*8–60.
The wedge is used to level the member. Determine the
horizontal force P that must be applied to begin to pushthe
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
2 m
1 m
500 N/m
A
B
C
P
5
809
8–61.
SOLUTION
Block B:
Solving,
Ans.P=39.6 lb
NA=80.5 lb
NB=82.5 lb
N¿=105.9 lb
+c©Fy=0;
NB cos 45° +0.3 NB cos 45° +0.3 N¿ cos 60° –N¿ sin 60° =0
:
+©Fx=0;
NB sin 45° –NB sin 45° +P–0.3N¿ sin 60° –N¿ cos 60° =0
+c©Fy=0;
0.3 NA–0.3 N¿ cos 60° +N¿ sin 60° –100 =0
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:
P=
39.6 lb
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
Since it is required that the wedge is on the verge to slide to the right, the frictional
force must act to the left on the top and bottom surfaces of the wedge and their
magnitude can be determined using friction formula.
Equations of Equilibrium:Referring to the FBD of the cylinder,Fig.a,
Referring to the FBD of the wedge shown in Fig.b,
Ans.x=0.01830 m=18.3 mm
=0–316.233(103)x4sin 10°
250 –5.25(103)x–0.35316.233(103)x4cos 10°©F
x=0;
:
+
N
2=16.233(103)x
N
2 cos 10° –0.35N
2 sin 10° –15(103)x=0+c©F
y=0;
Thus, (F
f)1=0.35315(103)x4=5.25(103)x
N
1=15(103)xN
1–15(103)x=0+c©F
y=0;
(F
f)2=0.35N
2
(F
f)1=mN
1=0.35N
1
ms=0.35. k 15 kN/m
P
B
Ans:
x=18.3
mm
811
8–63.
SOLUTION
From FBD (b),
Ans.P=2.39 kN
–2.841 sin 10° =0
P–0.3512.6252–0.3512.8412cos 10°:
+©F
x=0;
N
A=2.841 kN
N
Acos 10° –0.35N
Asin 10° –2.625 =0+c©F
y=0;
N
B–2.625 =0N
B=2.625 kN+c©F
y=0;
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
A
P
B
Equations of Equilibrium and Friction: Using the spring formula, Fsp = kx =
Ans:
P=2.39
kN
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.
cos a
(
1
–
ms
2
)
–
2ms
sin a
SOLUTION
cos a
(
1
–
m
s
2
)
–
2m
s
sin a
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
(2)
Solving Eqs. (1) and (2) yields:
From FBD (b):
Ans. P=4054 lb =4.05 kip
:
+©Fx=0;
P–0.3(3868.2) –4485.4 sin 15° –1794.1 cos 15° =0
+c©Fy=0;
NC+0.4 (4485.4) sin 15° –4485.4 cos 15° =0
NC=3868.2 lb
N=4485.4 lb
ND=2893.9 lb
+c©Fy=0;
N cos 15° –0.4N sin 15° –0.3ND–3000 =0
3000 lb
15°
P
A
B
D
Ans:
P=4.05
kip
8–66.
SOLUTION
(2)
Solving Eqs. (1) and (2) yields:
From FBD (b):
Ans. P=106 lb
:
+©Fx=0;
0.2(2905.0) +292.9 cos 15° –2929.0 sin 15° –P=0
+c©Fy=0;
NC–292.9 sin 15° –2929.0 cos 15° =0
NC=2905.0 lb
N=2929.0 lb
ND=475.2 lb
+c©Fy=0;
N cos 15° +0.1N sin 15° +0.2ND–3000 =0
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 .
Since the screw is being tightened, Eq. 8–3 should be used. Here,
;
; and .Since Mmust overcome
the friction of two screws,
Ans.
Note: Since , the screw is self-locking.fs7u
F=66.7 N
F(0.125) =2 [900(0.0125)tan(16.699° +3.643°)]
M=2[Wr tan(fs+u)]
M=F(0.125)fs=tan–1 ms=tan–1(0.3) =16.699°
tan–1c5
2p(12.5) d=3.643°
u=tan–1aL
2prb =
ms=0.3
200 mm
D
C
G
200 mm
B
A
Ans:
F=66.7
N
*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
;
and .Since Mmust overcome
the friction of two screws,
Ans.
Using the result of FCD and referring to the free-body diagram of member GAC
shown in Fig. a, we have
Ans.
Note: Since , the screws are self-locking.fs7u
F
G=674 N
©MA=0; 674.32(0.2) –F
G(0.2) =0
F
CD =674.32 N
50(0.125) =2[F
CD(0.0125)tan(16.699° +3.643°)]
M=2[Wr tan(fs+u)]
M=50(0.125)fs=tan–1ms=tan–1(0.3) =16.699°;
tan–1c5
2p(12.5) d=3.643°
ms=0.3.
200 mm
D
C
G
200 mm
B
A
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
Ans.W=7.19 kN
8010.52=W10.01252 tan121.80° +2.188°2
up=tan–1c3
2p112.52d=2.188°
fs=tan–110.42=21.80°
M=W1r2 tan1fs+up2
ms=0.4,
F=80 N
0.5 m
F
818
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
Since the screw is self locking. Ans.fs7up,
up=tan–1c3
2p112.52d=2.188°
fs=tan–110.42=21.80°
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 .
.Thus
Ans.
Note:Since , the screw is self-locking.f
s7
u
P=617 lb
120 =P(0.5) tan (16.699° +4.550°)
M=Wr tan (fs+u)
fs= tan –1ms= tan –1(0.3) =16.699°
2pr
2p(0.5)
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
Ans.F=5.38 lb
F(12) =600(0.5) tan (16.699° –4.550°)
M=Wr tan (fs–u)
fs= tan –1ms= tan –1(0.3) =16.699°; and W=600 lb. Thus
2pr
2p(0.5)
ms=0.3
6 in.6 in.
B
A
Ans:
F=5.38 lb
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
2p r;
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
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
Note: Since the screw is self-locking. It will not unscrew even if force Fis
removed.
Equations of Equilibrium:
Ans.M=48.3 N #m
1610.2910.032–M=0a+©MO=0;
fs7u,
F=1610.29 N
7=F10.0152tan 14.852° +11.310°2
15 mm
M
B
Ans:
M=48.3
N#m
824
*8–76.
SOLUTION
Ans.P=2851 N =2.85 kN
35 (0.250) =P(0.006) tan (11.98° +15.11°)
M=Wr tan (u+f)
u=tan–1a8
2p(6) b=11.98°
f=tan–1(0.27) =15.11°
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
125 mm
F
AB
Ans:
P=2.85 kN
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
Here,
Ans.=0.0637
ms=tan fswhere fs=u=3.643°
fs=tan–1ms
u=tan–1al
2prb=tan–1c4
2p1102d=3.643°.
A
Ans:
m
s=0.0637
826
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:
M=5.69 lb #in.
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:
F=1.98 kN
SOLUTION
Frictional Forces on Screw. Here u=tan–1
a
l
2pr
b
=tan–1
c
3
2p (10)
d
=2.7336°,
W=F
and f
s=
tan
–1
m
s=
tan
–1
(0.35)
=
19.2900°.
M=Wr tan (
u
+
f
s)
8=F(0.01) tan (2.7336°+19.2900°)
F=1977.72 N =1.98 kN
Ans.
Note: Since f
s >
u, the screw is “self-locking”. It will not unscrew even if the torque
is removed.
M
A