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Unformatted text preview: PROBLEM 3.142 A worker tries to move a rock by applying a 360—N force to a steel bar as
shown. (a) Replace that force with an equivalent forcecouple system at
D. (b) Two workers attempt to move the same rock by applying a vertical
force at A and another force at D. Determine these two forces if they are
to be equivalent to the single force of pan a. SOLUTION (:1) Have Have where (b) Have where 2F: M: II 360 N(—sin40°i — cos40“j) = —(231.40 N)i — (275.78 N)j = F
or F :360N 7 50°1
2M0: rB/D x R = M
raw : {(0.65 m)cos30°]i + [(0.65 m)sin30°]j
= —(0.56292 m)i + (0.32500 m) j
i j k
—0.56292 0.32500 0 Nm = [(155240 + 75.206)Nm]k
—231.40 —275.78 0
(230.45 N~m)k or M = 230 Nm ) 4 2MB: M = MD x Fr1 . rm = {(1.05 m)cos30°]i + [(1.05 m)sin30°] j = —(0.90933 m)i + (0.52500 m) j PROBLEM 3.142 CONTINUED i ' k
FA —0.90933 0.53500 0 Nm = [230.45 Nm]k
0 —1 0
or (0.90933FA)k = 230.45]:
. FA = 253.42N or FA = 253N 14
Have 2F: F = FA + FD —(231.40 N)i — (275.73 N)j = —(253.42 N)j + FD(—cosd9i — sin6j)
From i: 231.4ON = F300519 (1)
j: 22.36 N = FDsinB (2)
Equation (2} divided by Equation ( 1)
me = 0.096629
6 = 5.51930 or 6 = 5.52”
Substitution into Equation (1) FD = ﬂ = 232.48 N
cosS.5193° or F)D = 232 N 7 5.52°4 PROBLEM 3.152 To loosen a frozen valve, a force F of magnitude 70 lb is applied to the
handle of the valve. Knowing that 6 = 25°, Mx = —61 lbﬁ, and M: = —43 lbft, determine (9 and d. SOLUTION
Have 2M0: 134/0 x F = M0
where rm = —(4 in.)i + (11 in.)j — (d)k F = F(c056005¢i — sinBj + costin¢k)
For F = 70 lb, 9 : 25° F = (701b)J:(0.90631cos¢)i — 0.42262j + (0.90631sin¢)k]
i j k
M0 = (7011)) 4 11 —d in.
—0.9063ICOS¢ —0.42262 0.9063lsin¢
: (70 1b)[(9.9694sin¢ — 0.42262d)i + (ﬁ0.90631dcos¢ + 3.62525in¢) j
+ (1.69048 __ 9.9694cos¢)k]in. and MA = (701b)(9.9694sin¢ — 0.422625!) in. = —(61 lb’ft)(12in./ft) (1) M). = (70 1b)( ~0.90631dcos¢v + 3.6252sin¢$)in. (2) M: : (701b)(£.69048 — 9.9694cos¢) in. = —43lbﬁ(12in./ft) (3) PROBLEM 3.152 CONTINUED From Equation (3) ¢ = cos'[634'33 = 24.636°
697.86
or ((5 = 24.6° 4
From Equation (1)
d = [102290 2 34.577 in.
29.583 or d = 34.6 in. 4 _[ 15 in. PROBLEM 4.34 Neglecting friction and the radius of the pulley. determine (a) the
tension in cable ABD, (b) the reaction at C. SOLUTION A 7/;‘7jﬂ 30 lb 0'
00.4. 5 First note 15
= tan" — = 221520"
a [36] ,8 = tan"[£] = 36.870"
20 (a) From f.h.d. of member ABC
+‘) EMC = 0: {3010)(23 in.) — [Tsin 22.620°)(36 in.)
—(Tsin36.870°](20 in.) = 0
T = 32.500 lb or T = 32.51bd
(b) From f.b.d. of member ABC —+ 2F, = 0: C, + [32.500 lb}(00522.620° + cos 36.870°] = 0
. c, = —56.000 lb
or C, = 56.0001b —~
+1 2F), = 0: C), — 30 lb + (32.500 Ib)(sia122.620° + sin36.8'.'0°) = 0 c} = —2.0{)01lb
or Cy = 2.0001 lb I
Then c z W = (56.0); +(2.001)2 = 56.036 lb
and 9 = tan{3] = tan'] [in—J = 2.0454o 0; — 6.0 or C = 56.0 lb 7 2.DS°‘ PROBLEM 4.45
E_ ,_ *1 1m.— _,. 4.," A 2011) weight can be supported in the three different ways shown
Lz" #5 t—Séi: Encoaxglinfaiat the pulleys have a 4' 1n. radius, determine the reaction at A
SOLUTION
(a) (a) From f.b.d. 0MB
A:€AJ LEI1:0: A_,=0
i: +323..=0: AJ.—2010=0
’5 b“; or A). = 20.0 lb
and A = 2001014
+‘j EMA = 0: M1 —(2010)(1.5 ﬁ) = 0
MA. = 3001011
or M,‘ = 30.010034
2011: )Note: 4 mtgn'] = 0.33333 0
From £b.d. of AB
116?; 1.2550: A_.—201Ia=0
+7.b cl 20 lb or AI = 20.0 lb .—f 25‘. :0: AJ. —201b =0 or A}. = 20.0 lb Then A = 213+ Af. = {20.0)2 +(:10.0}2 = 28.284 lb A = 28.3 lb 4: 45°41
3: 2M,l = 0: MA, +(2010)(0.33333 a)
~(2010)(1.5 0 + 0.33333 11): 0
M_., = 300100
or M4 = 30.010034 ——————_—_—__—— PROBLEM 4.45 CONTINUED (5) From f.b.d. of A3
+ _ . _
(C) —a 2F; — 0. Ar  0
qua. =0: AI‘,—2(}lb—201b=0 or A}. = 40.011: and A = 40011:]!
+‘j 3M1 = 0: MA —[201b)(l.5 a — 0.33333 3)
—{2otb)(1.5 n + 0.33333 ﬂ) = 0
. M,' = 60.0]bﬁ L or MA : 60.01bﬁ )4 PROBLEM 4.54 A slender rod A3. of weight W. is attached to blocks A and B. which
move freely in the guides Show. The blocks are connected by an elasLic
cord which passes over a pulley at C‘ (a) Express the tension in the cord
in terms of W and 9. (2:) Determine the value of 8 for which the tension
in the cord is equal to 3 W. SOLUTION
5 (a) From f.b.d. of mdAB +3 EMC = 0: Home) + W[[%]cosa] * meosa) = o : Woosﬂ
2(co59 — sine?) Dividing both numerator and denominator by cos 9, “it I )
l—tanﬂ {b} For T = 3W, awe—g]— (1 —tan0) Itanﬂzl
6 9 = tan—{g} = 39.806“ or 9 = 39.3“‘ ...
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 Fall '09
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