*17–100.
A force of F = 10 N is applied to the 10-kg ring as shown. If
slipping does not occur, determine the ring’s initial angular
acceleration, and the acceleration of its mass center, G. Neglect
the thickness of the ring.
SOLUTION
45
30
0.4 m
G
A
C
F
17–101.
If the coefficient of static friction at C is μs = 0.3, determine
the largest force F that can be applied to the 5-kg ring,
without causing it to slip. Neglect the thickness of
the ring.
SOLUTION
45
30
0.4 m
G
A
C
F
17–102.
The 25-lb slender rod has a length of 6 ft. Using a collar of
negligible mass, its end A is confined to move along the
smooth circular bar of radius 3
22
ft. End B rests on the
floor, for which the coefficient of kinetic friction is m
B=0.4.
If the bar is released from rest when
u
= 30°, determine the
angular acceleration of the bar at this instant.
SOLUTION
+
25
A
B
3 2 ft
6 ft
u
Ans:
2
894
17–103.
The 15-lb circular plate is suspended from a pin
at A. If the pin is connected to a track which is given an
acceleration aA = 5 ft
>
s2, determine the horizontal and
vertical components of reaction at A and the angular
acceleration of the plate. The plate is originally at rest.
SOLUTION
S
+
ΣF
x=
m(a
G
)
x
; Ax=
15
32.2
(aG)x
+
c
Σ
Fy=m(aG)y; Ay–15 =
15
32.2
(aG)y
c
+ΣMG=IG
a
;
Ax(2) =
c1
2
a15
32.2 b
(2)2
d
a
a
G
=a
A
+a
G
>
A
aG=5i–2
a
i
(
+
c
)
(a
G
)
y
=0
(
S
+
)
(aG)x=5–2
a
Thus,
A
y
=15.0 lb
Ans.
Ax=0.776 lb
Ans.
a
=
1.67 rad
>
s
2
Ans.
a
G=
(a
G
)
x=
1.67 ft
>
s
2
G
A
a
A
2 ft
Ans:
A
y
=15.0 lb
Ax=0.776 lb
a
=
1.67 rad
>
s
2
895
*17–104.
1.5 ft
P
30
If P= 30 lb, determine the angular acceleration of the 50-lb
roller.Assume the roller to be a uniform cylinder and that
no slipping occurs.
a
=
7.44 rad
>
s
2
17–105.
1.5 ft
P
30
If the coefficient of static friction between the 50-lb roller
and the ground is determine the maximum force
Pthat can be applied to the handle, so that roller rolls on the
ground without slipping.Also, find the angular acceleration
of the roller.Assume the roller to be a uniform cylinder.
ms=0.25,
SOLUTION
17–106.
The uniform bar of mass m and length L is balanced in the
vertical position when the horizontal force P is applied to
the roller at A. Determine the bar’s initial angular
acceleration and the acceleration of its top point B.
SOLUTION
+
A
B
L
17–107.
Solve Prob. 17–106 if the roller is removed and the
coefficient of kinetic friction at the ground is μk.
SOLUTION
+ΣF
m(a
)
;
P–
m
c
+ΣMG=IG
a;
(
P–mkNA
)
L
2
=
a1
12
mL
2
b
a
+
c
Σ
Fy=m(aG)y;
NA–mg =0
Solving,
NA=mg
aG=
L
6
a
a=
6(P–
mk
mg)
mL
Ans.
a
B
=a
G
+a
B
>
G
(
S
+
)
aB=–
L
6
a+
L
2
a
aB=
La
3
aB=
2(P–
mk
mg)
m
Ans.
A
B
L
Ans:
6(P–
mk
mg)
899
*17–108.
SOLUTION
geometry
,A
lso, using
law of sines,,.Applying Eq. 17–16, we have
a
(1)
(2)
(3)
Kinematics:Assume that the semicircular disk does not slip at A,then .
Here, .
Applying Eq. 16–18, we have
Equating iand jcomponents, we have
(4)
(5)
Solving Eqs. (1), (2), (3), (4), and (5) yields:
Since , then the semicircular
disk does not slip.Ans.
F
f6(F
f)max =msN=0.5(91.32) =45.66 N
F
f=20.12 N N=91.32 N
a=13.85 rad>s2(aG)x=2.012 m>s2(aG)y=0.6779 m>s2
(aG)y=0.1470a–1.3581
(aG)x=0.3151a–2.3523
–(aG)xi–(aG)yj=(2.3523 –0.3151 a)i+(1.3581 –0.1470a)j
–(aG)xi–(aG)yj=6.40j+ak*(–0.1470i+0.3151j)–42(–0.1470i+0.3151j)
aG=aA+a*rG>A–v2rG>A
rG>A={–0.3477 sin 25.01°i+0.3477 cos 25.01°j}m ={–0.1470i+0.3151j}m
(aA)x=0
+cF
y=m(aG)y;N–10(9.81) =-10(aG)y
;
+©F
x=m(aG)x;F
f=10(aG)x
+10(aG)ysin 25.01°(0.3477)
+10(aG)xcos 25.01°(0.3477)
+©MA=©(Mk)A; 10(9.81)(0.1698 sin 60°) =0.5118a
u=25.01°
sin u
0.1698 =sin 60°
0.3477
rG>A=20.16982+0.42–2(0.1698) (0.4) cos 60° =0.3477 m
2(10)
The semicircular disk having a mass of 10 kg is rotating at
at the instant . If the coefficient of
static friction at Ais , determine if the disk slips at
this instant.
ms=0.5
u=60°v=4 rad>s
4 (0.4) m
3p
O
G
0.4 m
A
u
v
17–109.
SOLUTION
motion about point Ausing Fig. a,
a(1)
Kinematics: Since the culvert does not slip at A,.Applying the
relative acceleration equation and referring to Fig. b,
Equating the icomponents,
(2)
Solving Eqs. (1) and (2) yields
Ans.a=3 rad>s2
aG=1.5 m>s2:
aG=3–0.5a
aGi=(3 –0.5a)i+
C
(aA)n–0.5v2
D
j
aGi=3i+(aA)nj+(ak*0.5j)–v2(0.5j)
aG=aA+a*rG>A–v2rG>A
(aA)t=3m>s2
+©MA=©(Mk)A;0=125a–500aG(0.5)
The 500-kg concrete culvert has a mean radius of 0.5 m. If
the truck has an acceleration of , determine the
culvert’s angular acceleration.Assume that the culvert does
not slip on the truck bed, and neglect its thickness.
3m>s24m
0.5m
3m/s2
17–110.
The 15-lb disk rests on the 5-lb plate. A cord is wrapped
around the periphery of the disk and attached to the wall at
B. If a torque M = 40 lb
#
ft is applied to the disk, determine
the angular acceleration of the disk and the time needed for
the end C of the plate to travel 3 ft and strike the wall.
Assume the disk does not slip on the plate and the plate
rests on the surface at D having a coefficient of kinetic
friction of μk = 0.2. Neglect the mass of the cord.
SOLUTION
AB
C
D3 ft
M 40 lb ft
1.25 ft