Exam_2_solution_tan_copy

Exam_2_solution_tan_copy - Check off your section number...

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Unformatted text preview: Check off your section number 9:05 —— 9:55 am El 1 —— Sahar / Dr. Masters [1 2 — Matt / Dr. Masters Cl 3 —-Ryan / Dr. Masters Cl 4 — Dominik / Dr. Masters E MCH 11 : Statics Score (out of 2001 11:15 am ~ 12:05 pm El 5 «Brian / Dr. Masters EXAM 2 El 6 ~Dominik/ Dr. Masters El 7 —— Fan / Dr. Masters 1:] 8 -— Chandan / Dr. Masters October 24, 2005 1:25 -— 2:15 pm INSTRUCTIONS: El 9 ~Matt/ Dr. Masters [1 10 —— Chandan / Dr. Masters El 11 -— Ryan / Dr. Masters m 1. Read the problems carefully D 12 _ Brian / DR Masters 2. Your exam should contain 2 problems on separate pages and two pages of multiple choice concept questions. If something is missing, notify a proctor immediately. 2:30 -— 3:20 pm El 13 —— Dr. Huang 3. A Free Body Diagram must be shown and used to support your work in each equilibrium problem. Each sketch must be clear and accurate. 4. Show ALL work in a neat and logical order. Partial credit will only be given if the work and reasoning are shown clearly and legibly. Begin each problem by clearly indicating what you are trying to FIND. 5. Neatness counts. Points will be deducted for sloppy work. 6. Place answers in the answer blocks provided. Be sure to label your answer and include units. (write F = 45 lb, not just the number 45 as the answer) 7. Individual calculators only. Sharing of calculators is prohibited. For questions on the grading, write a detailed explanation of your question below, including a description of how the work you originally submitted supports your question. Return this to your instructor no later than Friday, November 4, 2005 and we will review your exam. / law Problem 1 (80 points) The bracket shown supports a concentrated load F = 3.5 kN and a triangular distributed load w = 20 kN / m. Assuming the bracket remains in static equilibrium under these loads, calculate the magnitude of the force exerted by each of the pins A and B on the bracket. «fig/m: F: 3.5 k/V) 03:20 kN/m, FWD: Ml J. lBl + 3.5/w/120HS’0 0°30) + /.z kAl/So+/S’0w30>mm =0 /Z‘/‘i MM 5 NO M A“) N5 = (0.93? k“ 22?} =0 =/4x —/(,.939)s7;730 Ax“ 3-970 k“ = —. ? k/u 0.42-3.5 fi/ /30 IA) = 3.7/ bk, 32MB :0 z /,2(2o)+3.5(/20> +/~/.307)//S’0(6° 30) —- {33/70y/80 541, z - 0. 35 / 05,0) by preventing the plate and shafi from turning. The weight of the assembly is carried entirely by the thrust bearing at A. 4 / —O?"ZL 7.— '3/\ 1A ’7': =/ M ‘ be "+"/‘ DC b0 V‘lo . V73 V75 gfifio = —21 x/Bxi‘v‘Bt?) q- (+3r~5;)x/—3oj") L ,3 A :3»— A gal. +(af—7§)X( flak) +/ZOJ“ lé-H Problem 3 (40 points) 3.1 Which one of the following things can be changed without changing the equivalent force~ couple system at Q? A) The magnitude of couple C Bil/me vertical location, y, of couple C ‘\ WEI) The magnitude of force F D) The horizontal location, x, of force F 3.2 There are three unknowns in this FBD (forces FA, FB, P). If you were given values for all the other parameters (a, b, h, M, and 9), which one of the following equilibrium equations would produce the value of P with just a single equation? A)ZFX=O B)2F,=0 C)ZMA=O B) Any one of the above FB 3.3 Which of the following figures is the correct FBD of this concrete slab? C) 3.4 Which one of following does NOT represent a set of INDEPENDENT equilibrium equations for the figure in problem 3.3 above? A)ZFX=O,ZFy=O,ZMA=O B)2FX=O,ZFy=O,ZMB=O C)ZFX=O,ZFy=O,ZMC=O D)ZFX=O,ZMA=O,2MC=O E); Fy=O,ZMA=O,ZMC=O KM F) ZMA=O,ZMB=O,ZMC=O ...
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This note was uploaded on 10/15/2008 for the course EMCH 211 taught by Professor Ventsel during the Spring '06 term at Penn State.

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Exam_2_solution_tan_copy - Check off your section number...

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