A s = (17.3sin 1360i + 17.3 cos 136° ]) km = (12.03 — 12.43) km. From station to plane, the position vector is P = (19.6si11153°i+ 19.6 cos 153°§+ 2.2012)1<m, 01‘ P = (8.90i — 17.5} + 2.2012)1<m. (a) To fly to the ship, the plane must undergo displacement D=S—P= (3.12i+5.02}— 22012) km . (b) The distance the plane rnust h‘avel is D=lD= (3.12)2 +(5.02)Z +(2.20)2 km: 6.31 km . Chapter 3 A 67 68 P3 .45 P3.46 P3.47 Vectors 41.0 km The hurricane's first displacement is [ )(300 h) at 60.00 N of W, and its second displacement is [25'0 km j(1.50 h) due North. With representing east and representing north, its total displacement is: (41.0 kYmcos 60.0°j(3.00 h)(—i) + (41.0 $501 60.0°j(3.00 11)} + (25.0 k$00.50 11)} = 61.5 l<m(—i) +144 km} with magnitude 11(615 km)Z +(144 km)2 2 157 km . A (a) E = (17.0 cm) c0s 27.0°i + (17.0 cm)sin 2.7.00] E = (15.13 + 7.723) cm A (b) F = —(17.0 cm)sin 27.0°i + (17.0 cm)c0s 2.7.00] F = (—7.721+15.1§) cm A (c) G = +(17.0 cm)sin 27.0°i + (17.0 cm) c0s 27.0°j FIG. P3.46 G: (+7.7zi+15.1§) cm Ax = —3.00, Ay = 2.00 A (a) A = A; + 14y]: —3.001 + 2.00} (b) [A]=\/Af +14; =\/(—3.00)Z +(200)Z = 3.61 me: i Z 2.00
(—3.00) x = —0.667, tan’1(—0.667) = —33.7° 6 is in the 2nd quadrant, s0 (9 = 180°+(—33.7°)= 146O . (c) Rx=0,Ry=—4.00,R=A+BthusBzR—A and B =R —A =0—(—3.00)=3.00,By_Ry Ay_ 4.00 2.00_ 6.00. x x x Therefore,B= 3003—600} . P3 .48 P3 .49 P3.50 Let +x 2 East, +y 2 North, (a) (b) (a) (b) Taking components along and , we get two equations: and e=tan41= 746° Nof E ‘Rlzwlxz +y2 = 470 km RX 2 40.0 cos 45.0°+30.0 cos 45.00 = 49.5
Ry = 40.0 sin 45.00—300 sin 45.0°+20.0 = 27.1 R = 49.53 + 27.1} W = (49.5)2 +(27.1)Z 2 56 .4 6 =tan1[£)= 28.7o Solving simultaneously, Therefore, 6.0011 — 8.0017 + 26.0 = 0 —8.00a + 3.0019 + 19.0 = 0. a = 5.00, b = 7.00 . 5.00A + 7003+ C = 0. Chapter 3 FIG. P3.49 69 70 Vectors
Additional Problems P3.51 Let Qrepresent the angle between the directions of A and B. Since
A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 1800—9, g, and The
magnitude of R is then R = 2A cos[§j. [Hint apply the law of cosines to the isosceles triangle and use the fact that B = A.] Again, A, —B, and D = A— B form an isosceles triangle with apex FIG_ P351
angle 6. Applying the law of cosines and the identity (1 — cos 6) = 2‘sinZ gives the magnitude of D as D = 2A sin[%j. The problem requires that R = 100D. Thus, 2A cos[§j 2 200A sin[%j. This gives tan[§j = 0.010 and (9 = 1.150 . P352 Let Qrepresent the angle between the directions of A and B. Since A and B have the same magnitudes, A, B, and R = A + B form an isosceles triangle in which the angles are 1800—9, g, and The 2 2
magnitude of R is then R = 2A cos[§j. [Hint apply the law of cosines to the isosceles triangle and use the fact that B = A. ] Again, A, —B, and D = A — B form an isosceles triangle with apex
angle 6. Applying the law of cosines and the identity FIG_ P352 (1 — cos 6) = 2‘sinZ gives the magnitude of D as D = 2A sin[§j. The problem requires that R = nD or cos[§j = nsin[€j giving 2
6 = Ztan’1 .
n Chapter 3 P353 (a) R = 2.00 ,R = 1.00 ,R = 3.00 x y Z (b) lRl = R: + R; + R3 = \/4.00 + 1.00 + 9.00 = J14.0 = 3.74 (c) cos19x =R—x=>19x =cos1[&]= 57.70 from ——x
W R
cos19 =R—y=>(9 =cos’1 £ = 7450 from ——y
y N y R ' cos19 =£=>19 =cos1[RZ]= 36.70 from +z >*P3.54 Take the xaxis along the tail section of the snake. The displacement from tail to head is 240 mi+ (420 — 240) mcos(180°—105°)i — 180 msin 75°} = 287 mi — 174 m}. Its magnitude is (287)2 + (174)2 m = 335 m. From 22 = dlStAﬁ, the time for each child's run is
' 335mh 1l<m 3600s
Inge: At  dIStance  ( )  101 s
v (12 km)(1 000 m)(1 h)
Olaf: At = = 126 s.
3.33 In
Inge wins by 126 —101 = 25.4 s .
>*P3.55 The position vector from the ground under the controller of the first airplane is r1 = (19.2 km)(cos 25°)i+(19.2 km)(sin 250)}+(0.8 @012
= (17.41 +8.11} + 0812) km. The second is at r2 = (17.6 km)(cos20°)i+ (17.6 km)(sin20°)i+(1.1 km)lA<
= (1651+ 6.02} + 1.112) km. Now the displacement from the first plane to the second is
r2 — r1 = (—0.863i — 2093 +0312) km with magnitude 1l(0.863)2+(2.09)2+(0.3)2 = 2.29 km . 72 Vectors >*P3.56 Let A represent the distance from island 2 to island 3. The displacement is A = A at 159°. Represent the displacement from 3
to 1 as B=B at 298° . We have 4.76 km at 37° +A+B=0. For xcomponents (4.76 km) cos 37°+A cos 159°+B cos 2980 = 0
3.80 km — 0.934A + 0.4693 2 0
B = —8.10 l<m+ 1.99A FIG. P356 For ycomponents, (4.76 km) sin 37°+A sin 159°+B sin 2980 = 0
2.86 km + 0.358A — 0.8833 2 0 (a) We solve by eliminating B by substitution: 2.86 km + 0.358A — 0.883(—8.10 l<m + 1.99A) = 0
2.86 l<m+ 0.358A + 7.15 km — 1.76A = 0 10.0 km 2 1.40A
A = 7.17 km
(b) B = —8.10 l<m+ 1.99(7.17 km): 6.15 km
>*P3.57 (a) We first express the corner's position vectors as sets of components
A = (10 m)cos50°i —(10 m)sin50°i = 6.43 mi —— 7.66 m}
B = (12 m)cos30°i —(12 m)sin30°i = 10.4 mi ——6.00 m}. The horizontal width of the rectangle is
10.4 m— 6.43 m = 3.96 In.
Its vertical height is 7.66 m—6.00 m: 1.66 In. Its perimeter is 2(3.96+1.66)m= 11.2m . (b) The position vector of the distant corner is Bxi +Ayi = 10.4 mi+ 7.66 mi = \/10.42 + 7.662 m at
7.66 m _
10.4 m _ tan’1 12.9 m at 36.40 . Chapter 3 73 P358 Choose the +xaxis in the direction of the first force. The total force, y\\ in newtons, is then
31 N 12.0%+31.0§—8.40§—24.0§= (3.603)+(7.00§) N . The magnitude of the total force is (3.60)Z +(700)2 N: 7.87 N (7 00) FIG. P358
and the angle it makes with our +xaxis is given by tan(9 = W,
(9 = 62.80. Thus, its angle counterclockwise from the horizontal is
35004—628O = 97.80 . P359 d1 : 100; y 6
012 = —3003 4—7 —\ x
013 = —150cos(30.0°)i — 150sin(30.0°)} = —130§ — 75.0} 014 = —200 cos(60.0°)i + 20051n(60.00)§= —1003 + 173}
R = d1 +dZ +d3 +d4 = (—130i— 2023) m
R = 1/(—130)2 + (—202)2 = 240 In
FIG. P359 45 = tan’1 = 57.20
130 6=180+¢= 237° dr d(4i+3}— 2153) P3.60 —_ =0+0—2°=—2.00 3
dt dt J ( m/S)J The position vector at t: 0 is 4i + 3}. At t: 1 s , the position is 4i + 1}, and so on. The object is
moving straight downward at 2 m/s, so d . .
—r represents 1ts veloc1ty vector . dt P361 v = vxi + vyj = (300 + 100 cos 30.0°)i+ (100 sin30.0°)§ v = (387i + 50.03) mi/h lvl 2 390 mi/h at 7.370 Not E 74 Vectors P3.62 (a) You start at pointA: r1 = rA = (30.0i — 20.03) m.
The displacement to B is r3 — rA = 60.0i + 80.0} — 30.03 + 20.0} = 30.03 + 100}. You cover half of this, (1503 + 50.03) to move to r2 = 30.03 — 20.0} + 15.03 + 50.0} = 45.03 + 30.0}. Now the displacement from your current position to C is rC rz— 10.03 10.0} 45.03 300}— 5503 40.0}. You cover onethird, moving to r3 = r2 + Ar23 = 4503 +300} +§(—550§ — 40.03) = 26.7i + 16.7}. The displacement from where you are to D is
rD — r3 = 40.03 — 30.0} — 26.7i — 16.7} = 13.33 — 46.73. You traverse onequarter of it, moving to r4 =r3 +£(rD —r3)= 26.7i+16.7i+ 1 Z(13.3}— 46.73) = 30.03 + 5.00}. The displacement from your new location to E is rE r4 = 70.03 : 60.0} 30.03 5.00} = 1003 : 55.0}
of which you cover onefifth the distance, —20.0i + 11.03, moving to r4 + Ar45 = 30.03 +5.00} — 20.03 + 11.0} = 10.03 + 16.0}. The treasure is at (10.0 m, 16.0 m) . (b) Following the directions brings you to the average position of the trees. The steps we took
numerically in part (a) bring you to 1 r +r
rA+_(rB_rA):[ A 3) 2 2
(r +r)
thent0(1‘A‘—1'B):1'C— A2 B :rA+rB+rC
2 3 3
(r +r+ )
thento (r “IE—Ht) I rD—AfMZTA‘l‘TB‘l—rcl—TD
3 ' 4 4
(r +r + +r)
and lastto (rA+rB+rC+rD) I IE_%ICD_IA+IB+TC‘FTD+TE
4 I 5 5 ' This center of mass of the tree dish‘ibution is the same location whatever order we take the
trees in. >*P3.63 P3.64 (a) (b) (a) (b) Chapter 3 75 Let T represent the force exerted by each child. The x 9’
component of the resultant force is TcosO + Tcos 120°+T cos 240° 2 T(1) + T(—0.5) + T(—0.5) = 0. The ycomponent is Tsm0+Tsin120+Tsin240 = 0+0.866T —0.866T = 0.
FIG. P3.63
Thus, 212:0. If the total force is not zero, it must point in some direction. When each child moves one
3600 N the same force, the new situation is identical to the old and the net force on the tire must still
point in the original direction. The contradiction indicates that we were wrong in supposing
that the total force is not zero. The total force must be zero. . Since each child exerts space clockwise, the total must turn clockwise by that angle, From the picture, R1 = ai+bi and Rd 2 \la2 +122 .
R2 = ai +bi+ clA<; its magnitude is 1lR12 +CZ =xla2 +122 +CZ. FIG. P3.64 76 Vectors P3 .65 Since A + B = 6.00},
we have
(Ax + Bx)i+ (Ay + By); = oi + 6.00}
giving
Ax +Bx :0 0r Ax =—Bx
and Ay + By 2 6.00.
Since both vectors have a magnitude of 5.00, we also have Ai+A§=B§+B§=5.002. From Ax = —Bx, it is seen that
A: 2 Bi
Therefore, A: +14: 2 3: +3: gives
A; = 35
Then, Ay = By and Eq. [2] gives
Ay = By 2 3.00. Defining 6as the angle between either A or B and the y axis, it is seen that A B
y =—y—&=0600 and 9:53.10. cos (9 = — —
A B 5.00 The angle between A and B is then ¢ = 249 = 1060 . FIG. P3.65 >*P3.66 P3.67 Chapter 3 77 Let Qrepresent the angle the xaxis makes with the horizontal. Since
angles are equal if their sides are perpendicular right side to right
side and left side to left side, Qis also the angle between the weight
and our y axis. The xcomponents of the forces must add to zero: —0.150 Nsin¢9 + 0.127 N = 0. (b) a: 57.90 FIG. P3.66 (a) The ycomponents for the forces must add to zero: +Ty —(0.150 N) cos57.9°= 0, Ty = 0.079 8 N . (c) The angle between the y axis and the horizontal is 90.00—579O = 32.10 . The displacement of point P is invariant under rotation of
2
the coordinates. Therefore, r = r/ and 1'2 = (1") or, )1 2 x2 + yz = (x’) + (y/)Z. Also, from the figure, 6 = (9 — 04 \ )X FIG. P3.67 Which we simplify by multiplying top and bottom by x cos 04 . Then, /  / .
x =xcosoz+ysmoz, y =—x81na+ycosa. ‘ NSWERS TO EVEN PROBLEMS P3.2 P3 .4 P3.6 P3.8 P3.10 P3.12 P3.14 (a) (2.17 m, 1.25 m); (—1.90 m, 3.29 m); P3.16 see the solution
(b) 4.55 m
P3.18 86.6 m and —50.0 m
(a) 8.60 m;
(134.47 In at _63_40; 4.24 m at 1350 P3.20 1.31 km north; 2.81 km east
(a) r at 180°—6; (b) 27 at 180°+6; (c) 37 at —6 P3.22 —25.0 m i+ 43.3 mi
14 km at 65° north of east P3.24 14.0 m/s at 113° west of north
(a) 6.1 at 112°; (b) 14.8 at 22° P3.26 788 mi at 480° north of east
9.5 N at 57° P3.28 7.21 m at 56.3°
79 m at 4° north of west P330 c = 7.30 cm i— 7.20 cm 3 78
P3.32 P3.34 P3.36 P3.38 P3 .40 P3 .42 P3 .44 P3.46 P3 .48 Vectors (a) 4.47 In at 63.40; (b) 8.49 In at 135°
42.7 yards 4.64 In at 78.60
(a) 10.4 cm; (b) 35.50
1.43 X 104 In at 32.20 above the horizontal —2201 + 57.6} = 227 paces at 165° (a) (3.121 ——5.023— 2.2012)1<m; (b) 6.31 km (a) (15.1i —— 7.723) cm; (b) (—7.72i —— 15.13) cm; (c) (+7.72i —— 15.13) cm (a) 74.60 north of east; (b) 470 km P3.50 P3.52 P3.54 P3.56 P3.58 P3.60 P3.62 P3.64 P3.66 a = 5.00, b = 7.00 2tan1[l)
n
25.4 s (a) 7.17 km; (b) 6.15 km 7.87 N at 97.80 counterclockwise from a
horizontal line to the right —2.00 In s C; its velocit vector
( J y (a) (10.0 In, 16.0 In); (b) see the solution (b) RZ =ai+bj+c12;R2)=\/a2 +112 +8 (a) 0.079 8N; (b) 57.90; (c) 32.10 (a) R1=a1+b§; ...
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 Fall '08
 DonaldShaw
 mechanics

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