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Unformatted text preview: PROBLEM 3.2 The force P is applied to the lever which controls the auger of a
snowblower. Determine the magnitude and the direction of the smallest
force P which has a 2.20— N m counterclockwise moment aboutA. — 561qu
a .
' .31 IE! lum For P to be a minimum, it must be perpendicular to the line joining points
A and B. '2 2
my: (86mm) +(122mm) 2149265111111 (I = ta 122 mm} = 54.819“ 86mm MA = VABP min 2.20 Nm [1000 mm]
149.265 mm 1 m
=14.7339N Pmin = 14.74N A 543° or Pmin =14.74 N 37 352° 4 PROBLEM 3.7 ._ A sign is suspended from two chains AE and BF. Knowing that the tension in BF is 45 lb, determine (a) the moment about A of the force
exert by the chain at B, (b) the smallest force applied at C which creates
the same moment about A. (0) Have
TB‘F: '15 IL {13) Have where MA = ram >< Tar Noting that the direction of the moment of each force component
about A is counterclockwise, 3"“ MA = XTBﬁ‘ + yTBFr
= (6.5 ﬂ)(451b}sin60° + (4.4 ft — 3.1&)(451b)cos60° = 282.56 [bit or MA = 233 lbﬁ ‘) 4 M 4 = r(‘‘A X (Ff) min For FC to be minimum. it must be perpendicular to the
linejoining points A and C. ' MA = d(F(‘l d = rm = (6.5 it)2 + (4.4 of : 7.3492 n 282.561bft = (7.3492 mm) min (1%)th : 35.999 lb ¢ = tan—{gi—E] = 34.0950 9 = 90“ e of; = 90° — 34.095° : 55905" or (mm = 36.0 lb 4: 55.904 PROBLEM 2.11 3 A rectangular plate is supported by three cables as shown. Knowing that
the tension in cable AC is 15 lb, determine the weight of the plate. The (vector) force in each cable can be written as the product of the
(scalar) force and the unit vector along the cable. That is, with E = (32 in.)i  (43 in.) j + (36 in.)k AB = (—32 in.)2 + (—43 in]: + (36 in.)2 = 68 in. E T . . . . .
TAB = n” = TME = 6814;]. [—(32 111.). ((48 m), + (36 m.)k] TAB : TAB(—0.4706i — 0.7059: + 0.5294k) C = (45 in.)i — (48 in.)j + (36 in.)k
AC = (45 in.)2 + (43 in.)2 + (36 in.)2 = 75 in. 7672; E _ 75 in[(45 in.)i — (48 in.) j + (36 in.}k] TAC = ”at = TAC TAC : TAC(0.60i — 0.64j + 0.48k) E = (25 in.)i — (4s in.)j — (36 in.)k AD 2 (25 in.)2 + (—48 in.)2 + (—36 in.)2 : 65 in. PROBLEM 2.113 CONTINUED AD T
T = T1 = T — = A”
’“3 "D ””3 AD 65 in. [(25 in.)i — (4s in.)j — (35 in.)k] TAD 2 TAD (0.3846i # 0.7385j — 0.5538k)
With W = Wj, out we have:
EF=O: TAB+TAC+TAD+Wj20 Equating the factors of i, j, and k to zero, we obtain the linear algebraic
equations: i: 4.4706135 + 0.6M“ — 0.38462“) = o (1)
j: 4.70591” — 0.642;“T 7 0.738511”) + W = o (2)
k: 052947;“, + 0.487“ m 0.5538150 = o (3) In Equations (1), (2) and (3), set TAC = 15 lb, and, using conventional methods for solving Linear Algebraic Equations (MATLAB or Maple,
for example), we obtain: TAB =136.01b
TAD : 143.0 lb W=2111b4 PROBLEM 2.1 30 A container of weight W is suspended ﬁ'om ring A, to which cables AC
and AE are attached. A force P is applied to the end F of a third cable
which passes over a pulley at B and through ring A and which is attached
to a support at D. Knowing that W = 1000 N, determine the magnitude
of P. (Him: The tension is the same in all portions of cable FBAD.) SOLUTION The (vector) force in each cable can be written as the product of the (scalar) force and the unit vector along
the cable. That is, with AB = —(0.78 m)i + (1.6 m)j + (0 m)k
AB = (—0.78111)2 + (1.6111)2 + (0)2 x 1.78m
E T . .
TAB = T143 = 1",“,E = ﬁ£—(0.78m)1+(i.6m)1+(0m)k] TAB : 11430043321 + 0.8989j + 0k) and
E = (0)1 +(1.6m)j +(1.2m)k
Ac = ,1(0m)3 + (1.6m)2 +(1.2 m)2 = 2m
K: 1" . . .
LC = T91.“ = ACE : ﬁlmll + (l.6m)]+ (1.2m)k]
T“. = TAC(0.8j + 0.6k)
and AD = (1.3 m)i + (1.5 m)j + (0.4 m)k 2 AD : (1.3111)2 + (1.6m)2 + (0.4 m) = 2.1m 11m = mm : an???) : if?“ (1.3 m)1 + (1.6 m)j + (0.4 m)k] TAD : TAD (0.61901 + 0.7619j + 0.1905k) PROBLEM 2.130 CONTINUED Finally,
XE : —(0.4 m)i + (1.6 m)j — (0.86m)k
AE = ‘i‘(—0.4m)2 + (1.6 m)2 + (—0.86111)2 = 1.86m
E T
.21“ :T—= A5 —.4' . °—.
TA) 1,15 )5 AE 1.86m[ (11 m)1+(16m)] (0 86m)k]
T“: = TAE(—O.2151i+ 0.8602j — 0.4624k)
With the weight of the container W = — J, at A we have: EF=0: TA3+TAC +TAD—Wj=0
Equating the factors of i, j, and k to zero, we obtain the following linear algebraic equations: 414382ng + 0.6190153 — 0.2151ng = 0 (1) 0.89892”, + 0.8% + 0.76191“) + (1.860er5 — W = 0 (2)
0.61:“. + 0.1905er , 0.4624245 = o (3) Knowing that W = IOOON and that because of the pulley system at E TAB : TAD : P,_ where P is the externally applied (unknown) force, we can solve the system of linear equations (1), (2) and (3) uniquely
for P. P=378N4 450 mm PROBLEM 3.6 j ‘ It is known that a vertical force of 800 N is required to remove the nail at
C from the board. As the nail ﬁrst starts moving, determine (a) the
moment about B of the force exerted on the nail, (b) the magnitude of the
force P which creates the same moment about B if a = 10°, (c) the
smallest force P which creates the same moment about B. (0) Have MB : rcmEu : (0.1m)(800N)
= 80.0 Nm or M8 = 80.0Nm)4 M3 = r/MPsinﬂ
6 = 90° — (90° k 70“) —a
: 90° — 20° — 10°
= 60°
80.0 Nm : (0.45 m)Psin 60°
P = 205.28N
or P = 205 N 4 (c) For P to be minimum. it must be perpendicular to the line joining
points A and B. Thus. P must be directed as shown. Thus M3 = dPrnin = rAfBPmiII
or 30.0 Nm : (0.45 m)P min Pmin =177.778 N or Pm :177.8N 4 20M _i PROBLEM 3.22 Before a telephone cable is strung, rope BAC is tied to a stake at B and is
passed over a pulley at A. Knowing that portion AC of the rope lies in a
plane parallel to the xy plane and that the tension T in the rope is 124 N,
determine the moment about 0 of the resultant force exerted on the
pulley by the rope. SOLUTION H Have M0 = Em x R [m
10" i where l'Axo =. (9 m)j+(1m)k
_ _ 5._—\
T1 / R=T1+2#2‘Bm'1
T] = {(124 N)coslo°]i — [(124 N)sin10°]j
2 4122.116 N)i u (21.532 NH 0 (1.5m) +(9m) +(1.3m) Xx =(20N)i—(120N)j+(24N)k V31»
2 [.5111 B 2 (124 N) . R = —(102.116N)i — (141.532 N)j + (24N)k
i j 1: —102.116 —l41.532 24 = (357.523 Nm)i — (102.116 Nm}j + (919.044 Nm)k or M0 = (358Nm)i — (102.1Nm)j + (919Nm)k < ...
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 Fall '08
 C.Trujilliano
 Statics

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