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ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): Please revew the followng statement: I certfy that I have not gven unauthorzed ad nor have I receved ad n the completon of ths exam. Sgnature: Instructor s Name and Secton: (Crcle Your Secton) Sectons: J Jones 9:30-10:20AM A Buganza 1:30-2:20PM B L 3:30-4:20PM J Jones Dstance Learnng INSTRUCTIONS Begn each problem n the space provded on the examnaton sheets. If addtonal space s requred, use the whte lned paper provded to you. Work on one sde of each sheet only, wth only one problem on a sheet. Each problem s worth 20 ponts. Please remember that for you to obtan maxmum credt for a problem, t must be clearly presented,.e. The only authorzed exam calculator s the TI-30IIS The allowable exam tme for Exam 1 s 70 mnutes. The coordnate system must be clearly dentfed. Where approprate, free body dagrams must be drawn. These should be drawn separately from the gven fgures. Unts must be clearly stated as part of the answer. You must carefully delneate vector and scalar quanttes. If the soluton does not follow a logcal thought process, t wll be assumed n error. When handng n the test, please make sure that all sheets are n the correct sequental order and make sure that your name s at the top of every page that you wsh to have graded. Problem 1 Problem 2 Problem 3 Total ME 270 Exam 1 Sprng 2017 Page 1

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): PROBLEM 1 (20 ponts 1A. The tenson n rope AC s 100 lbs. Determne the angle (θ) between cables AB and AC and the magntude of the projecton of the tenson n cable AC tenson n the drecton of AB. θ θ = Proj = (3 pts) ME 270 Exam 1 Sprng 2017 Page 2

ME 270 Sprng 2017 Exam 1 1B. The tenson n cable AB s 10kN. Determne the force vector NAME (Last, Frst): z T AB and the moment vector M D due to the cable about pont D. y x T AB = M D = (3 pts) ME 270 Exam 1 Sprng 2017 Page 3

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): 1C. Three of these four systems are equvalent. Whch system s not equvalent? What couple must be added to ths system to make t equvalent? System 1 2 3 4 (Crcle One) M = k kn-m (3 pts) 1D. Determne the x-centrod of the shaded area. If the 2-n radus hole was removed (.e., the block was sold), what qualtatve effect would ths have on the x-centrod? x = (3 pts) x-centrod would: Increase Reman Same Decrease (Crcle One) ME 270 Exam 1 Sprng 2017 Page 4

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): PROBLEM 2. (20 ponts) GIVEN: A tractor s lfted by three ndependent cables as shown n the fgure. Determne the tenson n the cables AB, AC, and AD f the tractor s weght s 80kN. FIND: a) Draw the free body dagram of the system of forces actng at pont A (3 pts) 1.25 1.25 b) Lst the cable force vectors as an unknown magntude multpled by a known unt vector (6 pts): T AB = ( + j + k ) T AC = ( + j + k ) = ( + j + k ) T AD ME 270 Exam 1 Sprng 2017 Page 5

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): c) Formulate the equatons of equlbrum. (3 pts): ΣF 0 = ΣF y = 0 = (1 pt) (1 pt) ΣF z = 0 = (1 pt) d) Solve the equatons of statc equlbrum for the magntude of the tenson n each cable. (6 pts): T AB = T AC = T AD = ME 270 Exam 1 Sprng 2017 Page 6

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): e) What s the maxmum load ths cable system can hold f the largest allowable tenson n any sngle cable s 60kN? W ma ME 270 Exam 1 Sprng 2017 Page 7

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): PROBLEM 3. (20 ponts) GIVEN: Frame ABCD s loaded as shown and s held n statc equlbrum by a fxed support at A. FIND: a) For the dstrbuted load shown, determne the magntude of the sngle-force equvalent and ts locaton measured from the y-axs. F eq = b) Draw the free body dagram of the system, lst all appled and reacton forces and moments, replace the trangular dstrbuted load wth a sngle equvalent load and mark ts locaton. (3 pts) ME 270 Exam 1 Sprng 2017 Page 8

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): c) Fnd the reactons at pont A. Express the results n vector form. (10 pts) F j (4 pts) A M A (6 pts) d) Qualtatvely descrbe the change n the magntude of the reactons at pont A f the 500 lb force s slghtly decreased by crclng the approprate trend. (3 pts) (F A ) x Increases Remans Same Decreases (Crcle One) (1 pt) (F A ) y Increases Remans Same Decreases (Crcle One) (1 pt) (M A ) Increases Remans Same Decreases (Crcle One) (1 pt) ME 270 Exam 1 Sprng 2017 Page 9

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): SOLUTIONS 1A. θ = 42.8 degrees Proj = 73.3 lbs 1B. T AB = 3.33 + 6.67j 6.67 k kn M D = 26.7 + 6.67j 6.67 k kn m 1C. System s 3 s not equvalent M = 80 k kn m 1D. x = 6.98 n x-centrod - Decreases 2A. Free body dagram 2B. T AB = T AB (0.524 + ( 0.327)j + ( 0.786) k ) T AC = T AC (0.524 + 0.327j + ( 0.786) k ) T AD = T AD ( 0.316 + 0j + ( 0.948) k ) 2C. F x = 0 = 0.524 T AB + 0.524 T AC 0.316 T AD F y = 0 = 0.327 T AB + 0.327 T AC F z = 0 = 0.786 T AB 0.786 T AC 0.948 T AD + 80kN 2D. T AB = 16.94 kn T AC = 16.94 kn T AD = 56.29 kn 2E. W ma 85.2 kn 3A. F eq = 300 lb 3 ft 3B. Free Body Dagram 3C. F A = ( 400 + 600 j ) lb M A = ( 800k )lb ft 3D. (F A )x Decreases (F A )y Decreases (M A ) Decreases ME 270 Exam 1 Sprng 2017 Page 10

ME 270 Sprng 2017 Exam 1 NAME (Last, Frst): ME 270 Exam 1 Equatons Dstrbuted Loads F = eq eq xf = 0 L w x dx 0 L Centrods y = y = In 3D, x da c da y da c da x A c A c x w x dx y A A x V c V Centers of Mass y = y = xcmρda x ρda ycmρda ρda y Buoyancy FB gv Flud Statcs p gh cm ρa ρa cm ρa F p Lw eq avg ρa ME 270 Exam 1 Sprng 2017 Page 11