An orthogonal cutting operation is being carried out in which uncut thickness is 0.010 mm, cutting speed is 130m/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)

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Solution

Given : t = 0.010 mm, v = 130 m/min
$α = 15^{\circ}, b = 6 m m, t_{c} = 0.015 m m$

$F_{c} = 60 N, F_{t} = 25 N$

$F = F_{c} sin α + F_{t} cos α$

$= 60 sin 15^{\circ} + 25 cos 15^{\circ}$

$= 39.6773 N$

Ratio of frictional energy to total energy

$= \frac{F}{F_{c}} . \frac{V_{c}}{V} = \frac{F}{F_{c}} (\frac{t}{t_{c}}) [∵ \frac{t}{t_{c}} = \frac{V_{c}}{V} = r]$

$= \frac{39.6773}{60} \times \frac{0.010}{0.015} = 0.4408$

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Given : t = 0.010 mm, v = 130 m/min α=15∘,b=6mm,tc=0.015mm Fc=60N,Ft=25N F=Fcsinα+Ftcosα =60sin15∘+25cos15∘ =39.6773N Ratio of frictional energy to total energy =FFc.VcV=FFc(ttc)[∵ttc=VcV=r] =39.677360×0.0100.015=0.4408