808
*8–60.
The wedge is used to level the member. Determine the
horizontal force P that must be applied to begin to pushthe
wedge forward. The coefficient of static friction between the
wedge and the two surfaces of contact is m
s=0.2.
Neglect
the weight of the wedge.
SOLUTION
Equations of Equilibrium and Friction. Since the wedge is required to be on the
Using this result and the FBD of the wedge, Fig. b,
2 m
1 m
500 N/m
A
B
C
P
5
809
8–61.
SOLUTION
Block A:
Solving,
The two blocks used in a measuring device have negligible
weight. If the spring is compressed 5 in. when in the position
shown, determine the smallest axial force Pwhich the
adjustment screw must exert on Bin order to start the
movement of Bdownward. The end of the screw is smooth
and the coefficient of static friction at all other points of
contact is .ms=0.3
60
45
k = 20 lb/in.
B
A
P
Ans:
810
8–62.
If P= 250 N, determine the required minimum compression
in the spring so that the wedge will not move to the right.
Neglect the weight of Aand B.The coefficient of static
friction for all contacting surfaces is Neglect
friction at the rollers.
SOLUTION
Free-Body Diagram:The spring force acting on the cylinder is .
Equations of Equilibrium:Referring to the FBD of the cylinder,Fig.a,
Referring to the FBD of the wedge shown in Fig.b,
F
sp=kx =15(103)x
ms=0.35. k 15 kN/m
A
P
B
10
811
8–63.
SOLUTION
From FBD (b),
Determine the minimum applied force Prequired to move
wedge Ato the right. The spring is compressed a distance of
175 mm. Neglect the weight of Aand B.The coefficient of
static friction for all contacting surfaces is
Neglect friction at the rollers.
ms=0.35. k=15kN/m
P
B
Ans:
812
*8–64.
If the coefficient of static friction between all the surfaces of
contact is
ms
, determine the force P that must be applied to
the wedge in order to lift the block having a weight W.
SOLUTION
Equations of Equilibrium and Friction. Since the wedge is required to be on
the verge sliding to the left, then slipping will have to occur at both of its contact
Solving Eqs. (1) to (4)
P
A
C
B
a
813
8–65.
Determine the smallest force Pneeded to lift the 3000-lb
load.The coefficient of static friction between Aand Cand
between Band Dis ,and between Aand B.
Neglect the weight of each wedge.
ms¿=0.4ms=0.3
SOLUTION
From FBD (a):
Solving Eqs. (1) and (2) yields:
3000 lb
15°
P
A
B
D
C
Ans:
814
8–66.
SOLUTION
From FBD (a):
Solving Eqs. (1) and (2) yields:
Determine the reversed horizontal force needed to pull
out wedge A.The coefficient of static friction between Aand
Cand between Band Dis ,and between Aand B
.Neglect the weight of each wedge.ms¿=0.1
ms=0.2
–P
3000 lb
15°
P
A
B
D
C
815
8–67.
If the clamping force at Gis 900 N, determine the horizontal
force Fthat must be applied perpendicular to the handle of
the lever at E.The mean diameter and lead of both single
square-threaded screws at Cand Dare 25 mm and 5 mm,
respectively.The coefficient of static friction is .
SOLUTION
ms=0.3
200 mm
E
D
C
G
200 mm
B
A
Ans:
816
*8–68.
If a horizontal force of F= 50 N is applied perpendicular to
the handle of the lever at E, determine the clamping force
developed at G.The mean diameter and lead of the single
square-threaded screw at Cand Dare 25 mm and 5 mm,
respectively.The coefficient of static friction is
SOLUTION
Since the screw is being tightened, Eq. 8–3 should be used. Here,
u=tan–1aL
ms=0.3.
200 mm
E
D
C
G
200 mm
B
A
Ans:
8–69.
The column is used to support the upper floor. If a force
is applied perpendicular to the handle to tighten
the screw, determine the compressive force in the column.
The square-threaded screw on the jack has a coefficient of
static friction of mean diameter of 25 mm, and a
lead of 3 mm.
SOLUTION
ms=0.4,
F=80 N
0.5 m
F
Ans:
8–70.
If the force F is removed from the handle of the jack
in Prob. 8–69, determine if the screw is self-locking.
SOLUTION
0.5 m
F
8–71.
If couple forces of F= 10 lb are applied perpendicular to the
lever of the clamp at Aand B, determine the clamping force
on the boards.The single square-threaded screw of the
clamp has a mean diameter of 1 in. and a lead of 0.25 in. The
coefficient of static friction is .
SOLUTION
Since the screw is being tightened, Eq. 8–3 should be used. Here,
ms=0.3
6 in.6 in.
B
A
820
*8–72.
If the clamping force on the boards is 600 lb, determine the
required magnitude of the couple forces that must be
applied perpendicular to the lever AB of the clamp at Aand
Bin order to loosen the screw. The single square-threaded
screw has a mean diameter of 1 in. and a lead of 0.25 in. The
coefficient of static friction is .
SOLUTION
Since the screw is being loosened, Eq. 8–5 should be used. Here,
ms=0.3
6 in.6 in.
B
A
821
8–73.
Prove that the lead lmust be less than for the jack
screw shown in Fig. 8–15 to be “self-locking.”
2
prms
SOLUTION
W
h
M
8–74.
SOLUTION
The square-threaded bolt is used to join two plates
together. If the bolt has a mean diameter of
and a lead of , determine the smallest torque M
required to loosen the bolt if the tension in the bolt is
.The coefficient of static friction between the
threads and the bolt is .ms=0.15
T=40 kN
l=3 mm
d=20 mm
M
Ans:
823
8–75.
The shaft has a square-threaded screw with a lead of 8 mm
and a mean radius of 15 mm. If it is in contact with a plate
gear having a mean radius of 30 mm, determine the resisting
torque Mon the plate gear which can be overcome if a
torque of is applied to the shaft. The coefficient of
static friction at the screw is Neglect friction of
the bearings located at Aand B.
mB=0.2.
7N#m
SOLUTION
15 mm
M
30 mm
B
824
*8–76.
SOLUTION
If couple forces of are applied to the handle of the
machinist’s vise,determine the compressive force developed in
the block. Neglect friction at the bearing A.The guide at Bis
smooth. The single square-threaded screw has a mean radius
of 6 mm and a lead of 8 mm, and the coefficient of static
friction is ms=0.27.
F=35 N
125mm
F
AB
Ans:
825
8–77.
The square-threaded screw has a mean diameter of 20 mm
and a lead of 4 mm. If the weight of the plate Ais5lb,
determine the smallest coefficient of static friction between
the screw and the plate so that the plate does not travel
down the screw when the plate is suspended as shown.
SOLUTION
A
8–78.
The device is used to pull the battery cable terminal C
from the post of a battery. If the required pulling force is
85 lb, determine the torque M that must be applied to the
handle on the screw to tighten it. The screw has square
threads, a mean diameter of 0.2 in., a lead of 0.08 in., and the
coeffi cient of static friction is ms=0.5.
M
Ans:
827
8–79.
Determine the clamping force on the board A if the screw
is tightened with a torque of
M=8 N #m.
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
M
A