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In a belt drive, the mass of the belt is 10 kg/m and its speed is 6 m/s. The drive transmits 20 KW of power. Determine the initial tension in the belt and the strength of the belt. The co-efficient of friction is 0.25 and wrap angle is 220°.

(a) 4096 N, 5762 N

(b) 3096 N, 4762 N

(c) 2096 N, 3762 N

(d) 1096 N, 2762 N

I have been asked this question at a job interview.

Asked question is from Mechanical Power Transmission in chapter Mechanical Power Transmission of Farm Machinery

1 Answer

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Correct option is (a) 4096 N, 5762 N

Best explanation: \(\frac{T1}{T2}\)=e^μθ=\(e^{0.25*220*(\frac{\pi}{180})}\) = 2.611

T1 = 2.611 T2

P = (T1-T2)*v

20*1000 = (T1-T2) * 6

T1 – T2 = 3333.33

2.611 T2 – T2 = 3333.33

T2 = 2069.11 N

T1 = 5402.44 N

Initial Tension = \(\frac{T1+T2}{2}+Tc=\frac{5402.44+2069.11}{2}\)+360 = 4095.77 N

Strength of the belt = total tension on tight side = T1 + Tc = 5762.44 N.

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