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Assistant Professor: Zhou Yufeng
N3.2-01-25, 6790-4482, yfzhou@ntu.edu.sg
http://www3.ntu.edu.sg/home/yfzhou/courses.html
CA Score Statistics
ME 10
ME 15
ME 17
PT 1
0-9
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0-9
0
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0-9
0
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0-9
0
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10-19
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10-19
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10-19
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10-19
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20-29
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20-29
2
8.3%
30-39
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30-39
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0%
30-39
1
3.3%
30-39
1
4.2%
40-49
4
13.8%
40-49
1
3.2%
40-49
3
10%
40-49
4
16.7%
50-59
5
17.2%
50-59
3
9.7%
50-59
5
16.7%
50-59
4
16.7%
60-69
6
20.7%
60-69
6
19.4%
60-69
2
6.7%
60-69 4
16.7%
70-79
5
17.2%
70-79
5
16.1%
70-79
5
16.7%
70-79
6
25%
80-89
5
17.2%
80-89
9
29%
80-89
5
16.7%
80-89 2
8.3%
90-99
2
6.9%
90-99
3
9.7%
90-99
3
10%
90-99 0
0%
100
2
6.9%
100
4
12.9%
100
6
20%
100
4.2%
1
1. A 2-kg block A passes over the highest point B of
the circular portion of the path ( =1.8 m) in the
vertical plane with a speed of 3 m/s.
(a) Calculate the normal force exerted by the path on
the block at B.
(b) Determine the maximum speed that the block
may have at point B without losing contact with
the path.
Draw the F.B.D and the K.D. of the particle, where any
vector is defined by (i) an arrow (unit vector) and (ii) a
signed magnitude. The equivalency of two vector
systems gives
(a)
mg  N =ma
where a 
(b)
v2

(1)
i.e.
N =mg ma
,so N =9.62 (N)
Set N=0 in equation (1), we have g  a 
v2

, so v  g  4.2(m / s)
2. Two packages are placed on a conveyor belt, which
is at rest. The coefficient of kinetic friction is 0.2
between the belt and package A, and 0.1 between
the belt and package B. If the belt is suddenly
started to the right and slipping occurs between the
belt and the packages, determine (a) the
accelerations of the packages, (b) the force exerted
by package A on package B.
Draw the F.B.D and the K.D. of the particles A and B, the
equivalency of two vector systems gives
Kinematic analysis:
Both blocks move in same acceleration a at the instant.
Kinetic analysis ( in i direction only):
For A: 0.2mAg  F = mA a
a = 1.417 (m/s2 )
For B: 0.1mBg + F = mB a
F = 13.07 (N)
3. A 15 kg block B is suspended from a 2.5 m cord attached
to a 20 kg cart A. Neglecting friction, determine (a) the
acceleration of the cart, (b) the tension in the cord,
immediately after the system is released from rest in the
position shown.




a

a

a
ˆ
B
A
B/ A
a A  a A (i )
Denote

t
n
t
where aB / A  aB / A  aB / A  aB / A  aB / A(25)
 aB / A (0.9063iˆ  0.4226 ˆj )
For block A: (linear motion)
T sin 25  mAaA  0.4226T  20a A
For block B: (curvilinear motion)
T (0.4226iˆ  0.9063 ˆj )  mB gˆj  mB [a A (iˆ)  aB / A (0.9063iˆ  0.4226 ˆj )]
(iˆ)  T (0.4226)  15[a A  (0.9063)aB / A ]
( ˆj )  T (0.9063)  15g  15(0.4226)aB / A
Solving equations above for T, aA and aB/A gives
T  117.6( N ), a A  2.49(m / s 2 ), aB / A  6.4(m / s 2 )
4. A system of light pulleys and inextensible cable connects bodies A, B and
C as shown. If the coefficient of friction between C and the support is 0.4,
what is the acceleration of each body? Take mA = 100 kg, mB = 300 kg
and mC = 80 kg.
Kinematic analysis
x A  2 xB  xC  C
 xA  2 xB  xC  0
Kinetic analysis
A: mA g  T  mA xA
B: mB g  2T  mB xB
xC
C: 0.4mC g  T  mC 
Now we have 4 equations for
4 unknowns. Solving the
simultaneous equations gives
T  94.88g  930.77( N )
xA  0.0512 g  0.502(m / s 2 )
xB  0.367 g  3.600(m / s 2 )
xC  0.786 g  7.711(m / s 2 )
5. The 2-kg collar C is free to slide in the vertical plane along
the smooth rod AB, which is welded onto collar A as
shown. The collar A is sliding along the vertical shaft with
acceleration 4 m/s2 upward. Determine the acceleration of
collar C , and the force acting on collar C by rod AB.
Kinematics:



a C  a A  a C / A  4 ˆj  a C / A eˆt
(1)
Kinetics:

 mgˆj  Neˆn  maC  m(4 j  aC / Aeˆt )
Dot (2) by eˆ t

aC / A  9.765  135

aC  7.49  157.2
Dot (2) by eˆ n

N  19.53135( N )
(2)