Question

In orthogonal turning of low carbon steel pipe with principal cutting edge angle of ${90}^o$, the main cutting force is $1000N$ and the feed force is $800N$. The shear angle is ${25}^o$ and the orthogonal rake angle is zero. Employing Merchant's theory, the ratio of friction force to normal force acting on the cutting tool is

• A. $1.56$
• B. $1.25$
• C. $0.80$
• D. $0.64$

Answer 1

The correct option is C $0.80$

Given,
Cutting force: $F_C=1000N$
Feed force : $F_T=800N$
Rake angle : $\alpha =0^o$
Shear angle: : $\varphi ={25}^o$
We know that, $\mu =\frac{F}{N}=\frac{F_{C}sin\alpha +F_{T}cos\alpha }{F_{C}cos\alpha -F_{T}sin\alpha }$

$=\frac{1000sin{0}^o+800cos{0}^o}{1000cos{0}^o-800sin{0}^o}$

$=\frac{800}{1000}=0.8$