NME (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: 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 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. Instructor s Name and Secton: Sectons: J. Slvers 8:30-9:30am B. Hylton 2:30-3:20pm J. Jones 11:30am-12:20pm J. Sepel 12:30-1:20pm M. Murphy 9:00-10:15am E. Nauman 9:30-10:20 am K. L 1:30-2:20pm J. Jones Dstance Learnng Problem 1 Problem 2 Problem 3 Total
NME (Last, Frst): PROBLEM 1 (20 ponts) Prob. 1 questons are all or nothng. 1(a) If the tenson n the cable s 400 N, determne the resultant force F actng on the pulley. Gve your answer n vector form. F = (2 ponts) In the followng dagram, the force vector F 1 les parallel to the x-y plane and s gven by F 1 = (- 100 120 j ) lb. Wrte down the poston vector ( r O ) for pont. Then, determne the moment of F 1 about pont O ( M ). O r O = (1 pont) M O = (2 ponts)
NME (Last, Frst): 1(b) Cable BC exerts a force F = ( 2.86 8.57 j + 4.28 k ) KN on the bar B at B. () Determne the unt vector u B that ponts from toward B. () Fnd the magntude F B of the projecton of F n the drecton of the unt vector u B. u B = (2 ponts) F B = (3 ponts) 1 (c) For the smply supported beam shown below, () Determne a sngle force equvalent F due to the dstrbuted load shown, () Fnd the locaton ( X ) of the equvalent force measured from pont. F = (2 ponts) X = (3 ponts)
NME (Last, Frst): 1 (d) Suppose that an area s defned as shown n the followng dagram between x = 0 and x = 2 ft. () Fnd the area of under the curve,. () Determne the centrod of the area n the x-drecton, X. = (2 ponts) X = (3 ponts)
NME (Last, Frst): PROBLEM 2. (20 ponts) GIVEN: hot ar balloon s suspended 3 feet above the ground, and s held n place by three cables, attached at rng. The balloon exerts an upward force on rng. Cable D runs parallel to the x-axs, and has a tenson of 84.0 lbs. FIND: a) On the axes provded, draw a free body dagram of rng. (4 pts) b) Wrte vector expressons for the forces n cables B and C n terms of ther unknown magntudes and ther known unt vectors. (6 pts) c) Usng your equatons of statc equlbrum, determne the magntude of the forces n cables B and C, and the magntude of the upward force from the balloon. (10 pts)
NME (Last, Frst):
NME (Last, Frst): PROBLEM 3. (20 ponts) GIVEN: Bar BC s loaded wth four forces as shown and s held n statc equlbrum by a fxed support at (Note Ths s not a pn jont at support.). FIND: a) On the artwork provded, complete the free body dagram. (4 pts) B 1 m C B C b) The two 200N forces shown form a force-couple. Determne the magntude and drecton of the force-couple n vector form. b) M C = 3 pts
NME (Last, Frst): c) Use your equatons of statc equlbrum to determne the magntudes of the reactons at the fxed support at. c) = x = y M = (3 pts) (3 pts) (5 pts) d) If the length of bar segment BC has shortened, what effect would ths have on the reactons at the fxed support at? (Increase, Decrease or Reman the Same) Crcle One (2 pts)
NME (Last, Frst): ME 270 Exam 1 Equatons Dstrbuted Loads F = eq eq 0 xf = L Centrods x = y = x = y = In 3D, w x dx 0 L xcd d y d c d x = x c c y x w x dx x V c V Centers of Mass x = y = x = y = xcmρd x ρd ycmρd ρd y cm ρ ρ cm ρ ρ
NME (Last, Frst): Solutons for Exam 1 1. F = 200 + 346.4j N O O M = 360-300j + 20k ft-lb r = 4 + 5j + 3k ft 1B. u B = 0.635 + 0.762j + 0.127k F B = -7.803 kn 1C. F = 4400N X = 3.45 m 1D. = 8 3 ft 2 X = 5 4 ft 4ˆ 3 T = T - j - kˆ 5 5 2. B B 2 ˆ 2 ˆ 1 ˆ T C = TC + j - k 3 3 3 2B. T C = 126 lb T B = 105 lb 2C. F B = 105 lb 3. FBD 3B. M C = 200k N-m 3C. X = -300 N 3D. Reman the same y = 350 N M = 37.5 N-m