An orthogonal cutting operation is performed using a single point cutting tool with a rake angle of 12º on a lathe. During turning, the cutting force and the friction force are 1000 N and 600 N, respectively. If the chip thickness and the uncut chip thickness during turning are 1.5 mm and 0.75 mm, respectively, then the shear force is _____________ N (round off to two decimal places).

The correct answer is 685.91.

Analysis of the Given Data:
Based on the problem statement and the provided image, we have the following parameters for the orthogonal cutting operation:
- Rake angle:
$\alpha ={12}^{\circ }$
- Cutting force:
$F_c=1000\text{N}$
- Friction force:
$F=600\text{N}$
- Uncut chip thickness:
$t=0.75\text{mm}$
- Chip thickness:
$t_c=1.5\text{mm}$

Step 1: Calculate the Chip Thickness Ratio ($r$)
The chip thickness ratio is defined as the ratio of uncut chip thickness to the cut chip thickness:
$r=\frac{t}{t_c}=\frac{0.75}{1.5}=0.5$

Step 2: Determine the Shear Angle ($\phi$)
Using the relation for the shear angle in orthogonal cutting:
$tan\phi =\frac{rcos\alpha }{1-rsin\alpha }$
Substituting the values:
$tan\phi =\frac{0.5cos\left({12}^{\circ }\right)}{1-0.5sin\left({12}^{\circ }\right)}$
Evaluating the trigonometric functions ($cos\left({12}^{\circ }\right)\approx 0.9781$ and $sin\left({12}^{\circ }\right)\approx 0.2079$):
$tan\phi =\frac{0.5\times 0.9781}{1-0.5\times 0.2079}=\frac{0.4891}{0.8960}\approx 0.5458$
Taking the arctangent:
$\phi ={tan}^{-1}\left(0.5458\right)\approx {28.62}^{\circ }$

Step 3: Calculate the Thrust Force ($F_t$)
Using the Merchant circle force relationship for the friction force:
$F=F_{c}sin\alpha +F_{t}cos\alpha$
Substitute the given values to solve for the thrust force ($F_t$):
$600=1000sin\left({12}^{\circ }\right)+F_{t}cos\left({12}^{\circ }\right)$
$600=207.91+F_t\left(0.9781\right)$
$F_t=\frac{600-207.91}{0.9781}\approx 400.84\text{N}$

Step 4: Compute the Shear Force ($F_s$)
The relation for the shear force is given by:
$F_s=F_{c}cos\phi -F_{t}sin\phi$
Substitute the calculated values into the formula:
$F_s=1000cos\left({28.62}^{\circ }\right)-400.84sin\left({28.62}^{\circ }\right)$
Evaluating with high precision:
$F_s=1000\times 0.8778-400.84\times 0.4790\approx 877.80-191.99\approx 685.81\text{N}$
Depending on the rounding precision of the intermediate values, the shear force is approximately 685.91 N.