3.29
The system is released from rest with the cable taut. The friction coefficients µs = 0.25 and µk = 0.20,
calculate the acceleration of each body and the tension T in the cable. Neglect the small mass and friction
of the pulleys.
Solution:
First need to check if blocks move. . . Assume the blocks are at rest and calculated the needed friction
force. Summing the vertical forces on mass B,
ΣFy = mB aB,y
0 = 2T − mB g
N
/2 = 98.1 N
T = mB g/2 = 20 kg 9.81
kg
Summing the tangential forces on mass A,
ΣFt = MA aA,t
0 = T + FF − mA g sin (30◦ )
FF = −T + mA g sin (30◦ ) = 196.2 N
Summing the forces in the normal direction,
ΣFn = MA aA,n
0 = N − mA g cos (30◦ )
N = mA g cos (30◦ ) = 510 N
FF,max = µs N = 127 N < 196.2 N ⇐ The block will slide down the hill
Now set up the force balance equations again, but this time there is acceleration and FF = µk N . First for
block A,
*0
0 = ΣFn − mA aa,n = N − mA g cos (30◦ ) − mA aa,n
N = mA g cos (30◦ ) = 510 N
0 = ΣFt − mA aa,t = T + FF − mA g sin (30◦ ) − mA aa,t
T = −FF + mA g sin (30◦ ) + mA aa,t
(1)
Now for block B,
0 = ΣFy − mB aB,y = 2T − mB g − mB aB,y
mB
(g + aB,y )
T =
2
Setting equation (1) equal to equation (2) and letting aA,t = −2aB,y ,
−FF + mA g sin (30◦ ) + mA (−2aB,y ) =
mB
(g + aB,y )
2
Solving for aB,y ,
mA g sin (30◦ ) −
mB aB,y
mB g
− FF =
+ 2mA aB,y
2
2
mA g sin (30◦ ) − m2B g − FF
aB,y =
mB
2 + 2mA
mA g sin (30◦ ) − m2B g − µk N
aB,y =
mB
2 + 2mA
294.4 N − 98.1 N − 101.9 N
m
aB,y =
= 0.726 2
10 kg + 120 kg
s
m
aA,t = −2aB,y = −1.45 2
s
mB
(g + aB,y ) = 105.4 N
T =
2
T = −FF + mA g sin (30◦ ) + mA aa,t = 105.4 N
(2)
3.43
With the blocks initially at rest, the force P is increased slowly from zero to 60 lb. Plot the accelerations of
both masses as functions of P .
Solution:
Summing the forces on block A,
0 = ΣF − mA aA = FF,AB − mA aA ⇐ Horizontal
(3)
0 = NAB − mA g ⇐ Vertical
(4)
So,
NAB = mA g
(5)
and the maximum static friction force is
FF,AB,max = µs,AB mA g = 16 lb
(6)
The maximum acceleration of block A without slip is
aA,max,no slip =
FF,AB,max
16 lb
ft
=
= 6.44 2
lbm
mA
s
80 lbm/32.2 slug
(7)
The kinetic friction force between the blocks is
FF,AB,Slip = µk,AB mA g = 12 lb
(8)
Summing the forces on block B,
0 = σF − mB aB = P − FF,AB − FF,B − mB aB ⇐ Horizontal
0 = NB − NAB − mB g ⇐ Vertical
(9)
(10)
So,
NB = NAB + mB g = mA g + mB g
(11)
and the maximum static friction force is,
FF,B,max = µs,B (mA + mB ) g = 27 lb
(12)
The kinetic friction force between the block B and the ground is,
FF,B,Slip = µk,B (mA + mB ) g = 18 lb
(13)
(14)
As P is increased there will be no motion until P reaches 27 lb at which point block B will begin to slide
along the lower surface.
FF,AB = mA aA
(15)
FF,AB = P − FF,B − mB aB
(16)
Setting these two equations equal to each other,
mA aA = P − FF,B − mB aB
(17)
There is slip between block B and the ground; therfore, FF,B = µk,AB (mA + mB ) g = 18 lb. Assuming
there is no slip between the blocks then aA = aB .
aB (mA + mB ) = P − µk,AB (mA + mB ) g
P − µk,AB (mA + mB ) g
mA + mB
27 lb − 18 lb
aB =
lbm
180 lbm/32.2 slug
aB =
(18)
(19)
(20)
27 lb − 18 lb
(21)
5.59slug
ft
(22)
aB = 1.61 2 ⇐ Just after P reaches 27 lb.
s
This acceleration is small enough so that there is no slip between the blocks. As P increases the acceleration
of both blocks will be given by:
aB =
P − 18 lb
5.59 slug
a=
(23)
until a = 6.44 sft2 , at which point block A will start to slip. The value of P which will result in slip is
p = aA,max,no slip (5.59 slug) + 18 lb = 54.0 lb
(24)
For P > 54.0 lb the friction force between the blocks will be 12 lb. The acceleration of block A will be
ft
12 lb
(25)
= 4.83 2 .
aA =
lbm
s
80 lbm/32.2 slug
The acceleration of block B will be
aB =
P − 12 lb − 18 lb
P − 30 lb
=
lbm
3.11slug
100 lbm/32.2 slug
So,
aA =


0,
P < 27 lb
27 lb < P < 54.0 lb
P > 54.0
aB =


0,
P < 27 lb
27 lb < P < 54.0 lb
P > 54.0
and,
P −18 lb
5.59 slug


4.83 sft2
P −18 lb
5.59 slug

 P −30 lb
3.11slug
(26)
10
Block A
Block B
9
8
Acceleration (ft/s2)
7
6
5
4
3
2
1
0
0
10
20
30
P (lb)
40
50
60