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An orthogonal cutting operations 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)
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- 0.4408
- 1.44
- 0.9938
- None of these
Correct Option: A
t = 0.010 mm
v = 130 m/min
α = 15°
b = 6 mm
tc = 0.015 mm
Fc = 60 N
Ft = 25 N
F = Fc sin α + Ft cos α
= 60 sin 15 + 25 cos 15
= 39.6773 N
Ratio of frictional energy to total energy
| = | . | = |  |  | ||||
| Fc | V | Fc | tc |
| = | × | = 0.4408 | ||
| 60 | 0.015 |
