+1 vote
in Farm Machinery by (38.9k points)
A simple band brake has the tight side of the band attached to a fixed pivot. The angle of wrap is 280° about a 450-mm diameter drum. A torque of 170 mm is sustained at 900 revs min^-1. If the co-efficient of friction is 0.2 the required belt tension at tight side of the belt will be _____

(a) 453.04N

(b) 549.12N

(c) 605.35N

(d) 1211.53N

The question was posed to me in semester exam.

Origin of the question is Numericals on Estimation of Tractor Power in division Tillage, Disc Plough & Numericals of Farm Machinery

1 Answer

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by (243k points)
selected by
Best answer
Right choice is (d) 1211.53N

To elaborate: \(\frac{T1}{T2}\) = e^μθ

\(\frac{T1}{T2}\) = e^0.2*280 * π/180

\(\frac{T1}{T2}\) = 2.657

Power = (T1-T2)v

\(\frac{2\pi NT}{60}\) = (T1-T2)*\((\frac{\pi DN}{60}) \)

2*170 = (T1-T2) * 0.45

T1-T2 = 340/0.45

T1 – \(\frac{T2}{2.657} = \frac{340}{0.45} \)

T1 = 1211.53 N.

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