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An orthogonal cutting operation is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15 and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________
(correct to two decimal places)
    Correct answer is '0.44'. Can you explain this answer?
    Verified Answer
    An orthogonal cutting operation is being carried out in which uncut th...
    F=FCsinα+FTcosα
    F=60sin15+25cos15
    F=39.67N….(ii)
    Ratio of frictional enery to Total energy will be,

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    Most Upvoted Answer
    An orthogonal cutting operation is being carried out in which uncut th...
    Given data:
    Uncut thickness (t0) = 0.010 mm
    Cutting speed (V) = 130 m/min
    Rake angle (α) = 15°
    Width of cut (b) = 6 mm
    Chip thickness (tc) = 0.015 mm
    Cutting force (Fc) = 60 N
    Thrust force (Ft) = 25 N

    Calculations:
    1. Shear plane angle (φ):
    The shear plane angle can be calculated using the formula:
    tan φ = (tc - t0) / (b * cos α)
    Substituting the given values:
    tan φ = (0.015 - 0.010) / (6 * cos 15°)
    tan φ = 0.005 / (6 * 0.9659)
    tan φ ≈ 0.000865
    φ ≈ 0.0496°

    2. Cutting velocity (Vc):
    The cutting velocity can be calculated using the formula:
    Vc = V / cos φ
    Substituting the given values:
    Vc = 130 / cos 0.0496°
    Vc ≈ 130 / 0.9998
    Vc ≈ 130.01 m/min

    3. Friction force (Ff):
    The friction force can be calculated using the formula:
    Ff = Ft - Fc
    Substituting the given values:
    Ff = 25 - 60
    Ff = -35 N

    4. Friction energy (Ef):
    The friction energy can be calculated using the formula:
    Ef = Ff * Vc * t0
    Substituting the given values:
    Ef = -35 * 130.01 * 0.010
    Ef = -45.0515 N.mm

    5. Total energy (Et):
    The total energy can be calculated using the formula:
    Et = Fc * Vc * t0
    Substituting the given values:
    Et = 60 * 130.01 * 0.010
    Et = 78.006 N.mm

    6. Ratio of friction energy to total energy:
    The ratio of friction energy to total energy can be calculated using the formula:
    Ratio = Ef / Et
    Substituting the calculated values:
    Ratio = -45.0515 / 78.006
    Ratio ≈ -0.5778

    Explanation:
    The ratio of friction energy to total energy is approximately -0.5778. However, the correct answer is given as 0.44. It seems there might be an error in the calculation or the given data. Please review the calculations and data provided to ensure accuracy.
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    An orthogonal cutting operation is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15 and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________(correct to two decimal places)Correct answer is '0.44'. Can you explain this answer?
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    An orthogonal cutting operation is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15 and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________(correct to two decimal places)Correct answer is '0.44'. 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 orthogonal cutting operation is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15 and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________(correct to two decimal places)Correct answer is '0.44'. 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 orthogonal cutting operation is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130 m/min, rake angle is 15 and width of cut is 6 mm. It is observed that the chip thickness is 0.015 mm, the cutting force is 60 N and the thrust force is 25 N. The ratio of friction energy to total energy is __________(correct to two decimal places)Correct answer is '0.44'. Can you explain this answer?.
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