Slutins--Ch. 1 (Math Review) CHAPTER 1 -- MATH REVIEW 1.1) Bth Parts a and b are straight vectr additin, graphical stle. Yu need a centimeter stick and prtractr t d them. a.) In this case, the velcit cmpnent cntributed b the rwer is a vectr directed straight acrss the river. It has a magnitude f 5 miles per hur. The velcit cmpnent cntributed b the river's mtin is directed dwnward. It has a magnitude f 2 miles per hur. Vectrial additin f the tw vectrs ields the net velcit f the bat during the crssing (see Figure 1). A prtractr 3 m velcit vectr f rwer (5 mi/hr) resultant velcit vectr 1 m FIGURE 1 velcit vectr f rwer (5 mi/hr) directin everne wants t g FIGURE 2 scale: appr. 1 cm/(mile/hur) velcit vectr f river (2 mi/hr) shuld be used t measure the angle. It turns ut t be -22 (minus because it is belw the reference line straight acrss the river). A centimeter stick measures the resultant at 5.4 centimeters. Multipling b the scaling factr, we get a net velcit vectr f 5.4 miles per hur. Nte: The Xer machine ever-s-slightl miniaturizes whatever it cpies. When I used the riginal t make m measurements, I gt the answer quted abve. If u use a centimeter stick and prtractr n the Xered cp, u will get an answer that is a bit ff. If u d the prblem n ur wn, u shuld get the same slutin as I did t a gd apprimatin. b.) The directin f an bject's net velcit is the same as the directin f the bject's net mtin. We want the net mtin t be alng the directin shwn in Figure I n page 22. The questin that is reall being asked is: "What bat-velcit vectr bat-angle relative t line directl acrss the river velcit vectr f river (2 mi/hr) actual (desired) velcit scale: appr. 1 cm/(mile/hur) 455
must eist if, when added t a water-velcit vectr, ields the net-desiredbat-mtin?" Clearl it must be upstream. Plaing with a cmpass and prtractr, the slutin is apprimatel 4 (see Figure 2 n the previus page) and the velcit is apprimatel 4 miles/hur. 1.2) This is als a prblem fr graphical vectr additin. a.) The vectrs invlved are shwn t the right. A prtractr was used t determine an angle f apprimatel 17 nrth f west. b.) The diagram is t scale. The resultant, measured b a centimeter stick, is apprimatel 5.6 centimeters. Multipling b the scaling factr ields (5.6)(2) = 112 miles. This is the net displacement. net directin net distance traveled scale: appr. 1 cm/(2 mile) 6 miles after westerl turn? FIGURE 3 3 8 miles nrth Nte: Once again, if u use a centimeter stick and the sketch in Figure 3 t fllw m steps in ding this prblem, u will prbabl find ur result and m result differing just a bit. I did the prblem ff the riginal 1 drawing. Xering slightl shrinks the material being cpied. 1.3) The graph f vectrs P and T are shwn in Figure 4. 1.4) Using the graph given in prblem- Figure II: in unit vectr ntatin: A = 2.25i +.5j. C = -.72i - j. in plar ntatin: B = 1.3 125. D = 2.6-2. -1 P = 7-1 FIGURE 4 1 =-6 T 456
Slutins--Ch. 1 (Math Review) Nte: I've made the prblem trick as far as the tw plar angles g. All angles must be measured either clckwise r cunterclckwise FROM THE + AXIS! 1.5) Each ais has its wn scale. If the scales had been the same, relative lengths and angles wuld have made sense. The aren't, hence the dn't. 1.6) This is all straight vectr manipulatin. Yu shuld be able t d this kind f thing in ur sleep! a.) -(1/3)A = (-1/3)((-8i + 12j) = (8/3)i - 4j. b.) -6E = (-6)((12 225 ) = (72 45 ). Nte: in plar ntatin, a scalar multiplies the magnitude f the vectr while a negative sign adds (r subtracts) 18 t the angle f the vectr. c.) A + B - C = (-8i +12j) + (i - 3j) - (5i + 6j -7k) = -17i + 3j + 7k. d.) A bit f trig and Figure 5 shw E's unit vectr equivalent. Written in unit vectr ntatin: E = -8.48i -8.48j. E = 12 cs 45 = 8.48 E = 12 sin 45 = 8.48 E =225 =45 Nte: The negative signs had t be put in manuall. E = 12 FIGURE 5 e.) A bit f trig and Figure 6 shws F's unit vectr equivalent t be: F = -.518i + 1.93j. Nte: Again, the negative signs had t be placed manuall. F = 2 sin 15 =.518 F = 2 F = 2 cs 15 = 1.93 =15, =15 FIGURE 6 457
f.) The Pthagrean relatinship, a bit f trig, and Figure 7 shws A's plar equivalent t be: A = 14.4 123.7. 2 2 1/2 A = [(-8) + (12) ] = 14.4 A 1-1 = tan (8/12) = 33.7 Nte: Tan -1 (8/12) = 33.7. Adding -1 that t 9 gives us the calculated 123.7. FIGURE 7 An alternate wa t d the prblem: take the actual cmpnents, minus signs and all, and determine tan -1 (A /A ). That is, tan -1 (12/-8) = -56.3. This is a furth quadrant angle, whereas ur vectr is a secnd quadrant angle. T adjust the situatin, we must add 18 t get the apprpriate 123.7 angle. Either wa will d. g.) The Pthagrean relatinship, a bit f trig, and Figure 8 shws B's plar equivalent t be: B = 5 216.9. Nte: Tan -1 (3/4) = 36.9. Added t 18 gives us the apprpriate 216.9. An alternate wa t d the prblem: take the actual cmpnents, minus signs and all, and determine the tan -1 (A /A ), r -1 = tan (3/4) = 36.9 B 2 2 1/2 B = [() + (-3) ] = 5 FIGURE 8 tan -1 (-3/) = 36.9. This is a first quadrant angle, whereas ur vectr is a third quadrant angle. T adjust the situatin, we must add 18 t get the apprpriate 216.9 angle. Either wa will d. h.) A. C = A C + A C + A z C z = (-8)(5) + (12)(6) + ()(-7) = 32. i.) D. E = D E cs φ = (7) (12) cs 75 = 21.74. 458
Slutins--Ch. 1 (Math Review) j.) i j k A B = -8 12-3 = i + j + [(-8)(-3) - (12)()]k = 72k. k.) i j k C B = 5 6-7 -3 = [(6)() - (-7)(-3)]i + [(-7)() - (5)()]j + [(5)(-3) - (6)()]k = -21i + 28j + 9k. l.) DE = D E sin φ (-k) = (7) (12) sin 75 = -81.1k. Nte: The -k directin came frm using the right-hand rule n DE. 1.7.) A dt prduct gives u a scalar equal t the magnitude f A times the magnitude f the cmpnent f B parallel t A. OR: It gives u a scalar equal t the magnitude f B times the magnitude f the cmpnent f A parallel t B. 1.8.) A crss prduct gives u a vectr whse magnitude is equal t the magnitude f A times the magnitude f the cmpnent f B perpendicular t A. OR: It gives u a vectr whse magnitude is equal t the magnitude f B times the magnitude f the cmpnent f A perpendicular t B. The directin f the crss prduct is perpendicular t the plane defined b A and B. 459
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