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A 37-kW tractor is required to pull a 3-furrow plough with an effective width of cut of 1050 mm in clay soil. Check if the plough and tractor are matched.

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13 weeks required to finish the ploughing activity.

Calculate the field area in square meters.

53 hectares * 10,000 square meters/hectare = 530,000 square meters

Calculate the working width of the plough.

2 furrows * 430 mm/furrow = 860 mm

Convert the working width to meters.

860 mm / 1000 mm/m = 0.86 m

Calculate the area covered by one pass of the plough.

0.86 m * 100 m = 86 square meters/pass

Calculate the number of passes required to plough the field.

530,000 square meters / 86 square meters/pass = 6,162 passes

Calculate the daily area ploughed.

0.86 m * 100 m * 8 hours/day = 6,880 square meters/day

Calculate the number of days to plough the field.

530,000 square meters / 6,880 square meters/day = 76.8 days

Adjust for working days per week.

76.8 days / 6 days/week = 12.8 weeks

Round to the nearest whole week.

12.8 weeks ≈ 13 weeks

Matching assessment:

The tractor and plough are not optimally matched because:

The tractor's power (44 kW) is significantly higher than the power required for a 2-furrow plough in loam soil. This suggests the tractor is underutilized.

The time required to plough the field (13 weeks) is longer than it would be with a better-matched plough.

Recommendations:

Consider using a 3-furrow plough to utilize the tractor's power more efficiently and reduce ploughing time.

Alternatively, consider using a smaller tractor with a 2-furrow plough to improve fuel efficiency and cost-effectiveness.

Question:-

A farmer wishes to plough a 53-ha field which has a loam soil using a 44 kW tractor and a 2

furrow plough. Each plough body makes a furrow 430 mm wide. The tractor operator works

an 8-hour day, and 6 days per week. Show if the tractor and plough are matched, and

calculate the calendar days required to finish the ploughing activity.

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