hw23 - The fcrce cf friction between the blcck of ice and...

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Unformatted text preview: The fcrce cf friction between the blcck of ice and the surface. r**Hu—~—fieem_mtnim#m_ l b. The kinetic coefficient Df eeefi fricticn Mk between the black flmh cf ice and the surface. saluticn The free—bcdy diagram includee the weight of the block W = 15(9.31} = 14?.15 M, thg are ‘ +e )33 = max: 'F ' 153 +T HF = me : .N — 14? 15 n D J’ .Y where ey = U since there is nc mcticn in the vertical directicn. Therefore Rewriting the x—ccmpcnent cf ecceleraticn using the chain-rule cf differentiaticn Where th. can'tlnt 5f intlgrltifln hll been chclln In that v I 15 min when “'0- ThenIr ifv-thlnx-Zflm, F i 4 H tntitlllili'l‘"""'""“"*""‘""'* an.‘ n IIIlI"" 'nd the kinetic centriciint cf frictien Ll III-Illicittiitu.littu.....¢‘ Mil. l1 utiiiii""' - ' k ll. frictien between the inclined The free-body diagram includee the weight ef the hicck H; the ncrmal fcrce H'and fricticn fcrce R exerted en the black by the enrifacelr the applied fcrce F; In terms cf cccrdinatee alcng and normal tn the surface, the equaticne cf meticn are and fi- 25 +ZZFx=max: 20-2551n20 —R=fia +5. 2? = me : H - 25 can 20 = 0 "ii... 1" y 1where a = 0 since there in ne metien in the directien marina]. ta #133, surface. Therefcre flag; .H = 23.4923 lb a. If R - G, then the x-ccmpcnent cf the equaticne cf mtien a = 14.?47 ft/e2 cenetant and integrating te get the velecity and pceitien given V'= 14.747: + Ci = 14_747t _ fine X = 7.3?35t2 + C: = 7_3735t2 ity gf the bleek after it led 10 m up the ineline. salutian The free—bad}! diagram includes the weight Bf “b 1. force Hand f ' * lock w, the norms. rlctlfln nab F=ZOOH farce R exerted on the hlfick by the surface, and + ee F. In terms ef e * the appllEd fer Dflrdlnatee an. I} 200 Gas 30 - 20(9.31} sin 30¢ — R = 20 a +3 EFX = max. D n +5 EFF = may: N - 20:19.81} nee 30 - EDD sin 30 = D where a = 0 since there is nn metien in the directien nermal tn the 1’ surface. Therefere N = 269.914 N If R = 0.10:? = 26.9914 N, then 2 a = 2.4057 m/e = eenetant mm" and integrating tn get the velocity and pflsitiflfl Eli-V55 v = 2.40531; + {:1 = 2.40571: X==1.2028t2 + c = 1.2023t from reet at x: 0. Then, when x ._ 15 m 15 = l'ZDZBtE ME. t = 3.53 a ..... . . . . . n”... .. .H... H . . . . . H...” and when x: 10 m 1” = 1.2023t3 t = 2-3334 5 1mg. = 6.94 m/a V = 2.4057(2.8334} W.F. 3115? E L.D. Eturgea required tfl reach an f lfl,flDD m. a. At lift—eff, m = 2.?5ilflfi} kg. _ a T- 33(10 } N, g = 9.81 m/e , and the initial. vertical acceleraticn cf the IDCkEt is fine. I I i - I I i i I i i ‘ ' i " i I I i I I I I I I I I i I l I I fl I I I I I I a = 2.19 m/EE e. Integrating the veleeity te get the peeition ef the til;- !""i.: 2 y = 1.095t m where the eenetant ef integration ie again zere einee y = D whjfi Then the racket will reach an altitude ef 10,000 111 when 10,000 a 1.095:2 "lli-‘II-Ili-II--.'-...----..l-*Ifi =41. J“ ' I lung time inL‘r::er~r.~-e1.Ir the rocket will " I ill. I . h " “'1 r :3,- r . a.-' ' . . ., I. . _ - ‘- al- I 7.1— I p . r h-" 3' r a "4 ."" r - -- :1... 43 l.- r E';.: _ -.'. ,- I'r!‘ ."fi - I cf the tee diagrams may be drawn fcr each cf the turn blccks. Chccsing to draw separate free-body diagrams results in the equaticns cf meticn a 30 +3 EFAX = mask: 30 am 35 + T - FA - 32.2 a fi = : N - 30 CD5 35 = 0 +2 EPA? maafiy A '52? - SOs' 35¢ T-F-ie + Bx - mfiaflx' 1“ B 32.2 E! + E = I - = Z FRY classy NE 50 ccs 35 D where the y—compcnents cf acceleration are both serc since mticn in the directicn ncrmal tn the surface and the r - ._-. acceleraticn are the same since the blacks were together. NA = 24.575 lb _F = H; - 49.953 1b finally, '- I I i I. .I .- l l I . .1“: Measuring the poeitien ef heth bleeke item the ceiling. the length at the rape em: he expressed I. -.._-1_ ‘ + a " .r _ _. A 5 ..._.: “teatime ef thin, 1- ":l' _I._-... .-.._' -. 14$. -:- {1' -'- I.- __ = -2|:"a J { 33} A d combining Eqs a and b gruee an +——2—Cia -21H?!+——-—.._.“?"r:l'{:II 22* = 2'3“ 32.2 A 32_2 (-2331 394 ftfez T 3A 5 1* ' ' ' ' i ' ' ' ' i e i e i I I I i i a i g n i 1 g i a y I i n u 1- II- . I I i - I I a e t I I n I t u i u i - I t I n i u u r u a g u i . a I n i p- H115: b If bleek E is replaced with a eenetant force of 120 1b, then the tens-m“ in the cable is just 1" = 120 lb and Eq. 3 gives the acceleration uf block A 200 - -& a 2(12‘31 ‘ ‘ 32.2 A 2 an = T i i I I I I I 1 l I I I e I I I i 1 I I I a 1 a e u n - a I i e i II fir. ...
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hw23 - The fcrce cf friction between the blcck of ice and...

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