8–119.
SOLUTION
A disk having an outer diameter of 120 mm fits loosely over
a fixed shaft having a diameter of 30 mm. If the coefficient
of static friction between the disk and the shaft is
and the disk has a mass of 50 kg, determine the smallest
vertical force Facting on the rim which must be applied to
the disk to cause it to slip over the shaft.
ms=0.15
*8–120.
The 4-lb pulley has a diameter of 1 ft and the axle has a
diameter of 1 in. If the coefficient of kinetic friction between
the axle and the pulley is m
k=0.20,
determine the vertical
force P on the rope required to lift the 20-lb block at
constant velocity. 6 in.
P
SOLUTION
–1
–1
8–121.
Solve Prob. 8–120 if the force P is applied horizontally to
the left.
6 in.
P
SOLUTION
–1
–1
8–122.
Then the radius of friction circle is
Determine the tension Tin the belt needed to overcome
the tension of 200 lb created on the other side. Also, what
are the normal and frictional components of force
developed on the collar bushing? The coefficient of static
friction is ms=0.21.
1.125 in.
2 in.
8–123.
If a tension force is required to pull the 200-lb
force around the collar bushing, determine the coefficient
of static friction at the contacting surface.The belt does not
slip on the collar.
T=215 lb
Frictional Force on Journal Bearing:The radius of friction circle is
and the coefficient of static friction is
Ans.ms=tan fs=tan 2.330° =0.0407
fk=2.330°
0.04066 =1 sin fk
rf=rsin fk
rf=0.04066 in.
1.125 in.
2 in.
Ans:
*8–124.
The uniform disk fits loosely over a fixed shaft having a
diameter of 40 mm. If the coefficient of static friction
between the disk and the shaft is m
s=0.15,
determine the
smallest vertical force P, acting on the rim, which must be
applied to the disk to cause it to slip on the shaft. The disk
has a mass of 20 kg.
150 mm
40 mm
P
SOLUTION
–1
–1
871
8–125.
The 5-kg skateboard rolls down the 5° slope at constant
speed. If the coefficient of kinetic friction between the
12.5-mm diameter axles and the wheels is
determine the radius of the wheels. Neglect rolling resistance
of the wheels on the surface. The center of mass for the
skateboard is at G.
SOLUTION
The effect of the forces acting on the wheels can be represented as if these forces are
acting on a single wheel as indicated on the free-body diagram shown in Fig. b.We have
Thus, the magnitude of R is
Thus, the moment arm of R from point O is
(6.25 sin 16.699 ) mm. Using these results and writing the moment equation about
point O,Fig.b, we have
a
Ans.r=20.6 mm
4.275(r)–49.05(6.25 sin 16.699° =0)+©MO=0;
°
fs=tan–1 ms=tan–1(0.3) =16.699°.
R=2Rx¿2+Ry¿2=24.2752+48.862=49.05 N
Ry¿=48.86 N48.86 –Ry¿=0©F
y¿=0;
Rx¿=4.275 NRx¿–4.275 =0©F
x¿=0;
N=48.86 NN–5(9.81) cos 5° =0©F
y¿=0;
mk=0.3,
250 mm
75 mm
300 mm
G
5
Ans:
r=20.6
mm
872
8–126.
The bell crank fits loosely into a 0.5-in-diameter pin.
Determine the required force Pwhich is just sufficient to
rotate the bell crank clockwise.The coefficient of static
friction between the pin and the bell crank is .ms=0.3
SOLUTION
Thus, the magnitude of Ris
We find that .Thus, the moment arm of R
from point Ois . Using these results and writing the moment
equation about point O,Fig. a,
a
Choosing the larger root,
Ans.P=42.2 lb
+©MO=0; 50(10) +2P2+70.71P+2500(0.25 sin 16.699°) –P(12) =0
(0.25 sin 16.699°) mm
fs=tan–1ms=tan–1(0.3) =16.699°
=2P2+70.71P+2500
R=2Rx2+Ry2=2(0.7071P)2+(0.7071P+50)2
+c©F
y=0; Ry–Psin 45° –50 =0Ry=0.7071P+50
:
+©F
x=0; Pcos 45° –Rx=0Rx=0.7071P
P
10 in.
12 in.
50 lb
45
Ans:
P=42.2
lb
873
8–127.
Thus, the magnitude of Ris
We find that the moment arm of Rfrom point Ois . Using these results
and writing the moment equation about point O,Fig. a,
a
Thus,
Ans.
m
s=tan
f
s=tan 22.35° =0.411
fs=22.35°
+©MO=0; 50(10) –41(12) –84.144(0.25 sin fs)=0
0.25 sin fs
R=2Rx2+Ry2=228.9912+78.9912=84.144 lb
+c©F
y=0; Ry–41 sin 45° –50 =0Ry=78.991 lb
The bell crank fits loosely into a 0.5-in-diameter pin. If
P= 41 lb, the bell crank is then on the verge of rotating
counterclockwise. Determine the coefficient of static
friction between the pin and the bell crank.
P
10 in.
12 in.
50 lb
45
Ans:
m
s=0.411
874
*8–128.
and Applying Eq. 8–11, we have
Ans.L78.8 lb
L260010.52
16.5
PLWa
r
r=a2.75
2b1122=16.5 in.a=0.5 in.=2600 lb,
The vehicle has a weight of 2600 lb and center of gravity at
G. Determine the horizontal force Pthat must be applied to
overcome the rolling resistance of the wheels.The
coefficient of rolling resistance is 0.5 in. The tires have a
diameter of 2.75 ft.
G
5ft
P
2ft
2.5 ft
Ans:
≈ 78.8 lb
875
8–129.
The tractor has a weight of 16 000 lb and the coefficient of
rolling resistance is a= 2 in. Determine the force Pneeded
to overcome rolling resistance at all four wheels and push it
forward.
Applying Eq. 8–11 with we have
Ans.PLWa
r
=
16000 a2
12 b
2
=1333 lb
W=16 000 lb, a=a2
12 b ft and r=2 ft,
2 ft
G
P
Ans:
P=1333
lb
876
8–130.
The handcart has wheels with a diameter of 6 in. If a crate
having a weight of 1500 lb is placed on the cart, determine
the force Pthat must be applied to the handle to overcome
the rolling resistance.The coefficient of rolling resistance is
0.04 in. Neglect the weight of the cart.
P
5
4
3
SOLUTION
Ans.P=25.3 lb
2.4 P=60 +0.024 P
P=Wa
r,4
5P=c1500 +Pa3
5bd(0.04)
3
+c©Fy=0; N–1500 –Pa3
5b=0
Ans:
P=25.3
lb
877
8–131.
The cylinder is subjected to a load that has a weight W.
If the coefficients of rolling resistance for the cylinder’s
top and bottom surfaces are and ,respectively,
show that a horizontal force having a magnitude of
is required to move the load and
thereby roll the cylinder forward. Neglect the weight of the
cylinder.
P=[W(aA+aB)]>2r
aB
aA
SOLUTION
a(1)
Since and are very small, . Hence, from Eq. (1)
(QED)P=W(aA+aB)
2r
cos fA–cos fB=1fB
fA
+©MB=0; P(rcos fA+rcos fB)–W(aA+aB)=0
+c©F
y=0; (RA)y–W=0(RA)y=W
:
+©F
x=0; (RA)x–P=0(RA)x=P
W
P
r
A
B
878
*8–132.
The 1.4-Mg machine is to be moved over a level surface
using a series of rollers for which the coefficient of rolling
resistance is 0.5 mm at the ground and 0.2 mm at the bottom
surface of the machine. Determine the appropriate
diameter of the rollers so that the machine can be pushed
forward with a horizontal force of P = 250 N. Hint: Use
the result of Prob. 8–131.
SOLUTION
Ans.d=38.5 mm
r=19.2 mm
250 =1400 (9.81) (0.2 +0.5)
2r
P=W(aA+aB)
2r
P
Ans:
d=38.5 mm
8–94.
841A
A minimum force of lb is required to hold the
cylinder from slipping against the belt and the wall.
Determine the weight of the cylinder if the coefficient of
friction between the belt and cylinder is and
slipping does not occur at the wall.
SOLUTION
(1)
Frictional Force on Flat Belt:Here, T1= 50 lb,
. Applying Eq. 8–6
Substitute this result into Eq. (1),
W= 9.17 lb Ans.
=50 e0.3(
p
6)=58.50 lb
T
2=T
1emb
b=a30°
180° bp=p
6 rad
–T
2 sin 30°(0.1 sin 30°) =0
ms=0.3
P=50
BA
0.1 ft
30
O
8–95.
841B
The cylinder weighs 10 lb and is held in equilibrium by the
belt and wall. If slipping does not occur at the wall,
determine the minimum vertical force Pwhich must be
applied to the belt for equilibrium. The coefficient of static
friction between the belt and the cylinder is ms=0.25.
(1)
Frictional Force on Flat Belt: Here,and Applying Eq. 8–6,
we have
(2)
Solving Eqs. (1) and (2) yields
Ans.
T
2=89.76 lb
P=78.7 lb
T
2=Pe0.251p>62=1.140P
T
2=T
1emb,
T
1=P.b=30° =p
6rad
–T
2sin 30°10.1 sin 30°2=0
BA
30°
O