Question

During orthogonal cutting with a tool of 10° rake angle, the cutting and thrust forces are 900 N and 275 N, respectively. The coefficient of friction on the rake surface of the tool is .............

Hint:

When calculating the coefficient of friction in orthogonal cutting, ensure you account for the cutting and thrust forces along with the rake angle. This is crucial in determining the cutting efficiency and tool wear.

Answer 1

In orthogonal cutting, the coefficient of friction \( \mu \) on the rake surface can be determined using the following relation:
\[ \mu = \frac{F_c \sin \alpha + F_t \cos \alpha}{F_c \cos \alpha + F_t \sin \alpha}, \] where:
- \( F_c \) is the cutting force (900 N),
- \( F_t \) is the thrust force (275 N),
- \( \alpha \) is the rake angle, which is given as 10°.
Substituting the given values into the equation:
\[ \mu = \frac{900 \sin 10^\circ + 275 \cos 10^\circ}{900 \cos 10^\circ + 275 \sin 10^\circ}. \] Now, calculate the trigonometric values:
- \( \sin 10^\circ \approx 0.1736 \),
- \( \cos 10^\circ \approx 0.9848 \).
Substituting these values into the equation:
\[ \mu = \frac{900 \times 0.1736 + 275 \times 0.9848}{900 \times 0.9848 + 275 \times 0.1736} = \frac{156.24 + 270.42}{885.32 + 47.74} = \frac{426.66}{933.06} \approx 0.51. \] Thus, the coefficient of friction lies between 0.49 and 0.53.