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Unformatted text preview: Problem] 824 Suppose that you want the bar of length
L to act as a simple brake that will allow the workpiece
A to slide to the left but will not allow it to slide to the
right no matter how large a horizontal force is applied
to it. The weight of the bar is H", and the coefﬁcient
of static friction between it and the workpiece A is #5.
You can neglect friction between the workpiece and the
surface it rests on.
Ca] What is the largest angle a for which the bar will
prevent the workpiece from moving to the right?
(b) If cr has the value determined in (a), what horizontal
force is required to start the workpiece A toward the
left at a constant rate? Solution: [a] To resist motion to the right no matter how large the horizontal
force requires a ver}r large friction force. For impending motion to the right. the sum of the moments about the bar hinge is
EM 2 +%Lsino — FNLsmo + choer = 0, where FN is the normal force and f is the friction force resisting
the impending motion.
Noting that f = grgFN. the sum of moments yields W sin Ct FN= manor—its coscq If the denominator vanishes. the normal force and hence the fric
tion force become as large as required. Thus sin a —ﬁt, one a =
0. from which g; = 12.31th.01' cs = tan—lute) [bl For impending motion to the left. the sum of moments about the
bar hinge is EM 2 +w [3)5ino —FNLsino— choscr:U__ where the friction force opposes the impending motion. Noting
FN = then W sino _ W sinct _ gs}?
2(sinct+msa) _ 4005c? _ 4 5'1: f: is the force required to start motion to the left. Problem 8.46 To obtain a preliminary evaluation of
the stahilir)r ol‘a turning car, imagine subjecting the sta
tionary car to an increasing lateral force F at the height
of its center of mass, and determine whether the car will
slip {skid} laterally before it tips over. Show that this will
be the case if blr'h := 2:45. l[Notice the importance of the
height of the center of mass relative to the width of the
car. This reﬂects on recent discussions ot'the stability of
sport utility,r vehicles and vans that have relatively high
centers of mass.) Solution: EQUILIBRIUM Eons:
2F}: F—fL—fR=D 2F”:
Elli}: Assume skid and tip simultaneously. N1, + NR — mg 2 U —hF + 6N3 — 3mg 2 0 fit =I—‘5NL~
fﬂ Zﬁlsll‘l'ﬂiskidll and My :0 (tip). fL = 3
fa =irsm9. The equilibrium eqns become
F = fa =irsNR =irsm9 and the moment eqn. uses 6:
—la:}.tsmg] + bﬁmg) — 51019 = U or f = Etta For TE I::v 2H5 . slip before tip ::2 2st“. tip before slip
big N low cm. relative to track width JuloIJ'10IJ'10I small N high cm. relative to track width Problem 8.74 The carjack is operated by turning the
threaded shaft at A. The threaded shaft ﬁts into a mating
groove in the collar at B, causing the collar to move
relative to the shaft as the shaft turns. As a result, points
3 and D move closer together or farther apart, causing
point C“ [where the jack is in contact with the car} to
move up or down. The pitch of the threaded shaft is
p = 5 mm, the mean radius of the thread is r = ll] mm,
and the coefﬁcient of kinetic friction between the thread
and the mating groove is 0.15. What couple is necessaryr
to turn the shaft at a constant rate and raise thejack when
it is in the position shown ifF = 6.13 kN‘? Solution: Isolate members BC." and ED. Assume that half the car load is carried by these members. The equilibrium conditions for
member BC are: Zr: =cI—BI :0. F
2M3 = 0.3 (E) — 0.150, = 0. These equations are solved for E zﬁzaasm
2 2 to obtain B: = 6.5 ML which is the force on the collar to be balanced
by the rotating threaded shaft. The pitch angle is 5
_ —1 _ o
ct—Lan (2ﬁ10)_4.55. The angle of kinetic friction is 9, = tail—1(U.15)= 3.530. The moment required to rotate the shaft at a constant rate is
M = (0.01)(6.5)tan(8.53° + 4.55“) = 0.0151kN m = 15.1 N m 300 mm —3 DD mat—— Problem 8.84 The mass of the suspended object is
4 kg. The pttlle}r has a IOU—mm radius and is rigidly,r
attached to a horizontal shaft supported by journal hear 25?"
ings. The radius ofthe horizontal shaft is 10 mm and the coefﬁcient of kinetic friction between the shaft and the ‘5
hearings is 0.26. What tension must the person exert on 100 mm
the hope to raise the load at a constant rate? Solution:
H : 0.1m
ﬁll, = 0.26 ShaftradjusOD] in
grkfshaftj = [3.26 tanlﬁk= 2m.
9;, =14.5?°
.11, = rFsinB;r
m=4k3 R:C.lm NK=U26 To Find F. we nrust ﬁnd the forces acting on the shaft.
ZFI: OJ. —Tc0525°=O (I) Shaft radius 0.0] m
K (shaft) : 0.26 Zr}: 09 —Tsmas°—mg=n(2;t
F = fog + 03 (3;.
2 M0: at" — ngl — M, = o (4;. M, ﬁrms. (5;. Unknowns: 0,: 03,. T. £11,, F
Solving. we get T = 40.914
Also. F =8'FGN.
M, 20.1’TONm
O: =3?.1N. o, = 56.5 N Problem 8.91) The shaft is supported by thrust bear—
ings that subject it to an axial load of 800 N. The coef—
ﬁcients of kinetic friction between the shaft and the left
and right bearings are 0.20 and 0.26, respectively. lWhat
couple is required to rotate the shaft at a constant rate? Solution: The Iq‘rbenring: The parameters are r0 = 38 rum. 1”; =0. e: 245°. in, 20.2.
and F =SDCIN. The moment required to sustain a constant rate of rotation is 2}.th J‘g — 3"?
3COSQ r3 —rl2 Mk”, 2 ) 25.73Nm. The right hearing: This is a ﬁatend hearing. The parameters are pig 2 0.26. r = 15 mm. and F = 800 N. The moment required to
sustain a constant rate of rotation is .trﬁght = 2mg“ = new m. The sum ofthe moments: AI = 5.?3 + ECG 2 ?.81 N rn Problem 8.99 The couple required to turn the wheel of the exercise bicycle is adjusted by changing the weight H". The coefﬁcient oi'kinetic friction hetween the Wheel and the belt is Ink. Assume the wheel turns clockwise. Ca] Show that the couple :‘lf required to turn the wheel
is M = n‘Rn _ 934“). {b} If H" = 40 lb and Ink 2 0.2, what iorce will the
scale S indicate when the bicycle is in use? Solution: Let :3 he the angle in radians of the belt contact with wheel. The tension in the top belt when the belt slips is Tapper =
We—“k'g. The tensioninthelowerbeit isle“ = TV. The moment applied to the wheel is
M = REED,” —Tuppﬂ.) = RWU — 9—"k’9j. This is the moment required to turn the wheel at aoonstant rate. The
angle .6 in radians is
7r 180
from which M = Hl’lr’il — e‘a'it‘k 1]. [hi The upper belt tension is :3 = 7r + (30 —15;. 1: ) =3_4o radians. Twp" 2409'3'4m'2l = 20.28113.
This is also the reading of the scale 5. ...
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This note was uploaded on 11/29/2010 for the course MECH&AE 101 taught by Professor Mal during the Fall '08 term at UCLA.
 Fall '08
 MAL

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