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A single reduction gear of 120 KW with a pinion of 150 mm pitch circle diameter and speed 550 rpm is supported in bearings on either side. Calculate the total load due to power transmission.

(a) 28.20 KN

(b) 35.21 KN

(c) 40 KN

(d) 45 KN

This question was addressed to me by my school teacher while I was bunking the class.

This interesting question is from Mechanical Power Transmission topic in chapter Mechanical Power Transmission of Farm Machinery

1 Answer

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The correct option is (a) 28.20 KN

The explanation: P=\(\frac{2πNT}{60} \)

T=\(\frac{120*1000*60}{2*π*550}\) = 2083.48 Nm  

Tangential load on pinion Ft = \(\frac{T}{r}=\frac{2083.48}{0.075}\)=27779.73 N

Total load due to power transmitted = \(\frac{Ft}{cos⁡θ} = \frac{27779.73}{cos⁡10}\) = 28.20 KN.

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