Question

In an orthogonal cutting test, the cutting force and thrust force were observed to be 1000 N and 500 N, respectively. If the rake angle of tool is ${11.31}^{\circ }$, the coefficient of friction in chip-tool interface will be [given sin 11.31 = 0.2]

• A. 1.27
• B. 0.63
• C. 0.32
• D. 0.78

Answer 1

The correct option is D 0.78

Given data:
$\sin {11.31}^{\circ }=0.2$
$\cos {11.31}^{\circ }=\sqrt{1-{\sin }^2{11.31}^{\circ }}$
$=\sqrt{1-(0.2{)}^2}=0.98$

$F=F_C\sin \alpha +F_t\cos \alpha$
$=F_C\times 0.2+F_t\times 0.98=69.\text{N}$

$\text{N}=F_C\cos \alpha -F_t\sin \alpha$
$F_C\times 0.98-F_t\times 0.2=880\text{N}$

$\mu =\frac{F}{N}=\frac{690}{880}=0.78$