A single point cutting tool with 0° rake angle is used in an orthogonal machining process. At a cutting speed of 180 m/min, the thrust force is 490 N. If the coefficient of friction between the tool and the chip is 0.7, then the power consumption (in kW) for the machining operation is
2.1

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Solution

The correct option is A 2.1
$\frac{F}{N} = \frac{F_{T}}{F_{C}} = 0.7$

$\Rightarrow \frac{490}{F_{C}} = 0.7$

$F_{C} = 700 N$

$Power = F_{C} \times V_{C} = \frac{700 \times 180}{60} = 2.1 k W$

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A single point cutting tool with 0° rake angle is used in an orthogonal machining process. At a cutting speed of 180 m/min, the thrust force is 490 N. If the coefficient of friction between the tool and the chip is 0.7, then the power consumption (in kW) for the machining operation is 2.1

The correct option is A 2.1 FN=FTFC=0.7 ⇒ 490FC=0.7 FC=700 N Power=FC×VC=700×18060=2.1 kW

A single point cutting tool with 0° rake angle is used in an orthogonal machining process. At a cutting speed of 180 m/min, the thrust force is 490 N. If the coefficient of friction between the tool and the chip is 0.7, then the power consumption (in kW) for the machining operation is
2.1

The correct option is A 2.1
$\frac{F}{N} = \frac{F_{T}}{F_{C}} = 0.7$

$\Rightarrow \frac{490}{F_{C}} = 0.7$

$F_{C} = 700 N$

$Power = F_{C} \times V_{C} = \frac{700 \times 180}{60} = 2.1 k W$