SOLUTION
:i.i
J
1.,
Free-Body
Diagram:
(a)+} 'fMB =0:
-Ay(3.6 ft) -(1461b)(1.44 ft) -(631b)(3.24 ft) -(90 Ib)(6.24 ft) = 0
or Ay = 2711bl ~
Ay = -271.10 Ib
(b) +) rMA = 0:
By(3.6 it) -(1461b)(5.04 it) -(631b)(6.84 it) -(90 Ib)(9.84 it) = 0
or Bv = 570Ib t ~
By = 570.10 Ib
PROPRIETARY MATERIAL. It> 2007 The McGraw-Hili Companies, .Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in anyform or by any means, without the prior written pelwission of the publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permi.~sion.
355
"
PROPRIETARY MATERIAL. f:>2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in anyfonn or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and
educators pennitted byJ\.(cGraw-Hill for their individual course preparation. If you are a st!/dcnt using this Manual, .vou are using it without permission.
1ti5
SOLUTION
Free-Body
Diagram:
.300"'M
Equationsof equilibrium:
+) rMA=0:
+
-x
-(330 N)(0.25 m) + Bsina(0.3 m) + Bcosa(0.5 m) = 0
1:F = 0:
Ax -Bsina
= 0
(1)
(2)
Ay -(330 N) + Bcosa = 0
(3)
(a) Substitution a = 0 into (1), (2), and (3) and solving for A and B:
B = 165.000N, Ax = 0, Ay = 165.0N
or A = 165.0N t. B = 165.0N t ~
(b) Substituting a = 9()° into (1), (2), and (3) and solving for A and B:
B = 275.00N, Ax = 275.00N,
(} = tan-l~
Ay = 330.00N
= tan-I 330 = 50.1940
Ax
m
:. A = 430 N ~ 50.2°, B = 275 N -~
PROPRIETARY MATERIAL. ~ 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part ofthis Manual may be displayed, reproduced
or distributed in any form or by any means, without the prior written permission ofthe publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
372
PROPRIET AR YMA TERIAL @ 2007 The McGraw-HilI Companies, Inc. All rights reserved. No part of this Ma~ual maybe displayed, reproduced
or distributed in any.form or by an}' means, without the prior written permission ofthe publisher, or used beyond the ,limited distribution to leachers and
educators pennitted by McGraw-Hill for their individual course preparation. If you are a stltdent using this ,'vfanual,:you are using it witholt! pennis.\'ion.
373
r"
1d"
::1:1)0
1R'
-,
-'""
PROBLEM 4.25
~J!!!;:;
,.,
A lever AD is hinged at C and is attachedto a control cable at A. If the
lever is subjectedto a 60-lb vertical force at B, detennine (a) the tension
in the cable, (b) the reaction at C.
contin/ied
PROPRIETARY ,'fATERIAL. @2007The McGraw-HilI Companies, Inc. All rights reserved. No part ofthis Mfnuul maybe displayed, reproduced
or di.\"tributed in any.form or b_van}' means, ~ithout the prior written permi.l'sion ofthe publisher, or used beyond thr limited distribution to teachers and
educutors permitted by,\1cGraw-Hill.for.t/~!r individual course preparation. If you are a student u.I'ing this Manua(, you are using it ~'ithout permi.l'sioll.
PROPRIETARY ,WATERIAL. IQ 2007 The McGraw-Hili Companies. Inc. All rights reserved. No part ofthi~ Manual may be displayed, reproduced
or distributed in anyform or b.y any means, lvithout the prior lI'rinen permission ofthe publisher, or used beyond the limited distlibution to teachet~ and
educators pennitted b}. McGraw-Hill.for their individual course preparation.if.vou are a student using this Manual, you are using it ..'ithout permission.
389
SOLUTION
(a)
Free-Body Diagram:
+ EF = 0:
-x
E x =0
Ey -16.2 kN -5.4 kN -18 kN = 0
orE=39.6kNt~
Ey= 39.6 kN
ME + (16.2 kN)(4.8 m) + (5.4 kN)(2.6 m) -(18 kN)(1.5 m) = 0
ME = -64.8 kN.m
orME=64.8kN°m)~
PROPRIETARY MATERIAL. ~ 2007 The McGraw-HilI Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any fonn or by any means, without the prior written pennission of the publisher, or used beyond the limited distribution to teachers and
educator.~permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission.
416
I~.
!~
YJ
PROBLEM 4.59
~
and T
A vertical load P is applied at endB ofrodBC. The constantof the spring
is k and the spring is unstretchedwhen 0 = o. (a) Neglecting the weight
of the rod, expressthe angle 0 correspondingto the equilibrium position
in tenDSof P, k, and I. (b) Oetennine the value of 0 correspondingto
equilibrium if P = 2kl.
kx
2klsin~
2
or 0
PROPRIETARY MATERIAL. () 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in anyfo1m or by any means, without the prior written pe1mission ofthe publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hillfor their individual course preparation./fyou are a student using this Manual, you are using it without permission.
428
SOLUTION
Free-Body
Diagram:
~
,?=\~'""
'
-~
p
/
~
0.$'"
'=.,
-~
~
2"
~~
Express all forces in tenDSof rectangularcomponents:
rA = (2.4 m)i
rB = (1.8 m)j
AD = -(2.4 m)i + (0.3 m)j + (1.2m)k
BE = -(1.8 m)i + (0.6 m)j -(0.9 m)k
WA = -(440 N)j
WB = -(440 N)j
Then
AD = TAD~(-2.4)2
-2.4i + 0.3j + l.2k
8
+ (0.3)2+ (1.2)2 = -9T~
--'"
T
BE
~
-1BE-
BE
-'1'
-.iDE
-1.8i + O.6j -O.9k
-6
~(-1.8)2+(O.6)2+(-O.9f
---AnI
T
7
1
+ 9T~
.2
+
T
-ADJ
7
+ ~TAJJk
.3
T
--AD""
'-
Ir
7
PROPRIETARY MATERIAL
11::12007
The McGraw-Hill Companies,ffic. All rights reserved, No part of this Manual may be displayed, reproduced
or distributed in anyform or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course preparation. If }'Ou are a student using this Manual, ,YOUare using it without permission.
k"':;~~:'"'
522
PROBLEM 4.117 CONTINUED
LMc = 0:
r.4 x TAD+ rB x TBE+ rA x WA + rB x WB
0
or
0
Equating the coefficients of the unit vectorsto zero:
.9.6
J:
9
T,W+-7-1.BE
5.47'
0
1848= 0
k
or J:w
1890N ~
TSE
2610
N..
Forceequations:
Cx
0,
or C.,. 3920.0N
44ON
0, or c"
0,
or
Cv
.76.657 N
279.99N
c
(3920N)i
(76.7 N)j + (280 N)k
PROPRIETARY l\1ATERIAL. '£' 2007 The McGraw-Hili Companies, Inc. All rights reserved. No pa!.t ofthis.iManual may beliisplayed. reproduced
or di.l.tribllted in any {O/l11or by any means. without the prior written pe/lI1ission of the publisher. or ILsedbeyon~the /muted distribution to teachers and
f,ducators permitted b..v,WcGraw-Hillfor thei,' individual course preparation. lfyou aloea .~tudentu.~ingthi.~lvlanu4/. .vou are IL'iingit without permi.~~ion.
523
/"
~'t
~
SOLUTION
Free-Body
Diagram:
'1
28",.
10;~.
-,",Co,
74;".
/~/
~.
b
--
-"
Go
;If 1."
~
"!
~"t~~
~O\¥\.
; ~.
Expressforces in terns of their rectangularcomponents:
-40i + 74 j -32k
BG = ~BG--~
~(-40)2 + (74)2 + (-32)2 -BG
TBG= TBG-BG
-BH
TBH -TBHBii
-30i+60j-60k
-TBH ~(30)2 + (60)2 + (-6~)2
(-""4:5"1
20 .
+
( 1.
2.
37
45
~k)
45
2
= TBH 31 + 3J -3k
)
P = -(751b)j
AD
}..AD = "AD" =
80i -60j
~(80Y
+ :-6~)2 = O.8i -O.6j
PROPRIETARY MATERIAL. ~ 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproduced
or distributed in anyform or by any means, Without the prior written permission of the publisher, or used beyond/he limited distribution to teachers and
educators permitted by McGraw-Hill for their individual course preparation. If y'ou are a stu4ent u.~ingthis Manual. you are using it without permission.
S78