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A motor car has a wheel base of 2.73 m and pivot centre of 1.065 m. the front and rear wheel track is 1.217 m. What will be angle of outside lock and turning circle radius of the outer front and inner rear wheels when the angle of inside lock is 40°?

(a) 34.6°

(b) 42°

(c) 32.4°

(d) 29°

This question was addressed to me during an online interview.

My question is based upon Numericals on Estimation of Tractor Power in chapter Tillage, Disc Plough & Numericals of Farm Machinery

1 Answer

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Correct option is (c) 32.4°

For explanation: cot φ – cot θ = \(\frac{c}{b} \)

cot φ = \(\frac{c}{b}\) + cot θ

cot φ = \(\frac{1.065}{2.743}\) + cot⁡ 40°

cot φ = 0.388 + 1.19175 = 1.579

cot φ = 1.58m

φ = cot^-1 (1.58) = 32.4°.

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