.
,~
PROBLEM
8.3
Detennine whetherthe to-kg block shown is in equilibrium, and find the
magnitude and direction of the friction force when P = 40 N and
(J = 20°.
PROPRlfTARY
MATERIAL. ~ 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part <?fthis Manual ma.vbe displayed. reproduced
or distribllted in anyform or by any means, ~'ithout the prior written permission of the publisher. or used beyond'the limited distribution to teachers and
educators permitted bv ,'vlcGra»'-Hill for their individual cour,~epreparation. If.}'OU are a student using this Manual. .vou (lre u.~ingit without permi~~ion.
1189
PROBLEM 8.6
Knowing that the coefficientof static friction betweenthe 20-kg block
andthe incline is 0.30,detemlinethe smallestvalueof (J for whichthe
blockis in equilibrium.
20 kg
220N
SOLUTION
FBD Block:
y
22~~/,,:;:~
~9~
x
') '.,~~/
,,/
'"
For 9min motion will impend up the incline, so F is downward and
F = .usN
/
, IFy=0:
N -(220 N)sin8 -(196.2 N) cos35°= 0
'/
'V'~f: \'1
/
EFx=O
(1)+ (2):
F = .usN = 0.3(220 sine + 196.2 cos35°)N
(1)
(220 N)cos8 -F -(196.2 N) sin35° = 0
(2)
0.3(220sinO+ 196.2cosO)N
= (220 cosO)N-(196.2sin35°)N
or
220cosO-66sinO = 160.751
Solvingnumerically:
()= 28.~ ~
PROPRIETARY ,HA TERIAL. @ 2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed. reproduced
or distributed in any form 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 you are a stude~t ~ing this Manual. you are using it without permission.
1192
...~
/\
200
PROBLEM
8.46
Two 80 wedges of negligible weight are used to move and position a
530-lb block. Knowing that the coefficient of static friction is 0.40 at all
surfaces of contact, detenI1inethe magnitude of the force P for which
motion of the block is impending.
PROIrRlET ARY MA TERIAL. (!:)2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of ~hisMfnual mav be displayed. reproduced
or distributed in anyfonn or by any means, without the prior written permission of the publisher, or used beyond th~ limited distribution to teachers and
educators permitted b.v McGra,,'-Hill.for their individual course preparation. If you are a student using this Ma~ual, .¥ou are lL~ingit ..ithout permi~.~ion.
1233
.
i--
PROBLEM
p
8.50
The elevation of the end of a steelbeam supportedby a concretefloor is
adjusted by means of the steel wedgesE and F. The base plate CD has
beenwelded to the lower flange of the beam, and the end reaction of the
beam is known to be 90 KN. The coefficient of static friction is 0.30
between the two steel surfaces and 0.60 between the steel and the
concrete. If the horizontal motion of the beam is prevented by the force
Q, determine (a) the force P required for impending upward motion of
the beam, (b) the correspondingforce Q.
PROPRIETARY MATERIAL. () 2007 The McGraw-Hili Companies, Inc. All rights reserved. No part of this ,'./anual 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 coursepreparation. If you are a student using this Manual. you are using it without permission.
1237
r"'
PROBLEM 8.104
A 120-kg block is supported by a rope whi~h is wrapped It times
around a horizontal rod. Knowing that the co~fficient of static friction
betweenthe rope and the rod is 0.15, determin~the range of values of P
for whichequilibriumis maintained.
kg
SOLUTION
!}!:: 020 kj L1' Sf m/sz.
p
"'-
=- 1/.77,.2
pI
f3 = 1.5 turns = 31Cfad
For impending motion of W up
p=
WeJisf3=(1177.2 N)e(O.15)31r
= 4839.7N
For impending motion of W down
-
P = We-JIs!3= (1177.2N)e,-{O.15)3ir'
i
I
= 286.3 N
I
~or equilibrium
286N ~ P ~ 4.84kN..
I
r
PROPRIETARY MATERIAL. It) 2007 The McGraw-Hili Companies,lnc. All rights reserved. No part ofthif !fanual may be displayed, reproduced
or di.vtributed in an)'form or by an)' means, without the prior written permission of the publisher, or u.vedbe)'on~ the limited distribution to teachers and
educators permitted b\-'lWcGraw-Hillfor their individual cow',vepreparation. !f)'OU are a student using this Mam1al, you are using it K'ithout permission.
1.295
~
SOLUTION
FBD motor and mount:
Impendingbeltslip: cw rotation
Tz = 7ie.us/3= 7ieO.407r
= 3.5l367i
( EMD = 0:
(12 in.)(1751b)-(7 in.)~ -(13 in.)1i = 0
2100 lb = [(7 in.)(3.5136) + 13 in.]11
1i = 55.858lb,
~ = 3.513611 = 196.2631b
FBD drum at B:
rMB = 0
MB -(3in.)(196.2631b -55.8581b) = 0
MB = 4211b.in.
~
r=3in.
(Compareto 857 lb. in. using V-belt, Problem 8.130)
PROPRIETARY MATERIAL. tC>2007 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced
or distributed in any fonn or by any means, mthout the prior written permission of the publisher, or used beyond the limited distribution to teachers and
educators permitted by McGraw-Hillfor their individual course preparation.lfyou are a student using this Manual, you are using it without permission.
1298
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