The rake angle of a cutting tool is 15°, shear angle 45° and the cutting velocity 35 m/min. What is the velocity of chip along the tool face?

28.57 m/min

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27.38 m/min

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25.32 m/min

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23.56 m/min

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Solution

The correct option is A 28.57 m/min
We know,

$\frac{V}{cos (ϕ - α)} = \frac{V_{C}}{sin ϕ}$
$∴ V_{C} = \frac{35 \times sin 45^{\circ}}{cos 30^{\circ}} = 28.57 m / min$

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Similar questions

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Q. If a is the rake angle of a cutting tool,

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Linked Data:
A $ϕ 40 m m$ job is subjected to orthogonal turning by $+ 10^{o}$ rake angle tool at $500 r e v / m i n$ . By direct measurement during the cutting operation, the shear angle was found equal to $25^{o}$ .

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Common Data:
An orthogonal operation is carried out at $20 m / m i n$ cutting speed, using a cutting tool of rake angle $15$ degree. The chip thickness is $0.4 m m$ and uncut chip thickness is $0.2 m m$

The chip velocity (in $m / m i n$ ) is

Q. Common Data:
An orthogonal operation is carried out at

20 m / m i n

cutting speed, using a cutting tool of rake angle

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The shear angle (in degrees) is

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The rake angle of a cutting tool is 15°, shear angle 45° and the cutting velocity 35 m/min. What is the velocity of chip along the tool face?

The correct option is A 28.57 m/minWe know, Vcos(ϕ−α)=VCsinϕ ∴VC=35×sin45∘cos30∘=28.57m/min

The correct option is A 28.57 m/min
We know,

$\frac{V}{cos (ϕ - α)} = \frac{V_{C}}{sin ϕ}$
$∴ V_{C} = \frac{35 \times sin 45^{\circ}}{cos 30^{\circ}} = 28.57 m / min$