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:
SOLUTION
Frictional Forces on Screw. Here u=tan–1
al
2pr
b
=tan–1
c3
2p (10)
d
=2.7336°,
M
A
829
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
Since the screws are being tightened, Eq. 8–3 should be used. Here,
Referring to the free-body diagram of wedge Bshown in Fig. ausing the result of T,
we have
u= tan –1aL
2prb= tan –1c7.5
2p(12.5) d=5.455°;
mœ
s=0.15.
ms=0.2,
A
B
250 mm
15 15
C
830
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
Referring to the free-body diagram of wedge Cshown in Fig. a, we have
and W= T= 4166.68N.Since
Mmust overcome the friction of two screws,
fs=tan–1ms=tan–1(0.15) =8.531°; M=P(0.25);
mœ
s=0.15.
ms=0.2,
A
B
250 mm
15 15
C
Ans:
831
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°
SOLUTION
Frictional Force on Flat Belt: Here, and .
Applying Eq. 8–6, we have
T
2=250(9.81) =2452.5 NT
1=F
F
832
*8–84.
SOLUTION
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
Ans:
833
8–85.
SOLUTION
Since the cow is on the verge of moving,the force it exerts on the rope is
and the force exerted by the man on the rope is T1.Here,.Thus,
b=2(2p)=4prad
T
2=250 lb
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
Ans:
834
8–86.
SOLUTION
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
Ans:
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
SOLUTION
b=p
2
A
Ans:
836
*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,
SOLUTION
15
B
30
Ans:
8–89.
SOLUTION
Block A:
Plate B:
Solving Eqs. (1)–(5) yields
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:
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
T
2=T
1emb,
T
1=T¿.T
2=Tb=u
2,
A
T
θ
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
SOLUTION
Frictional Force on Flat Belt: If the normal force between the man and the boat is
equal to zero, then, and . Applying Eq. 8–6, we have
The least number of half turns of the rope required is turns.Thus
Equations of Equilibrium: From FBD (a),
8.980
p=2.86
T
2=500 lbT
1=130 lb
A
C
B
840
*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
Frictional Force on Flat Belt. Here b=270°=
3
p
2
rad.
700 mm
60 mm
25 mm
C
300 mm
B
M
A
D
P
841
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
Frictional Force on Flat Belt. Here b=270°=
3
p
2
rad.
T
D=
T
A
emb
A
842
*8–96.
SOLUTION
Equations of Equilibrium and Friction: Since the block is on the verge of sliding
up or down the plane, then, If the block is on the verge of sliding
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
If the block is on the verge of sliding down the plane, and
T
2=28.28 lb.T
1=W
T
2=W.T
1=42.43 lb
b=45° +90° =135° =3p
4
F=msN=0.2N.
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:
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
Ans:
M=75.4
N#m
844
8–98.
SOLUTION
FBD of a section of the belt is shown.
Proceeding in the general manner:
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
Wheel:
*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
Frictional Forces on Flat Belts: When the cord pass over peg D, b=180°=p rad
and T2=P. Applying Eq. 8–6, T2=T1 emb, we have
P=T1
e0.1 p T1=0.7304P
Equations of Equilibrium: From FDB (b),
+
c
ΣFy=0; NB–98.1=0 NB=98.1 N
From FDB (b),
µ
µ
µ
P
300 mm
400 mm
AC
D
B
D = 0.1
C = 0.4
B = 0.4
A = 0.3
µ
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,
SOLUTION
a
Ans: