This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: EGM 2511 Exam 1 Spring 2009 NAME (Last, First) Section: 1576 Or 1578 or 6583 (circle one)
SHOW YOUR WORK  BOX 0R UNDERLINE ANSWERS — FBDs REQUIRED ' Problem 1. (33 pts) Three cables are used to tether a balloon as shown. The balloon exerts an upward
vertical force P on point A where the cables are tied. The tension force in cable is knoWn to be 259 N. a) Write unit Vectors RAB, RAG, and LAD in the AB, AC and AD directions. (fractions OK)
b) Find direction cosines and direction angles for the line AC. _
c) Calculate the tension force in Cable AC required to keep the system equilibritim. (1) Suppose it is also known that the vertical component of the force in Cable AD is exactly — 461.49 N.
What is the force P caused by the balloon? 5.6 m 5.) FM M) ayséw grit.2 .azérezrx {it '
Q :2 Mstfaijéjm” cps e3: :: Jﬂm 93—. ‘75"? _ .9 . 
1: "t "a glﬁ a . . O
Cos 9% x $13.. = . $t75é7’x ﬁts
Q. 5‘ C55“ .ﬁé z: “+0 A
f a...) 55 gazezgﬂﬁﬁégﬁf'qms wax.15 cs ‘fjiN. A) $223 :2 P— may egzzmagémmﬁe Pmlﬂ’gl‘z7 «v APB/RM m EGM 2511 Exam 1 Spring 2009 NAME (Last, First) Section: 1576 or 1578 or 6583 (circle one) ‘ ' i 0
SHOW YOUR WORK — BOX OR UNDERLINE ANSWERS—FBDsREQ V I' ' " '  Problem 2. (33 pts) The cable AB has tension equal to 540 N. Considering only the force acting at A,
a) Find the equivalent forcecouple system that would be acting at point C (give answers in
standard vector form). As part of your answer, draw your forcecouple system as vectors on the
original ﬁgure. b) Find the magnitude and sign of the component of the moment caused by the force at A about
the line from D to C. ' y ‘ a? Ab: aha/q, 4w?) x43: F :21 6W 1.4124 mag. m2?
AB 1%; W rat; ' :':'— H126 322+ 285’.é7"3f— 432.7é6k \ ﬁg: «PM; A
z ..... 2 g k
63‘ \/ I 4412.22 283% “9132,? 3 . ~\
A76 =5 — memos; 4az7e~933430mw
4 .
520: —étqz,., Maegan: 2.322: kNm 53 SE : 4055217”? 3 be: 2.5
ﬁéc‘: — "° Mm = in» 57¢ : amass as 4.9;: mm EGM 2511 Exam 1 Spring 2009 NAME (Last, First) Section: 1576 or 1578 or 6583 (circle one)
SHOW YOUR WORK — BOX 0R UNDERLINE ANSWERS — FBDs REQUIRE F R E l g S
UIL ‘ P OBLEMS '
Problem 3. (33 pts) The lightweight door shown below is suppo ed by a s1ngl:rhz::geé;1¥he hing :
has broken and is no longer connected) and the doorstop DE. Doorstop DE is parallel to the yz plane, has
negligible weight, and can be considered to have pinned or ball—and—socket connections at D and E.
Neglecting the weight of the door and knowing that the applied force vector at C is F =  20i — 25j  45 k lb,
a) Find the support reaction force at E, and ‘ f b) Find the ﬁve support reaction components at
(Answers to a) and b) will be in scalar form — do not give vectors as answers). (Hint: For DE, think 3—45) y a.) ngj =+ esgaépgpgzzow ) we 714115 ca ﬂit!
to.) 25¢: Awsz ,3 Aweigh __ .2 Mg’ =9 + 5x gmamlé’) ta + 45 (/0>== 0) a» : M55331: leggy
2F «Ev/)3 ‘— 25 +gig7q.4I/£):0) A33133527g: gig}.
I 2 M23 5‘ "t 52 “? iﬂowig) 32—} a. 25 engage) =0
. La» coat—.m ~ ﬁt; @410? 45 rgtgwyczwgﬁp 9 AW gym a; @in EGM 2511 Exam 1 Spring 2009 NAME (Last, First) Section: 1576 or 1578 or 6583 (circle one)
SHOW YOUR WORK — BOX OR UNDERLINE ANSWERS ~ FBDs REQUIRE F R E q: I:  A / Mme: WWW  ._ ‘ f. Us Problem 3. (33 pts) The lightweight door shown below is suppo‘ ed by a single A , : 
has broken and is no longer connected) and the doorstop DE. Doorstop DE is parallel to the yz plane5 has negligible weight, and can be considered to have pinned or ball—and—sOcket connections at D and V
Neglecting the weight of the door and knowing that the applied force Vector at C is.F = — 201 — 25 lb, a) Find the support reaction force at E, and b) Find the ﬁve support reaction components at '
(Answers to a) and b) will be in scalar form — do not give vectors as answers). (Hint: For DE, 3~45) 51.) amps :«17’5[3é)+5§p£:[3ff)=0 ) DE: 77.4% 4:: Lip,
inFf/iﬁ’wiﬂgszéﬁéﬁ' . H
ng' = a 5x + gimmwsﬂo ’ eta/aka) ax: M55382; Mil: m
2F :: A3 —~ 25 +%(7q,41/3):0) A3: #3335qu ,ﬂéjb.
2 w = » C; + iﬂwwﬁt} ﬂ 2513Mw§200®=0
5 ngmw :: Apﬁgk ‘
in; Engrt 45 @{rnwﬁme 3 Ag: 25w; «x w ' any“ ...
View
Full
Document
This note was uploaded on 09/06/2011 for the course EGM 2511 taught by Professor Jenkins during the Fall '08 term at University of Florida.
 Fall '08
 Jenkins
 Statics

Click to edit the document details