748
8–1.
Determine the maximum force
P
the connection can
support so that no slipping occurs between the plates.There
are four bolts used for the connection and each is tightened
so that it is subjected to a tension of 4 kN.The coefficient of
static friction between the plates is .
SOLUTION
Free-Body Diagram:The normal reaction acting on the contacting surface is equal
to the sum total tension of the bolts.Thus,When the plate is
Equations of Equilibrium:
N=4(4) kN =16 kN.
ms=0.4
P
P
2
P
2
Ans:
749
8–2.
SOLUTION
Equations of Equilibrium:
a
Ans.
NC=1622.22 lb =1.62 kip
+©MC=0 2N
B (9) +400(2.5) –7500(5) –600(12) =0
The tractor exerts a towing force Determine
the normal reactions at each of the two front and two rear
tires and the tractive frictional force Fon each rear tire
needed to pull the load forward at constant velocity.The
tractor has a weight of 7500 lb and a center of gravity
located at . An additional weight of 600 lb is added to its
front having a center of gravity at .Take 0.4. The
front wheels are free to roll.
ms=GA
GT
T
=400 lb.
4 ft
3 ft
5 ft
2.5 ft
A
CB
T
F
GA
GT
750
8–3.
0.15 m
A
G
B
0.9 m
0.6 m
10 kN
1.5 m
SOLUTION
Equations of Equilibrium: The normal reactions acting on the wheels at (Aand B)
are independent as to whether the wheels are locked or not. Hence, the normal
reactions acting on the wheels are the same for both cases.
a
N
B=42.316 kN =42.3 kN
N
A11.52+1011.052–58.8610.62=0+©MB=0;
The mine car and its contents have a total mass of 6 Mg and
a center of gravity at G. If the coefficient of static friction
between the wheels and the tracks is when the
wheels are locked, find the normal force acting on the front
wheels at Band the rear wheels at Awhen the brakes at
both Aand Bare locked. Does the car move?
ms=0.4
751
*8–4.
the bed of the truck. If the loaded bin has a weight of 8500 lb
and center of gravity at G,determine the force in the cable
needed to begin the lift. The coefficients of static friction at
Aand Bare and respectively.Neglect
the height of the support at A.
mB=0.2,mA=0.3
SOLUTION
a
+©MB=0; 8500(12) –N
A(22) =0
G
12 ft10 ft B
A
30
Ans:
8–5.
The automobile has a mass of 2 Mg and center of mass at G.
Determine the towing force F required to move the car if
the back brakes are locked, and the front wheels are free to
roll. Take m
s=0.3
.
SOLUTION
Equations of Equilibrium. Referring to the FBD of the car shown in Fig. a,
Ans:
F
0.75 m
30
0.3 m
0.6 m
G
A
C
B
1.50 m
1 m
8–6.
The automobile has a mass of 2 Mg and center of mass at G.
Determine the towing force F required to move the car.
Both the front and rear brakes are locked. Takem
s=0.3
.
SOLUTION
Equations of Equilibrium. Referring to the FBD of the car shown in Fig. a,
S
+
ΣFx=0;
FA+FB–F cos 30°=0
(1)
Ans:
F
0.75 m
30
0.3 m
0.6 m
G
A
C
B
1.50 m
1 m
754
8–7.
SOLUTION
To hold lever:
T
h
e
bl
oc
k
b
ra
k
e cons
i
sts of a p
i
n-connecte
d
l
ever an
d
friction block at B.The coefficient of static friction between
the wheel and the lever is and a torque of
is applied to the wheel.
Determine if the brake can hold
the wheel stationary when the force applied to the lever is
(a) (b) P=70 N.P=30 N,
5N#mms=0.3,
200 mm 400 mm
P
150 mm O
B
A
5N m
50 mm
Ans:
755
*8–8.
The block brake consists of a pin-connected lever and
friction block at B.The coefficient of static friction between
the wheel and the lever is , and a torque of
is applied to the wheel. Determine if the brake can hold
the wheel
stationary when the force applied to the lever is
(a) , (b) .P=70 NP=30 N
5N#mms=0.3
SOLUTION
To hold lever:
200 mm 400 mm
P
150 mm O
B
A
5N m
50 mm
Ans:
756
8–9.
The pipe of weight Wis to be pulled up the inclined plane of
slope using a force P.If Pacts at an angle ,show that for
slipping sin( ), where is the angle
of static friction; .u=tan–1 ms
ua +u)>cos(f–uP=W
fa
SOLUTION
+a©Fy¿=0;
N+P sin f–W cos a=0
N=W cos a–P sin f
P
α
φ
757
8–10.
SOLUTION
P=W(sin a+tan u cos a)
cos f+tan u sin f
+Q©Fx¿=0;
P cos f–W sin a–tan u (W cos a–P sin f)=0
+a©Fy¿=0;
N+P sin f–W cos a=0
N=W cos a–P sin f
Determine the angle at which the applied force Pshould
act on the pipe so that the magnitude of Pis as small as
possible for pulling the pipe up the incline.What is the
corresponding value of P? The pipe weighs Wand the slope
is known. Express the answer in terms of the angle of
kinetic friction, .u=tan–1 mk
a
fP
α
φ
758
8–11.
SOLUTION
a)
W
3sin 45° +N–200 =0+c©F
y=0;
Determine the maximum weight Wthe man can lift with
constant velocity using the pulley system, without and then
with the “leading block” or pulley at A.The man has a
weight of 200 lb and the coefficient of static friction
between his feet and the ground is ms=0.6.
(a)
45°
C
B
C
B
(b)
w
A
w
Ans:
759
*8–12.
T
h
e
bl
oc
k
b
ra
k
e
i
s use
d
to stop t
h
e w
h
ee
l
from rotat
i
ng
when the wheel is subjected to a couple moment . If the
coefficient
of static friction between the wheel and the
block
is , determine the smallest force Pthat should be
applied.
ms
M0
SOLUTION
a
+©MC=0; Pa –Nb –msNc =0
O
M
0
P
a
c
b
r
C
8–13.
If a torque of is applied to the flywheel,
determine the force that must be developed in the hydraulic
cylinder CD to prevent the flywheel from rotating. The
coefficient of static friction between the friction pad at B
and the flywheel is .
SOLUTION
Free-BodyDiagram:First we will consider the equilibrium of the flywheel using the
free-body diagram shown in Fig. a.Here,the frictional force must act to the left to
Equations of Equilibrium: We have
a
N
B=2500 N0.4 N
B(0.3) –300 =0+©MO=0;
FB
ms=0.4
M
=
300 N
#
m
30
0.6 m
60 mm
A
D
B
C
1 m
Ans:
761
8–14.
The car has a mass of 1.6 Mg and center of mass at G.If the
coefficient of static friction between the shoulder of the road
and the tires is determine the greatest slope the
shoulder can have without causing the car to slip or tip over
if the car travels along the shoulder at constant velocity.
ums=0.4,
SOLUTION
Tipping:
a
–Wcos u12.52+Wsin u12.52=0+©MA=0;
θ
A
B
G
5ft
2.5 ft
8–15.
The log has a coefficient of static friction of with
the ground and a weight of 40 lb/ft. If a man can pull on the
rope with a maximum force of 80 lb, determine the greatest
length lof log he can drag.
ms
=0.3
SOLUTION
Equations of Equilibrium:
+c©Fy=0;
N–40l=0
N=40l
80 lb
BA
l
Ans:
763
*8–16.
SOLUTION
Free – Body Diagram. Since the weight of the man tends to cause the friction pad A
to slide to the right, the frictional force FAmust act to the left as indicated on the
free – body diagram of the ladder,Fig. a. Here, the ladder is on the verge of slipping.
T
hus,.
F
A=msN
A
The 180-lb man climbs up the ladder and stops at the position
shown after he senses that the ladder is on the verge of
slipping.Determine the inclination of the ladder if the
coefficient of static friction between the friction pad Aand the
ground is .Assume the wall at Bis smooth.The center
of gravity for the man is at G.Neglect the weight of the ladder.
ms=0.4
u
G
A
B
10 ft
u
8–17.
The 180-lb man climbs up the ladder and stops at the position
shown after he senses that the ladder is on the verge of
slipping.Determine the coefficient of static friction between
the friction pad at Aand ground if the inclination of the ladder
is and the wall at Bis smooth.The center of gravity for
the man is at G.Neglect the weight of the ladder.
u=60°
SOLUTION
Free – Body Diagram. Since the weight of the man tends ot cause the friction pad A
Equations of Equilibrium.
+c©F
y=0; N
A–180 =0N
A=180 lb
G
B
10 ft
3ft
Ans:
8–18.
The spool of wire having a weight of 300 lb rests on the
ground at B and against the wall at A. Determine the force
P required to begin pulling the wire horizontally off the
spool. The coefficient of static friction between the spool
and its points of contact is m
s=0.25.
SOLUTION
Equations of Equilibrium. Referring to the FBD of the spool shown in Fig. a,
S
+
ΣFx=0;
P–NA–FB=0
(1)
Frictions. It is required that slipping occurs at A and B. Thus,
Solving Eqs. (1) to (5),
A
B
O
3 ft
1 ft
P
8–19.
The spool of wire having a weight of 300 lb rests on the
ground at B and against the wall at A. Determine the
normal force acting on the spool at A if P
=
300 lb.
Thecoefficient of static friction between the spool and the
ground at B is m
s=0.35.
The wall at A is smooth.
SOLUTION
Equations of Equilibrium. Referring to the FBD of the spool shown in Fig. a,
A
B
O
3 ft
1 ft
P
767
*8–20.
The ring has a mass of 0.5 kg and is resting on the surface of
the table. In an effort to move the ring a normal force P from
the finger is exerted on it. If this force is directed towards the
ring’s center O as shown, determine its magnitude when the
ring is on the verge of slipping at A. The coefficient of static
friction at A is m
A=0.2
and at B, m
B=0.3
.
SOLUTION
FA=FB
Ans:
75 mm
O
B
P
60
A