+1 vote
in Farm Machinery by (38.9k points)
What is the width of belt to transmit 15 HP to a pulley 35 cm in diameter. Pulley makes 1600 rev/min, coefficient of friction between belt is 0.22 and angle of contact is 20°. Maximum tension in belt not to exceed 8 kg per cm width.

(a) 7.79cm

(b) 9.79cm

(c) 3.49cm

(d) 8.79cm

I had been asked this question in an international level competition.

The above asked question is from Power Transmission System in chapter Ignition and Power Transmission of Farm Machinery

1 Answer

+2 votes
by (243k points)
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Best answer
The correct choice is (d) 8.79cm

Easy explanation: \(\frac{T1}{T2}\) = e^0.22*3.66=e^0.8065

T1 = T2 * 2.24

15*746 = \(\frac{[π*35*1600(T1-T2)]}{100*60} \)

2.24 T2-T1 = 381.63

T2 = \(\frac{381.63}{1.24}\) = 307.76N

T1 = 307.76*2.24=689.38N

Width of the belt = \(\frac{689.38}{8*9.8}\) = 8.79cm.

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