(^678) „„„„„ A Textbook of Engineering Mechanics
Let the belt leave the larger pulley at E and G and the smaller pulley at F and H as shown in
Fig. 33.7. Through O 2 draw O 2 M parallel to FE. From the geometry of the figure, we find that O 2 M
will be perpendicular to O 1 E.
and sin α=^12
rr
l

• ...(where α is the angle MO 2 O 1 in radians.)
  Since the angle α is very small, therefore
  sin α=^12
  rr
  l


• 
  

  α= ...(i)
  ∴ Length of arc JE= 1
  2
  r
  ⎛⎞π
  ⎜⎟+α
  ⎝⎠
  ...(ii)
  Similarly, length of arc FK = 2
  2
  r
  ⎛⎞π
  ⎜⎟+α
  ⎝⎠
  ...(iii)
  and EF=MO 212 =+l^22 –(r r)
  2
  l 1– rr^12
  l
  ⎛⎞+
  = ⎜⎟
  ⎝⎠
  Expanding this equation by binomial theorem,
  EF=
  2
  1 –^112 .....
  2
  rr
  l
  l
  ⎡⎤⎛⎞+
  ⎢⎥⎜⎟+
  ⎢⎥⎣⎦⎝⎠

  
  

  2

  
  

• ()^12
  2

rr

l

l

+

...(iv)

We know that the length of the belt,

L= Length of arc GJE + EF + Length of arc FKH + HG

= 2 (Length of arc JE + EF + Length of arc FK) ...(v)

Substituting the values of length of arc JE from equation (ii), length of arc FK from equation

(iii) and EF from equation (iv) in this equation,

L=

2

12

12

()

2–

222

rr

rl r

l

⎡⎤⎛⎞ππ+ ⎛⎞

⎢⎥⎜⎟+α + + ⎜⎟+α

⎣⎦⎝⎠ ⎝⎠

=

2

12

11 2 2

()

2–

222

rr

rrl r rx

l

⎡⎤ππ+

⎢⎥+α+ + +

⎣⎦

=

2

12

12 1 2

()

2( ) ( ) –

22

rr

rr r r l

l

⎡⎤π +

⎢⎥++α++

⎣⎦

=

2

12

12 1 2

()

()2( )2–

rr

rr r r l

l

+

π+ +α + +

Substituting the values of

rr 12

l

+

α= from equation (i),

L=

2

12 12

12 12

2( ) ( )

() ()2–

rr rr

rr rr l

ll

++

π+ + + +

=

22

12 12

12

2( ) ( )

() 2–

rr rr

rr l

ll

++

π+ + +

=

2

12

12

()

()2

rr

rr l

l

+

π+ + +