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**Unformatted text preview: **CHAPTER 3 22. Angles are given in ‘standard’ fashion, so Eq. 3-5 applies directly. We use this to
write the vectors in unit-vector notation before adding them. However, a very different-
looking approach using the special capabilities of most graphical calculators can be
imagined. Wherever the length unit is not displayed in the solution below, the unit meter should be understood. (a) Allowing for the different angle units used in the problem statement, we arrive at E=3.73i+4.703
é:1.451+3.733
” —5.2oi+3.003
E+F+G+H=128i +6.60]. :2
n (b) The magnitude of the vector sum found in part (a) is «((1.28 m)2 + (6.60 m)2 = 6.72 m. (c) Its angle measured counterclockwise from the +x axis is tan—1(6.60/1.28) = 79.0°. ((1) Using the conversion factor 7: rad = 180° , 790" = 1.38 rad. 1e 1’s trip (0.50 m east or 0.5 i) and 28. Let [T represent the first part of Beet
ded voyage (1.6 In at 50° north of east). For represent the ﬁrst part of Beetle 2’s trip inten their respective second parts: E is 0.80 m at 30° north of east and D is the unknown.
The ﬁnal position of Beetle 1,is 2+1} = (0.5 m)i+(0.8 m)(cos30° i+sin30° 3) = (1.19 m) i+(o.40 111)}. The equation relating these is A + E = (j + D , where C‘=(1.60 m)(cos50.0°i+sin50.0°:i)=(1.03 m)i+(1.23 m)3
(a) We ﬁnd D = ENE—6 =(0.16 m)i+(—0.83 m)3, and the magnitude is D = 0.84 m. (b) The angle is tan—1(—0.83/ 0.16) =—79°, which is interpreted to mean 79° south of east (or 11" east of south). 33. Examining the ﬁgure, we see that 6—; + g + = 0, where Z i B . (a) | 3 x 3 | = (3.0)(40) = 12 since the angle between them is 90°. (b) Using the Right-Hand Rule, the vector 31 x 5 points in the ix} = k , or the +2 direction. —> —> —>
b (c)| 3x 3|=1 3x(—E; — b)|=|——(a>< )1: 12. (d) The vector —& x 13 points in the —i x} z —k , or the - z direction. (e)] b’x c’1=1b’x(—a’ — b’)|= 1—(3x3)1=1(3x 13’)|=12. (f) The vector points in the +2 direction, as in part (a).‘ 37. We apply Eq. 3-30 and Eq.3-23. If a vector-capable calculator is used, this makes a
good exercise for getting familiar with those features. Here we brieﬂy sketch the method. (a) We note that 5x5 = —8.0i+5.03+6.0k. Thus,
a-(E x a) =(3.0)'(—8.0)+ (3.0)(5.0)+(—2.0) (6.0): —21.
(b) We note that 5 + '5 =1.0i — 2.0}+ 3.012. Thus, 5-(Z;+E) = (3.0) (l.0)+(3.0) (—2.0)+(—2.0) (3.0) = —9.0.
(0) Finally, 5x03 + a) = [(3.0)(3.0)—(—2.0)(—2.0)] i+[(—2.0)(1.0)—(3.0)(3.0)]j
+[(3.0)(—2.0)—(3.0)(1.0)] 12
= 5i — 113' — 912 '
38. Using the fact that
iXJzk,‘ j‘xkzi, kxi=j
we obtain ‘ 2A >< 3:2 (2.00i+3.003—4.00k)x(—-3.00i+4.003+2.00k)=44.0i+16.03+34.0k. Next, making use of we have 36-(2AXB) =3 (7.00i—8.003)~(44.0i+16.0] +34.012)
=3[(7.00)(44.0)+(—8.00)(16.0)+(0)(34.0)]= 540. ...

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- Spring '08
- Richardr
- mechanics, Work