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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} fourwheel 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 =FriNA +Nsi=
_+. err; = :[ﬁLﬂg FA + FB = me
H.8ﬂmg=mﬁ a = Geog = 0.311(931 mlsi) = 7.343 mr'sz [it] Riserwhee! 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 — {Ilam3} : [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 SCIlb cabinet is mounted on casters That allow it to move freelyr
{,u = D} on the ﬂoor. If a 2511:: 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 impend ) ; 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 lﬂib 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; = lﬂdiOET + 5.?ﬂ4l (3} Saba001(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 fan21 _ 1
= 16 I=Em(o"+e2}
f, =i 4ﬂﬂlb .2 2
12 —32‘2 ms: [1:45.45 ft} +{3.e ft} ] = 53.5093 1be~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} = (53.5o931htt32)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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 Summer '06
 

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