Question

During an orthogonal machining operation on mild steel, the results obtained are:
$t_1=0.25mm,t_2=0.75mm,\alpha =0{}^{\circ },$
$F_c=900N,F_t=450N$
What is the coefficient of friction between the tool and the chip?

• A. 0.25
• B. 0.3
• C. 0.5
• D. 0.7

Answer 1

The correct option is C 0.5

We know,

$F=F_c\sin \alpha +F_t\cos \alpha =450N$

$N=F_c\cos \alpha -F_t\sin \alpha =900N$

$\mu =\frac{F}{N}=\frac{450}{900}=0.5$