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Unformatted text preview: Problem I (25 Points) Given two vectors, A (1, 2, “2), and B (6, 3, 2), ﬁnd: a) A unit vector paraiiei to B (7)
b) The angle between A and B (10)
c) llprojABll (8) Solution: a)B= [6,3,2], 1IB=./(62+32+22):JE§=7 2m 6, 3, 2 , ,
Unit vector paraileitoB= [ 7 ], or [%!+—:J+%k] 5 pts b) To ﬁnd the angle betwen A and B, we evaiuate AB
AB : llA'!.llBl].cos (8), 6 is the angle between A and B 2 pt A.B= (1x6+(2)x3 +(~2)x2)=w4 3pts
llAll=x/(12+(m2)2+(2)2 =~f§=3 23:
or we can com ute'é m[~]—i—E'—g~k]
’ p A 3 3j 3 ’
cosB=[§i+§~j+i2—k].[ling—g~k]m 2—3—1 :~.19
7 7 7 3 3 3 7 7 21
—4 —4
e :__..=__=w,19 2 t
COS” 7x3 21 p5
, 6 ""—“ 300.930 lpt,0 ifacute angle
. AB A "41—2—2 [4,8,8]
ww _ : ___ :0 7 2 5 5 t
°“’“°“ nAu liAlt Jim] 9 p5 llprojAB” =% (“4): + 82 + 82 )m g, or 1.33 3 pts Problem 2 (25 Points) As Shown in the: ﬁgure at {he riggmn block A. is
supported by two cables which are securely
attached at ring P. One ()FLhBSC cables is 6 ft in
length and has its other end fastened main31y in
tlw Wall 211 (I). This other cable: passes over a
iiictionlﬁss pulley and is then connected to block
8. The weight of block A is W4 2 SOIbS,
distance y = 3 ft. and the system is in static
equilibrium.
3) Draw the free body diagrams; needed to solve
this problem. (5 points)
b) Determ'im: the weight. ofblock B necessary to
maintain the equilibrium gecmieiry Shawn. (10 points)
9) Determine the tension in cable OF. (H) pointS)
F 80‘ ' A F” D 8
Fr g A 'l v‘wT‘E; i 13 2 ‘ I’I‘JWI’S
J ('35 1' you! we 5130»: THE' gsoavuiji'ﬁy m? "NH? Sysrgm 32165749 W €9*$f'vmr(y2) :30" ' in“: li*'Wcs‘D C v 11 u Kim) TAM ‘P = {@3ch W (,9 :: Flaw; mapn xii? :; (~ gaggyf+ SIM3OJ;}‘1F;(:VSZ3.BDI' + gawzzia‘ﬁ) 4‘03 = 35 lbi®
09’s" . 3 29?: ~ 0.9mm. +0.‘?JSF2 ‘: 0 Hal» ;, W "37865 F}? L331“? u“@
#3: 0:500'711 0403:er ~§o 0 [H mm gags‘rn‘I/ITJNG (we) mm {W} WELD;
F. , '" z. C“
0.5‘U05vﬁj + O'VIOEVFR “g0 “‘7' 0’97“? F2 “’0 I!
C 19;? c.) bMEEETlTVIT'N£‘ Fix R.er wru (1} Wifij a: ,2“? p; : (Loggﬂs’simi) : soc1y
Page2of4 Problem 3 {25 Paints) Use the (Pause—Jordan elimination method to solve the ‘ibllewing system. of Einear equations:
)51 + 8x3 '2 1 x] +2x2 +3x3 x 2
2x, +5x2 +3x3 m 3 :1.) Write the augmented matrix (4 peints) b) Use elementary row Operations Ee obtain {he reduced r0w~eehefen form of the augmented
matrix (E8 pein‘ts). 0) Write the seiutien ferthe three unknowns (3 points) Note: No credit will be given for Paris (b). and (C) unless ALL intermediate work is shown For Part {b} I Augmented Matrix Action taken hem Jreviuus matrix 151 rew remains unchanged
0 2 ~— 5 i Add negative of 1“ row to 2.“(1 row
0 5 "#13 I Multipty J. by 1St row and add to 3rd row _ ISl rew remains unchanged 1
0 I ~% Muitiply 2'“1 row by 1/2
h 1‘ lMumpy '39 by 2nd row and add to 3“1 row 1”" row remains unchanged
0 l 0 8 Muitipiy 2'“1 row by 1/2
0 0 1 3 Muftipy $512 by 2"d mw and add to 3” row Solution: x; :23, x1 "2 8, x e e CD Page ‘E 0'!" 1 Problem 4 (25 points) Given:
The tension in cable AB is 777 N and in cable
BC is 990 N.
Deﬁne the x—y—z coordinate system to be
centered at point 0 as shown, with associated
unit vectors (i, j, k)
Detezmine:
it. the force vector acting on point B by
cable BC“ {9 points)
2.. the force vector acting on point B by I _ H
cable AB” {9 points) ‘ 3 the vector resultant ofthe forces exerted
at point B by the two cables (AB and
CB) ('7 points)
Give all answers using the x»yz coordinate
system and associated unit vectors, (i, j, k),
being careful to supply all physical units
when applicabic‘ )m m A 5: Ed”
O’?¢”’g¢€&c is W " (WWW 62X 231. 3’.qu s. [22.14 (m)
W a; mam» (mew (A221 0’” M u 9.?m A
,0.?3 ,. 3.4%721: (m)
. ' A far‘5 '“" “tr e :2? 342:4 (WW) \
@é‘q" 5’4 6A 5A 2/ ‘ im+ﬂf.¢)z¥{72) 5’“
m W‘WW
IL  W4
"h ...t " » A. .2342 (“M _
m:— M at 4:0 fg‘y)’c7z. (/Zﬂaitﬁﬂ04); WE" s (ﬂaw): vii/‘5’ "' j Page 4 GM ...
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 Fall '09
 SPILKER

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