While turning a 60 mm diameter bar, it was observed that the tangential cutting force was 3000 N and the feed force was 1200 N. If the tool rake angle is 32°, then the coefficient of friction is nearly (may take sin 32° = 0.53, cos 32° = 0.85 and tan 32° = 0.62)

Concept:

From Merchant cycle, 

F = F_C sin α + F_T cos α 

N = F_C cos α - F_T sin α

Thus \(\;\mu = \frac{F}{N} = \frac{{{F_c}\;sin\alpha \; + \;{F_{t\;}}cos\alpha }}{{{F_c}\;cos\alpha \; - \;{F_{t\;}}sin\alpha }}\)

Also, μ = tan β or β = tan^-1 μ

Where, F = friction force in N, N = normal force in N, F_C = cutting force in N

F_T = thrust force in N, μ = coefficient of friction, and α, β = rake angle and friction angle respectively. 

Calculation: 

Given, α = rake angle = 32°, Cutting force F_C = 3000 N, Feed force F_t = 1200 N

F = F_C sin α + F_T cos α = 3000 × sin 32° + 1200 × cos 32° = 2610 N

N = F_C cos α - F_T sin α = 3000 × cos 32° - 1200 × sin 32° = 1914 N

\(\mu = \frac{F}{N} = \frac{{2610}}{{1914}} = 1.36\)