(^684) „„„„„ A Textbook of Engineering Mechanics
From the geometry of Fig. 33·10, we find that
∠ HO 2 K=∠ JO 1 E
2
θ

∴∠ MO 1 O 2 180 –
2
θ
=°
and ∠O 1 MO 2 =90°
∴ cos ∠ MO 1 O 2 11 12
12 1 2
OM OE EM r r
OO O O l
++
== =
0·225 0·1
0·1667
1·9 5

• 
  

  ==
  ∴∠ MO 1 O 2 = 80·4°
  i.e. 180 – 80·4
  2
  θ
  °=°
  or 180 – 80·4 99·6
  2
  θ
  =° °= °
  ∴θ= 2 × 99·6° = 199·2° Ans.
  Power transmitted by the belt
  We know that speed of the belt,
  1 0·45^200 4·71 m/s
  60 60
  dN
  v
  π π× ×
  == =
  and θ= 199·2° = 199·2° ×
  180
  π
  = 3·477 rad
  We know that^1
  2
  2·3 log
  T
  T
  ⎛⎞
  ⎜⎟⎜⎟
  ⎝⎠
  =μθ = 0·25 × 3·477 = 0·8693
  1
  2
  log
  T
  T
  ⎛⎞
  ⎜⎟⎜⎟
  ⎝⎠

  
  

  0·8693
  0·3780
  2·3

  
  

  or
  2
  1000
  2·388
  T
  ⎛⎞=
  ⎜⎟
  ⎝⎠
  ...(Taking antilog of 0·3780)
  ∴ T 2 =
  1000
  2·388
  = 418·8 N
  We know that power transmitted by the belt,
  P=(T 1 – T 2 )v = (1000 – 418·8) × 4·71 = 2740 W = 2.74 kW Ans.
  33.15.CENTRIFUGAL TENSION
  We have already discussed that the belt continuously runs over both the pulleys. It carries
  some centrifugal force in the belt, at both the pulleys, whose effect is to increase the tension on both,
  tight as well as the slack sides. The tension caused by the centrifugal force is called centrifugal
  tension. At lower speeds, the centrifugal tension is very small and may be neglected. But at higher
  speeds, its effect is considerable, and thus should be taken into account.
  Consider a small portion AB of the belt as shown in Fig. 33·11.