A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity ratio of 0.5. The center-to-center distance is 9 ft. The angular speed of the small pulley is 1750 rev/min as it delivers 2 hp. The service is such that a service factor Ks of 1.25 is appropriate. (a) Find Fc, Fi, (F1)a, and F2, assuming operation at the maximum tension limit. (b) Find Ha, nfs, and belt length.

A) The values of  Fc, Fi, ( F1)a  and F2

B ) The values of  Ha, nfs  and belt length

Given data :

Angular velocity ratio of large pulley = 0.5

width of polyamide ( b ) = 6 inches = 0.1524 m

pulley width ( d ) = 2 inches = 0.0508 m

center to center distance ( L ) = 9 ft =  2.7432 m

angular speed of small pulley ( n ) = 1750 rev/min

power of small pulley ( Hnom ) = 2 hp = 1491.4 Watts

Ks = 1.25

γ = 9500 N/m³

g = 9.81 m/s²

t = 0.0013 mm

A) Determine the values of Fc, Fi, F1a and F2

w = γ * b * t

   = 9500 * 0.1524 * 0.0013 = 1.88 N/m

V = = ( π * 0.0508 * 1750 ) / 60 = 4.65 m/s  

T = 10.17 Nm ( calculated value )

i) Fc =  ---- ( 1 )

Insert values into equation ( 1 )

Fc = ( 1.88 * 4.65² ) / 9.81

    = 4.15 N

ii) Fi =  - Fc ----- ( 2 )

where ( F1 )a = 0.1524 * 6000 * 0.7 * 1 = 640 N

             F2 = ( F1 )a - =  640 -  = 239.6 N

back to equation ( 2 )

Fi = [ ( 640 + 239.6 ) / 2 ] - Fc

   =  435.65 N

iii) ( F1 )a = b * Fa * Cp * Cv

              = 0.1524 * 6000 * 0.7 * 1  = 640 N

iv) F2 = ( F1 )a -

         = 640 -  = 239.6 N

B) Find the values of Ha, nfs and belt length

i) Ha = 1491 * 1.25

        = 1863.75 W

ii) nfs =

iii) belt length ( dip ) =  

                                 = ( 2.7432 * 1.88 ) / 435.65  = 11.8 mm

Hence we can conclude that the values of the variables in the question is as listed above

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a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2  = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches  = 0.0508 m

n = 1750 rpm

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

b)

dip =

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