Chapter 3 Kinematics in Tw Dimensins; Vectrs
Vectrs and Scalars Additin f Vectrs Graphical Methds (One and Tw- Dimensin) Multiplicatin f a Vectr b a Scalar Subtractin f Vectrs Graphical Methds Adding Vectrs b Cmpnents Prjectile Mtin Prjectile Mtin Is Parablic Relative Velcit
Adding Vectrs b Cmpnents An vectr can be expressed as the sum f tw ther vectrs, which are called its cmpnents. Usuall the ther vectrs are chsen s that the are perpendicular t each ther. V x V In the figure abve, the vectr pints at an angle f 3 with respect t the +x directin (in fact, it can pint at an directin θ). The cmpnent is called the prjectin f n the x axis with magnitude V x given b the line C, while the cmpnent is the prjectin f n the axis with magnitude V given b the line B. Using the parallelgram methd fr vectr additin: (3.4)
Adding Vectrs b Cmpnents Frm nw n, we will dente the magnitude f a vectr b the fllwing expressin: (3.5) (3.6) Where the smbl represents the mdule f the vectr magnitude., r, in ther wrds, its We can als write, and Nte that the eq. 3.6 fllws frm the Pthagrean therem. V x V V V x
Adding Vectrs b Cmpnents Example 3.4: Using the figure belw, shw that if and are tw vectrs in a plane defined b tw perpendicular axes x and, and is the resultant f the additin f these tw vectrs, then: (3.7) x Slutin develped n the blackbard Equatin 3.7 states that the resultant f the sum f tw vectrs in a tw dimensinal plane is equivalent t the sum f their cmpnents n each f the x and axes.
Adding Vectrs b Cmpnents Equatin 3.7 als applies when bth vectrs start awa frm the rigin f the crdinate sstem:
Adding Vectrs b Cmpnents If the cmpnents are perpendicular, the can be fund using trignmetric functins. In the figure, the vectr makes and angle θ with the x axis. Nte: We define (b cnventin) θ with respect t the +x directin: -θ θ sin θ = sine f theta = cs θ = csine f theta = (3.8) (3.9) (3.1) tan θ = tangent f theta =
Adding Vectrs b Cmpnents Yu can use ur calculatr t cmpute sin, cs and tan. If u want t find the angle, u can use the functins sin -1, cs -1 and tan -1 fr equatins 3.8, 3.9 and 3.1, respectivel. We write: If u calculatr des nt have these functins, it shuld have an ke called invert r inv r smething similar. Trignmetric functins can be added, multiplied, etc when needed. Yu can find several functins and identities in Appendix A f the text bk. (3.8) (3.9) (3.1) Three ver imprtant trignmetric identities are the fllwing:
Adding Vectrs b Cmpnents Using equatins 3.8 3.1, we can write the cmpnents f the vectr as: (3.11) (3.12) (3.8) (3.9) (3.1)
Adding Vectrs b Cmpnents Prblem 3.1 (textbk): Three vectrs are shwn in Fig. 3 32. Their magnitudes are given in arbitrar units. Determine the sum f the three vectrs. Give the resultant in terms f (a) cmpnents, (b) magnitude and angle with the x axis. Fig. 3-32
1. Three vectrs are shwn in Fig. 3 32. Their magnitudes are given in arbitrar units. Determine the sum f the three vectrs. Give the resultant in terms f (a) cmpnents, (b) magnitude and angle with the x axis. A B C x x x = = = = 44. cs 28. 38.85 A 44. sin 28. 2.66 = = = = 26.5 cs 56. 14.82 B 26.5 sin 56. 21.97 = = = = 31. cs 27. C 31. sin 27 31. r (a) r r ( A + B + C ) ( ) ( A r + B r + C r ) ( ) = 38.85 + 14.82 +. = 24.3 = 24. x = 2.66 + 21.97 + 31. = 11.63 = 11.6 (b) r r r A + B + C ( ) ( ) 2 2 = 24.3 + 11.63 = 26.7 11.63 θ = tan = 25.8 24.3 1
Adding Vectrs b Cmpnents Prblem 3.13 (textbk): Fr the vectrs given in Fig. 3 32, determine (a) (b) (c) r r r A B + C, r r r A + B C, r r r C A B. Fig. 3-32
r r r r r r 13.Fr the vectrs given in Fig. 3 32, determine (a) A B + C, (b) A + B C, and r r r (c) C A B. (a) A B C x x x r = = = = 44. cs 28. 38.85 A 44. sin 28. 2.66 = = = = 26.5 cs 56. 14.82 B 26.5 sin 56. 21.97 = = = = 31. cs 27. C 31. sin 27 31. r r ( A B + C) = ( ) + = x r r r ( A B C) ( ) 38.85 14.82. 53.67 + = 2.66 21.97 + 31. = 32.31 Nte that since the x cmpnent is psitive and the cmpnent is negative, the vectr is in the 4 th quadrant. r r r 2 2 1 32.31 A B + C = ( 53.67) + ( 32.31) = 62.6 θ = tan = 31. belw + x axis 53.67
r (b) r r + = 38.85 + 14.82. = 24.3 ( A B C) ( ) r x r r + = 2.66 + 21.97 31. = 73.63 ( A B C) ( ) r r r 73.63 A + B C = ( 24.3) + ( 73.63) = 77.5 θ = tan = 71.9 24.3 2 2 1
(c) r r r =. 38.85 14.82 = 24.3 ( C A B) ( ) r r r x ( C A B) = 31. 2.66 21.97 = 73.63 Nte that since bth cmpnents are negative, the vectr is in the 3 rd quadrant. r r r 2 2 1 73.63 C A B = ( 24.3) + ( 73.63) = 77.5 θ = tan = 71.9 belw x axis 24.3 Nte that the answer t (c) is the exact ppsite f the answer t (b).
Prjectile Mtin Prjectile mtin is a generalizatin f purel vertical mtin (falling bjects) we studied in chapter 2. Nw we will cnsider the mtin in tw dimensins (can be easil generalized t three dimensins). An example f prjectile mtin is that f a tennis ball, speeding bullets, etc.. In this chapter, we will nt cnsider hw an bject is set in mtin, but nl its mtin after it start mving and befre it lands r is caught.
Prjectile Mtin Prjectile mtins can be apprached using the methd f vectr cmpnents as described befre. We can analze the mtin b studing its vertical and hrizntal cmpnents separatel (Galile was the first t describe prjectile mtin using this methd). In this figure, a ball rlls ff the end f a table with an initial velcit V x at an instant f time t = in the hrizntal directin. Furthermre, its velcit pints in the directin f the mtin at an instant f time t, and is tangent t its path.
Prjectile Mtin We can then analze the mtin in the x and directin separatel using the equatins f mtin we btained in chapter 2: hrizntal mtin (2.1) (2.12) (2.13) vertical mtin (2.16) (2.17) (2.18)
Prjectile Mtin directin In the vertical directin, we have the acceleratin, g, f gravit. The initial velcit is zer (V = ). The crdinate is chsen t be psitive upward and zer at the rigin selected as where the pint the ball start experiencing vertical mtin as depicted in the figure. Then, at t = we have: = V = a = g (a pints dwnward) Using equatin 2.17 we have: (3.13) Using eq. 2.16: (3.14)
Prjectile Mtin x directin In the hrizntal directin, there is n acceleratin. The initial velcit is V x. The crdinate x is chsen t be eastward and zer at the rigin selected as the pint where the ball start experiencing vertical mtin as depicted in the figure. Then, at t = we have: x = initial velcit = V x a = Using equatin 2.12 we have: (3.15) Using eq. 2.1: (3.16) The velcit is cnstant in the hrizntal directin.
Prjectile Mtin Example 3.5: Shw that an bject prjected hrizntall will reach the grund in the same time as an bject drpped verticall frm the same height (use the figure belw). The nl mtin that matters here is the vertical ne since (displacement in directin). Equatin 3.13 gives the displacement fr bth balls since the are subject t similar initial cnditins, namel: = V = a = g (a pints dwnward) Therefre applies in bth cases at an instant f time. Thus, the tw ball reach the grund tgether.
Prjectile Mtin If an bject is launched at an initial angle f θ with the hrizntal, the analsis is similar except that the initial velcit has a vertical cmpnent. Yu can use equatins 3.11 and 3.12 and then develp the prblem in the ver same wa as n the previus slides.
Prjectile Mtin Prblem 3.21 (textbk) : A ball is thrwn hrizntall frm the rf f a building 45. m tall and lands 24. m frm the base. What was the ball s initial speed? Slutin develped n the blackbard
Prblem 3.21 (textbk) : A ball is thrwn hrizntall frm the rf f a building 45. m tall and lands 24. m frm the base. What was the ball s initial speed? Chse dwnward t be the psitive directin. The rigin will be at the pint where the ball is thrwn frm the rf f the building. In the vertical directin, v = 2 a = 9.8 m s. = and the displacement is 45. m. The time f flight is fund frm appling Eq. 2.17 t the vertical mtin. ( ) 2 2 2 2 45. m 1 1 = + v t + a t 45. m = 2 2 ( 9.8 m s ) t t = = 3.3 sec 2 9.8 m s The hrizntal speed (which is the initial speed) is fund frm the hrizntal mtin at cnstant velcit: x = v t v = x t = 24. m 3.3 s = 7.92 m s x x
Prjectile Mtin Prblem 3.31 (textbk) : A prjectile is sht frm the edge f a cliff 125 m abve grund level with an initial speed f 65. m/s at an angle f 37.º with the hrizntal, as shwn in Fig. 3 35. (a) Determine the time taken b the prjectile t hit pint P at grund level. (b) Determine the range X f the prjectile as measured frm the base f the cliff. At the instant just befre the prjectile hits pint P (c) Find the hrizntal and the vertical cmpnents f its velcit (d) Find the magnitude f the velcit (e) Find the angle made b the velcit vectr with the hrizntal. (f) Find the maximum height abve the cliff tp reached b the prjectile. Slutin develped n the blackbard
Chse the rigin t be at grund level, under the place where the prjectile is launched, and upwards t be the psitive directin. Fr the prjectile, (a) v = 65. m s θ = 37. a = g = 125 v = v sinθ The time taken t reach the grund is fund frm Eq. 2.17, with a final height f. = + v t + a t = 125 + v sin θ t gt 1 2 1 2 2 2 1 ( g ) 2 ( )( ) 2 2 1 v sinθ ± v sin θ 4 g 125 2 39.1± 63.1 t = = = 1.4 s, 2.45 s = 1.4 s 2 9.8 Chse the psitive sign since the prjectile was launched at time t =.
Chse the rigin t be at grund level, under the place where the prjectile is launched, and upwards t be the psitive directin. Fr the prjectile, (b) v = 65. m s θ = 37. a = g = 125 v = v sinθ The hrizntal range is fund frm the hrizntal mtin at cnstant velcit. ( θ ) ( ) ( ) x = v t = v cs t = 65. m s cs 37. 1.4 s = 541 m x
Chse the rigin t be at grund level, under the place where the prjectile is launched, and upwards t be the psitive directin. Fr the prjectile, (c) v = 65. m s θ = 37. a = g = 125 v = v sinθ At the instant just befre the particle reaches the grund, the hrizntal cmpnent f its velcit is the cnstant The vertical cmpnent is fund frm Eq. 2.16: x ( ) v = v θ = = cs 65. m s cs 37. 51.9 m s 2 ( ) ( )( ) v = v + at = v sinθ gt = 65. m s sin 37. 9.8 m s 1.4 s = 63.1m s
Chse the rigin t be at grund level, under the place where the prjectile is launched, and upwards t be the psitive directin. Fr the prjectile, (d) v = 65. m s θ = 37. a = g = 125 v = v sinθ The magnitude f the velcit is fund frm the x and cmpnents calculated in part c) abve. 2 2 x ( ) ( ) 2 2 v = v + v = 51.9 m s + 63.1m s = 81.7 m s
Chse the rigin t be at grund level, under the place where the prjectile is launched, and upwards t be the psitive directin. Fr the prjectile, (e) v = 65. m s θ = 37. a = g = 125 v = v sinθ The directin f the velcit is (see slide number 7 f this lecture) v θ = tan = tan = 5.6 v 51.9 and s the bject is mving. 1 1 63.1 x 5.6 belw the hrizn
Chse the rigin t be at grund level, under the place where the prjectile is launched, and upwards t be the psitive directin. Fr the prjectile, (f) v = 65. m s θ = 37. a = g = 125 v = v sinθ The maximum height abve the cliff tp reached b the prjectile will ccur when the - velcit is, and is fund frm Eq. 2.18. ( ) v = v + 2 a = v sin θ 2 g 2 2 2 2 m ax ( ) 2 2 2 2 v sin θ 65. m s sin 37. m ax 2 = = = 2 g 2 9.8 m s ( ) 78.1 m