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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 40º. The required specific draft (or unit draft) in kPa is ____________
Correct answer is between '117,118'. Can you explain this answer?
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An inclined blade cutting tool of 250 mm width is operating at 200 mm ...
To calculate the cutting force, we need to find the component of the normal load that acts parallel to the cutting edge. This can be done using trigonometry.
The component of the normal load parallel to the cutting edge is given by:
Force_parallel = Normal_load * sin(tool_lift_angle)
Force_parallel = 1000 N * sin(40°)
Force_parallel = 1000 N * 0.6428
Force_parallel = 642.8 N
Now, we can calculate the cutting force per unit length of cutting edge using the given value:
Cutting_force_per_unit_length = 20 N/mm
Next, we can find the total cutting force by multiplying the cutting force per unit length by the width of the cutting tool:
Total_cutting_force = Cutting_force_per_unit_length * Width
Total_cutting_force = 20 N/mm * 250 mm
Total_cutting_force = 5000 N
Finally, we can calculate the total force required to overcome the soil-metal friction:
Friction_force = Total_cutting_force * coefficient_of_soil_metal_friction
Friction_force = 5000 N * 0.3
Friction_force = 1500 N
The total force required to operate the inclined blade cutting tool is the sum of the cutting force and the friction force:
Total_force = Cutting_force + Friction_force
Total_force = 5000 N + 1500 N
Total_force = 6500 N
The component of the normal load parallel to the cutting edge is given by:
Force_parallel = Normal_load * sin(tool_lift_angle)
Force_parallel = 1000 N * sin(40°)
Force_parallel = 1000 N * 0.6428
Force_parallel = 642.8 N
Now, we can calculate the cutting force per unit length of cutting edge using the given value:
Cutting_force_per_unit_length = 20 N/mm
Next, we can find the total cutting force by multiplying the cutting force per unit length by the width of the cutting tool:
Total_cutting_force = Cutting_force_per_unit_length * Width
Total_cutting_force = 20 N/mm * 250 mm
Total_cutting_force = 5000 N
Finally, we can calculate the total force required to overcome the soil-metal friction:
Friction_force = Total_cutting_force * coefficient_of_soil_metal_friction
Friction_force = 5000 N * 0.3
Friction_force = 1500 N
The total force required to operate the inclined blade cutting tool is the sum of the cutting force and the friction force:
Total_force = Cutting_force + Friction_force
Total_force = 5000 N + 1500 N
Total_force = 6500 N
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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 soilcutting force per unit length of cutting edge is 20 N mm-1. The tool lift angle is 40º. The required specific draft (or unit draft) in kPa is ____________Correct answer is between '117,118'. Can you explain this answer?
Question Description
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 soilcutting force per unit length of cutting edge is 20 N mm-1. The tool lift angle is 40º. The required specific draft (or unit draft) in kPa is ____________Correct answer is between '117,118'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 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 soilcutting force per unit length of cutting edge is 20 N mm-1. The tool lift angle is 40º. The required specific draft (or unit draft) in kPa is ____________Correct answer is between '117,118'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 soilcutting force per unit length of cutting edge is 20 N mm-1. The tool lift angle is 40º. The required specific draft (or unit draft) in kPa is ____________Correct answer is between '117,118'. Can you explain this answer?.
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 soilcutting force per unit length of cutting edge is 20 N mm-1. The tool lift angle is 40º. The required specific draft (or unit draft) in kPa is ____________Correct answer is between '117,118'. Can you explain this answer? for GATE 2024 is part of GATE preparation. The Question and answers have been prepared according to the GATE exam syllabus. Information about 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 soilcutting force per unit length of cutting edge is 20 N mm-1. The tool lift angle is 40º. The required specific draft (or unit draft) in kPa is ____________Correct answer is between '117,118'. Can you explain this answer? covers all topics & solutions for GATE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 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 soilcutting force per unit length of cutting edge is 20 N mm-1. The tool lift angle is 40º. The required specific draft (or unit draft) in kPa is ____________Correct answer is between '117,118'. Can you explain this answer?.
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