evised 1/10/008 (110-TEST 6) More on Vecors I. esolving a vecor ino is horizonal and verical componens: In secion.5, ou learned how o find he magniudes of he horizonal and verical vecors, V and V, respecivel, of an given vecor. The vecors V and V, called, respecivel, he horizonal and verical componens of V, are deermined as follows: V = V cos, V, = V sin where is he sandard - posiion (no erence) angle of V. (Noe: The - and - componens in his handou are rounded for he sake of illusraion onl. hen ou do problems ourself, ou should be caul o keep all calculaor digis unil rounding he final answer.) Eample 1: Find he horizonal and verical componens of a vecor V wih he magniude 66 and sandard posiion angle 1. Soluion: V = 66 cos 1= -531, V = 66 sin 1= -33. Noe ha he negaive value for V and V lec he fac ha he - and - componens are direced in he negaive direcions of he - and - aes respecivel. Finding hese values is called "resolving V ino is - and - componens." Noe also ha no diagram need be drawn o solve such a problem. N 6 138.8 S.8 F B E 31 = 650 LB. Eample : A wind blows a 6.0 miles per hour a a heading of 31. Find is componens. Soluion: If we consider our compass direcions o be aligned along he - and - aes in he radiional manner (see lef), we firs noe ha a heading of 31 corresponds o a sandard-posiion angle of 138. Thus, V = 6.0 cos 138 = -46.1, V = 6.0 sin 138 = 41.5. Inerpreing hese resuls in erms of compass direcions, we find ha he wind has a componen of 46.1 miles per hour owards he wes and a componen of 41.5 miles per hour owards he norh. Eample 3: A 650 lb. ruck is parked on a ramp inclined.8wih he horizonal. Find he componen of is weigh direced down he ramp. Soluion: The phsical siuaion illusraed a lef where he 650 lb. weigh is direced down (oward he cener of he earh) and he force F is o be deermined. One wa o solve he problem is o se up our recangular coordinae ssem so ha F will be parallel o one of he aes as shown in he second diagram. Since B is complemenar o.8, B= 67., and he sandard-posiion angle of he weigh vecor in our coordinae ssem is = 180 + B = 47.. e have se up our coordinae ssem so ha F = = cos = 650 cos 47. = -4 lbs, where he negaive simpl indicaes he negaive - direcion in our coordinae ssem. Thus, he answer is 4 lbs. II. Adding vecors: In secion.5, ou found he magniude and he erence angle,, of he direcion of a vecor from is - and - componens as follows: V V V an V V e can find he sandard - posiion angle of he vecor if we use he above in connecion wih he signs of V and V. Eample 4: If a plane heads due wes a 10 miles per hour and if a 50 mile per hour wind blows from due souh, find he acual veloci of he plane wih respec o he ground. The veloci V p impared o he plane b he plane and he veloci V w impared o he plane b he wind are he - and - componens of is resuling veloci V wih respec o he ground. If we align he recangular coordinae ssem and he compass direcions in he usual wa (as in Eample ) we have V = -10 (since due wes is he negaive - direcion) and V = 50 (since he wind is owards due norh). 1
Thus, V V V 10 50 16 ; and, an V V 50 10 giving = 13.4. Since V < 0 and V Y > 0, is a second quadran angle. Theore, he veloci of he plane is 16 miles per hour, 13.4 N of and, hence, a a heading of 83.4. Finall, we can combine he problems of resolving vecors ino componens and recombining componens, as in he previous eample, o add an wo or more vecors. The sum is called he resulan. Eample 5: Find he resulan of he hree vecors A, B, C shown below wih A = 444, B = 388, and C =. Y Soluion: Y Y C C Y 70.0 58.0 B 0 A B 0 C Firs we resolve each of he hree vecors ino heir - and - componens, being caul o use he appropriae sandard -posiion angles. Then, since he - and - componens are in he same or opposie direcions, we can add hem direcl o find and. Vecor -componen -componen A 444 cos 0 = +444 444 sin 0 = 0 B 388 cos 38 = -06 388 sin 38 = -39 C cos ll0 = -76 sin 110 = +09 +16-10 Finall, we obain 10 16 10 0; and an, 16 36.5 Q IV, 33.5 (he sandard-posiion angle) Eample 6: A plane ravels 550 miles per hour a a heading of 345 and hen 84 miles a a heading of 0. Find is displacemen from is saring poin. Soluion: Once we realize ha we are finding he resulan of wo displacemen vecors, here is no need o draw a diagram, ecep possibl o conver he headings of 345 and 0 o heir corresponding sandard-posiion angles of 105 and 70 respecivel. Vecor -componen -componen A 550 cos 105 = - 14 550 sin 105 = +531 B 84 cos 70 = + 88 84 sin 70 = +791 +146 +13 Finall, we obain 13 146 13 1330; and an, 83.7 83.7 146 Theore, he plane is 1330 miles from is saring poin a a heading of 6.3 (corresponding o he sandard-posiion angle of 83.7 ). B Y A Y 0
Eample 7: Find he resulans of he forces F and G given below in erms of magniudes and sandardposiions angles: F = 543 lbs, F =173 G = 445 lbs, G = 07 Soluion: = F + G = 543 cos 173 + 445 cos 07 = -935 Y = F Y + G Y = 543 sin 173 + 445 sin 07 = -136 136 = = = 935 + 136 = 945 lbs; and an q = =, q = 8.3, q = 188.3 935 Eample 8: If he wo forces in Eample 7 ac on an objec, wha hird force is needed o keep i from moving? Soluion: hen forces ac on an objec and i remains moionless, we sa ha he forces are in equilibrium; ha is, heir sum is he vecor wih magniude zero. Since we alread have added he forces F and G in Eample 7 o find heir resulan, he hird force mus be equal in magniude and opposie in direcion for he sum of all hree o be zero. Theore, he answer is 945 lb. wih a direcion of 8.3. If we had no found he resulan of F and G alread, we would sar b finding he sum of heir - and - componens o be -935 and -136 respecivel (as we did in Eample 7.) Then he - and - componens of he hird force would be heir opposies +935 and +136, respecivel, for heir oals o be 0. Finall, we would find he hird force from hese componens as usual. Ofen a problem involving vecors ma be solved hrough more han one approach. The following is such a problem: Eample 9: A sign hangs from wo ropes as shown below. If he ension in he lef rope is 150 lbs, find he ension in he righ rope and he weigh of he sign. Soluion 1: 150 lbs 50 150 lbs Sign 50 70 40 T 40 130 40 T Soluion 1: Since he sign is in equilibrium, he sum of he horizonal componens mus be zero; ha is: 150 cos 130 + T cos 40 + cos 70 = 0 T cos 40 = -150 cos 130 150 cos 130 T 16 lbs cos 40 Also, he sum of he verical componens mus be zero: 150 sin 130 + 16 sin 40 + sin 70 = 0 150sin130 16sin sin 70 40 196 lbs. T 40 50 150 lbs Soluion : 50 50 40 Soluion : e add vecor T o he 150 lb. ension vecor geomericall as in he hird diagram, forming a riangle. Since he sign is in equilibrium, he resulan mus be equal o in magniude, ha is, =. Since he inerior angle formed a he origin mus be complimenar o 50, i equals 40 ; and hus, he remaining inerior angle, formed b T and mus be 50. (h?) e now have enough informaion o use he Law of Sines o find T and = : T 150, T 16 lbs. and 196 sin 40 sin 90 sin 50 lbs. 3
Eercises: 1. Find he resulan of he following vecors: 30 45.4 53.0 155 a) A = 56.0, A = 76.0 b) A = 1.9, A = 36. c) B = 4.0, B = 00.0 B = 96.7, B = 11.5 C = 6.9, c = 143.4. A 100 lb block ress on an incline plane which makes an angle of 0 wih he horizonal. Find he force perpendicular o he plane. 3. A ship sails 15.0 km on a heading of 00, hen 30.0 km on a heading of 350. How far and a wha heading mus i ravel o reurn o por? 4. A 10 lb gmnas hangs b her hands from a high horizonal bar. Her hands are separaed so ha her hands form 65 angles wih he bar. Find he ension in her arms. (Noe: Each arm suppors half her weigh.) 5. Find he ension in he ropes supporing he 0 lb sign below. 1 30.8 8 f 5 f 5 f 0 lb 6. A boa which ravels 9.00 miles per hour in sill waer aims 50.0 upsream agains a curren running 4.00 miles per hour. Find is resulan veloci. Answers: 1. a) = 47.0, = 101.0 4. 66. lbs b) = 50., = 50.3 c) = 35, = 148.3 5. 183 lbs. 94.0 lbs 6. 6.47 miles per hour, 6.6 upsream 3. 18.6 km a 146. 4
MATH 110 TEST #6 SOLUTIONS TO VECTO HANDOUT 1A A 56 a 76 A 13.55 A Y 54.34 B 4 a 00 B -.55 B Y -8.1-9 Y 46.13 =, 9 46.13 47. 0 q \ æ 1 46.13ö - = an ç» 79.0, q in Quad II çè 9 ø q = 180-79 = 101 1B Componen Y A 1.9 a 36. -1. -18. B 96.7 a 11.5 94.8 19.3 C 6.9 a 143.4-50.5 37.5 3.1 38.6 = 3.1 38.6 50. 38.6 an 1 50.3, in Quad I 3.1 50.3 1C Componen A 155 a 37 13.8 93.3 B 30 a 135.4-15.0 1.1 C 1 a 39. Y -108.6-18.1-199.8 13.3 (nd Quad) -199.8 13.3 34. 8 13.3 an 1 31.7, in Quad II 199.8 180-31.7 148.3 5
() 0 50 100 lb F Y = 100 lb sin 50 = - 94 lb F Y = 94 lb (3) B Componens: A B 15 km heading 00 A A 15 cos 50 15 sin 50-5.13-14.10 Using sandard posiion angle of 50 30 km heading 350 30 cos 100 30 sin 100-5.1 9.54 Using sand posiion angle of 100 B B 10.34 15. 44 = 18.6 = resulan (adding) -10.34 15.44 ( nd Quad) = an -1 15.44 56. 10.34 56. 56. 146. Since is in QII, he angle o reurn o por would be direcl opposie, wih a erence angle of 56. in QIV, ha is, a heading of 90 + 56. = 146.. 6
(4) 65 sin65 = 60 65 60 60 =» sin65 66.» 66. 10 (5) 8 5 3 4 5 an = ¾, 36.9 following eercise 4 36.9 36.9 110 0 lbs sin 36.9 = 110 110 = 183 sin 36.9 7
(6) B Componens: 9 B B = 9cos 50 = 5.785 B = 9 sin 50 = 6.894 C = 4 cos 70 = 0 C = 4 sin 70 = - 4.000 = 5.785 =.894 4 50 C = + = 6.47 mph IV.894 an q = = =.5003 5.785 \ q q = 6.6, q in Quad I = 6.6 8