Spring-2012-Exam-1-Solution

Spring-2012-Exam-1-Solution - NAME ME 270 — Spring 2012...

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Unformatted text preview: NAME: ME 270 —- Spring 2012 Examination No. 1 Please review the following statement: I 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 white lined paper provided to you. Work on one side of each sheet only, with only 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, is. o The coordinate system must be clearly identified. 0 Where appropriate, free body diagrams must be drawn. These should be drawn separately from the given figures. 9 Units must be clearly stated as part of the answer. a You must carefully delineate vector and scalar quantities. lf 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. lnstructor’s Name and Section: Section 1: J. Silvers 1:30 — 2:20 pm Section 2: J. Jones 9:30 — 10:20 am Problem 1 Problem 2 Problem 3 Total NAME: ME 270 -—- Spring 2012 PROBLEM 1 (20 points) -=- Prob. 1 questions are all or nothing. PROBLEM 1A. (5 points) FIND: The force vector of cable TAB = 220T+54.93 - 330EN. Determine the unit vector u0A . Calculate the magnitude of the projection of force TAB in the direction of bar 0A. “a re “:- i‘” a y Limp: ‘3’ “A. 2 W 2.: Q, “i” ‘ \rom W M t "‘ Tm ° Hop: 3(2-10? 4-54.3-330'12) (c.5553 +0.83; (<3 e a +* , 400% “1 0.55155; +0.83l? : 5 uOA (Znts) 144- Mew (in div-radian offloai'l-t in 01% ( 3 pts) E 1m: PROBLEM 1B. (5 points) FIND: The weight of cylinder C is 5OON. On the coordinate frame provided, draw the free body diagram of ring E. A Determine the magnitude of the weight of cylinder A (WA) ' to hold the assembly in the position shown. Determine the magnitude of the tension in cable TED. “I: W Eazc 3* “lab ‘3" 500 @3350" l ESE: 2 “WA +‘ 500 3‘56 Tab NAME: ME 270 — Spring 2012 PROBLEM 1C. (5 points) FIND: Cable AB is attached to the tree at point A and to the tractor at B and carries a tension of 2 kN. Write force vector TAB . Determine the moment vector about the base of the tree (point 0) due to cable TAB. (i.e., MO). mo .3 m, a?“ 201:] “0.4561?” H.3gf—l3 { » 1b PROBLEM 1D. (5 points) 5, 3 100 g 4 , , FIND: Replace the two forces and couple shown by an equivalent force and couple acting at base A. Express these in vector form. (Hint: This is not a static equilibrium problem). A 300 lb ft (flag 2 23:6 80? 4005' +522? 2: - 30? «(90%; lbs. * (I'm :zm : +aooE+ 80m;— (90(4)? fig s: 1.0;th— NAME: ME 270 — Spring 2012 PROBLEM 2. (20 points) GIVEN: A cylinder weighing 1200 N is hung from three cables as shown and is in static equilibrium. FIND: a) In the space provided, draw a free body diagram of attachment A. (4 pts) b) Write vector expressions for cables TAB, TAG, and TAD in terms of their unknown magnitudes, and their known unit vectors. (6 pts) c) Using your static equilibrium equations, determine the magnitude of tension in cables AB, AC and AD. (10 pts) ZF=6 ZFY‘O 4 1 OZTABfigTAC_§]—:1D 2w 3 2 OngAC_§TAD sto 2 0=§TAD—1200 31200 TAD: (2 ) Plug into y equation 3 2 O=gTAC 5 2 TAC = 2000 N Plug into x equation 4 1 0:TAB_§TAC ~§TAD TAB = §(2000)+%(1800) TAB = 2200 N NAME: NAME: ME 270 — Spring 2012 PROBLEM 3. (20 points) GIVEN: Boom AB is secured to the wall by a pin joint at point A, and supported by a cable at point B. The boom supports a vertical load of BOON at point D. The loads are attached by collars, which do not move. Neglect the weight of the boom. FIND: a) On the sketch provided below, draw a free body diagram of the boom. (4 pts) b) Determine the magnitude of the tension in the cable BC. (10 pts) c) Determine the magnitudes of the reactions at point A. (6 pts) b) ZMA =0, inIE 0=—E (1.500330)+§T(2.5$in30)+%T(2.5cos30) T: PI(1.500330) :(2.58in30)+%(2.500s30) 1' “l ‘3 x 0) 2F. =0 O=Ax—0.8T Ax =0.8(452) 1 AK A H 2F. =0 7 0=A —Fl+0.6T y A y (800) —0.6(452) A =529 N y ...
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