Subtopic: Calculation of performance parameters – field capacity, efficiency, performance index, application rate and losses
Question Statement:
A tractor drawn five-tine sweep type cultivator is set at an angle (load angle) of 45° with the soil surface during its operation. The sweep of each tine experiences a normal force of 960 N on its surface. If the coefficient of soil-metal friction is 0.3, the total draft requirement of the cultivator is ________ kN.
Official Answer Key:
4.35 to 4.45
Step-by-Step Solution & Explanation
For a tillage tool inclined at load angle α, the horizontal (draft) component per tine is:
D₁ = N sin α + μN cos α = N(sin α + μ cos α)
= 960 (sin 45° + 0.3 cos 45°) = 960 (0.7071 + 0.2121) = 960 × 0.9192 = 882.5 N.
For 5 tines: D = 5 × 882.5 = 4412 N = 4.41 kN — within the official range 4.35–4.45.
Common error: omitting the frictional term μN cos α, which gives 3.39 kN.
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