In an orthogonal machining operation, the cutting and thrust forces are equal in magnitude. The uncut chip thickness is 0.5 mm and the shear angle is 15°. The orthogonal rake angle of the tool is 0° and the width of cut is 2 mm. The workpiece material is perfectly plastic and its yield shear strength is 500 MPa. The cutting force is _________ N (round off to the nearest integer).

The correct answer is 2732.

1. Identify the given parameters from the problem description:
- Uncut chip thickness, _$t$ = 0.5 mm
- Width of cut, _$b$ = 2 mm
- Shear angle, _$\varphi$ = 15°
- Orthogonal rake angle, _$\alpha$ = 0°
- Yield shear strength of the workpiece material, _{${\tau }_s$} = 500 MPa = 500 N/mm²
- The cutting force, _{$F_c$}, and the thrust force, _{$F_T$}, are equal in magnitude: _{$F_c=F_T$}.

2. Determine the friction angle (λ):
In Merchant's circle diagram, the friction force (F) and normal force (N) on the rake face are related to the cutting and thrust forces by the following relations:
$$F=F_c\sin \alpha +F_T\cos \alpha$$
$$N=F_c\cos \alpha -F_T\sin \alpha$$
Given that the rake angle is _{$\alpha =0°$}:
$$F=F_c\sin (0°)+F_T\cos (0°)=F_T$$
$$N=F_c\cos (0°)-F_T\sin (0°)=F_c$$
Since _{$F_c=F_T$}, it follows that:
$$F=N$$
The coefficient of friction _$\mu$ at the tool-chip interface is defined as:
$$\mu =\frac{F}{N}=1$$
The friction angle _$\lambda$ is related to _$\mu$ by:
$$\mu =\tan \lambda =1\Rightarrow \lambda ={\tan }^{-1}(1)=45°$$

3. Calculate the shear force (F_s):
The area of the shear plane (A_s) is given by:
$$A_s=\frac{b\times t}{\sin \varphi }$$
Using the shear strength _{${\tau }_s$}, the shear force _{$F_s$} is:
$$F_s={\tau }_s\times A_s=\frac{{\tau }_s\times b\times t}{\sin \varphi }$$
Substituting the given values:
$$F_s=\frac{500\times 2\times 0.5}{\sin (15��)}=\frac{500}{0.258819}\approx 1931.85\text{N}$$

4. Calculate the cutting force (F_c):
From Merchant's force relations, the relationship between the cutting force _{$F_c$} and the shear force _{$F_s$} is:
$$F_c=F_s\times \frac{\cos (\lambda -\alpha )}{\cos (\lambda -\alpha +\varphi )}$$
Substitute _{$\lambda =45°$}, _{$\alpha =0°$}, and _{$\varphi =15°$} into the equation:
$$F_c=\frac{500\times 2\times 0.5}{\sin (15°)}\times \frac{\cos (45°-0°)}{\cos (45°-0°+15°)}$$
$$F_c=1931.85\times \frac{\cos (45°)}{\cos (60°)}$$
Using the values of trigonometric functions:
- _{$\cos (45°)=\frac{1}{\sqrt{2}}\approx 0.707107$}
- _{$\cos (60°)=0.5$}
Substitute these values:
$$F_c=1931.85\times \frac{0.707107}{0.5}$$
$$F_c=1931.85\times 1.414214\approx 2732.05\text{N}$$

Rounding off to the nearest integer, we get:
F_c ≈ 2732 N