Question

In an orthogonal cutting with a tool of rake angle \(0^\circ\), the value of the cutting force is two times of the thrust force. The coefficient of friction is ................. (round off to 1 decimal place).

Hint:

For orthogonal cutting with a tool of rake angle \(0^\circ\), if the cutting force is twice the thrust force, the coefficient of friction is the ratio of the cutting force to the thrust force.

Answer 1

- In orthogonal cutting, the relationship between the cutting force (\(F_c\)) and the thrust force (\(F_t\)) can be expressed as: \[ F_c = 2 F_t \] - The cutting force is related to the friction force (\(F_f\)) by the following equation: \[ F_f = \mu F_n \] Where \(\mu\) is the coefficient of friction and \(F_n\) is the normal force. For the given scenario with a rake angle of \(0^\circ\), the normal force \(F_n\) is approximately equal to the thrust force \(F_t\). \[ F_f = \mu F_t \] Since \(F_c = 2 F_t\), the coefficient of friction \(\mu\) can be calculated as: \[ \mu = \frac{F_c}{F_t} = 2 \] Thus, the coefficient of friction is \(0.5\).