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Unformatted text preview: 986* A rope attached to a ZZO—kg
block passes over a fixed
drum as shown in Fig. P9—86.
If the coefficient of
friction between.the rope and
the drum is 0.30, determine
(a) The minimum force ﬁ that
must be used to keep the
block from falling. (b) The minimum force § that
must be used to begin to
raise the block. Fig.P986
SOLUTION
From a freebody diagram
for the block: ‘Y’
+T2Fy=T—mg=0
T = mg = 220(9.807) = 2158 N
W3
From Eq. 9—16b:
_ MB _ o.3o(n/2) _ T2 — Tie — Tie — 1.6020'1‘1 (a) During lowering of the block: T2 = 2158 N
T2 2158 Pmin = T,1 = m = m = 1347.1 N g 1347 N Ans. (b) During raising of the block: T1 = 2158 N P , = T = 1.6020T1 = 1.6020(2158) = 3457 N ¥ 3460 N Ans. min 2 9—89 The electric motor shown
in Fig. P989 weighs 30
lb and delivers 50 in.lb
of torque to pulley A of a
furnace blower by means of
a V—belt. The effective
diameters of the 36
pulleys are 5 in. The
coefficient of friction
is 0.30. Determine the
minimum distance a to
prevent slipping of the
belt if the rotation of
the motor is clockwise. SOLUTION
. _ ___Ji___., __Jlﬁ¥L__ _
For a pulley' “enh ' sin («/2) ‘ sin (36/2) “ 0'9708
u 5 = 0.9708(n) = 3.050
enh
From Eq. 9—17: T2 = T1e3'050 = 21.11T1 From a freebody diagram
for the blower pulley: + D 2MB = Tarp — TirP — cA 21.11T1(2.5)
— T1(2.5)  50 = 0
T = 0.9945 lb T = 21.11T1 21.1l(0.9945) = 20.99 lb From a freebody diagram
for the motor: + C EMA = T2(8 + rP) + T1(8  pp) — WM(a)
= 20.99(10.5) + 0.9945(5.5) — 30a = 0 7.529 in. E 7.53 in. Ans. 9:
u 990 The band brake of Fig P990 is used to control
the rotation of a drum. The coefficient of friction
between the belt‘and the
drum is 0.35 and the weight
of the handle is 15 N. If
a force of 200 N is applied
to the end of the handle,
determine the maximum torque
for which no motion occurs
if the torque is applied
(a) Clockwise. (b) Counterclockwise. SOLUTION 180 + 30 360 3.665 rad. (211') B: 0.35(3.665) = 1.2828 #3 = From Eq. 916b: . 8
T =Te =Te1282 = 2 1 1 3.6067T1 (a) From a freebody diagram
of the handle when the
rotation of the drum is 200N clockwise:
+ C EMA = 200(675) — 15(225) — T2(75) = 0
T2 = 1755.0 N
T
l _ 2 _ 1755.0 _ 1
T1 ' 3.607 ‘ 3.606 ' 486‘6 N
Tmax = (T2  T1)R = (1755.0  486.6)(0.150) = 190.3 N'm Ans
T, 200);
(b) From a free—body diagram
of the handle when the
rotation of the drum is
counterclockwise:
WSantl
T2 = 1755.0 N n
5 \s N
T2 = 3.6067T1 = 3.6067(1755.0) = 6329.8 N
T = (T2  T1)R = (6329.8 — 1755.0)(0.150) = 686 N'm Ans. 996* A 100—N crate is sitting on
a ZOO—N crate as shown in
Fig. P996. The rope that
joins the crates passes
around a frictionless
pulley and a fixed drum.
The coefficient of
friction is 0.30 at all
surfaces. Determine the
minimum force F that must
be applied to the ZOON
crate to start motion. SOLUTION L2“ ,1“
From a free—body diagram .T r
of the 100—N crate: g
I
. . . . m "I
For impending sllding: No )ooN 3 )ooN
+ T XFy = An — 100 = 0 __
n B
A = 100 N R «3’ Bn
" "n
A =uA =0.3(100> =30N u TVqu
f n For impending tipping: + C 2MB = T(2.5)  100(0.6) = 0
T = 24 N < 30 N (crate tips) 'yi From Eq. 9—16b: T2 : T1eu6 = 24eo.3o(2N/3) = 45 N From a freebody diagram
of the ZOO—N crate: 5" cn=300N + —e ZFX = P — Af  Cf  T2 = 0 P , = Ar + Cf + T2 = 24 + 0.3(300) + 45 = 159.0 N Ans. Min ...
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 Spring '10
 jones
 Statics

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