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Unformatted text preview: rr——1 12191. The man pulls the boy up to the tree limb C by
walking backward. If he starts from rest when xA = 0 and
moves backward with a constant acceleration aA =
0.2 m/sz, determine the speed of the boy at the instant
y” = 4 m. Neglect the size of the limb. When .13; = 0, ya: 8 m, so that A‘ and B are coincident, i.e., the rope is 16 m
long. Position  Coordinate Equation : Using the Pythagorean theorem Io‘dcunn‘ne
I“. we have!“ = «1} +81. Thus. I=Illc+y§
16: xi+81+y5 y. =15—‘fxii64 m Time Derivative . Taking the ﬁmcdaivaﬁvc of 5441;} where uA = %
I
andvs =%,wchavc ‘
_dy, _ 1.4 05"
v, — — ——  ——
d3 xﬁ +64 ‘1'
X
”a =—_._"‘__v [2] ﬁ/xi +64 A Anne inswmy, = 4 m_. from Eq.[l], 4 £16—‘/x}+64, xA = 8.944 m. The velocity
of the man anhat instantw: beobmined. u: = (adj ”(an [a 5mg]
1:} = 0+ 2(O.2)(8.944~0)
DA = 1.891m/s Substitute \hc above results into Eq.[2~] yiclds 8.944
u, =—~——————(L891)=—L4lmis= 1.41 m/s T Ans V 33442 + 64 Note : The negative sign indicates that. velocity unis in thc opposite direction to dual of
positive y, . \ 12205. At the instant shown car A is traveling with a
velocity of 30 m/s and has an acceleration of 2 m/s2 along
the highway. At the same instant B is traveling on the
trumpet interchange curve withva speed of 15 m/s, which
is decreasing at 0.8 m/sz. Determine the relative velocity
and relative acceleration of B with respect to A at this
instant. V8 = YA + "BM 15 cosﬁO°i+ l5 sin60°j = 30i+ (VB/A )x i+ (vs/A )yj 15 c0560° = 30 + (VB/A ), A
15 5111600 = 0+ (VB/A ),
(”BM )2: = 22.5 = 22.5 m/s (~— 5' (VB/A )y = 12.99 m/s T vB/A = s/ (22.5)2 + (12.99)2 = 26.0 m/s Ans W” (a. a); 0’ 9M 1.. /. 5106 m4: 83 = 3A + 33/4
o.s coséO°i— 0.8 sin60°j +. 0.9 mam —o.9 00560°j = 2i+ (am ),i+ (am ), j AMI/F4 L
,0 8 cosﬁO“ +0.9 smso° = 2+ (am )x ' —0. 8 sin60° —0.9 eos60° = (dB/A )y 43,, = J(1..6206)2+(1.1428)2 = 1.98 mfsz Ans 1.1428
*1 __ = 0
(1.6206) 35.2 7 Ans (am ), = 4.6206 ids2 = 1.6206 111152 <— (am )y = —1.l428 £2132 = 1.1428 m/s2 .t ¢~ mu 3742, A car or mass m is traveling at a slow velocity
yo. If it is subjected to the drag resistance of the wind,
which is proportional to its velocity, i.e., FD = k7),
determine the distance and the time the car will travel
before its velocity becomes 0.5%. Assume no other
frictional forces act on the car. eXE=max; ~ﬁb=m—— (1) m :\
Setv = 0.5 120, r = I 111(2) 1‘ = 0.693 ~ Ans From Eq.(1), —kvdx = mvdv ~J:kdx = Iv mdv kx'= new—vs)
Atv = 0.5 v0, 1 = $0.5m l
x = 0531:10— ’ Ans ‘ 119 1326. At the instant shown the IOOlb block A is
moving down the plane at 5 ft/s while being attached to
the SOlb block B. If the coefﬁcient of kinetic friction is
M = 0.2, determine the acceleration of A and the
distance A slides before it stops. Neglect the mass of the
pulleys and cables. Block A : 3 100
y'ZF} = max; TA ‘02NA +100[§)=(3_2§)a‘ 4
+‘\>:Fy =ma,; M100(§)=0 105 N? Thus, ‘ ’
TA —44= —3.1056aA (1) 0.1%, Block B : A] +TEI'} =ma,; 7b50=(§1:O—2)aa 1};~50=l.553a5 (2)
PulleysaECandD:
+T2F,=o; 22;—22;,=0 TF2} (3) Kinematics : 5015 5.4 +2.31 :1
50+(SDSB)#I’
sc+d+‘sD =d'
Thus. 0A =_25C ‘ T3 T3
2aD=a3 ac = ‘ap, so that a; = a5 (4)
Solving Eqs. (l)(4) :
‘ 11‘ = a3=—l.288 ml
71 = 1;; =48.0 lb
Thus, aA =i.291ft/s2 Ans (+/) v2=vﬁ+24c(s—so)
0 = (5)2 + 2(—.1.288)(s — 0) s =9.70ft Ans *13—56; The jet plane is traveling at a constant speed of
1000 ft/s along the curve y = 20(10’6)x2 + 5000, where x
and y are in feet. If the pilot has a weight of 180 lb,
determine the normal and tangential components 0f the
force the seat exerts on the pilot when y = 10 000 ft. , = 2000")? +5000 10000 =20(1o")x2 +5000 x215811ﬁ
5? = r2119: 40(10'6)xi = 0 63246
dx x=15811 '
9:32.31°
dzy —6
—=40 10
a, ( )
1) [14%) T 11+(063246)31‘;
=—~————=  =41.4 . .
9 ﬂ [40006)] _ 130(33):: 4:2 ' A E‘Fn = man; E. — 18000532.31° = [%J[:%) P; = 2871b Ans
«72:17, = 1nd,; F. —180sin32.31° =0 E = 96.215 Ans ...
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 Spring '06
 Keil

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