lecture2 - . ME 221 I: RECTANGUkAfl Con/(Po NE r13 O F A...

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Unformatted text preview: . ME 221 I: RECTANGUkAfl Con/(Po NE r13 O F A Foams age NV ~63;th CAkLE 3\RECT10N Closuv ANGkFS E = \Fl-Wez ) FE: \fiwea) E’WFTW-e-ZA F‘= Frmeyffiflwefifi‘flmmezk Bose- VE‘CFOQ 2F 7" '1'“ 894933+€va82 /' A ' /. i“. .x’ fg‘flék f" ME '''''''''' I. FNHISEFJIN‘. 1~ DIM .3 e; =_L' 1—._3_ UNIT VEC—TOIQ ' . 5 5- BASL: .TX TOPS PRINTING College Station M ffffffffff I, L’fl.‘ lllllllll NIL. K |._ TOPS PRINTING College Station 1 3 “é =E.S‘\1 2 +;’7&‘i~3 + .385$3\F\ _ .LENCn—FRO'F- iCuNIT‘f) 69%: W (5:1))93: QM“ ‘(ncsb efcécw ) 9v=w°~+ ) %= ,3 0° 6125079 ° RESUKT'ANT' OF’ VECTORS Fences NEG—TOR ADDVWON) ‘ ‘3 I? .5 s- A F =C¢o¢ +804d 3ND5 -" . A A, G, =(5_C503£ +3 (503 d )AES A 5 A 55 G=Q+oL+BoJ )Mas Lb5 lab; _/oo 1, +- |1£_2} ’91"? ,1 RE3UKTANT FoRCE‘ OF MANYF’orLCL VEQToIZS \S Q /\ FE =—. gaff) Fr L€~FU\S + (i F2) XL ESOLTfi MAGywlTU be ‘6? Wm if (E Fr? +Ci Raf; + g F239: (BIAS? VECTOR e KE‘SUl-JTAmT ‘5' é}: (gm/3t +LzFa‘QS +£inV§ (1 FN- + {ZEN + ké F23? .TX TOPS PRINTING College Station C" (" P y I I u ., . _ . ‘ .I A ’ —i X x I V V 1' -. I I . 1000‘” -What are the horizontal and vertical components of the force shown ? -Express the force in Cartesian Vector form(i,j, k) F} = (/DOOLB)MQB\O°) ) Fir—QWOLQCW 106°) F1 = Qooogafi.zua\‘ J pfi=_aowkb\ G? 3?) A A F=BHQZ —- 0/3? +041 MECHa'HIt'AI. FVIJVEERING x '— I: o z «I U) o a: 3 '5 U TOPS PRINTING For the force shown in I Fig. P2—61 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. F=15001b Fig. P2-61 SOLUTION 2 2 2 (a) d = /(—4> + (12) + (7) = /209 = 14.457 FY = F cos 6x = 1500(—4//209> = —415.03 lb 2 -415 lb Ans. Fy = F cos 6y = 1500(12//209> = 1245.09 lb 3 1245 lb Ans. Fx = F cos 6: = 1500(7//209) = 726.30 lb a 726 lb - Ans. (b) F = —415 i + 1245 j + 726 E lb Ans. («I (N _ _ _. .. .. .. __ ~What are the horizontal and vertical 6 I - components of the force shown ? 52"..- I .. _ _ _ _ ‘ - i: _ __ Ill-LLB Express the force in Cartesian Vector form (i,j, k) F;=-COOLL§ 5:3 - F5: QOoLbL‘é \ 3'3. '2. J 1 __7_ +5 JED-+3 =—éOOLb.—£ - = 090kb(_1 3' 5.93 \5'.83 - F' =—§IH.S‘L‘|=§‘R 3 =3oeLbs! F: —SW.SL|D 2 +303ka This eye—hook, with the loads applied as shown, is attached to the concrete by a number of concrete bolts that will each pull out of the concrete at 490 Lbs . Determine how many such bolts are required so that the bolts will not pull out. Assume a) that all bolts share the load evenly and b) that the X component of forces on the bolts can be ignored form this calculation. This eye—hook, with the loads applied as shown, is attached to the concrete by a number of concrete bolts that will each pull out of the concrete at 490 Lbs . Determine how many such bolts are required so that the bolts will not pull out. Assume a) that all bolts share the load evenly and b) that the X component of forces on the bolts can be ignored form this calculation. Total force Y =2415.24 N x 490|b =2415.24lbs N: 4.9 Since can’t buy .9 of a bolt Number reguired = 5 M'FI'HANII‘AI. ENUl‘fl-‘EIIWG TOPS PRINTING College Station ,TX Determine the magnitude R of the resultant and the y angle 9x between the llne Fz=mom F _”0w of action of the resultant 3— F,=1200w and the x—axis for the four F4=5201b forces shown in Fig. P2-77. Fm.Pz77 SOLUTION PU 1| 6 F1 cos 91 + F2 cos 62 + F3 cos 9 + F4 cos 3 4 H 1200(-4/5) - 900 cos 700 + 750 cos 30° + 520(12/13) = ~412.82 lb m H y F1 sin 61 + F 1200(3/5) + 900 sin 700 + 750 sin 600 + 520(5/13) = 2415.24 lb R = R: + 12:: /(-412.82)2 + (2415.24)2 = 2450.27 lb z 2450 lb Ans. sin 9 ' 6 ' G 2 2 + F3 51n 3 + F.1 Sln 4 H R - -1 ._v - -1 m _ o o 9x — tan Rx — tan _412.82 — 99.699 3 99.7 Ans. /~ fi = 2450 lb 5 80.30 Ans. .I /’ 1?-_ y :3 l L 7 a a; . /f_fi_...—’g_ [liar :' Jr‘- ’1 \ I ’v‘ (7 —" .._.— _, -3 ' 1 , l (5 O °Determine the set of three components (x, y, z) °Determine the three direction cosines \F/ = 973 Lb _ 3‘) F}: - (AS—35in 07) I: =(3 99»I3{r\b>(- 7 07) y; a: KWSLDC my?) W W Wfifir) =\[_EL__ 11,-- I”?! FEZTFQZ ‘1‘th I733? +Q7ffl)?’ -r-’9f|( .9 “)1 CMQG1)_—_L112:9 )awke‘é) a. men?) 804.6,!) MECHANICAL E‘HUIVEI'ZII I“? 65/ TOPS PRINTING Col lege Station .TX Determine the x and y scalar c0mponents of the force shown in Fig. P2-52. SOLUTION For the force shown in Fig. P2—56 (a) Determine the x, y, and z scalar components of the force. (b) Express the force in Cartesian vector form. . y SOLUTION (a)flny = F cos m = 900 cos 35° = 737.2 N{ Fz = F sin ¢ = 900 sin 35° = 516.2 N a 516 N Ans. Fx = ny cos a = 737.2 cos 66° = 368.6 N a 369 N Ans. 5V = ny sin e = 737.2 sin 60° = 638.4 N z 638 N Ans. (b) \F 369 i + 638 j + 516 E N Ans. MECHANIC“; ENGINEERINli r x g k 5.; P-o E : EN :15 mo 1 9 O: h a U Determine the x and y scalar components of the force shown in Fig. P2—46. SOLUTION Determine the x and y scalar components of the force shown in Fig. P2—48. SOLUTION F: 1650N Fig. P2-46 ...
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This note was uploaded on 01/31/2010 for the course MEEN 221 taught by Professor Mcvay during the Fall '08 term at Texas A&M.

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lecture2 - . ME 221 I: RECTANGUkAfl Con/(Po NE r13 O F A...

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