3 - has 8.7 H 9.368 as u use 2950 03.mEZom i moo amewdl n...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: has 8.7 H 9.368 as u use 2950 03 .mEZom i moo amewdl n omnmoo 333 ea .55 Save u «I u 0mm. .5 833' a» 03 49 can :vmcozaacm E EconomEOu _. Ea _ wczazcm o 35.0 n v.8 ET 33 o o + at n x _ _ k\0k X Us: + a?» H ®> mm a.» mnemxu 8.x :8 a? 3 .92 m8 833 + :2 E 3.3de u 0 0mm Em Nd 9mm. 80 Nd Q03 0 o e. .q _ ITO“ Q\U.—X QQS+¢>N @> m_ G kc DUE? 2:. omém H n £25 90 32 2: wEbn—n—a am "Egan—cm Sm an 2: .«o 562m; 3395 2: we 5:3 is: N 8 a2 2: o. w5>oE mm & 052m 2: 95 0mm H m t 62.: E2 been E :32? ESP? 9: $2200 3%: E9395 in [he x-y are Problem 17.170 Points B and C city vectors of arms AB and BC are (0,13 = —0.5k (rad/s) and mgr Determine the velocity of point C. plane. The angular vclo 2.0k (rad/s). The vector 0 o Solution jsin 15") = 734.” — 196.7j (mm). 760(icos [5° rB/A The vector rc/B = 900(icos 50° +jsin 50“) = 5785i + 689.4j (mm), The velocity of point B is «Ms >< l'B/A = V8 V3 = ~98,35i ~ 367.Ij (mm/s). The velocity of point C V(' = VB + (03c X 1773 i — 367.lj + (2)(k x (578.5i + 689.4j)). —98.3 J 0. .I 5 + .l 9 00 7 3 _ U 7. 6 3 _ "M 3 00 0., V(' - 1477.2i + 790j (mm/s) Problem 17.162 If the crankshaft AB is turning in the counterclockwise direction at 2000 rpm (revolutions per minute), what is the velocity of the piston? Solution: The angle of the crank with the vertical is 45°. The Equate expressions for vB and separate components: angular velocity of the crankshaft is —296.2 = 4.7961030 2 a) = 2000 (l) = 209.44 rad/s. 60 —296.2 —_- uc — l.4l4w3c. The vector location of point B (the main rod bearing) r3 = 2(—-isin 45° +jcos 45°) = |.4l4(—i +j) in. The velocity of point B (the main rod bearing) is Solve: vc = —383.5j (in/s) = —32j (ft/s) (03c = -—6l.8 rad/s. i j k vB=wxr3M=L4l4w 0 0 l —l l O = —296.2(i +j)(in/s). From the law if Sines the interior angle between the connecting rod 2 5 and the vertical at the piston is obtained from —,—— = , , from smb’ stn45° which 25'n 45° 0 = sin"( '5 )=16.43°. The location of the piston is rc = (2 sin 45° + 5cos(-I)j = 6.21j (in.). The vector mm = r3 — rc = —t.414i— 4.796j (in.). The piston is constrained to move along the y axis. In terms of the connecting rod the velocity of the point B is i j k VB = VC + (03c X l‘B/c = vcj + 0 0 war. 4.414 —-4.796 0 = vcj + 4.7%de — l.4l4w3c j (in/s). v“= Irma}, (4.: 363m w: mu“ w— Hf“ - “- —-::I' __,6. id'rwa I; - “Ear 1:“. '— — FER-‘1'?"- Er" i r‘mgfi'fi gal-5:13: r'a'gr' VIE-Lek”; IU'A : FEMS EFF "F {if—Ella“. g + rmfiaaimhé‘ {Emembw Ham-F I'r 5:519 HIE-”J! 1'41: FIE-flea at; n_n1,' M “WP. 53" i! Flare}. bun-I'- walls-H .I'n'h; 'H‘E fin} = [fir ran-mm (:4 may]. 5—“.— i‘ P5313 35;? 4- rec-5““ [I'll + 641.3555; —..--5. "fl-31;] .— IFMI'] J 9.95;; lit-1p. gird-lb _ .rlnf“ L.— -'a {,1le aft: = I . ' r .- “ca-D ragl'fl-dj-‘fie‘gr + {HQQEQE -- f‘E] Qfifljln-EIFJ . :1 Ir. 415;] Ala-u _ '5' . r-L-J. =fifiip._ liq-Lad": [/JI“ EJ3EBJIEI¢I 7' Fair-Hal: [Efj =fi 3-; —,1.'= j:'-Efl.mflsmg _ F¢5"'a:|’?r' "LE-“63.1159 + @[K FE'fififl*L'5 3}]3-15; _ réqfl MEMQEE .. -::I ._ _ 5.35ng 5213 T 'rfigssdsmflw‘ FIE] Affidavit? ffi’gflg" F'lfififlflafl"? grflfllé— P53 35mg? IE: TWI’Jflf/Hifilfij'] -“ r GflfiiSJr-fifi‘jneg? £15 ‘.' Ffiéarm’jnfl 4 f5lm": “5555] EL :1? gfl’l‘ :Eié-Pfé' Jim: ”MES - :Qréfifilmfl ' 9369;122- ' “'3’“: {I F'1'53‘5'r-23J3J3E’ If ' I .d" i -‘ *i' +[ gr'fi'fl'jlfldfrfld FEJJJ'I’TIS' — Fyflgjhfib E'Efi '_ Fri-E EtEflE-H'JEE LE; _ _‘v" _ film’s , but £3“ 1; "33M" = (31333 T»; v “I , . = "F : “'5' I -_— 5,35 3; £le - —g-| #13 h-jhg-Il Fain-h {9:3 _ |I____ L" _ _.. ____ ”51' 49% -= 5.4. 4‘3 gr 1;?fo diffs; {HEP} f} L _ .—" ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern