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828
*8–80.
If the required clamping force at the board A is to be 2 kN,
determine the torque M that must be applied to the screw
to tighten it down. The square-threaded screw has a mean
radius of 10 mm and a lead of 3 mm, and the coefficient of
static friction is m
s=0.35.
Ans:
8.09 N #m
SOLUTION
Frictional Forces on Screw. Here u=tan-1
al
2pr
b
=tan-1
c3
2p (10)
d
=2.7336°,
W=2000 N
and f
s=
tan
-1
m
s=
tan
-1
(0.35)
=
19.2900
°
.
M=Wr tan (
u
+
fs
)
=2000 (0.01) tan (2.7336°+19.2900°)
=8.09 N #m
Ans.
M
A
8–81.
If a horizontal force of P= 100 N is applied perpendicular
to the handle of the lever at A, determine the compressive
force Fexerted on the material. Each single square-
threaded screw has a mean diameter of 25 mm and a lead of
7.5 mm. The coefficient of static friction at all contacting
surfaces of the wedges is and the coefficient of
static friction at the screw is
SOLUTION
; and ,
where Tis the tension in the screw shank.Since Mmust overcome the friction of
two screws,
Ans.
Referring to the free-body diagram of wedge Bshown in Fig. ausing the result of T,
we have
(1)
(2)
Solving,
Using the result of Nand referring to the free-body diagram of wedge Cshown in
Fig. b, we have
Ans.F=11563.42 N =11.6 kN
2(6324.60) cos 15° -230.2(6324.60) sin 15°4-F=0+c©F
y=0;
N¿=5781.71 NN=6324.60 N
N¿+0.2N sin 15° -N cos 15° =0+c©F
y=0;
4015.09 -0.2N¿-0.2N cos 15° -N sin 15° =0©F
x=0;
:
+
T=4015.09 N =4.02 kN
25 =23T(0.0125) tan (8.531° +5.455°)4
M=2[Wr, tan(fs+u)4
W=Tfs=tan-1ms=tan-1(0.15) =8.531°; M=100(0.25)=25 N #m
mœ
s=0.15.
ms=0.2,
A
B
250 mm
15 15
C
8–82.
Determine the horizontal force Pthat must be applied
perpendicular to the handle of the lever at Ain order to
develop a compressive force of 12 kN on the material.Each
single square-threaded screw has a mean diameter of 25 mm
and a lead of 7.5 mm. The coefficient of static friction at all
contacting surfaces of the wedges is and the
coefficient of static friction at the screw is
SOLUTION
Using the result of Nand referring to the free-body diagram of wedge Bshown in
Fig. b, we have
Since the screw is being tightened, Eq. 8–3 should be used. Here,
and W= T= 4166.68N.Since
Mmust overcome the friction of two screws,
Ans.P=104 N
P(0.25) =234166.68(0.0125) tan (8.531° +5.455°)4
M=23Wr tan (fs+u)4
fs=tan-1ms=tan-1(0.15) =8.531°; M=P(0.25);
u=tan-1cL
2prd=tan-1c7.5
2p(12.5) d=5.455°;
T=4166.68 N
T-6563.39 sin 15° -0.2(6563.39) cos 15° -0.2(6000) =0©F
x=0;
:
+
N¿=6000 N
N¿-6563.39 cos 15° +0.2(6563.39) sin 15° =0+c©F
y=0;
N=6563.39 N
mœ
s=0.15.
ms=0.2,
A
B
250 mm
15 15
C
8–83.
Acylinder having a mass of 250 kg is to be supported by the
cord which wraps over the pipe. Determine the smallest
vertical force Fneeded to support the load if the cord
passes (a) once over the pipe,,and (b) two times
over the pipe,.Take .ms=0.2b=540°
b=180°
*8–84.
SOLUTION
Applying Eq. 8–6, we have
a) If
Ans.
b) If
Ans.F=16 152.32 N =16.2 kN
F=2452.5e0.2(3 p)
T
2=T
1emb
b=540° =3prad
F=4597.10 N =4.60 kN
F=2452.5e0.2 p
T
2=T
1emb
b=180° =prad
Acylinder having a mass of 250 kg is to be supported by the
cord which wraps over the pipe. Determine the largest
vertical force Fthat can be applied to the cord without
moving the cylinder.The cord passes (a) once over the pipe,
, and (b) two times over the pipe,.Take
.ms=0.2
b=540°b=180°
F
833
8–85.
SOLUTION
and the force exerted by the man on the rope is T1.Here,.Thus,
Using this result and referring to the free - body diagram of the man shown in Fig. a,
Since , the man will not slip, and he will
successfully restrain the cow.
F6F
max =ms¿N=0.3(180) =54 lb
:
+©F
x=0; 37.96 -F=0F=37.96 lb
+c©F
y=0; N-180 =0N=180 lb
T
1=37.96 lb
250 =T
1e0.15(4p)
T
2=T
1emsb
b=2(2p)=4prad
A 180-lb farmer tries to restrain the cow from escaping by
wrapping the rope two turns around the tree trunk as
shown. If the cow exerts a force of 250 lb on the rope,
determine if the farmer can successfully restrain the cow.
The coefficient of static friction between the rope and the
tree trunk is , and between the farmer’s shoes and
the ground .ms
œ=0.3
ms=0.15
834
8–86.
SOLUTION
Yes, just barely. Ans.
F
max =0.8 (185) =148 lb 7136.9 lb
F=136.9 lb
:
+©F
x=0; 136.9 -F=0
N=185 lb
+c©F
y=0; N-185 =0
T
2=T
1emb =100 e0.2 p
2=136.9 lb
b=p
2
The 100-lb boy at Ais suspended from the cable that passes
over the quarter circular cliff rock. Determine if it is
possible for the 185-lb woman to hoist him up; and if this is
possible, what smallest force must she exert on the
horizontal cable? The coefficient of static friction between
the cable and the rock is , and between the shoes of
the woman and the ground .ms
œ=0.8
ms=0.2
A
835
8–87.
The 100-lb boy at Ais suspended from the cable that passes
over the quarter circular cliff rock. What horizontal force
must the woman at Aexert on the cable in order to let the
boy descend at constant velocity? The coefficients of static
and kinetic friction between the cable and the rock are
and , respectively.mk=0.35ms=0.4
*8–88.
Theuniform concrete pipe has aweight of 800 lb and is
unloaded slowly from the truck bed using the rope and skids
shown. If the coefficient of kinetic friction between the rope
and pipe is determine the force the worker must
exert on the rope to lower the pipe at constant speed. There
isapulley at B,and the pipe does not slip on the skids.The
lower portion of the rope is parallel to the skids.
mk=0.3,
Ans.T1=73.3 lb
T2=T1emb, 203.466 =T1e(0.3)(195°
180°)(p)
b=180° +15° =195°
T2=203.466 lb
15
B
30
837
8–89.
SOLUTION
Block A:
Aca
bl
e
i
s attac
h
e
d
to t
h
e 20-
k
g p
l
ate B, passes over a f
i
xe
d
peg at C, and is attached to the block at A. Using the
coefficients of static friction shown, determine the smallest
mass of block Aso that it will prevent sliding motion of B
down the plane.
30
B
A
m
A
0.2
m
C
0.3
C
m
B
0.3
Ans:
mA=2.22 kg
838
8–90.
The smooth beam is being hoisted using a rope which is
wrapped around the beam and passes through a ring at Aas
shown. If the end of the rope is subjected to a tension Tand
the coefficient of static friction between the rope and ring is
determine the angle of for equilibrium.ums=0.3,
SOLUTION
Equation of Equilibrium:
Frictional Force on Flat Belt: Here, and Applying Eq. 8–6
we have
(2)
Substituting Eq. (1) into (2) yields
Solving by trial and error
The other solution, which starts with T' = Te0.3(0/2) based on cinching the
ring tight, is 2.4326 rad = 139°. Any angle from 99.2° to 139° is equilibrium.
Ans.u=1.73104 rad =99.2°
e0.15 u=2 cos u
2
2T¿cos u
2=T¿e0.15 u
T=T¿e0.31u>22=T¿e0.15 u
T
2=T
1emb,
T
1=T¿.T
2=Tb=u
2,
2=0T=2T¿cos u
2
A
T
θ
Ans:
u=92.2°
839
8–91.
The boat has a weight of 500 lb and is held in position off the
side of a ship by the spars at Aand B.A man having a weight
of 130 lb gets in the boat, wraps a rope around an overhead
boom at C,and ties it to the end of the boat as shown. If the
boat is disconnected from the spars,determine the minimum
number of half turns the rope must make around the boom
so that the boat can be safely lowered into the water at
constant velocity.Also,what is the normal force between the
boat and the man? The coefficient of kinetic friction
between the rope and the boom is .Hint:The
problem requires that the normal force between the man’s
feet and the boat be as small as possible.
ms=0.15
The least number of half turns of the rope required is turns.Thus
Use half turns Ans.
Equations of Equilibrium: From FBD (a),
From FBD (b),
Frictional Force on Flat Belts: Here,.Applying Eq. 8–6, we have
Ans.N
m=6.74 lb
N
m+500 =(130-N
m)e0.15 (3 p)
T
2=T
1emb
b=3prad
+c©F
y=0; T
1+N
m-130 =0T
1=130 -N
m
+c©F
y=0; T
2-N
m-500 =0T
2=N
m+500
n=3
8.980
p=2.86
b=8.980 rad
500 =130e0.15b
T
2=T
1emb
A
C
B
Ans:
n=3 half turns
Nm=6.74 lb
*8–92.
Determine the force P that must be applied to the handle of
the lever so that B the wheel is on the verge of turning if
M
=
300 N
#
m. The coefficient of static friction between
the belt and the wheel is m
s=0.3.
SOLUTION
3
p
700 mm
60 mm
25 mm
C
300 mm
B
M
A
D
P
8–93.
If a force of P
=
30 N is applied to the handle of the lever,
determine the largest couple moment M that can be resisted
so that the wheel does not turn. The coefficient of static
friction between the belt and the wheel is m
s=0.3.
700 mm
60 mm
25 mm
C
300 mm
B
M
A
D
P
SOLUTION
3
p
842
*8–96.
SOLUTION
up the plane [FBD (a)],
If the block is on the verge of sliding down the plane [FBD (b)],
Frictional Force on Flat Belt: Here, rad. If the block
is on the verge of sliding up the plane, and
Ans.
If the block is on the verge of sliding down the plane, and
Ans.W=13.95 lb =13.9 lb
28.28 =We0.3
A
3p
4
B
T
2=T
1emb
T
2=28.28 lb.T
1=W
=86.02 lb =86.0 lb
W=42.43e0.3
A
3p
4
B
T
2=T
1emb
T
2=W.T
1=42.43 lb
b=45° +90° =135° =3p
4
T
2+0.2135.362-50 sin 45° =0T
2=28.28 lbQ+©F
x¿=0;
N-50 cos 45° =0N=35.36 lba+©F
y¿=0;
T
1-0.2135.362-50 sin 45° =0T
1=42.43 lbQ+ ©F
x¿=0;
N-50 cos 45° =0N=35.36 lba+©F
y¿=0;
Determine the maximum and the minimum values of
weight Wwhich may be applied without causing the 50-lb
block to slip.The coefficient of static friction between the
block and the plane is and between the rope and
the drum Dmœ
s=0.3.
ms=0.2,
45°
W
D
Ans:
W=86.0 lb
W=13.9 lb
843
8–97.
Granular material, having adensity of is
transported onaconveyor belt that slides over the fixed
surface,having acoefficient of kinetic friction of
Operation of the belt is provided by amotor that supplies a
torque Mto wheel A.The wheel at Bis free to turn, and the
coefficient of static friction between the wheel at Aand
the belt is If the belt is subjected to apretension
of 300 Nwhen no load is on the belt, determine the
greatest volume Vof material that is permitted on the belt
at any time without allowing the belt to stop.What is the
torque Mrequired to drive the belt when it is subjected to
this maximum load?
mA=0.4.
mk=0.3.
1.5 Mg>m3,
SOLUTION
100mm
100mm
BA
M
m
k
0.3 m
A
0.4
Ans:
M=75.4
N#m
V=0.171 m3
844
8–98.
SOLUTION
Replace ,
Using this and , the above relations become
Combine
Integrate from
to
we get,
Q.E.D
T
2=T
1e
¢
mb
sin a
2
≤
u=b,T=T
2
u=0, T=T
1
dT
T=mdu
sin a
2
Tdu=2adN sin a
2b
dT =2mdN
(dT)(du):0
dF =mdN
cos du
2by 1,
sin du
2by du
2
©F
y=0; -(T+dT) sin du
2-Tsin du
2+2dN sin a
2=0
Show that the frictional relationship between the belt
tensions, the coefficient of friction , and the angular
contacts and for the V-belt is .T2=T1emb>sin(a>2)
ba
m
T
2
T
1
Impending
motion
b
a
845
8–99.
SOLUTION
The wheel is subjected toatorque of If the
coefficient of kinetic friction between the band brake and
the rim of the wheel is determine the smallest
horizontal force Pthat must be applied to the lever to stop
the wheel.
mk=0.3,
M=50 N #m.
400 mm
M
C
P
Ans:
P=53.6
N
846
*8–100.
Blocks A and B have a mass of 7 kg and 10kg, respectively.
Using the coeffi cients of static friction indicated, determine
the largest vertical force P which can be applied to the cord
without causing motion.
SOLUTION
µ
µ
µ
P
300 mm
400 mm
AC
D
B
D = 0.1
C = 0.4
B = 0.4
A = 0.3
µ
Ans:
P=223 N
847
8–101.
The uniform bar AB is supported by a rope that passes over a
frictionless pulley at Cand a fixed peg at D.If the coefficient
of static friction between the rope and the peg is
determine the smallest distance xfrom the end of the bar at
which a 20-N force may be placed and not cause the bar
to move.
mD=0.3,
CD
Ans:
x=0.384 m
