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Unformatted text preview: PROBLEM 8.2 Determine Whether the block shown is in equilibrium, and ﬁnd the #9030
”4020 MN magnitude and direction of the friction force when (9 =35° and
P = 400 N4
SOLUTION
FBD block:
u map A/
\
/ t
\ 4
/
35°
5 __ .L *
7‘00 N E
N
4’ Assume equilibrium: \ 2F" = o: N — (1000 N)cos35° — (400 N)sin35° = 0 N = 1048.6 N /' 2F, = 0: F — (1000 N)sin35° + (400 N)cos35° = 0
F = 246N = Feq F : ,usN = (0.3)(104s.6 N) = 314N max Fm < Fmax 0K equilibrium { ‘. F=246N/4 ,\ PROBLEM 8.15 A 48kg cabinet is mounted on casters which can be locked to prevent
their rotation. The coefﬁcient of static friction between the ﬂoor and each
caster is 0.30. Knowing that h : 640 mm, determine the magnitude of
the force P required for impending motion of the cabinet to the right (a) if
all casters are locked, (b) if the casters at B are locked and the casters at A
are free to rotate, (c) if the casters at A are locked and the casters at B are
free to rotate. 480 mm SOLUTION
FBD cabinet: Note: For tipping, NA = E4 2 0 Q 2MB : o: (0.24 m)W — (0.64 m)Pt = 0 PHP 2 0,375W in
(a) All casters locked: Impending slip: F A = ySNA, F” = ,uSNB
TEF}, =0: NA+NBW=0 NA+NB =W
So FA + F" = ”KW ~>ZFX=01 P—FA—FBzo P=FA+FB=ySW P = 0.3(470.88 N) or P =141.3 N 4
(P = 0.3W < Ptip 0K)
W 2 4S kg(9.81 m/SZ) (b) Casters atA free, so FA = 0
= 47088 N Impending slip: F3 : ,uSNB
,ub : 0.3 ~2Fx=0: PFB=0 PZFBzﬂA'NB NB:—“' C EMA = 0; (0.64 m)P + (0.24 m)W ~ (0.48 m)NB = 0 8P+3W—66%:0 P=0.25W (P : 0.25W < p m 0K) P 2 0.25(470.88 N) P = 117.7 N 4 PROBLEM 8.15 CONTINUED
(c) Casters atB free, so FB = 0
Impending slip: F A = ”SNA ”212:0: P—FA=0 P:FA=,uSNA N=—=—
A #c 03 P P
(2MB = o; (0.24 m)W — (0.64 m)P — (0.48 m)NA = 0 3W—8P—6%=0 P=O.10714W=50.45N (P<P up 0K) P=50.5N{ PROBLEM 8.32 The 25kg plate ABCD is attached at A and D to collars which can slide
on the vertical rod. Knowing that the coefﬁcient of static friction is 0.40
between both collars and the rod, determine whether the plate is in
equilibrium in the position shown when the magnitude of the veﬂical force applied at E is (a) P z 0, (b)P = 80 N. SOLUTION
(a) P = 0; assume equilibrium:
FBD plate:
10W
5‘ to: (EMA = o: (0.7 m)ND — (1 m)W = 0 ND = T
E —>2Fx=0: ND—NA=0 NA=ND=—107W (FA )mx 2 IiiNA (FD)max = IusND 20pr
W = 25 kg(9'81N/kg) So (121+ Farm = ”ANA + ND) = 7 =1.143W
=245.25N (25:0: FA+FD—W=0
FA + FD = W < (FA + Fame“ OK. Plate is in equilibrium 4
(b) P = 80 N; assume equilibrium:
{EMA = o: (1.75 m)P + (0.7 m)ND — (1 m)W = 0 or ND=W—1.75P
0.7
mzpr=o; ND_NA:0 NDzNA=W (Ir/1%“ = #SNA (F3)max : #:NB So (FA + F3) 2 0.4M =120.29 N m 0.7
(213:0: FA+FD—W+P=O
FA+FD=W—P=165.25N (FA + FD)equ“ > (FA + FD)“: Impossible, so plate slides downward 4 ...
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 Spring '08
 Jenkins
 Statics

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