PHYS
homework 3 solutions

# homework 3 solutions - CHAPTER 3 22 Angles are given in...

• 2

This preview shows pages 1–2. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

This is the end of the preview. Sign up to access the rest of the document.

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. ...
View Full Document

• Spring '08
• Richardr

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern