In a orthogonal cutting process the tool used has rake angle of zero degree. The measured cutting force and thrust force are 500 N and 250 N, respectively. The coefficient of friction between the tool and the chip is

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Solution

$∵ tan β = \frac{F γ / F_{C} + tan α}{1 - (tan α) F γ / F_{C}}$

$∵ α = 0$

$tan β = \frac{F γ}{F_{C}} = \frac{250}{500} = 0.5$

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= z e r o

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Common Data:
In an orthogonal machining operation:
Uncut thickness $= 0.5 m m$
Cutting speed $= 20 m / m i n$
Rake angle $= 15^{o}$
Width of cut $= 5 m m$
Chip thickness $= 07 m m$
Thrust force $= 200 N$
Cutting force $= 1200 N$
Assume Merchant's theory.

The coefficient of friction at the tool-chip interface is

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In a orthogonal cutting process the tool used has rake angle of zero degree. The measured cutting force and thrust force are 500 N and 250 N, respectively. The coefficient of friction between the tool and the chip is

∵tanβ=Fγ/FC+tanα1−(tanα)Fγ/FC ∵α=0 tanβ=FγFC=250500=0.5

$∵ tan β = \frac{F γ / F_{C} + tan α}{1 - (tan α) F γ / F_{C}}$

$∵ α = 0$

$tan β = \frac{F γ}{F_{C}} = \frac{250}{500} = 0.5$