Engineering Mechanics

(Joyce) #1

(^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.



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