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In a belt drive the mass of the belt is 1 kg/m and its speed is 6m/s. The drive transmits 9.6 KW of power. Determine the strength of the belt when the coefficient of friction is 0.25 and wrap angle is 220°.

(a) 2629.78 N

(b) 2404.73 N

(c) 2003.09 N

(d) 1829.78 N

I had been asked this question in an internship interview.

This key question is from Power Transmission System topic in portion Ignition and Power Transmission of Farm Machinery

1 Answer

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Correct answer is (a) 2629.78 N

For explanation: Tc = mv^2 = 1*6^2 = 36 N


\(\frac{T1}{T2}=e^{0.25*220*\frac{\pi}{180}}\) = 2.6102

P = (T1-T2) * V

9.6*10^3 = (T1-T2) * 6

T1-T2 = 1600

T1 = 2593.78 N

T2 = 993.78 N

Strength = T1 + Tc = 2629.78 N

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