Question
An orthogonal cutting operations is being carried out in which uncut thickness is 0.010 mm, cutting soeed 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)
Answer :
0.44
t = 0.010 mm
V = 130 m/min
α = 15°
b=6mm
tc = 0.015 mm
Fc = 60N
Ft = 25N
F = Fc sinα + Ftcosα = 60sin15° + 25cos15° = 39.6773
Ratio of frictional energy to total energy = (F/Fc)*(Vc/V) = (F/Fc)*(t/tc) (∵ t/tc = Vc/V =r)
= (39.6773/60) * (0.010/0.015) = 0.4408
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