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Unformatted text preview: PROBLEM 8.102 Blocks A and B are connected by a cable that passes over support C.
Friction between the blocks and the inclined surfaces can be neglected.
Knowing that motion of block B up the incline is impending when W; = 16 lb, determine (a) the coefﬁcient of static friction between the rope and the support, ([3) the largest value of W B for which equilibrium is
maintained. (Hint: See Problem 8.128.) SOLUTION FBD A: FBD B: 2015 T ”4/” IS W. _
A <’ w i
, ’ /
’1 y, a ,1 8
t I“ .53 \
”a ”if! g ,\ l
_’~_3P_. _:?_._x 7G.
N , />:F, =0: TA—201bsin30°=0 “'5
\ 2ny = 0: WBsin30° — TB = 0 TA : 10 lb
W
T8 = 7”
From hint, ﬂ = 60" = grad regardless of shape of support C
. v . Wlg
(a) For motion of B up mohne when W], = 16 1b, TB = 7 : 8 lb
and ,usﬂ = 1na or #5 = 1an = i Inim— = 0.213086
TB 5 TB 7: 8 lb #5 = 0.213 ‘ (b) For maximum WE, motion of B impends down and TB > TA So TB 2 new = (101b)e°2‘3°W”3 = 12.500113
Now WE = 2TB
So that W” = 25.0 lb 4 PROBLEM 8.109 A ﬂat belt is used to transmit a torque from pulley A to pulley B. The
radius of each pulley is 3 in., and a force of magnitude P = 2251b is
applied as shown to the axle of pulley A. Knowing that the coefﬁcient of
static friction is 0.35, determine (a) the largest torque which can be transmitted, (b) the corresponding maximum value of the tension in the
belt. SOLUTION
FBD pulley A: Impending slipping of belt: T2 = 7'16””i
T2 = Tle0'35” = 3.00287]
~~> ZFX =0: T1 +T2 4251b = 0
T1(1+ 3.0028) = 2251b or T1 = 56.211113
T2 = 3.00287] or 72 = 168.7911:
(a) Q EMA = 0: MA + (6 in.)(Tl — T3) = 0 or MA = (3 in.)(168.79 1b — 56.21113)
max. torque: M/1 = 338 lbin. 4 (h) max. tension; T2 = 168.8 lb 4 (Compare with MA = 638 lbin. with Vbelt, Problem 8.131.) PROBLEM 8.116 Bucket A and block C are connected by a cable that passes over drum B.
Knowing that drum B rotates slowly counterclockwise and that the
coefﬁcients of friction at all surfaces are #3 = 0.35 and ,uk = 0.25,
determine the smallest combined weight Wof the bucket and its contents
for which block C will (a) remain at rest, (b) be about to move up the
incline, (c) continue moving up the incline at a constant speed. SOLUTION
FBD block: / 2F" = 0; NC — (2001b)cos30° = 0; N = 1006 1b 3:; 200 lb
W \ 2F, = 0: TC — (2001b)sin30° ; FC = 0
71
/_ (“342: TC = 100 lb i FC (1)
(L ﬁt, \3‘\ 1': where the upper signs apply when FC acts \ (a) For impending motion of block \, FC \, and FC = A‘NC = 0.35(100J§1b) = 35J§ lb
So, from Equation (1): TC = (100 — 35$) lb But belt slips on drum, so TC = [Ii/AH“? WA = [(100 — 35x5)1b] {“5831} WA = 23.311: 4
(b) For impending motion of block \, FC \ and FC : #3Nc : 35% lb From Equation (1): TC = (100 + 355) lb Belt still slips, so WA : TCe’W’ = [(100 + 356) lb]e”'25(g%) WA = 95.11b4 PROBLEM 8.116 CONTINUED
(0) For steady motion of block \, FC \, and FC = ,ukNC = 25x5 lb
Then, from Equation (1): T = (100 + 25/3) lb. Also, belt is not slipping on drum, so
, 2,»
WA = Tee’ﬂs” = [(100 + 256) lb]e °'”( 3) WA = 68481b 4 ...
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 Spring '08
 Jenkins
 Statics

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