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hw8 - PROBLEM 15.174 Pin P is attached to the wheel shown...

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Unformatted text preview: PROBLEM 15.174 Pin P is attached to the wheel shown and slides in a slot cut in bar 30. The wheel rolls to the right without slipping with a constant angular ‘- V 1“” / velocity of 20 rad/s. Knowing that x 2 480 mm when 6 2 0, determine 3 (a) the angular acceleration of the bar and (b) the relative acceleration of pin P with respect to the bar for the given data. 3\ l)’ {3‘13”}; 'k” if; r‘LngVll) mm ‘W 3:10 shim». 6 2 90°. SOLUTION 5’ Coordinates. i ”17/; I, *6 } Y4 : (x/l)0 + V8, (VA 2 I” 1‘99“ 7 '§_*‘“ r3 2 0, )8 2 I j. l xci : X42 .VL' : 0 #0 xp 2 )01 + esin 6. yl) 2 r + ecosfi Data: ()04)0 2 480 mm 2 0.48 m r 2 200mm 2 0.20 m r e=140mm=0.14m 6 : 90° xp : 0.48 + (0.20)(%j + (0.14) = 0.93416 m, [5‘ = 0 8 Velocity analysis. . V a)“ 2 20 rad/s 2, (0,3,) 2 cow) 0 : [m -+ l + [mac 1] : [(0,2)(20) 2. ]+ [(0.14)(20) l] = [4 in./s me- ]+ [2.8 in./s l] vp, = [Wm H : [0934166030 1], vm. = u ~—~ Use VP 2 VP: + VP”; and resolve into components. 3;: 4 2 u u 2 4.00 m/s. +1: 2.8 = 0.93416wBD (1)30 = 2.9973 rad/s) . . - - _ _ \ Acceleianon analysw. 004C — 0, (130 — 0581) / 2 aA : 0, am 2 rij : (0.14)(20) = 56 m/s2 2.. 2 3P 2 a"? + am, 2 56 m/s «- PROBLEM 15.174 CONTINUED ap/ : [xpagn i J + [x1020 *’“ 1 : [0.93416a30 l ] +[(0.93416)(2.9973)2 2. ] : [0-93416aoo 1} [8.3923 m/s2 ... 1 aP/F = it W Coriolis acceleration 2%,)24 = (2)(2.9973)(4.00) = 23.978 m/s3 3 Use 3;, : apv + ap/F + 2a),;u l and resolve into components. 2;: —56 = 8.3923 + o, a : 47.6 m/sl, +1: 0 : 0.93416aBU + 23.978 (280 = —25.7 rad/sZ 0‘51) : 25.7 rad/s2 VS, 4 am 2 47.6 m/SZ «— 4 11,. m“ PROBLEM 15.177 Disks \ l [/1 ‘ H”: .1) m} The Geneva mechanism shown is used to provide an intermittent rotary " / , motion of disk S. Disk D rotates with a constant counterclockwise angular 5/ , ' velocity u) n of 8 rad/s. A pin P is attached to disk D and can slide in one 0‘ i , of six equally spaced slots cut in disk S. It is desirable that the angular \ g velocity of disk S be zero as the pin enters and leaves each of the six ” ' mi 1) slots; this will occur if the distance between the centers of the disks and “hm 0" ””5 the radii of the disks are related as shown. Determine the angular velocity I -> in“ ~+ and angular acceleration of disk S at the instant when ¢ : 150°. SOLUTION Geometry. P Law of cosines. / '7 ‘" ‘30:, r2 = 125° + 2.502 — (2)(1.25)(2.50)cos30° 3 Fa 30° . .3 8 : 150;“. r 1.54914111. Law of Sines. sm ’6 = sm 30 1.25 r ,8 : 23794" Let disk S be a rotating frame of reference. Q 2 (Us 3, Q = as) Motion of coinciding point P' on the disk. VP, = rwS =1.54914w5 h ,6 a... = —aSk x rm — wérp/U = [154914623 F\ fl] + [1.54914w§ 7 fl] Motion relative to the frame. VP/s=“7fl aP/S=ll7,3 Coriolis acceleration. 2‘05“ \< 3 v,. = v,,. + vm = [154914605 h 13] + [u 7 ,8] 3P 2 ap + am + 260814 \ : [1.549140% h ,6] {1.549141}; 7 o] + [a 7 ]+ [26051: \ ,8] Motion of disk D. (Rotation about B) vP = (BP)a)D = (l.25)(8) = 10 in./s 7 30° aP = [(BP)aD 7 60°] + [(BP)a)§ <1; 30°] : 0 + [(1.25)(8)2 wq 30°] = 801n/s2 st: 30° PROBLEM 15.177 CONTINUED Equate the two expressions for VP and resolve into components. h fl: 154914603 = IOCos(30° + B) _ 10cos53.794° 1.54914 (US 2 3.8130 rad/s cos 2 3.81rad/s 3 4 7 [3: u =10sin(30° + ,8) = 1051n53,794° = 8.0690 in./s Equate the two expressions for ap and resolve into components. K ,6: 1.54914a5 — 2wSu = 805in(30° + ,6) _ 8051n53.794° + (2)(3.8130)(8.0690) _ 1.54914 (15 = 81.4 rad/s2 as = 81.4 rad/s2 3 4 C4) “213' ImaIS : [#Arfiha§§ No“ Loam centroid . C .‘ ZAfZW ) v— : [QC/Mlfi efleoc‘d’x 05 , 7 ,flfl/awoh :— %{%’a ”fig—wag) {a : gmoéaz); 9mg. 23:1:ch , 7) 1 Inc 1: 7M!) 6"” 3—21 Sou/476071) IMI>Z$6 ’1‘ Mdf/(E) 1—. m 0 th/o”é (6%? 5'21 <13 I5»? = 1565’ + ”04965 '5 lea’f-m (4M ~dfim) , / 145%";[email protected] (94;), we Aux/e PROBLEM 16.6 For the truck of Sample Prob. 16.1, determine the distance through which the truck will skid if (a) the rear-wheel brakes fail to operate, (b) the front-wheel brakes fail to operate. SOLUTION (a) [frear—whce/ brukesfili/ to operate: ”m? §*\~WM_ 1 2 w (W, l ‘7 Q “#426 , 14% NI),(12 ft) — W(5 ft) = 1115(4 ft) W 1N8:5W+lic_l 12 3g _ W _ J; 2F : Zvalt-rr: FA. : ma. [MAIN/H : Ea ’ 5 1 W W, : —a ‘12 3 g , g s \ w 0.699(1‘7 1(322 my) a : “/ a — 12.227 fi/s1 « 1— 0.233 (Jnf/(‘Wm/y acce/eralcd motion 12- : v}, + lax 0 = (30 11/53) — 2(12227 ft/sl)x 6 8 It 4 Ii 1.6) (/7) [fl'fi‘onflwhccl brakes/21H m ()pcrale.‘ 5w «a :1 1 5 ”472:9 : W (JV N _ J l m " W AMI” “~65? {921W 32:22 All-1 1W;T/Vp h?» 4’17: 2M“ : 2(11“) W(7 ft) ; N ,(12 ft) : 1775(411) e 11‘ I N‘. : 1w AL: 12 3 g PROBLEM 16.13 A uniform circular plate of mass 6 kg is attached to two links AC and BD v; 1) of the same length. Knowing that the plate is released from rest in the ~ _ fl position shown, in which lines joining G to A and B are, respectively, C“ "T?“ ‘ horizontal and vettical, determine (a) the acceleration of the plate, (:5) the l ' :3 tension in each link. r I‘ T'y, ) i y ,5 ( SOLUTION (a) Acceleration '11” 2F : EFL)”: mgeos75° : Inc—1 c7 : geos75° : 2.53901m's3 2.54m/s3 / 15° 4 E (/7) Tension in vac/7 link if, 2MB : ZlM/ilcn'; (FICOS750)I‘ + (FI sin 750))” : (ansin750)/‘ F,(cos75° + sin 75°) : (6 kg)(2.539 m/sz)sin 75° If, : 12.0146 F1 : 1201 N Tension 4 +lx 75° SF 2 ER”: F, + F,; — mgsin75O : 0 6.009 lb + F,f — (6 kg)(9.81 n1,/'s:)sin75° = L 14/; : 44.839 N FM) 2 44.8 N Tension 4 PROBLEM 16.16 At the instant shown the angular velocity of links BE and CF is 6 rad/s counterclockwise and is decreasing at the rate of 12 rad/s3. Knowing that the length of each link is 15 in. and neglecting the weight of the links, determine (a) the force P, (b) the corresponding force in each link. The weight of rod AD is 15 lb. SOLUTION Links: V II 60‘ 6 feral/3' a” : V0}: : (1.25 ft)(6 rad/s)2 : 45 rad/s3 -‘ [‘de ’ 7 v 9/ A A? ”A a, : m: : (1.25 ft)(12 rad/5‘) : 15 rad/3‘ a” V: /5/,2:/ 257% 69;: Bur AD is in Iranslalion F/ZE v I C- * 5(1’5172115431’! 1 1" 2M“. : 2(M( F“, sin 60°( 15 in.) — FM-sin600(151n.) X i )Cff' I FM? : FR/r +1 2M” 2 2W”) ell ' F”.- sin 600(30 in.) , (15 lb)(15 in.) : mur sin 300(15 in.) + ma” sin 600(15 in.) / , q - '3 '3 F”.- (25.981 in.) , 2251b-in. : (ijfl 5 in.)(15 rad/s“ sin30° + 45 rad/5’ sin 60°) 32.2 ft/s‘ , F“,- : 21.15861b or F”. = FM 2 21,21bT4 PROBLEM 16.16 CONTINUED 2F : HF)“: (Fm; + F” )cos60° + P : —ma, cos30° + Irmwcos60° cos 30° + 45 rad/s: COS 60°] \, 21.15861b+P:[ 15” fllsrad S- 322 11/53 ‘ 21.15861b + P 2 4.4299 lb P : —16.7287 lb or P 216.731b +n 4 ...
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