An inclined blade cutting tool of 250 mm width is operating at 200 mm cutting depth. The normal load on the tool and the coefficient of soil-metal friction are 1000 N and 0.3, respectively. The soil cutting force per unit length of cutting edge is 20 N mm⁻¹. The tool lift angle is 40°. The required specific draft (or unit draft) in kPa is ______
Official Answer Key:
117 to 118
Step-by-Step Solution & Explanation
Total horizontal draft on the blade is taken as the sum of the soil-cutting resistance along the edge and the friction resistance on the tool face: soil-cutting force = (force/length)×width = 20×250 = 5000 N; friction force = μ×N = 0.3×1000 = 300 N. Draft = 5000+300 = 5300 N. Specific (unit) draft = Draft/(width×depth) = 5300/(250×200) = 0.106 N/mm² = 106 kPa. Discrepancy note: the official accepted range is 117–118 kPa. Introducing the given 40° lift angle into the friction term (e.g. resolving the normal-load friction force through the lift angle, μ·N/cos40°, or an earth-moving/Reece-type formulation combining cutting and adhesion terms) shifts the result closer to the key but we were unable to exactly reproduce 117–118 kPa from the data as transcribed; the lift angle's precise role in the intended force resolution could not be confirmed from the scanned question. We show the direct summation method (106 kPa) and flag the official key value for scoring.
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