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Unformatted text preview: Witia ,czua, :0 Mar /\C PROBLEM 6.126 400 mm The press shown is used to emboss a small seal at E. Knowing
that P = 250 N, determine (a) the vertical component of the force
exerted on the seal, (19) the reaction atA. ir_
SOLUTION '
FBD Stamp D: (S a Q “Fa'u‘ \‘X’C’ka 50 20 u = 0: E —FBD c0520° = 0, E =FBD cos20° um
FBD ABC: ';
éT ,
1‘7
‘ LEM}, = 0: (0.2 m)(sin30°)(FBD cos 20°) + (0.2 m)(cos 30°)(FBD sin 20°)
1 — [(0.2 m) sin 30° + (0.4 m)coslS°](250 N) = 0
F BD = 793.64 N C
and, from above,
E = (793.64 N) cos 20°
(a) E = 746 N 4 “~25, = 0: A, —(793.64 N)sin20° = 0
A, = 271.44 N—» , T213. = 0: A), + (793.64 N)c0520°  250 N = 0 A), =495.73Nl
so (b) A=565N=q613°4 .2. _ 3 m. .}
AHE/‘ﬁmltvﬁvla COR"ti UR: ‘(t‘kC/i' Cur “30 ‘5 ‘3‘ I E q rd“) 0 A . * v' Q). ‘l’O dh’tuu ‘01—) F N Frs‘r s 30 m Ovﬁa and \ WUOJH  ﬂ— 1&1 Sﬁﬁmc'vﬂ ‘Px{=‘Cxx “Nd 3% = gee, + a? Cab\ste/w Ag CMQ‘T a membw) '3 A, 3y 3% SMMMz—ﬁj PM }= ssx\ “Wad A“: 6‘1
Aggx EFH 1: S’OOT ;AY :0 Arm: an in = Bx as 480 \to (‘5’) “Die MC” VP.”
B ' I“ ‘
, ‘1 *eacﬂﬁuw ﬁn"
JKA (5 he...) ragarwﬂ
3 T
353*} Z r
L g ME PymwrF (9)11)“ (nur3 ;) + '3» (a‘r ‘5) =0
z] gutQC.) 1» \‘J F 1 lbs 1300
I x I h “use; *0
F ‘ —..q~€«o
iv) in“ i X .   . ‘b
F“ 4‘10 F '5.
F =
x FxrEx —_9X :0
M440téﬁe+3xq+80 :C}
Ex:rc\?ol
\b(_~—>>
Er  _
y TEV 16
‘m ‘1' Eva—+00 _;O
Eyre \v)
“ch PROBLEM 6.149 l_____________i ‘P——]U iII,—————>§‘——]0 in.
a i 1
V ' L I _ ‘ __ Knowing that the frame shown has a sag at B of a =1in., determine
‘ the force P required to maintain equilibrium in the position shown. SOLUTION
We note that AB and BC are twoforce members ﬂ, (an... iow 99
..._ w *4
Free body: Toggle 9x _._.——‘""" a.
P 
By symmetry: Cy = 3 I
Q _ i
E 10in. a
szncvhzzﬁ
a ' a 2 a Z Free body: Member CDE
+j)ZME = 0: Cx (6 in.)  Cy (20 in.) — (50 1b)(] 0 in.) = 0 55%) — 5(20) = 500
a 2 30
P ——10 =500
(a J i
For a=1.0in.
P[3—19—10j=500
20P=500 P:°5.01bi< 240 lb 05 = 035
V m. = 0,25 PROBLEM 8.1 ark—— SOLUTION Harem {Yaw—w,  Assume eguilibrium: Maximum friction force: l5
\KEFX = O: F+(2401b)sin 25°—(1501b)cos 25° :0 F =+34.5181b *7! BF}. = 0: N—(2401b)cos 25°—(1501b)sin 250 = 0 N = +280.91 1b
Fm = ,uSN = 0.35(280.91 1b) = 98.3191b Since F < Fm, (b) it: Mare, '1'V\d'~.+ ("J I? ‘2 Hera, ! PROBLEM 8.11 SOLUTION
(a) Free body: 20kg block W1 2(20 kg)(9.81 111/53): 196.2 N
F] = ySNI = 0.4(196.2 N) = 78.48 N LZFr—O: T—F1 =0 T=F1 278.48N
Free body: 30kg block W2 = (30 kg)(9.81m/sz)= 294.3 N
N2 2196.2 N + 294.3 N = 490.5 N 5 “we “PP‘ F:=#SN2:0.4(490.5N)=196.2N
JLZF=O: P—Fl—FZ—T=0 P = 78.48 N +1962 N+78.48 N =353.2 N Free body: Both blocks
Blocks move together W = (50 kg)(9.81 m/sz) =490.5 N
in =0: P—F=0 P = ,uSN = 0.4(490.5 N) = 196.2 N A F = 34.5181bi\ Determine whether the block shown is in equilibrium and ﬁnd the magnitude and
P direction of the friction force when 19 = 25° and P = 150 1b. N = 280.911b / Block is in equilibrium 4
F = 34.5 lb\ 4 The coefﬁcients of friction are ,us =O.4O and ,uk =O.30 between all
surfaces of contact. Determine the smallest force P required to start the
30kg block moving if cable AB (a) is attached as show, (b) is removed. P=353N‘*—< w=m~$77
Lm '~*
fag/v fly:¥2>.r~ % > PROBLEM 8.105
The coefﬁcient of static friction between block B and the horizontal
surface and between the rope and support C is 0.40. Knowing that
mA 2 12 kg, determine the smallest mass of block B for which equilibrium
is maintained.
SOLUTION
Support at C: FBD block B:
773
(31' $+u~rn : rad 1—6 $953
fr;
Z; = h”! '3 rs
WA = mg = (12 kg)g
12F), =0: NB —WB =0 or NB =WB
x Impending motion: FB = ,uSNB = 0.4NB = 0.4 WE
—>ZFX =0: FE —TB =0 or TB =FB =O.4WB
At support, for impending motion of WA down:
’ WA = TBeﬂvﬂ
so TB = WAe‘ﬂsﬂ = (12 kg)g*‘~°‘”’“'2 = (6.4019 kg)g
NOW W KZ'B—z g 0.4 0.4
W l .
sothat m3: B=M :1“; 2.7“, mB=l6.00kg4
g g m, ...
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 Spring '06
 ZEHNDER

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