*8–40.
If determine the minimum coefficient of static
friction at Aand Bso that equilibrium of the supporting
frame is maintained regardless of the mass of the cylinder C.
Neglect the mass of the rods.
u
=30° C
LL
uu
SOLUTION
8–41.
If the coefficient of static friction at Aand Bis
determine the maximum angle so that the frame remains
in equilibrium, regardless of the mass of the cylinder.
Neglect the mass of the rods.
the free-body diagram shown in Fig.a.
Equations of Equilibrium:We have
Ans.u=31.0°
tan u=0.6
F
BC sin u–0.6(F
BC cos u)=0©F
x=0;
:
+
N
B=F
BC cos uN
B–F
BC cos u=0+c©F
y=0;
u
ms
=0.6, C
LL
uu
© 2016 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently
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8–42.
The 100-kg disk rests on a surface for which m
B=0.2.
Determine the smallest vertical force P that can be applied
tangentially to the disk which will cause motion to impend.
SOLUTION
0.5 m
B
A
P
791
8–43.
Investigate whether the equilibrium can be maintained. The
uniform block has a mass of 500 kg, and the coefficient of
static friction is m
s=0.3.
SOLUTION
A
800 mm
200 mm
3
45
600 mm
B
*8–44.
The homogenous semicylinder has a mass of 20 kg and mass
center at G. If force P is applied at the edge, and r
=
300mm,
determine the angle
u
at which the semicylinder is on the
verge of slipping. The coefficient of static friction between
the plane and the cylinder is m
s=0.3.
Also, what is the
corresponding force P for this case?
G
u
P
r
4r
3p
SOLUTION
8–45.
SOLUTION
Post:
Assume slipping occurs at C;
a
Ans.
(O.K.!)(FB)max =0.4(533.3) =213.3 N 7121.6 N
FB=121.6 N
NC=811.0 N
P=355 N
+c©Fy=0; 3
5P+NC–533.3 –50(9.81) =0
:
+©Fx=0; 4
5P–FB–0.2NC=0
+©MC=0; –4
5P(0.3) +FB(0.7) =0
FC=0.2 NC
NB=533.3 N
T
h
e
b
eam AB
h
as a neg
li
g
ibl
e mass an
d
t
hi
c
k
ness an
d
i
s
subjected to a triangular distributed loading. It is supported
at one end by a pin and at the other end by a post having a
mass of 50 kg and negligible thickness. Determine the
minimum force Pneeded to move the post.The coefficients
of static friction at Band Care and
respectively.
mC=0.2,mB=0.4
2m 400 mm
800 N/m
C
B
300 mm
A
P
4
3
5
8–46.
T
h
e
b
eam AB
h
as a neg
li
g
ibl
e mass an
d
t
hi
c
k
ness an
d
i
s
subjected to a triangular distributed loading. It is supported
at one end by a pin and at the other end by a post having a
mass of 50 kg and negligible thickness. Determine the two
coefficients of static friction at Band at Cso that when the
magnitude of the applied force is increased to
the post slips at both Band Csimultaneously.
P=150 N,
Post:
a
Ans.
Ans.mB=FB
NB
=51.429
533.3 =0.0964
mC=FC
NC
=68.571
933.83 =0.0734
FC=68.571 N
:
+©Fx=0; 4
5(150) –FC–51.429 =0
FB=51.429 N
+©MC=0; –4
5(150)(0.3) +FB(0.7) =0
NC=933.83 N
+c©Fy=0; NC–533.3 +150a3
5b– 50(9.81) = 0
NB=533.3 N
2m 400 mm
800 N/m
B
300 mm
A
P
4
3
5
795
8–47.
B
AC
D
SOLUTION
diagram of the crates shown in Figs. aand b.
Equations of Equilibrium. Referring to Fig. a,
(1)
Also, by referring to Fig. b,
(2)
Solving Eqs. (1) and (2), yields
Ans.
F
CD =8.23 lb
u=16.3°
+Q©F
x¿=0; 0.35(150 cos u)–F
CD –150 sin u=0
a+©F
y¿=0; N
B–150 cos u=0N
B=150 cos u
+Q©F
x¿=0; F
CD +0.25(200 cos u)–200 sin u=0
a+©F
y¿=0; N
A–200 cos u=0N
A=200 cos u
Crates Aand Bweigh 200 lb and 150 lb, respectively.They
are connected together with a cable and placed on the
inclined plane. If the angle is gradually increased,
determine when the crates begin to slide.The coefficients
of static friction between the crates and the plane are
and .mB=0.35mA=0.25
u
u
Ans:
u=16.3°
796
*8–48.
Two blocks A and B, each having a mass of 5 kg, are
connected by the linkage shown. If the coefficient of static
frictionat the contacting surfaces is m
s=0.5
, determine the
largest force P that can be applied to pin C of the linkage
without causing the blocks to move. Neglect the weight of
the links.
Ans:
23.0 N
SOLUTION
Equations of Equilibrium. Analyze the equilibrium of Joint C Fig. a,
+
c
ΣF
y
=0;
FAC sin 30°–P cos 30°=0
F
AC =2
3 P
S
+
ΣFx=0;
F
BC –
P sin 30°
–
(
2
3P) cos 30°
=
0
FBC =2 P
Referring to the FBD of block B, Fig. b
+a ΣFx=0;
2 P cos 30°–FB–5(9.81) sin 30°=0
(1)
+Q ΣF
y=
0;
NB–2 P sin 30°–5(9.81) cos 30°=0
(2)
Also, the FBD of block A, Fig. C
S
+
ΣFx=0;
23P cos 30°– FA=0
(3)
+
c
ΣF
y
=0;
N
A–2
3P sin 30°
–
5(9.81)
=
0 (4)
Friction. Assuming that block A slides first. Then
FA=
m
s NA=0.5 NA
(5)
Solving Eqs. (1) to (5)
P=22.99 N =23.0 N
Ans.
NA=68.96 N
FA=34.48 N
FB=15.29 N
NB=65.46 N
Since
FB6 (FB)max =
m
s NB=0.5(65.46) =32.73 N
, Block B will not slide.
Thus, the assumption was correct.
P
30
30
30
A
C
B
797
8–49.
The uniform crate has a mass of 150 kg. If the coefficient of
static friction between the crate and the floor is m
s=0.2
,
determine whether the 85-kg man can move the crate. The
coefficient of static friction between his shoes and the floor
is m
′
s=0.4
. Assume the man only exerts a horizontal force
on the crate.
SOLUTION
2.4 m
1.2 m
1.6 m
798
8–50.
The uniform crate has a mass of 150 kg. If the coefficient of
static friction between the crate and the floor is m
s=0.2
,
determine the smallest mass of the man so he can move the
crate. The coefficient of static friction between his shoes and
the floor is m
′
s=0.45
. Assume the man exerts only a
horizontal force on the crate.
Ans:
m=66.7 kg
SOLUTION
2.4 m
1.2 m
1.6 m
8–51.
Beam AB has a negligible mass and thickness, and supports
the 200-kg uniform block. It is pinned at A and rests on the
top of a post, having a mass of 20 kg and negligible thickness.
Determine the minimum force P needed to move the post.
The coefficients of static friction at B and C are m
B=0.4
and m
C=0.2
, respectively.
SOLUTION
1.5 m 1.5 m
C
B
0.75 m
1 m
AP
4
3
5
800
*8–52.
Beam AB has a negligible mass and thickness, and supports
the 200-kg uniform block. It is pinned at A and rests on the
top of a post, having a mass of 20 kg and negligible thickness.
Determine the two coefficients of static friction at B and atC
so that when the magnitude of the applied force is increased
to
P=300 N,
the post slips at both B and C simultaneously.
Ans:
m
B=0.105
m
C=0.138
SOLUTION
1.5 m 1.5 m
C
B
0.75 m
1 m
AP
4
3
5
801
8–53.
Determine the smallest couple moment that can be applied
to the 150-lb wheel that will cause impending motion. The
uniform concrete block has a weight of 300 lb. The
coefficients of static friction are m
A=0.2
, m
B=0.3
, and
between the concrete block and the floor, m
=0.4
.
1 ft
5 ft
B
A
1.5 ft
M
Ans:
M=55.2 lb #ft
SOLUTION
802
8–54.
AB
C
G
2.5 m
0.25 m
2.5 m
u
Determine the greatest angle so that the ladder does not
slip when it supports the 75-kg man in the position shown.
The surface is rather slippery, where the coefficient of static
friction at Aand Bis .
SOLUTION
Equations of Equilibrium:Referring to the free-body diagram shown in Fig.b,
we have
(1)
(2)
Dividing Eq. (1) by Eq. (2) yields
Ans.
Using this result and referring to the free-body diagram of member AC shown in
Fig. a, we have
a
Since , end Awill not slip.Thus,
the above assumption is correct.
F
A6(F
A) max =msN
A=0.3(607.73) =182.32 N
N
A=607.73 NN
A+133.66 cos
¢
33.40°
2
≤
–75(9.81) =0+c©F
y=0;
F
A=38.40 NF
A–133.66 sin
¢
33.40°
2
≤
=0©F
x=0;
:
+
F
BC =133.66 NF
BC sin 33.40°(2.5) –75(9.81)(0.25) =0+©MA=0;
u=33.40° =33.4°
tan u>2=0.3
F
BC cos u>2=N
B
N
B–F
BC cos u>2=0+c©F
y=0;
F
BC sin u>2=0.3N
B
F
BC sin u>2–0.3N
B=0©F
x=0;
:
+
ms=0.3
X
Ans:
u=33.4°
803
8–55.
The wheel weighs 20 lb and rests on a surface for which
mB=0.2. A cord wrapped around it is attached to the top
of the 30-lb homogeneous block. If the coeffi cient of static
friction at D is mD=0.3, determine the smallest vertical
force that can be applied tangentially to the wheel which
will cause motion to impend.
SOLUTION
1.5 ft
1.5 ft
C
DB
A
P
3 ft
Ans:
P=13.3
lb
804
*8–56.
The disk has a weight Wand lies on a plane which has a
coefficient of static friction . Determine the maximum
height hto which the plane can be lifted without causing the
disk to slip.
m
SOLUTION
25
z
x
y
2a
a
h
25
805
8–57.
SOLUTION
Ans.
a
Ans.d=1.50 ft
+©MO=0; 200(d)–100(3) =0
:
+©F
x=0; P–100 =0;
P=100 lb
F
max =0.5 N=0.5(200) =100 lb
The man has a weight of 200 lb, and the coefficient of static
friction between his shoes and the floor is
Determine where he should position his center of gravity G
at din order to exert the maximum horizontal force on the
door.What is this force?
ms=0.5.
d
3ft
G
Ans:
P=100
N
d=1.50 ft
806
8–58.
Determine the largest angle that will cause the wedge to
be self-locking regardless of the magnitude of horizontal
force Papplied to the blocks.The coefficient of static
friction between the wedge and the blocks is .
Neglect the weight of the wedge.
SOLUTION
Equations of Equilibrium:Referring to Fig.a, we have
Using the requirement we obtain
Ans.u=33.4°
N tan u>2…0.3N
F…0.3N,
F=N tan u>2
2N sin u>2–2F cos u>2=0+c©F
y=0;
ms=0.3
X
PP
u
Ans:
u=33.4°
807
8–59.
a
From FBD (b),
Ans.
Since a force is required to pull out the wedge, the wedge will be self-locking
.snAnehw P=0.
P70
P=5.53 kN
–0.35113.142=0
P+12.78 cos 80° –0.25112.782cos 10°:
+©F
x=0;
N
B=13.14 kN
N
B–12.78 sin 80° –0.25112.782sin 10° =0+c©F
y=0;
N
A=12.78 kN
–6.00122–16.0152=0
N
Acos 10°172+0.25N
Asin 10°172+©MD=0;
If the beam AD is loaded as shown, determine the
horizontal force Pwhich must be applied to the wedge in
order to remove it from under the beam.The coefficients of
static friction at the wedge’s top and bottom surfaces are
and respectively. If is the
wedge self-locking? Neglect the weight and size of the
wedge and the thickness of the beam.
P=0,mCB =0.35,mCA =0.25
3m
AP
10°
4 kN/m
C
B
4m
D
Ans:
P=5.53
kN
yes