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hw14 - PROBLEM 16.5 meing that the eoeFfieient 01“...

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Unformatted text preview: PROBLEM 16.5 meing that the eoeFfieient 01“ static friction between the tires and the road is (1.30 for the eutemehile shown, determine the maximtun possible acceleration on a level reed, assuming {a} four-wheel drive, (31] rear- wheel drive, {-3} fi'ent‘wheel drive. SULUTIDH {a} F our—whee! drive: +TEF}. =e: N_4+NH— = FJ+FB =FiNJ+FkNB =Fr-iNA +Nsi= _+. err; = :[fiLflg FA + FB = me H.8flmg=mfi a = Geog = 0.311(931 mlsi) = 7.343 mr'sz [it] Riser-whee! drive: +3 2MB = E[M3}cfi: {l m}W 43.5 unlit:1 = —[0.5 m)me HI = {MW + Heme E1 = ##Nfi = e.ee{e.4W + Emma} = H.32mg + fl.lfim§ E(F‘ FA =ma IL“: {1.32mg + 0.16mi = mg? [L32g = 0.343 32 D. (9.31 nvsi] = 3.1371 mars? PROBLEM 15.5 CONTINUED ‘ [c] From—whee! drr'w: l T | EMA = EMA)”; {2.5 n1].i"-"E — {1.5 m}W = —{r.:.5 mm | NL: = MW — 02an Thus: F'H = #thr = D.Sfl[fl.6W — {Ila-m3} : [MEmg — lllfimfi -+h Ea. = 2(3) FB = ma :r't' ' 13.4ng — 13.164151 = m5 0.43;: = 1.165 a = E(9.31 mfsz} = 4.0593 [[1st 1.15 - u: E = 4.0fi mfg: -—~ *1 PROBLEM 1E? A SCI-lb cabinet is mounted on casters That allow it to move freelyr {,u = D} on the floor. If a 25-11:: feree is applied as ehown. determine {a} the acceleration of the cabinet, {Er} the range of values of h for which the cabinet will net tip. SDLUTIDN I. {a} Acceleration [a] For tipping to imp-end ) ; A = :2: Hf EMF = E[M3}m: {251th -{501b}{12 in.) = mE{36 in.) 25h = em] - [25}{36} I: = en in. Ferfipping to impend ‘j; 3 = [I +( 2M3f = Ewajcfi: [25]b]fi+[fifllb]{12in.] ”15(35} or h=12in. cabinet will not tip for 12 in. 1: h 1: 6D in. 1 __._. ___———-_-.-—- PROBLEM 16.23 In order tr: determine the mass moment rat" inertia efa flyWheel of radius 1.5 ft, 0 lfl-ib block is attached to a wire that is wrapped areund the flywheel. The black is released and is observed t0 fall 12 ft in 4.5 s. To eliminate bearing Friction term the mmputatiens, a second block 01' weight 4t] [13 is used and is observed to fall 12 ft in 2.8 0. Assuming that the moment {if the couple due tr: fiietien remains constant, determine the mass moment {if inertia ef the flywheel. SDLU'I'IDN Kinematics Kinetics Halli—:Mfig " +3: 2M3 = E[Mfl'}efi': [m_4g]r — M! = Ft)! + [mflir —a mAET—Mj = I:J+ mAar 1 . 1 Casel: y=120. 1:055; }’=Ealt‘:>l2fl=%a{4.55}' 0...: 1.1002 0153 Fremtl] {2010105 0] — M_,- = F[ 1.1032 0101] +[ 2010 . 1.1032101 1.50 1.50 32.2 fits—M S )[ i 30— 30,. = 0.200131T +1.1042 (3:15:21 y=12tL 1:205; y=%cg3:>12n=%a_{2.ssf a._= 3.0012 1052 1400105 01—0. = F[M][fl 1](3.0012 0.153)“; n) 1.5 0 32.2 015* 60 —- M; = lfldi-OET + 5.?fl4l (3} Saba-001(2) from [3] 30 = 1.25002:— + 4.5000 .7 = 20300111101:2 or .7 = 20.310113.2 4 PROBLEM 1E.53 A S—m beam of mass 2110 kg is lowered by means of two cables unwinding from overhead cranes. he the beam approaches the ground, the crane operators apol},r brakes to slow the unwinding motion. Knowing that the deceleration of cable A is 5 mii's2 and the deceleration of cable 3 is iii inn-.2. detelmine the ieneien in each cable. SOLUTION Kinelica: e Me'- of I ' G‘- 8 Pi _—,>| #5311: [+— K Emmerics: ad=ofi +41: snee31= os nnie2 T +{4m}rx e = 1.125 naniieZ j a = a3 +1.5e = 0.5 rnie2 + [1.5 m}[i.125 radfsz} a = 2,1315 inie2 2 l 1 "' I=— L =— Zflflkg 5m =41fi.fi?k -m' 12 IE( H } g ,_ \‘i'. l +(' EMgzfingjufi: rA[4m]—{2ookg}[gj me[i.5m]+ia i144 in} — [zoo kg][9.8i n‘ii's2](1.5 m] = [zoo kg}(2.ia'i5 miei]{i.5 m} + (4115.6? kg~m2](l. 125 radio!) L=iGlTN or TA=IDITN4 PROBLEM 15.53 CONTINUED ] f :15}. = z[:«;.]cfi: it. + 1'}, —{2m kg]{g] = ma 101m + T3 = [2m kg}(9.31rnf22] + [2cm kg}(2.] 37's m6) | I}, =1332.5N or T3 =1383N «I I PROBLEM 15.53 CONTINUED ] f :15}. = z[:«;.]cfi: it. + 1'}, —{2m kg]{g] = ma 101m + T3 = [2m kg}(9.31rnf22] + [2cm kg}(2.] 37's m6) | I}, =1332.5N or T3 =1383N «I I L . PROBLEM 16.54 The 490—“: crate shown is lowered by means of two overhead eranee‘. Knowing that at the instant shown the deceleration of cable A is 3 fits‘ and that of cable B is l W52... determine the tension in each cable. SOLUTION Kit/reflex: W=mg fan-21 _ 1 = 16 I=Em(o"+e2} f, =i 4flfllb .2 2 12 —32‘2 ms: [1:45.45 ft} +{3.e ft} ] = 53.5093 1b-e~s°- +1.23? zzmjm; TA +TB—4flflzflfi t1} 3 t;f + T5 = 4oo[1+ 5] 3 +1: EMU =E{Mo} [TA —Tfi}[1.3}=fa eIT: {rd — rfi}[1.ee} = (53.5o931h-tt-32)a t2} Ktnematt‘ott: a=5+ra {0'th = E +1341; 3 = E +1.30: {3} [nth = a —1.Ea'; 1: a — Lost or = 1155556 mots.2 PROBLEM 15.54 CONTINUED TA + T3 = 40:3[1 + i] = 424.34 lb TA — T5 = (32.5051b43](0.55555 radial) 1",, = 2214492 n;- TA = 22111: 1 T3 = 203.390? 1b T3 = 24331]: i ...
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