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Review2Solutions - 1 ‘ 4 u i I 7-83‘ Determine all...

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Unformatted text preview: 1 ' ‘ 4 u i I 7-83‘ Determine all forces acting on member BCD of the _ linkage shown in Fig. 137—83. - 40 1b Eff: - 400050 +6“ - 4M, :0 [5x ' 690.4 ~35c4. = Zéf—y éF‘J: £114 4051/?30b—fij‘7 :0 fl, = Mm, - 2.0-: 40,4.“T m N0. 5505 ' Engineer’s Computation Pad __L#_# «'2 '. 7-91 * Determine all forces actihioifix member ABCD of the H ‘1 . f frame shown in Fig. P7-91. T 3 . fl _——- —- r43 . ‘3 ' i - SQ .7, _fi$ ]\ . .———- (a ' -'-.'s.'-3“-;.-..'-'5v.:s «1113-31. {p a b T _. .3... 1: g, [4 F a 2 .<;__ TSIb - - Engineel’s Computation Pad No. 5505 __ + L-mm—T-W @fl’lfi :0 = 12F-7.r(24) - # 3 7r *9 56M? : ’43 ~7r=7fl4e -'7r #:- F' 7275“)““0 4%” = Fat/33:70 aflzumfl W k3 . E ‘ i 7-88“ A block with a mass of 150 kg is supported by a ca— ‘ ble which passes over a ISO-mm diameter pulley that is at- - tached to a frame as shown in Fig P7- 88. Determine ail forces acting on member BCD of the frame. a ’77, Cal. 65 I472” 2’5" Cx— H710“ 3‘; Ca; ‘ [38'31'0—b- 2/333? Cq-H‘Il—mzam 6 Cg: [4724-603 ={97J7U - N0. 5505 w Engineers Computation Pad 705 ng= 1393(500) “/97ré/0‘9" 3K(”’m) 414900)190000 ( . 5k , 13mm)... /9?J‘(wo)_ _3,54 A, .,.. 53,5: 3 2237A!” 07 " /?7J‘Nf ‘ a , 7- 9“, Determine all forces acting on member ABE of the 1 1 Me shown' In Fig P7 90. ' - [SON :GJ/nfl : lJ?[3°O)-D[3oo) Immm W No. 5505 Engineer’s Computation Pad Fig. 97-90 D = 1.51) lo '2‘ r [355: ‘0: Kat/0793,30 :r—y Ev M W m) CL... E" waD [60 TC: +£Z/fl76 *—€ 76%) + /¢a(aoo) ‘ ' - C1,. Malawi)... 442; #‘F 5'7 *500—1JD :0 > (W 7 Z/éw‘fly 14/JvflqJO’0 £1 : 50M 4 ""‘ 6 = ash-Aw =300Nf‘ :7 + E ”W 9/5 c = £2 {‘00) vméw) 5!: = 30021) :52“! 1 E; - 8k “ZS-9’0 l . {3.2 .g g): .._/JD.- 300—50 {$05,149. F133: | L; 7-98 Determine all forces acting on member ACE of the frame shown in Fig. P7-98. The diameter of the pulley at E " is 120 mm. The mass of body Wis 100 kg. ZIFV ‘ Ev " Q 2543’)" q?! qgl flis ’5 2" For“ W56!" Ed: l‘f—"éfZ—J"J Engineers Computation Pad What forces are. exerted on forces on the pliers? the bolt at E in Fig. 6.40 As. a result of the ISO-N {Kg/Mp 3 '- I“) (230) +flfi33Z3o) 57.7130): D H52‘% '6’“ i 1 i . '- . The tractor boom supports the uniform mass of 500 Kg ’ in the‘bucket which has a center of mass at G. Determine the force in cylinder ~ ' . .CD and the resultant force at pin .F. The load is supported equally on each side of the tractor by a similar mechanism. 2 i I A Gil-kg cabinet is mounted on casters which can be locked to‘pre— vent their rotation. The coeflicient of friction £59.30. If' it = 800 mm. determine the magnitude of the force F required to move the cabinet to the right (a) if all casters are locked, (b) if the casters at B are locked and the casters at A are free to rotate. ' 3. The coefliclents of friction are p5 = 0.25 and pk = 0.20 between all surfaces of contact. Determine the smallest force I? required to start block D moving if (a) block C is restrained by cable AB as shown, (11) cable AB is removed. 4; ., The coefficients of friction'between the block and the incline are [15 = 0.30 and a}: = 0.2.5. Determine whether the block is in equilibrium and find the magnitude and direction of the friction force when P ='150 N. 5. A hand brake is used to control the speed of a flywheel as shown. The coefficients of fiiction are as = 0.3 and pk = 0.25. What torque should be applied to the flywheel to keep it rotating clockwise at a constant speed when P = 10 lb? ME 224L—MECHANICS LAB WORKSHEET 9 Page 2 6. Find the centroid of the C-shaped area with respect to the x and y axes shown. 7. Find the centroid of the plane area with respect to the x and y axes shown. 8. Find the centroid of the'Fshaped area with respect to the x and y axes shown. 9. Find the coordinates of the centroid of the shaded area with respect to the x and y axes. 51B SHEEN?! MW SWEETE‘ 2M}! EHIEII'IT‘i Afi-flflfl 17124433 22‘ -'fl (EL-41L @ ‘ _ . . I" LG LomATED [Harlan The tractor boom supports the-unifbnn mass of 500 Kg in the'bucket which has a clutter of mass at G. Datemfine the for-(SB in each cylinder AB and CD and the resultant force; at pins E and F. The load is supportad equally on each side ofthe tractor by a. similar machanism. .. — ., ' _ , 02m 1.25111 :3th ‘Ffifi Feb map ,fiaéw #3?” . Pm; E #Hub (3: . - . Mm .i M $ ,. fl} r3, ‘5: “if"? EMF: £0 24—238 (4%,?) =- 3; .13;erij #— {gig Fm; gas T735 « W MW..W mmfiM . “5:934 FM-fi r2641: -* [3734. [2? ; 4:32;: {2724/2/42 . _,,_._....-M..—vn -fir..._»m..;—w=m....r-.m—__» gyfgm. - _ '. )3 . F a? {’96. :: 753;?952 r" {é—i i K :9 : Fx FE‘ éfiwflr FOL- u—:-:—.r-.,_' — - "Wa- flag? wwww—J—Hu— #19“ “LE“...«erb—a; ”44“.) war: rfiiér:/ ( __ . '31“: w, :3 .fi ,L 1.32;” "2769 J_ r: ‘41}? ”fl “9' mm?) ) Mfiwwmflmwun F -awmr:w‘w :« 2F’2’WQEV‘J¢ if "1 2! :réreié“: fit,» 51 H. $95715 rut-('57!) HUD ADM 07L} , 39b}... Cyélwsreni. W {scab—J. NU. that 0| :1... :m E» “1’ Qfiflmmflm Engineer's Computation Pad “/ @A- +M;>:/aw P: ('5 (’50) (7-94“) r—j P MW)? iii } GALE 3‘4 ’33. 9951’? {M 0, #601,253“ 77P9 Fggg'r“ (gm—4 fié—éé! F? b B 5NL~ ‘ - “ Ffl—f" __._.-——-. . If ng. 57219 :94“ g A125: Loam . £3,214,419 - . '19 anNg—agmw_9flfiPr_a g 600 Mg. =§Q¢M+EBEP N fi,§‘LQ-§—-é§- T. a. .5? f” A g N0. 937 811E Engineer‘s Computaiion Pad 600/05 FL 2; Q“ My; 19 LR; vi V . ‘2’” *jx'iF =0 .‘7 .. ML "Mg, NO. 55:}! 51 IL": WWEm'fi-JEN Engineer's Camputation Pad (UL )1 .2 ML .1. ME: 7:. (3137 +_{47LE .2. 2,45”; ’5'" .zr(24r37: «2:33.233— N0. 93:” 6'! 'I 1:: fl “3 fifi‘WEWEW Engineer's Computation Pad Fer—D: A) (E: $3de: 1; m1 mun. _______________. 19.3 min-Q ébua 0F Ji‘razdnép Face“ : ' (\vawacosch—arsfis‘mzfsb N: 4&53 1- 9;.2 : 75L: BaM/‘V NeeaQecfiIS-QU‘ figux _ (Ema-:14. FML Wm; -_[‘_fi LwM z/MM T—(OJEEé’éZLl) 3c: 4 'nga’i ELDCL Vs; (M NO. 93! U1 1|: E9 @WEWEW _ Engineers Computation Pad :en- t semcammmmm , Example 811 WWWWMWNWWWN;M%WWMW$W - mtt‘ilfmii Wm . , Examplesrlz SolutiOn Centroid of 0 Composite Area with a Hole Problem Statement Let the C—shaped area of Fig. E8103 be represented by two parts (1 and 2) as shown in Fig. E811. Part 1 is a rectangular area without the hole, and part 2 is a square area corresponding to the hole. Determine the centroid of the C—shaped area by considering parts 1 and 2. . Figure £8.11 From Fig. E831, we can tabulate the sums of the moments of the areas of parts 1 and 2 with respect to the x and y axes, as shown in Table 13.8.11. Note that the sums in Table E8.11 are the same as the sums in Table E810. Thus, the centroid coordinates are, as in Example 8.10, F 25500 x— 1500 —17.0mm __4osoo_ 32—1500 “27.0mm The centroidal coordinates have the same signs as those in Example 8.10, relative to the x and 3: reference axes. Also, the area of the hole (part 2 in Fig. E8.ll) is taken as nega- tive, and, therefore, the moments of that area are also negative {see Eq. (8.21)]. Table £8.11 Centroid soordinates of a composite area with a hole AREA cam-note DISTANCE MOMENT our AREA PA RT [mmzl [mm] [m3] _____—————*———‘—r———‘_"—‘_"_“ i A: xi Yr APT! Ar}: 1 2 400 20 3O 48 000 72 000 2 —900 25 35 —22 500 -31 500 Sums 1 500 —— -— 25 500 40 500 Centroid of the Cross Section of 0 Structural Member Problem Statement Along, flat sheet of steel, 0.5 in thick, is bent into the shape shown 1n cross-section in Fig. E8.12a. Determine the centroid coordinates (E, i) of the cross sec— tion. SOLUTION Composite Pa rte. The plate 15 divided into three segments as shown in Fig. 9—185. Here the area of the small rectangle © 15 considered ‘negative” since it must be subtracted from the larger one Q). Moment Arms. The centroid of each segment is located as indicated ' in the figure. Note that the 32' coordinates of ® and C3) are negative. Summertime. Taking the data from Fig. 948b, the calculations are tabulated as follows: Segment A (1’12) §‘(ft) 37m) SEA (1‘?) 37A (ft3) W 1 §(3)(3) = 4.5 1 1 4.5 4.5 2 (3)(3) = 9 -—1.5 1.5 —13.5 13.5 3 —(2)(1) = —2 —2.5 2 s —4 M = 11.5 23m = 74 2331 = 14 H__mm—_n______ Thus, BIA —4 X“ -- E—A'H m — —0. 348 ft Ans — _ EM _ .21 _ y— E A 11' 5 — 122ft Ans. NOTE. If these results are plotted m Fig 9—18, the location of point C seems reasonable. 9.3 COMPOSITE BODIES 481 (b) 92L 3.9,,— is; y; A“ a: :LS’ 2%‘2965 75: ‘29” (9 Qzflzg} 291/250 0&7 ‘7'” 7 f— iji‘dl - é * W a: fijfiefiii " :2; £3739 ...
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