8–102.
50 mm
A
B
D
45°
30°
M= 0.8 N⋅m
The belt on the portable dryer wraps around the drum D,
idler pulley A, and motor pulley B. If the motor can develop
a maximum torque of determine the
smallest spring tension required to hold the belt from
slipping.The coefficient of static friction between the belt
and the drum and motor pulley is ms=0.3.
M=0.80 N #m,
SOLUTION
8–103.
SOLUTION
Blocks A and B weigh 50 lb and 30 lb, respectively. Using
the coeffi cients of static friction indicated, determine the
greatest weight of block D without causing motion.
A
B
C
D
m0.5
mBA 0.6
mAC 0.4
20
*8–104.
From FBD (b),
a(2)
Frictional Force on Flat Belt: Here,Applying Eq. 8–6,
we have
(3)
Solving Eqs. (1), (2), and (3) yields
Ans.
T
1=42.97 N T
2=110.27 N
M=3.37 N #m
T
2=T
1e0.3p=2.566T
1
T
2=T
1emb,
b=180° =prad.
M+T
110.052–T
210.052=0+©MO=0;
The 20-kg motor has a center of gravity at Gand is pin-
connected at Cto maintain a tension in the drive belt.
Determine the smallest counterclockwise twist or torque M
that must be supplied by the motor to turn the disk Bif
wheel Alocks and causes the belt to slip over the disk. No
slipping occurs at A.The coefficient of static friction
between the belt and the disk is ms=0.3.
50 mm
M
50 mm
150 mm
100 mm
B
C
AG
8–105.
A
B
E
C
uu
A 10-kg cylinder D, which is attached to a small pulley B,is
placed on the cord as shown. Determine the largest angle
so that the cord does not slip over the peg at C.The cylinder
at Ealso has a mass of 10 kg, and the coefficient of static
friction between the cord and the peg is
SOLUTION
In the case where cylinder Eis on the verge of ascending, and
Here,,Fig. b.Thus,
ln
Solving by trial and error, yields
In the case where cylinder E is on the verge of descending, T2= 10(9.81) N and
T1= . Here,.Thus,
ln
Solving by trial and error, yields
Thus, the range of at which the wire does not slip over peg Cis
Ans.
u
max =38.8°
24.2° 6u638.8°
u
u=0.6764 rad =38.8°
(2 sin u)=0.1ap
2+ub
0.1 ap
2
+ ub
10(9.81) =49.05
sin u e
T
2=T
1e m sb
p
2+u
49.05
sin u
u=0.4221 rad =24.2°
0.5
sin u=0.1ap
2+ub
0.1 ap
2
+ ub
49.05
sin u=10(9.81) e
T
2=T
1e ms
b
p
2+uT
1=10(9.81) N.
T
2=T=49.05
sin u
sin u
ms=0.1.
X
8–106.
Applying Eq. 8–6, we have
Equations of Equilibrium: From FBD (a),
a
Ans.
From FBD (b),
Thus, the spring stretch is
Ans.x=F
sp
k=1143.57
4000 =0.2859 m =286 mm
F
sp –21578.712=0F
sp =1143.57 N:
+©F
x=0;
M=50.0 N #m
M+571.7810.12–1500 +578.1210.12=0+©MO=0;
T=571.78 N
500 +T=Te0.2p
T
2=T
1emb
T
1=T.
A conveyer belt is used to transfer granular material and
the frictional resistance on the top of the belt is
Determine the smallest stretch of the spring attached to the
moveable axle of the idle pulley Bso that the belt does not
slip at the drive pulley Awhen the torque Mis applied.
What minimum torque Mis required to keep the belt
moving? The coefficient of static friction between the belt
and the wheel at Ais ms=0.2.
F=500 N.
0.1 m 0.1 m
F= 500 N
k=4kN/m
A
B
M
8–107.
The collar bearing uniformly supports an axial force of
P
=
5 kN. If the coefficient of static friction is m
s=0.3
,
determine the smallest torque M required to overcome
friction.
s=0.3
200 mm
150 mm
P
M
*8–108.
The collar bearing uniformly supports an axial force of
P
=
8 kN. If a torque of M
=
200 N
#
m is applied to the
shaft and causes it to rotate at constant velocity, determine
the coefficient of kinetic friction at the surface of contact.
SOLUTION
200 mm
150 mm
P
M
855
8–109.
The floor-polishing machine rotates at a constant angular
velocity. If it has a weight of 80 lb. determine the couple
forces Fthe operator must apply to the handles to hold the
machine stationary.The coefficient of kinetic friction
between the floor and brush is Assume the brush
exerts a uniform pressure on the floor.
mk=0.3.
SOLUTION
Ans.F=10.7 lb
F(1.5) =2
3(0.3) (80)(1)
M=2
3mPR
2ft
1.5 ft
Ans:
F=10.7
lb
8–110.
The double-collar bearing issubjectedto an axial force
Assuming that collar Asupports 0.75Pand
collar Bsupports 0.25P,both with auniform distribution of
pressure,determine the maximum frictional moment M
that may be resisted by the bearing.Take for both
collars.
ms=0.2
P=4 kN.
SOLUTION
Ans.=16.1 N #m
M=2
5(0.2)
¢
(0.03)3–(0.01)3
(0.03)2–(0.01)2(0.75) (4000) +(0.02)3–(0.01)3
(0.02)2–(0.01)2(0.25) (4000)
≤
M=2
3msP
¢
R3
2–R3
1
R2
2–R2
1
≤
A
B
30 mm
10 mm
20 mm
P
M
857
8–111.
The double-collar bearing is subjected to an axial force
P
=
16 kN. Assuming that collar A supports 0.75P and
collar B supports 0.25P, both with a uniform distribution of
pressure, determine the smallest torque M that must be
applied to overcome friction. Take m
s=0.2
for both collars.
P
30 mm
50 mm
A
B
M
75 mm
100 mm
Ans:
M=237 N #m
858
*8–112.
The pivot bearing is subjected to a pressure distribution at
its surface of contact which varies as shown. If the
coefficient of static friction is determine the torque M
required to overcome friction if the shaft supports an axial
force P.
m,
SOLUTION
Thus, Ans.M=0.521 PmR
=1.454p0R2
=4p0R2a1–2
pb
=p0
B
1
A
p
2R
B
2cos apr
2Rb+r
A
p
2R
B
sin apr
2Rb
R
0
R12p2
P=L
A
dN =LR
0
p0acos apr
2RbrdrbL2p
0
du
=0.7577mp0R3
=mp0
¢
16R3
p2
≤
cap
2b2
–2d
=mp0
B
2r
A
p
2R
B
2cos apr
2Rb+
A
p
2R
B
2r2–2
A
p
2R
B
3sin apr
2Rbd0
R12p2
=mp0LR
0ar2cos apr
2RbdrbL2p
0
du
M=L
rmp0cos apr
dF =mdN =mp0cos apr
2RbdA
P
M
2R
r
R
Ans:
M=0.521 P
m
R
859
8–113.
SOLUTION
Ans.=2msP
R2cos u
R3
3=2msPR
3 cos u
M=LrdF =LmsrdN =2msP
R2cos uLR
0
r2dr
dN =pdA =P
pR2
¢
2prdr
cos u
≤
=2P
R2cos urdr
P=ppR2p=P
pR2
P=Lpcos udA =Lpcos u
¢
2prdr
cos u
≤
=2ppLR
0
rdr
dr
The conical bearing is subjected to a constant pressure
distribution at its surface of contact. If the coefficient of
static friction is determine the torque Mrequired to
overcome friction if the shaft supports an axial force P.
ms,
P
M
R
Ans:
M=
2
ms
PR
3 cos u
860
8–114.
T
h
e4–
i
n.-
di
ameter s
h
aft
i
s
h
e
ld i
nt
h
e
h
o
l
esuc
h
t
h
at t
h
e
normal pressure acting around the shaft varies linearly with
its depth as shown. Determine the frictional torque that
must be overcome to rotate the shaft. Take ms=0.2.
SOLUTION
The differential area (shaded)
Ans.
=
1440p(0.2)
=
905 lb
#
in.
=1440pm lb #in.
T=2LdF =2LmdN =80pm L6
0
xdx
dN =pdA =10x(4pdx)=40pxdx
dA =2p(2)dx =4pdx
M
6in.
60 lb/in
2
Ans:
T=905
lb #in.
8–115.
200 mm
F
A
D
P
100 mm
125 mm
150 mm
S
M0.8 N m
C
SOLUTION
a
Ans.P=118 N
+©MF=0; 88.525(0.2) –P(0.15) =0
P¿=88.525 N
4.00 =2
3(0.4) (P¿)a(0.125)3–(0.1)3
(0.125)2–(0.1)2b
M=2
3mP¿aR3
2–R3
1
R2
2–R2
1b
M=26.667(0.150) =4.00 N #m
The plate clutch consists of a flat plate A that slides over the
rotating shaft S. The shaft is fixed to the driving plate gear B.
If the gear C, which is in mesh with B, is subjected to a
torque of M = 0.8N # m, determine the smallest force P, that
must be applied via the control arm, to stop the rotation.
The coefficient of static friction between the plates A and D
is ms = 0.4. Assume the bearing pressure between A and D to
be uniform.
*8–116.
The collar fits loosely around a fixed shaft that has a radius
of 2 in. If the coefficient of kinetic friction between the shaft
and the collar is determine the force Pon the
horizontal segment of the belt so that the collar rotates
counterclockwise with a constant angular velocity. Assume
that the belt does not slip on the collar; rather, the collar
slips on the shaft. Neglect the weight and thickness of the
belt and collar.The radius, measured from the center
of the collar to the mean thickness of the belt, is 2.25 in.
mk=0.3,
SOLUTION
Equilibrium:
Hence
a
Ans.P=13.8 lb
+©MO=0; –a2P2+202b(0.5747) +20(2.25) –P(2.25) =0
R=2R2
x+R2
y=2P2+202
:
+©Fx=0; P–Rx=0Rx=P
+c©Fy=0; Ry–20 =0Ry=20 lb
rf=2 sin 16.699°=0.5747 in.
fk=tan–1mk=tan–10.3 =16.699°
20 lb
P
2in.
2.25 in.
863
8–117.
Equilibrium:
Hence
a
Ans.P=29.0 lb
+©MO=0; a2P2+202b(0.5747) +20(2.25) –P(2.25) =0
R=2R2
x+R2
y=2P2+202
:
+©Fx=0; P–Rx=0Rx=P
+c©Fy=0; Ry–20 =0Ry=20 lb
rf=2 sin 16.699° =0.5747 in.
The collar fits loosely around a fixed shaft that has a radius of
2in. If the coefficient of kinetic friction between the shaft and
the collar is determine the force Pon the horizontal
segment of the belt so that the collar rotates clockwise with a
constant angular velocity.Assume that the belt does not slip
on the collar; rather,the collar slips on the shaft. Neglect the
weight and thickness of the belt and collar.The radius,
measured from the center of the collar to the mean thickness
of the belt, is 2.25 in.
mk=0.3,
P
2in.
2.25 in.
Ans:
P=29.0
lb
8–118.
The pivot bearing is subjected to a parabolic pressure
distribution at its surface of contact. If the coefficient of
static friction is ,determine the torque Mrequired to
overcome friction and turn the shaft if it supports an axial
force P.
ms
Ans.=8
15 msPR
M=LrdF =LmsrdN =2msP
pR2L2p
0
duLR
0
r2
¢
1–r2
R2
≤
dr
dN =pdA =2P
pR2
¢
1–r2
R2
≤
(rdu)(dr)
P=pR2p0
2p0=2P
pR2
P=LpdA=Lp0
¢
1–r2
R2
≤
(rdu)(dr)=p0L2p
0
duLR
0
r
¢
1–r2
R2
≤
dr
P
p
0
pp
0
(1)
r
2
––
R
2
M