In an orthogonal cutting test on mild steel, the following data were obtained
Cutting speed : $40 m / m i n$
Depth of cut : $0.3 m m$
Total rake angle : $+ 5^{o}$
Chip thickness : $1.5 m m$
Cutting force : $900 N$
Thrust force : $450 N$
Using Merchant's analysis, the friction angle during the machining will be

{26.6}^{o}

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{31.5}^{o}

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45^{o}

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{63.4}^{o}

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Solution

The correct option is B ${31.5}^{o}$
The coefficient of friction
$μ = \frac{F}{N} = \frac{F_{c} s i n α + F_{T} c o s α}{F_{c} c o s α - F_{T} s i n α}$

$= \frac{900 s i n 5^{o} + 450 c o s 5^{o}}{900 c o s 5^{o} - 450 s i n 5^{o}}$

$= \frac{526.72}{857.35} = 0.614$

Friction angle
$∴ β = t a n^{- 1} (μ) = {31.5}^{o}$

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Q. In an orthogonal cutting test, the following observation were made
Cutting force

= 1200 N

Thrust force

= 500 N

Tool rake angle

= z e r o

Cutting speed

= 1 m / s

Depth of cut

= 0.8 m m

Chip thickness

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Friction angle during machining will be

Common Data:
In an orthogonal machining operation:
Uncut thickness $= 0.5 m m$
Cutting speed $= 20 m / m i n$
Rake angle $= 15^{o}$
Width of cut $= 5 m m$
Chip thickness $= 07 m m$
Thrust force $= 200 N$
Cutting force $= 1200 N$
Assume Merchant's theory.

The coefficient of friction at the tool-chip interface is