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 ( μk ) A = 0.30 and ( μk ) B = 0.32, determine the
initial acceleration of each box.
SOLUTION
Assume that aB > a A so that the normal force NAB between the boxes is zero.
A:
ΣFy = 0: NA − WA cos 15° = 0
or
NA = WA cos 15°
Slipping:
FA = ( μk ) A NA
A:
= 0.3WA cos 15°
ΣFx = m A a A : FA − WA sin 15° = m A a A
0.3WA cos 15° − WA sin 15° =
or
WA
aA
g
a A = (32.2 ft/s 2 )(0.3 cos 15° − sin 15°)
or
= 0.997 ft/s 2
B:
B:
ΣFy = 0: N B − WB cos 15° = 0
or
N B = WB cos 15°
Slipping:
FB = ( μk ) B N B
= 0.32WB cos 15°
ΣFx = mB aB : FB − WB sin 15° = mB aB
or
or
0.32WB cos 15° − WB sin 15° =
WB
aB
g
aB = (32.2 ft/s 2 )(0.32 cos 15° − sin 15°) = 1.619 ft/s 2
aB > a A Ÿ assumption is correct
a A = 0.997 ft/s 2
15° W
a B = 1.619 ft/s 2
15° W
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319
PROBLEM 12.16 (Continued)
Note: If it is assumed that the boxes remain in contact ( NAB ≠ 0), then assuming NAB to be compression,
a A = aB
and find (ΣFx = ma) for each box.
A:
0.3WA cos 15° − WA sin 15° − N AB =
WA
a
g
B:
0.32WB cos 15° − WB sin 15° + N AB =
WB
a
g
Solving yields a = 1.273 ft/s 2 and NAB = −0.859 lb, which contradicts the assumption.
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320
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/s2, determine (a) the horizontal force between the tires
and the ground, (b) the force between the boulder and the
pallet.
SOLUTION
aT = 1 m/s 2
Kinematics:
a A = a B = 0.5 m/s 2
5000
= 155.28 slugs
32.2
200
mA =
= 6.211 slugs
32.2
1000
mB =
= 31.056 slugs
32.2
mT =
Masses:
Let T be the tension in the cable. Apply Newton’s second law to the lower pulley, pallet and boulder.
Vertical components
:
2T − (m A + mB ) g = (m A + mB ) a A
2T − (37.267)(32.2) = (37.267)(0.5)
T = 609.32 lb
Apply Newton’s second law to the truck.
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321
PROBLEM 12.17 (Continued)
Horizontal components
(a)
: F − T = mT aT
Horizontal force between lines and ground.
F = T + mT aT = 609.32 + (155.28)(1.0)
F = 765 lb W
Apply Newton’s second law to the boulder.
Vertical components + :
FAB − mB g = mB aB
FAB = mB ( g + a) = 31.056(32.2 + 0.5)
(b)
FAB = 1016 lb W
Contact force:
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322
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 μs = 0.20 and μk = 0.15. If P = 0, determine
(a) the acceleration of block B, (b) the tension in the cord.
SOLUTION
From the constraint of the cord:
2 x A + xB/A = constant
Then
2v A + vB/A = 0
and
2a A + aB/A = 0
Now
a B = a A + a B/A
Then
aB = a A + ( −2a A )
or
aB = − a A
(1)
First we determine if the blocks will move for the given value of θ . Thus, we seek the value of θ for
which the blocks are in impending motion, with the impending motion of A down the incline.
B:
ΣFy = 0: N AB − WB cos θ = 0
or
N AB = mB g cos θ
Now
FAB = μ s N AB
B:
= 0.2mB g cos θ
ΣFx = 0: − T + FAB + WB sin θ = 0
T = mB g (0.2cos θ + sin θ )
or
A:
A:
ΣFy = 0: N A − N AB − WA cos θ = 0
or
N A = (m A + mB ) g cos θ
Now
FA = μ s N A
= 0.2( mA + mB ) g cos θ
ΣFx = 0: − T − FA − FAB + WA sin θ = 0
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323
PROBLEM 12.18 (Continued)
T = m A g sin θ − 0.2(m A + mB ) g cos θ − 0.2mB g cos θ
or
= g[ mA sin θ − 0.2(m A + 2mB ) cos θ ]
Equating the two expressions for T
mB g (0.2cos θ + sin θ ) = g[m A sin θ − 0.2(mA + 2mB ) cos θ ]
8(0.2 + tan θ ) = [40 tan θ − 0.2(40 + 2 × 8)]
or
tan θ = 0.4
or
or θ = 21.8° for impending motion. Since θ < 25°, the blocks will move. Now consider the
motion of the blocks.
ΣFy = 0: N AB − WB cos 25° = 0
(a)
B:
or
N AB = mB g cos 25°
Sliding:
FAB = μk N AB = 0.15mB g cos 25°
ΣFx = mB aB : − T + FAB + WB sin 25° = mB aB
or
T = mB [ g (0.15cos 25° + sin 25°) − aB ]
= 8[9.81(0.15cos 25° + sin 25°) − aB ]
= 8(5.47952 − aB )
A:
(N)
ΣFy = 0: N A − N AB − WA cos 25° = 0
or
N A = (m A + mB ) g cos 25°
Sliding:
FA = μk N A = 0.15(m A + mB ) g cos 25°
ΣFx = m A a A : − T − FA − FAB + WA sin 25° = m A a A
Substituting and using Eq. (1)
T = m A g sin 25° − 0.15( mA + mB ) g cos 25°
− 0.15mB g cos 25° − m A (− aB )
= g[ mA sin 25° − 0.15(m A + 2mB ) cos 25°] + m A aB
= 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40aB
= 91.15202 + 40aB
(N)
Equating the two expressions for T
8(5.47952 − aB ) = 91.15202 + 40aB
or
aB = −0.98575 m/s 2
a B = 0.986 m/s 2
(b)
We have
or
25° W
T = 8[5.47952 − (−0.98575)]
T = 51.7 N W
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324
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
μ s = 0.20 and μk = 0.15. If P = 40 N
, determine (a) the
acceleration of block B, (b) the tension in the cord.
SOLUTION
From the constraint of the cord.
2 x A + xB/A = constant
Then
2v A + vB/A = 0
and
2a A + aB/A = 0
Now
a B = a A + a B/A
Then
aB = a A + ( −2a A )
or
aB = − a A
(1)
First we determine if the blocks will move for the given value of P. Thus, we seek the value of P for which
the blocks are in impending motion, with the impending motion of a down the incline.
B:
ΣFy = 0: N AB − WB cos 25° = 0
or
N AB = mB g cos 25°
Now
FAB = μ s N AB
B:
= 0.2 mB g cos 25°
ΣFx = 0: − T + FAB + WB sin 25° = 0
A:
T = 0.2 mB g cos 25° + mB g sin 25°
or
= (8 kg)(9.81 m/s 2 ) (0.2 cos 25° + sin 25°)
= 47.39249 N
A:
or
ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0
N A = (m A + mB ) g cos 25° − P sin 25°
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325
PROBLEM 12.19 (Continued)
Now
FA = μ s N A
or
FA = 0.2[(mA + mB ) g cos 25° − P sin 25°]
ΣFx = 0: − T − FA − FAB + WA sin 25° + P cos 25° = 0
−T − 0.2[(m A + mB ) g cos 25° − P sin 25°] − 0.2mB g cos 25° + mA g sin 25° + P cos 25° = 0
or
or
P(0.2 sin 25° + cos 25°) = T + 0.2[(m A + 2mB ) g cos 25°] − m A g sin 25°
Then
P(0.2 sin 25° + cos 25°) = 47.39249 N + 9.81 m/s 2 {0.2[(40 + 2 × 8) cos 25° − 40 sin 25°] kg}
P = −19.04 N for impending motion.
or
Since P, < 40 N, the blocks will move. Now consider the motion of the blocks.
ΣFy = 0: N AB − WB cos 25° = 0
(a)
B:
or
N AB = mB g cos 25°
Sliding:
FAB = μk N AB
= 0.15 mB g cos 25°
ΣFx = mB aB : − T + FAB + WB sin 25° = mB aB
T = mB [ g (0.15 cos 25° + sin 25°) − aB ]
or
= 8[9.81(0.15 cos 25° + sin 25°) − aB ]
= 8(5.47952 − aB )
(N)
A:
ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0
or
N A = (m A + mB ) g cos 25° − P sin 25°
Sliding:
FA = μk N A
= 0.15[(m A + mB ) g cos 25° − P sin 25°]
ΣFx = m A a A : − T − FA − FAB + WA sin 25° + P cos 25° = m A a A
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326