1
KINETICS OF PARTICLES
5kg B

A
10kg
125N
Knowing that the coefficient of friction is 0.30 at all surfaces of contact, determine (a) the
acceleration of plate A, (b) the tension in the cable. (Neglect bearing friction in the pulley)
 Kinematics
yt
xB
B

A
l
xt
xA
-
Assume cable is inextensible and, therefore of constant length
 xA + xB = constant
aA = -aB ……………….……. (i)
 Motion of Plate B
-
Free-body diagram
WB = 5(9.81)N = 49.05N
T
FB
T
NB
Fy = 0: NB – 49.05 = 0  NB = 49.05 N
Friction: FB = 0.3NB = 0.3(49.05) = 14.72 N
Fx = MB aB: FB-T = 5 aB
 T = 14.72 – 5aB ………………(ii)
 Motion of Plate A
- Free-body diagram
NB = 49.05N
FB = 14.72N
WA = 10(9.81)N = 98.1N
125N
T
FA
NA
Fy = 0 : NA – 49.05 – 98.0 = 0
 NA = 147.15 N
Friction: FA = 0.3 NA = 0.3(147.15) N
FA = 44.15 N
Fx = mAaA: 125 – T – 14.72 – 44.15 =
10aA………….(iii)
With (ii), (iii) becomes 125 – (14.72 – 5aB) – 14.72 – 44.15 = 10aA
2
51.41 + 5aB = 10A
aB = -aA from (i)  a A 
51.41
 3.43 m / s 2
15
From (ii) T  14.72  5(3.43)  31.87 N
(a) Disk about to slide to the right
-
X
Free-body diagram
mg
60o
a
F
Y
N
Fx = max: 0.2 N = m a cos 60 ………………… (i)
Fy = may: mg – N = m a sin 60 ………………..(ii)
Solve (i) and (ii) simultaneously  a  0.297 g  2.91 m/s2
(b) Disk about to slide to the left
Y
Free-body diagram
mg
a
60o
F
X
N
Fx = max: 0.2 N = m a cos 60 ………………… (iii)
Fy = may: N - mg = m a sin 60 ………………..(iv)
Solve (iii) and (iv) simultaneously  a  0.612 g  6 m/s2
In a manufacturing process, disks are moved from one elevation to
another by the lifting arm shown; the coefficient of friction
between a disk and the arm is 0.20. Determine the magnitude of
the acceleration for which the disks slide on the arm, assuming the
acceleration is directed (a) downward as shown, (b) upward.
3
For the pulley arrangement shown below, with s =
0.25 and k = 0.20, calculate the acceleration of each
block and the tension in the cable. Neglect the small
mass and friction of the pulleys.
Kinematics
 Assume cable is inextensible
 xA – 2yB = constant  x A  2y B  0
 aA = 2aB …….. (i)
Free-body diagrams
T
60g
T
T
A
F

B
N
30o
20g
 Check if block A slips
For no slipping, Fx = 0 : 60 g sin 30 – T – F = 0
 F = 30(9.81) – T = 294.3 – T
Also Fy = 0 : 2T – 20 g = 0  T = 10(9.81) = 98.1 N
 F = 294.3 – 98.1 = 196.2 N
Friction: Fmax = sN = 0.25 (60 g cos 30) = 127.4 N
F > Fmax

A slips down inclined plane.
 Motion of A
Fx = mAaA : 60 g sin 30 - T – F = 60 aA  294.3 – T – F = 60 aA …………(ii)
Fy = 0 (A does not move in the y direction): N – 60 g cos 30 = 0  N = 509.7
N…(iii)
Friction: F = kN = 0.2(509.7) = 101.9 N ………(iv)
with (iv), (ii) becomes 294.3 – T – 101.9 = 60 aA  60 aA + T = 192.4 ……..(v)
4
 Motion of B
Fy = mBaB: 2 T – 20 g = 20 aB  T – 10 aB = 98.1 ………..(vi)
Rearrange (i), (v) and (vi) 
aA – 2aB + (0)T = 0
60 aA + (0)aB + T = 192.4
(0)aA – 10 aB + T = 98.1
- solve the above using Cramer’s rule
aA 
0
2
0
192.4
0
1
98.1
1
2
0
60
0
1
0
aA 
 10 1

0(0  10)  (2)(192.4  98.1)  0(192.4  0
1(0  10)  (2)(60  0)  0(600  0)
 10 0
188.6
 145 m / s 2
130
1
0
0
60 192.4 1
aA 
0
98.1
1
1
2
0
60
0
1
0
aA 

1(192.4  98.1)  0(60  0)  0[60(98.1)  0]
130
 10 1
94.3
 0.725 m / s 2
130
5
T
1
2
0
60
0
192.4
0
 10
98.1
1
2
0
60
0
1
0

1(0  192.4)  (2)[( 60)(98.1)  0]  0(600  0)
130
 10 1
T
13696
 105.4 m / s 2
130
A 30-kg crate rests on a 20-kg cart. The coefficient of static friction between the crate and cart is
0.25. If the crate is not to slip with respect to the cart, determine (a) the maximum allowable
magnitude of P, (b) the corresponding acceleration of the cart.
30 kg
y
P–F=0
P=F
20kg
P
x
Crate: (Assume slipping impending)
W = 30-kg
30 a
=
F = 0.25N
Fy = 0  N = 30g
F = 0.25(30g) = 7.5g
N
FA = max  7.5g = 30 a
ax = 0.25g  a = 2.45 m/s2
50g
=
P
50 a
Fx = ma  P = 50(0.25g) = 12.5g
P = 122.6 N
Check on assumption of slipping
Fmax = SN = 0.25 x 30g = 7.5g
P – F = 0  F = 12.5g  F > Fmax  crate slips
Crate & Cart
6
A 2 kg ball revolves in a horizontal circle as shown at a constant speed
of 1.5 m/s. Knowing that L = 600 mm, determine (a) the angle  that
the cord forms with the vertical, (b) the tension in the cord.
 Free-body diagram of ball
y
T

x
2g
-
ball does not move in the y-direction
 Fy = 0: Tcos = 2(98.1) = 19.62 ………… (i)
-
Motion of ball in the x-direction
2V 2
2(1.5) 2

Fx = max: T sin  
0.6 sin  0.6 sin 
or Tsin2 = 7.5 ……(ii)  T = 7.5/sin2
substitute for T in (i) 
7.5
cos   19.62
sin 2 
sin2 = 1 – cos2  7.5 cos = 19.62 – 19.62cos2
cos 2  
7.5
19.62
cos  
 0  cos 2   0.3823 cos   1  0
19.62
19.62
cos  
 0.3823  (0.3823) 2  4
2
cos  
0.3823  2.0362
2
cos 1 
1.6539
 0.82695  1  34.2
2
cos 1 
2.4185
 1.20925
2
T
7.5
 23.7 N
(sin 34.2) 2
(impossible ) 
  34.2
7
Block A slides along a plane
inclined at an angle  to the
horizontal. The block is initially at a
height, h, above a point B.
Determine the velocity of the block
at B if it starts from rest.
W
A

h

B
Solution (i) (Case of no friction)
- Free-body diagram of block
Principle of Work and Energy
Y

UiB =TB - Ti
UiB (initial position)
Ti = 0 (Block starts from rest)
X
NP
F = NP
UiB = W sin  
h
 Wh
sin 

1W 2
VB  Wh  VB  2gh
2 g
Solution (ii) (Friction considered)
Block A does not move in direction  inclined plane  Fy = 0: NP – W cos  = 0
UiB = ( W sin   W cos )

h
 Wh  Wh cot 
sin 
1W B
V2  Wh  Wh cot   VB 2gh (1   cot )
2 g
For  = 90  cot  = 0  V =
2gh
as in solution (i)
For  = 0 , V =
2gh
For real values of V, 2gh  2gh cot     tan 
A 20 kg package is projected up a 20 incline with an initial
velocity of 12 m/s. The coefficient of friction between the
incline and the package is 0.15. Determine (a) the maximum
distance the package will move up the inclined plane; (b) the
velocity of the package when it returns to its original position.
8
(a) Free-body diagram of block
Principle of Work and Energy
196.2N
20o
F = NA
Y
NA
X
UAB = TB - TA
TA =
1
1
mV A2  (20)(12) 2  1440 J
2
2
At maximum distance up the incline VB = 0  TB = 0
UAB = -(196.2 sin 20 + 0.15 NA)d
(Force and displacement in opposite directions)
Package does not move  to inclined plane  Fy = 0
NA – 196.2 cos 20 = 0  NA = 196.2 cos 20
 -(196.2) (sin 20 + 0.15 cos 20)d = 0 – 1440
d  15.2 m
(b) Free-body diagram
196.2N
20o
F = 0.15NB
NB
Principle of Work and Energy
UBA = TA - TB
 TA = UBA 
1
(20)VA2  15.2(196.2)(sin 20  0.15 cos 20)
2
VA2  59.96 
VA  7.74 m / s