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HW2Solutions - r-r——1 12-191 The man pulls the boy up...

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Unformatted text preview: r-r——1 12-191. 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—‘fxi-i-64 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, . \ 12-205. 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 13-26. At the instant shown the IOO-lb block A is moving down the plane at 5 ft/s while being attached to the SO-lb 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 ‘0-2NA +100[§)=(3_2§)a‘ 4 +‘\>:Fy =ma,; M-100(§)=0 105- N? Thus, ‘ ’- TA —44= —3.1056aA (1) 0.1%, Block B : A] +TEI'} =ma,; 7b-50=(§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+(SD-SB)#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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