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: PROBLEM 2.70  A 350“: load is wpponed by the repemdpulley arrangement show.
Knowing [hat a = 35”; determine {a} the angle ﬁt (1:) the magnitude of
the force P which should be exerted on the free end of the rope to
maintain equilibrium. (Him: The tension in the rope is the same on each
side of a simple pulley. This can be proved by the methods of Chapter 4.} s
P . .
3‘ J
i A "
a: @
i
3501b
SOLUTION
FM'BW ““5"“ 1’ “"93 A _+ EFI = o: ZPsinﬁ — Pcos25° = 0
Hence:
(a) sinﬂ = £93525" or ,6 = 24.2“!
(b) +1 25,20: 2Pcosﬁ+Psin35°+350 lb = 0
Hence: 2Pc0524.2° + Psin35°  350 lb 2 O
or P = 145.97 lb P = 146.0 lb 4 PROBLEM 2.111 A trmmlission tower is held by three guy wires attached to a pin an! and
anchored by bolts at B. C, and D. If the tension in wire AB is 3.6 kN.
determine the vertical force P exerted by the tower on the pin at A. The force in each cable can be Written as the product of the magnitude of
the force and the unit vector along the cable. That is, with A .= (13m}i  (30111)] + (54:11)]; .. AC = (13 m]’ + Hem]: + (5.4m}1 = 35.4m
E T _ _
TK = n.“ = FACE = ﬁﬁls m):  {30 111)] + (5.4 m_}k] TAU = adumsﬁ _ 0,34751+ 0.1525k) and ﬁ=—(ﬁm)i(3um}j+(7.sm}u AB = (—6m):+{—30 m): +{7.5m): = 3LSm AB T . .
T” 2 T1.” = THE = ﬁ[—(6m)l — [30 111)] + (75 Ink] 1“ = Tﬂ[—{}.1905i— 0.95245 + 023mm N Finally E = —{5m]i — (30111))"(222111JR
.41) = J{5 131]: + [—30 m)2 + [—22.2 m):  313 m
Tm = HAD = ring : _..._.THD [[5m)i —{30m]j — [22.2 m)k] AD 37.8 m
Tm = Tm(0.158?i  0.79373' — 0.581%.)
With P = P]. 31A:
£1720: 'l"m +TAC +TAD+Pj =0 13qu the factors of L j, and II to zero, we obtain the linear algebraic
equations:  ' i: 4.1905235 + 0.503er — 0.1537er = o (1) j: 4.952414, .7 0.34751}:  0.19371'“ + P = 0 (2]
In Equations (1). (2} and (3}. set T” = 3.6 kN. and. using conventional methods for solving Linear Algebraic Equations (MATLAB 01' Maple.
for example). we obtain: TE : 1.963 m Tm = 1.959 W r=6.66kNT4 PROBLEM 3.21 Before 1112 trunk of a large tree is felled. cables AB and BC are attached as
shown. Knowing d1at.thc tension in cables AB and BC are TI? N and
990 N, nespectiwiy. determine the moment about 0 of the resultant force
exerted on the tree by the cables at B. SOLUTION
Have Mu = ram} x F3
where r50: [3 4 111”
F3 7' T43 + T3:
 0.9 '— 3.4 ' 7.2
Tm =kmna = ( “1,: ( Thq , “mm? H}
[09]" + (8.4)” + {7.2)‘ m
5.1 ' .4 ' .
TM : 1N7“: WW—WWEW N ,
J[5.1)' + [8.4j‘ +[121‘ In
F:[ [63.[0Nli saw}; pawns] [{SION}i[S40N}j+[120N}k]
= [447N)i 41423 N]j+[624 Njk
i j k
and Mo: 0 3‘4 0 Nm=[5241.6Nm]l{3?54.SN»m]k
44? 4423 524 or Me = (524mm): (3.75kNm)k 1 PROBLEM 3.49 ' To liﬂ a heavy crate. a man uses a block and tackle attached to the bottom
ofan Ibeam. at hka. Knowingthat the moments about theyami : axes
of the force exerted at B by portion A3 of the rope axe. respectively,
100 lbﬂ and 400 Ibﬂ , dctm'mine the distance a. SOLUTION
Bsscdon M0 = 11m x T“ where Mo = M1i+ Myj + M;
= MJ1+(100 [bﬂ}j—(400ib~ﬁ]k w = Eﬁﬁii+(4ﬁ}i TM 2 Marty
(6ft]i — {12 an  (0)1:
= —d—T£‘
821
i j k
MJ+100j400k= 6 4 a Si
6 —12 —a ”A
= iH‘iay + {60)} — {96M
dim '
. . . mo
From jcoeﬂimem. lDUdm = 607',“ or TM 2 6—d“ {1)
a
. 400
From kcoefﬁcnmt: 4100:!” = —96TRA or TM = Eds“ [2}
Eqmting Equations (Hand [2) yields a = 10096}
6(400) or 0:4.00114 ...
View
Full Document
 Fall '09
 K

Click to edit the document details