In an orthogonal machining with a single point cutting tool of rake angle $10^{\circ}$, the uncut chip thickness and the chip thickness are $0.125$ mm and $0.22$ mm, respectively. Using Merchant’s first solution for the condition of minimum cutting force, the coefficient of friction at the chip-tool interface is _______ (round off to two decimal places).

edited

chip thickness ratio (r) = to\tc = .125/.22 = .568

$tan(\Phi ) = rcos\alpha /(1- rsin\alpha ) = .568 cos (10)/ (1- .568 sin(10))= 31.78$

for minimum cutting power merchant’s first Equation is given by:-

$2\phi +\beta -\alpha = 90$

$we get \beta = 36.44$

$coefficient of friction (\mu ) = tan\beta \Leftrightarrow \mu= 0.73$ Ans.
Note:- Gate answer is between (0.72 – 0.76)
by (140 points) 3