engr200-t1-f2009-answers

engr200-t1-f2009-answers - ENGR 200 TEST 1 Spring 2009 Name...

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Unformatted text preview: ENGR 200 TEST 1 Spring 2009 Name 1) Determine the smallest force F1 such that the resultant of the three forces, F1, F2, and F 3 is vertical. Also find the angle Ct at which F1 should be applied. (From Engineering Mechanics Statics, by Michael Plesha, Gary Gray and Francesco Constanzo, Mc Graw Hill, 2010) F; = 200113 . £5 {74 00 :“L- r ~~“ M 2 Caqu mac» (4)550”, ~ (F, Sir/xx F6“ MMWUM F; 3/1494 2:!” :7 pygmy E : when eels? : 5+???“ ficolellj QuT c613 /0" Cat/Wavi- .[ .Mffiflhc‘) own?" "i 0 our (90° (tut/00119 qfls m yrceaevrrmj cg‘chu-zfi) K2. fF‘F :0 eK. n0 Gin&(-L 011 cm“); may“ dew“ A E, :0 K {\luT a. Kaouweh’tevd” c519- lazuli/EM 59 Zoo H¢>) _ 9 c [M ‘Z Byuuvlfi (I? D‘s (go “‘5’, “WWW” tin 0%“ 6 AW“ til—122+ "‘1 n4qu Euwuv “5/7 —- Wi/LUH? *' (5)“) ’3'th (/10 (Jecg tuth YGU 0?»? 0V 'yio C(Q3bC3/\ (SPEC/>17”) “HEW me ENGR 200 Ryan Franz FALL 2009 EXAM 1 - PROBLEM 2 SOLUTION The systems shown are all free body diagrams so no more drawing is needed. A coordinate system should be drawn. This solution assumes a right handed coordinate system with counter-clockwise rotations positive. It should be noted that all systems have the same dimensions. The most common point to take moments about was the bottom right corner of the structure. Full credit is still available if another point was chosen and pointed out in the problem. SYSTEM-A l i | SYSTEM D SUM Fx (lbs) 300 — 100 ‘ 100 + 100 = 200 i = 200 SUM Fy (lbs) 100 — 100 o o 100 — 100 = i=0 SUM Ma (ft—lbs) -(1oo)(4) + (100) (2) ‘ ‘ (10o)(4) — (100)(6) = -200 II I N O O For two systems to be statically equivalent, the sum of the force and moment vectors of system 1 must equal those of system 2. Therefore: system A is staticall (equivalent to System D. a" w»: a: a ENGR 200 FALL 2009 EXAM 1 - PROBLEM 2 GRADING SCALE The following eight items were scored on a scale of O — 1 points: The following four items were given 0.5 points if present: 1) 2) Z17X,SYSTEMA ZFX 2F X,SYSTEM C 2F X,SYSTEMD ,SYSTEMB 2M SYSTEM A 2M 2M SYSTEMB SYSTEM C 2M SYSTEM D Ryan Franz The student received 0.5 points if the equation was set up properly and 0.5 points if equation was evaluated correctly. Ifthe equation was set up improperly due to incorrect use of negative signs, but would have been correct if the negative signs were fixed, the student received 0.5 points. 2F? = 21% 2M1 = 2M2 Answer 1: System A = System D Answer 2: System B = System C Units were not considered for points, since the “answer” did not involve units. The sum of the forces of the y-direction was considered trivial and not included in the grade. Points were not deducted for not drawing a coordinate system or indicating where moments were being taken about if this information could be inferred from the student’s solution. Some observations on how scores were determined: 1) If the student thought statically equivalent meant "in equilibrium” but still did all the work, a total of 8 points was available. 2) If the student thought statically equivalent only involved forces/moments and did not analyze the systems for moments/forces, 5 points were available. 3) If the student only did the work for A+D or B+C and showed no work for the other system, 5 points were available. 4) lf the student somehow did all the work, correctly identified the meaning of static equivalence, and still only managed to see that one pair was statically equivalent, 9.5 points were available. 5) If the student wrote ”B=C" and absolutely nothing else, they received 3 points as a “compromise” between the 0.5 points they probably should have received and the 5.5 points they could argue for. Justification: Emphasis was placed on the student showing work. The student was tested indirectly on their ability to determine a moment from a force and perpendicular distance, their ability to identify a couple, and their ability to sum forces and moments. If a student did halfthe problem, they earned half the points. Showing all work and misinterpreting the results earned the student 8 points. The problem cannot be identified as a "trick problem with two solutions” as showing ALL work and only identifying one answer was worth 9.5 points. ENGR 200 TEST 1 Spring 2009 Name 3) Determine the components of a unit vector normal to plane ABC. (Modified from: Engineering Mechanics Statics, by Michael Plesha, Gary Gray and Francesco Constanzo, Mc Graw Hill, 2010) (‘l A}? my 12-0 : ifs/>57”; +7 {My ~- We {.rgfi c» (.2 (ii/x1} 32 v 3A g2» ...
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This note was uploaded on 09/18/2011 for the course ENGR 200 taught by Professor Mullen during the Spring '08 term at Case Western.

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engr200-t1-f2009-answers - ENGR 200 TEST 1 Spring 2009 Name...

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