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Unformatted text preview: SOLUTEON PROBLEM 3.24 A wooden board AB, which is used as a temporary prop to support a
small roof, exerts at point A of the roof 3 228 N force directed along BA.
Determine the moment about C of that force. MC = rAFC 3‘ FHA
”I = (0.96 m)i  (0.12 m)j + (0.72 m}k F321 : 132F321 —(0.1 m)i + (1.8 m)j — (0.6 mm:
({(0.1)2 + (1.8)2 +(0.6)2 m : “(12.0 N)i + (216 N)j , (7? N)“ (228 N} i j k
‘. MC = 0.96 —U.12 0.72 Nm
—12.0 216 —72 = —(146.38 Nm)i + (60.480 Nm)j + (205.92 Nm)k or MC 2 —(146.9 Nm)1 + (60.5 Nm)j + (206 N»m)k 4 PROBLEM 3.91 Two workers use blocks and tackles attached to the bottom of an Ibeam
to liﬁ a large cylindrical tank. Knowing that the tension in rope AB is
324 N, replace the force exerted at A by rope A3 with an equivalent
forcecouple system at E. SOLUTION
Have 2F: Tm = F where TAB = 1HT” (0.75 m)i — (6.0 111)] + (3.0 m)k( 324 N}
6.75 m TM, = 36 N(i 7 si + 4k) so that F = (36.0 N)i—(288 N)j+(144.0N)k 4 Have 2M5: ”EXT” =1“
1 j It
or (7.5m)(36N)0 1 0:M
1 —8 4 M = (270Nm)(4i — k) or M = (1030 Nm)i — (270 Nm)k 4 PROBLEM 3.28 In Problem 3.21, determine the perpendicular distance from point 0 to
cable BC. Problem 3.2]: Before the trunk of a large tree is felled, cables AB and
BC are attached as shown. Knowing that the tension in cables AB and BC are 777 N and 990 N, respectively. determine the moment about 0 of the
resultant force exerted on the tree by the cables at B. Have lMo l = reed where d = perpendicular distance from 0 to line BC. M0 = r310 x TBC ram : 8.4mj
TM.  lacTnc 7 (5.1111): — (8.4111); + (1.2 m)k (990 N)
2 2 1
Jon) +(3.4) +0.2) m
: (510 N)i — (840 N)j + (120 N)k
i j k
M0 = o 8.4 0 =(1008Nm)i—(4284Nm)k
510 —840 120 2 2
and mo]: Janos) +(4234) = 4401.0 Nm 4401.0Nm = (990 N)a’ d : 4.4454 In or d =4.45m1 PROBLEM 3.73 Knowing that P = 0, replace the two remaining couples with a single
equivalent couple, specifying its magnitude and the direction of its axis. M=M1+M2 where M] = rm, x Pm
r03 2 (0.96 m)i v (0.40 m)j PIC : 2(100 N)k i j k
M, = 0.96 —0.40 0 = (40 Nm)i+(96 Nm)j
0 O —100
Also, M2 = ram " P25 rDM : (0.20 m)j — (0.55 m)k
P25 = K5055 = —(0.48 m)i + (0.55 m)k (146 N) (0.48)2 + (0.55)2 m = 7(96 N)i + (110 N)k i j k
M2 = 0 0.20 4155 Nm
—96 0 110 = (22.0Nm)i + (52.3 Nm)j + (19.2 Nm)k PROBLEM 3.73 CONTINUED
and M =[(40Nm)i +(96Nm)j]+[(22.0Nm)i
+(52.8Nm)j + (19.2 Nm)k] = [62.0 N.m}i + (148.8 Nm)j + (19.2 Nm)k [Mg : M3 + M; + M3 = {62.0)2 +(14S.8)2 + (19.2)2 =162.339Nm or M 2162.3Nm4 M 62.0i + 148.8j + 19.2k 71 = _ a
‘M‘ 162.339 : 0.38192i + 0.91660j + 0.118271k
cosﬁx = 0.38192 a, = 61547"
or 9x = 67.5” 4
c053), : 0.91660 6y : 23.566°
or 9), : 23.6” 4
(30592 = 0.118271 Hz = 83.208° or 92 2 832° 4 SOLUTION Have where PROBLEM 3.75 Knowing that P = 5 lb, replace the three couples with a single equivalent
couple, specifying its magnitude and the direction of its axis. M=M4+M7+M5 i j k
M4 = er XE“; = —10 0 o lbin. = (401bin.)j
o o 4
i j k 7
MT = rm x Fm = —5 3 0 [El lbin. = 0.76835(21i+ 35j) lbin.
—5 3 7
(See Solution to Problem 3.74.)
i j k
M5 = rcM x FSC = 10 —6 7 lbin. 2 —[351bin.)i+(501bin.)k
0 5 o M = [(151353 — 35)i + (40 + 26.892)j+(50)k]1bin.
= —(18.86471bAin.)i +(66.892 1bin.)j+ (501bin.)k M = JME + ME. + M3 : (13.3647)2 + (66.392)2 + (50)2 = 85.6181bin. or M 2 85.61b'in.4 M —13.8647i + 66.892j + 50k 1. = 7 = : 4.2203“ + 0.78129j + 0.58399]:
M 85.618
cosﬁlx = 4122034 9x =102.729° or a, :102.7° 4
0056}, : 0.78129 0), = 38.621° or By : 38.6° 4 case: = 0.53399 a. : 54.2w or 92 = 54.3“! A ...
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 Fall '08
 C.Trujilliano
 Statics

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