In an orthogonal machining with a single point cutting tool of rake angle $10^{o}$ , the uncut chip thickness and the chip thickness are $0.125 m m$ and $0.22 m m$ respectivley. 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).

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Solution

$α = 10^{o}$
$t = 0.125 m m$
$t_{c} = 0.22 m m$
$r = \frac{t}{t_{c}} = \frac{0.125}{0.22} = 0.5682$
$t a n ϕ = \frac{r c o s α}{1 - r s i n α} = \frac{0.5682 c o s 10^{o}}{1 - 0.5682 s i n 10^{o}}$
or $ϕ = {31.83}^{o}$
Using Merchant's thoery.
$ϕ = 45^{o} + \frac{α}{2} - \frac{β}{2}$
$or, {31.83}^{o} = 45^{o} + \frac{10^{o}}{2} - \frac{β}{2}$
$or, β = {36.34}^{o}$
$μ = t a n β = t a n {36.34}^{o}$
$= 0.7356 = 0.74$

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α=10o t=0.125mm tc=0.22mm r=ttc=0.1250.22=0.5682 tanϕ=rcosα1−rsinα=0.5682cos10o1−0.5682sin10o or ϕ=31.83o Using Merchant's thoery. ϕ=45o+α2−β2 or,31.83o=45o+10o2−β2 or,β=36.34o μ=tanβ=tan36.34o =0.7356=0.74