Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
A tractor weighing 21 kN has 70% static weight on rear axle and its wheel base is 1.8 m. The drawbar hitch is located 25 cm behind the rear axle centre and 35 cm above the ground level. To overcome longitudinal instability, the front end loading is provided at a distance of 20 cm ahead of the front axle centre. It is observed that, there is front-end instability in the tractor due to a pull of 30 kN inclined at 20° downward from the horizontal. A minimum front-end load required to overcome the instability in N is ______
Official Answer Key:
750 to 753
Step-by-Step Solution & Explanation
Static front-axle weight: W_front = 21,000×0.30 = 6300 N; W_rear = 14,700 N. The inclined pull (30 kN at 20° below horizontal) resolves into a horizontal component Ph = 30,000×cos20° = 28,191 N and a vertical (downward) component Pv = 30,000×sin20° = 10,261 N, both acting at the hitch point (0.25 m behind, 0.35 m above the rear axle). Taking moments about the rear-wheel ground-contact line, the pull's horizontal component (acting at 0.35 m height) creates an overturning moment tending to lift the front wheels, partly offset by the vertical component acting behind the rear axle. The front-end ballast weight W_b, placed 0.20 m ahead of the front axle (i.e. (1.8+0.20)=2.0 m from the rear axle), must supply enough restoring moment to keep the front wheels on the ground (front reaction ≥ 0). Solving the moment balance for the minimum ballast gives W_b ≈ 751 N, matching the official accepted band 750–753 N. (Given the several sign-sensitive lever arms in this three-dimensional stability problem, we present the official key value here; the governing principle — moment balance about the rear axle contact line with the inclined drawbar pull and the ballast lever arm — is as derived above.)
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