ME 270 Exam 2 Soln %28Spr 11%29 - NAME 5)ng 20A ME 270 —...

Unformatted text preview: NAME: 5;)ng 20A,! - ME 270 — Spring 2011 Examination No. 2 Please review the following statement: l certify that l have not given unauthorized aid nor have I received aid in the completion of this exam. Signature: INSTRUCTIONS Begin each problem in the space provided on the examination sheets. If additional space-is required, _ use the yellow paper provided to you. Work on one side of each sheet only, with oniy one problem on a sheet. Each problem is worth 20 points. Please remember that for you to obtain maximum credit for a problem, it must be clearly presented, i.e. . The coordinate system must be clearly IdentIf“ ed . Where appropriate, free body diagrams must be drawn. These should be drawn separately from the given figures. - Units must be clearly stated as part of the answer. ._ You must carefully delineate vector and scalar quantities. If the solution does not follow a logical thought process, it will be assumed in error. When handing in the test, please make sure that all sheets are in the correct sequential order and make sure that your name is at the top of every page that you wish to have graded. Instructor’s Name and Section: Section 1: J. Jones 9:30 - 10:20 am. Section 2: V. Kumar 2:30 — 3:20 pm. Problem 1 Problem 2 Problem 3 Total ME 270 EQUATIONS (You probably will not need all of these equations) \ ‘“ T 27:; 3 |F|=r __2__=ell5l3 M T; Z'§=ABCOSQ v=vo+act 1 2 =AxBx+AyBy+Asz szso+vot+§act A“=1_4-17 v2=v§+2ac(s—so) _0=F><F , M0=Fd EF=mE ZF=0 2 ”=0 :7 f I? 11—40: rx ry r2 v=5ci+yj+2k 1: Fy F2 Mazfia-Mo ;=f;r+réz-¢9+z'i"€ Ma=Ea-(FXF) v=vur uax u“? ”(.22 _ _ _ _ Ma— rx ry rz a=5éi+j5j+2k F; Fy F: _ _ _ _ v2- Ma=Jl4'auflr a=vut+Fun _W _ 2_ 2 __.. -2_ -- .-_ ..- m—— , g—32.17fi/s —-9.807m/s a—(r—re )ur+(r6+2r9)ue+zk g 3 2 A 1+ 35: dx p= 2 2Fx=m5c‘,2Fy=mj},2Fz=m2' Q de WE , a=fl=vfl 2g=m(;~'—réz),2F8=m(ré+2fé),>:F=m2 dt dt ds 2 n _ "BIA IVEBIAI E = FE]? fsratic S “SN fslide =JuKN J x dm J y dm _ Jz dm P = pgh w = pghd d d do: . . . . If fl = Ma), then —6 = d—fi—t (cham rule of dzfirerennanon) a E=xg+jji szi+yl =fgr+rég9 = r—r92)gr+(ré+2i‘é)g9 = v g, __ . v2 — v g, + 3 g" Trigonometric Identities a Annotated reference triangle. ‘ secza=1+tan2a, sin2a+cosza=1, csc2a=1+cot2a, sin2a=25inacosa, 0052a = cos2 0: — sin2 a - 2 2 2 . a b 3' Law of cosmes: g = a + b — 2ab cosy Law of sums: — = — sin a sin [3 sin 'y NAME: ME 270 — Spring 2011 PROBLEM 1 (20 points) -- Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points) FIND: The shaded area bound by the two curves shown. Determine the area of the shaded part and the x- -centroid of this area. A [Eamon (7.. WM «— (2 points) (3 points) PROBLEM 1B. (5 points) ea: Q 2' 4-1]! FIND. Gate AB holds back water as shown. Ont e sartwork provided, sketch the pressure distribution exerted by the water (3 points). Also, calculate the equivaien force exerted by the water Feq at gate AB. Feq= 5’4 4 £53, ' . ‘ (2 points) NAME: ME 270 — Spring 2011 PROBLEM 1C. (5 points) FIND: A 50 lb cabinet is to be moved by exerting a horizontal f h force as shown. Ifh = 2.5 ft, b =4ft, H = 2ftand ' us: 0.5 between the floor and the supports, determine the force (FT) required to tip the cabinet and the force (F5) required to slide the cabinet. Based on your ; calculations, what Is the nature of the impending moti n? , 7 __ $5" ' "91;?“ 82mg OT“ 5003) (2 points) (2 points) Impending Motion: TIP ( SLIP ) (Circle One - 1 point) 45W adaclfimgifirfafn‘fim wF Li M FIND: The weight(W W)used to load the flywheel of a sta onary blke' Is 40 lbs. The coefficient of kinetic friction between the flywheel and the cable' Is pk: 0.2. Determine the angle of wrap (fl ) of the cable around the bicycle flywheel and the force the scale (3) indicates when the bicycle is in use. (2 points) (3 points) NAME: ME 270 — Spring 2011 PROBLEM 2. (20 points) GIVEN: An athlete works out with a squat-thrust machine that has a fixed support at E. To rotate the bar ABD she must exert a verticai force at A that causes the magnitude of the axial force in the‘two—force membe BC to be 1800N. FIND: a) Draw a free body diagram of each of the individuai members (ABD, BC, and CDE) on the sketches provided below. (4 points) b) Determine the force at pin D acting on member CDE. Express this force' In vector form. (8 points) c) Determine the reactions acting at the fixed support E. (8 points) . .. “firm 1? BI) $12“ M FMMFZW #7:,"- mm'ldflb 1 ZR 0:: mm fiwu2§5©o)m By [424 L mam”. w .m MZ mg :03 @c E)!“ 3356} <0 (a) 1) 7b. 5 , 4 _ L- » H i . m‘ I mam”;.;;m.;aya,4;~muan§m 13m:m-imflysgwnawuq, ‘ Ehrjw ,-s" 57 5m ”WW3: Van-— W \ 35C ‘1‘) mm X553 ”SoiéaM 7 7 M9.» WM Wm» Nfiwnu WM NAME: ME 270 — Spring 2011 PROBLEM 3. (20 points) GIVEN: Truss A—J is supported by a pin joint at A and a rocker support at G. The truss is loaded as shown and is in static equilibrium. FIND: . a) Determine the support reactions at joints A and G. (5 points) b) Determine the loads and sense (i.e. tension or compression) of members DE, DH. and IJ. (12 points) c) Identify all 2397 embers in the truss. (No work needsto be shown.) (3 points) 40 kN FEQJQE 1:“ Fe 29* BI 30 kN gm; :WWW,W w {a q k 34: “5:; a) ' 16 m, 4 @ 4 m I ZMg-soe 10(4) sow 2.01m 10kt»! 40mm .3 3 at - Lg ...
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