Engineering Mechanics

(^674) „„„„„ A Textbook of Engineering Mechanics
also cause some forward motion of the belt without carrying the driven pulley with it. This is called
slip of the belt, and is generally expressed as a percentage.
The result of the belt slipping is to reduce the velocity ratio of the system. As the slipping of
the belt is a common phenomenon, thus the belt should never be used where a definite velocity ratio
is of importance (as in the case of hour, minute and second arms in a watch).
Let s 1 % = Slip between the driver and the belt,
s 2 % = Slip between the belt and the follower,
Now velocity of the belt, passing over the driver per minute,
v= 11 – 11 1
100
s
ππ×dN dN
= 11 1–^1
100
s
dN
⎛⎞
π ⎜⎟
⎝⎠
...(i)
Similarly, πd 2 N 2 = –1–^22
100 100
ss
vv v
⎛⎞
×=⎜⎟
⎝⎠
Substituting the value of v from equation (i),
πd 2 N 2 = 11 1–^12 1–
100 100
ss
dN
⎛⎞⎛⎞
π×⎜⎟⎜⎟
⎝⎠⎝⎠
or
21 1 2
12
1– –
100 100
Nd s s
Nd
⎛⎞
= ⎜⎟
⎝⎠
... Neglecting^12
100 100
⎛⎞ss+
⎜⎟
⎝⎠×
112
2
1–
100
dss
d
⎡ ⎛⎞+ ⎤
= ⎢⎥⎜⎟
⎣⎦⎝⎠
1
2
1–
100
d s
d
= ⎛⎞
⎜⎟
⎝⎠
...(where s = s 1 + s 2 i.e., total percentage of slip)
Note: If thickness of the belt is considered, then
21
12
1–
100
Ndt s
Ndt

• ⎛⎞
  == ⎜⎟


• ⎝⎠...(where t is thickness of the belt)
  Example 33.3. An engine shaft running at 120 r.p.m. is required to drive a machine shaft by
  means of a belt. The pulley on the engine shaft is of 2 meters diameter and that of the machine shaft
  is of 1 meter diameter. If the belt thickness is 5 mm, find the speed of the machine shaft when (i) there
  is no slip, and (ii) there is a slip of 3%.
  Solution. Given: Speed of the engine shaft (N 1 ) = 120 r.p.m; Diameter of the pulley on the
  engine shaft (d 1 ) = 2m; Diameter of the machine shaft (d 2 ) = 1m; Thickness of the belt (t) = 5 mm
  = 0·005 m and slip (s) = 3%.
  (i) Speed of the machine shaft when there is no slip
  We know that speed of the machine shaft,
  N 2 =
  1
  1
  2
  2 0·005
  120
  10·005
  dt
  N
  dt


• 
  
  
  
  • ×=×
    ++
    = 239·4 r.p.m. Ans.
    (ii) Speed of the machine shaft when there is a slip of 3%
    We know that speed of the machine shaft,
    N 2 =
    1
    1
    2
    1–
    100
    dt s
    N
    dt
  
  
  
  


• ⎛⎞
  × ⎜⎟


• ⎝⎠ =
  2 0·005 3
  120 1 –
  1 0·005 100


• ⎛⎞
  × ⎜⎟


• ⎝⎠
  =232 r.p.m. Ans.