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Question

In an orthogonal cutting test, the following observation were made
Cutting force =1200N
Thrust force =500N
Tool rake angle =zero
Cutting speed =1m/s
Depth of cut =0.8mm
Chip thickness =1.5mm

Friction angle during machining will be

A
22.6
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B
32.6
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C
57.1
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D
67.4
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Solution

The correct option is A 22.6
FC=12000N,FT=500N;α=0
VC=1m/s,d=b=0.8,t2=1.5mm
tan(βα)=FTFC
β=α+tan1(FTFC)=0+tan1(5001200)=22.62o

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