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PROBLEM 12.14 (Continued)
PROBLEM 12.15
Each of the systems shown is initially at rest.
Neglecting axle friction and the masses of the pulleys,
determine for each system (a) the acceleration of
block A, (b) the velocity of block A after it has
moved through 10 ft, (c) the time required for block
A to reach a velocity of 20 ft/s.
PROBLEM 12.15 (Continued)
PROBLEM 12.16
Boxes A and B are at rest on a conveyor belt that is initially at
rest. The belt is suddenly started in an upward direction so
that slipping occurs between the belt and the boxes. Knowing
that the coefficients of kinetic friction between the belt and
the boxes are
( ) 0.30
kA
and
( ) 0.32,
kB
determine the
initial acceleration of each box.
B: 0: cos15 0
yBB
FNW
or
cos 15
BB
NW
Slipping:
()
0.32 cos 15
BkBB
B
FN
W
:sin15
xBBBB BB
Fma FW ma
or 0.32 cos 15 sin 15
B
BB B
W
WW a
g
or
22
(32.2 ft/s )(0.32 cos15 sin 15 ) 1.619 ft/s
B
a
BA
aa
assumption is correct
2
0.997 ft/s
A
a
15°
2
1.619 ft/s
B
a
15°
PROBLEM 12.16 (Continued)
PROBLEM 12.17
A 5000-lb truck is being used to lift a 1000 lb boulder B that
is on a 200 lb pallet A. Knowing the acceleration of the truck
is
1 ft/s
2
, determine (a) the horizontal force between the tires
and the ground, (b) the force between the boulder and the
pallet.
PROBLEM 12.17 (Continued)
PROBLEM 12.18
Block A has a mass of 40 kg, and block B has a mass of 8 kg.
The coefficients of friction between all surfaces of contact are
0.20
s
and
0.15.
k
If P
0,
determine (a) the
acceleration of block B, (b) the tension in the cord.
B
A: 0: cos 0
yAABA
FNNW
or
()cos
AAB
Nmmg
Now
0.2( ) cos
AsA
AB
FN
mmg
0: sin 0
xAABA
FTFFW
PROBLEM 12.18 (Continued)
8[9.81(0.15cos 25 sin 25 ) ]
8(5.47952 ) (N)
B
B
a
a
0: cos25 0
yAABA
FNNW
or
()cos25
AAB
Nmmg
Sliding:
0.15( ) cos 25
AkA AB
FN mmg
:sin25
xAA AABA AA
Fma TFF W ma
Substituting and using Eq. (1)
sin 25 0.15( ) cos 25
0.15 cos 25 ( )
[ sin 25 0.15( 2 ) cos 25 ]
9.81[40 sin 25 0.15(40 2 8)cos 25 ] 40
91.15202 40 (N)
AAB
BAB
AABAB
B
B
Tmg m mg
mg m a
gm m m ma
a
a
Equating the two expressions for T
8(5.47952 ) 91.15202 40
BB
aa
or
2
0.98575 m/s
B
a
2
0.986 m/s
B
a
25
(b) We have 8[5.47952 ( 0.98575)]T
or
51.7 NT
PROBLEM 12.19
Block A has a mass of 40 kg, and block B has a mass of 8 kg.
The coefficients of friction between all surfaces of contact are
0.20
s
and
0.15.
k
If
40 NP
, determine (a) the
acceleration of block B, (b) the tension in the cord.
0: sin 25 0
xABB
FTFW
or
2
0.2 cos 25 sin 25
(8 kg)(9.81 m/s )(0.2 cos 25 sin 25 )
47.39249 N
BB
Tmg mg
A:
0: cos 25 sin 25 0
yAABA
FNNW P
or
()cos25sin25
AAB
Nmmg P
