Document

Assistant Professor: Zhou Yufeng
N3.2-01-25, 6790-4482, yfzhou@ntu.edu.sg
http://www3.ntu.edu.sg/home/yfzhou/courses.html
1. Two blocks A and B, of mass 12 kg and 15 kg, respectively,
hang from a cable that passes over a pulley of negligible
mass. The blocks are released from rest in the position
shown and block B is observed to strike the ground with a
velocity 1.4 m/s. Determine (a) the force exerted by the
cable on each of the two blocks, (b) the work done by the
friction torque during the interval.
(a) For Particle A
1
1
2
 TA  mA g s  2 mA v  TA  12 g15.  2 (12)14. 2  TA=125.56 (N)
For Particle B
1
1
2


15
g

T
1
.
5

(15)1.4 2  TB= 137.35 (N)
 mB g  TB  s  2 mB v 
B
2
(b) For the massless pulley
(TB  TA)s  | Mf  | = 0
| Mf  |= 17.685 (Nm)
Note: the work done by the friction torque should be negative.
Alternatively, we may consider System = A + pulley + B
2. The coefficients of friction between the load and the flatbed trailer shown
are s =0.4 and k =0.35. Knowing that the speed of the rig is 90 km/h,
determine the shortest time in which the rig can be brought to a stop if the
load is not to shift.
Knowing that the trailer and the load A are moving together:


v1  90iˆ(km / h)  25iˆ(m / s), v2  0
Consider the load A

F f   s ( mA g )i
Using Principle of Impulse and Linear Momentum



I12  L2  L1



 F f t  mA v 2  mA v1




(
m
g
)

t
i


m
v

s
A
A 1

t 
25
 6.371( s)
0.4 g
3. The 50-N block is original at rest on the smooth surface. It is acted upon
by a radial force of 10 N and a horizontal force of 35 N, always directed at
30 from the tangent to the path as shown. Determine the time required to
break the cord, which requires a tension T = 150 N. What is the speed of
the block when this occurs? Neglect the size of the block for the
calculation.
Considering all the forces acting on B at the moment when
the cord is broken (at position 2), and using Newton’s second
law in normal direction, we have
F
n
 man  m22 r  2  4.4753(rad / s)  v  5.37(m / s)
Applying Principle of Angular Impulse and Momentum,
The moment of the inertia of the particle about O is
50
I O  mr 2  1.22  7.3394(kgm2 )
g
 ( ang) 

I12  H O 2  H O1   M O t  I O2  0
 35 cos 30(1.2)t  7.3394(4.4753)  t  0.903
P2.17 A rod AB of length 600 mm
is pinned at B on a disk of radius r
= 200 mm. End A is pin joined to a
bar OA, which is of length 600 mm
and connected to a fixed pivot at O.
The disk rolls without slipping on
the ground. At the position shown,
point G has a speed of v = 100
mm/s to the right and zero
acceleration. Determine angular
velocity and angular acceleration of
bar OA at the position shown.
  
vB  v A  vB / A
BA


 0, vB  vA

vB  150iˆ
 AO



v A   AO  rAO   AO rAO iˆ
vA
vB


 0.25rad / s
rAO rAO

 
aB  a A  aB / A

n
(150  100) 2 ˆ
aB  aB  
j  25 ˆj
100
2
n
vA ˆ
150 2 ˆ
aA  
j
j  37.5 ˆj
rAO
600



a At   AO  rAO   AO rAO iˆ


t

aB / A  aB / A   B / A  rBA   B / A rBA60
iˆ : 0   AO rAO  aBt / A cos 60
ˆj : aBn  a An  aBt / A sin 60
  AO
a An  aBn

 0.012rad / s 2
rAO tan 60