Under orthogonal cutting condition, a turning operation is carried out on a metallic workpiece at a cutting speed of 4 m/s. The orthogonal rake angle of the cutting tool is 5°. The uncut chip thickness and width of cut are 0.2 mm and 3 mm, respectively. In this turning operation, the resulting friction angle and shear angle are 45º and 25°, respectively. If the dynamic yield shear strength of the workpiece material under this cutting condition is 1000 MPa, then the cutting force is ______________ N (round off to one decimal place).

The correct answer is 2573.40.

Step-by-Step Explanation:

1. Identify the given parameters from the problem:
- Cutting speed, V = 4 m/s
- Rake angle, α = 5°
- Uncut chip thickness, t = 0.2 mm
- Width of cut, w = 3 mm
- Friction angle, β = 45°
- Shear angle, ϕ = 25°
- Dynamic yield shear strength of the workpiece material, τ_s = 1000 MPa = 1000 N/mm²

2. Calculate the shear force (F_s):
The shear force is the force acting along the shear plane, defined by the formula:
$$F_s={\tau }_s\times \frac{w\times t}{\sin \varphi }$$
Substituting the given values into the equation:
$$F_s=1000\times \frac{3\times 0.2}{\sin ({25}^{\circ })}$$
Let's compute the value of the term:
$$F_s=1000\times \frac{0.6}{0.422618}\approx 1419.72\text{N}$$

3. Relate the cutting force (F_c) and the shear force (F_s) using Merchant's Circle relations:
From Merchant's force circle diagram, the relationship between cutting force and shear force is given by:
$$\frac{F_c}{F_s}=\frac{\cos (\beta -\alpha )}{\cos (\varphi +\beta -\alpha )}$$
Therefore, the cutting force is expressed as:
$$F_c=F_s\times \frac{\cos (\beta -\alpha )}{\cos (\varphi +\beta -\alpha )}$$

4. Calculate the angles:
- Numerator angle: $\beta -\alpha ={45}^{\circ }-5^{\circ }={40}^{\circ }$
- Denominator angle: $\varphi +\beta -\alpha ={25}^{\circ }+{45}^{\circ }-5^{\circ }={65}^{\circ }$

5. Compute the Cutting Force (F_c):
$$F_c=1000\times \frac{3\times 0.2}{\sin ({25}^{\circ })}\times \frac{\cos ({40}^{\circ })}{\cos ({65}^{\circ })}$$
Substitute the trigonometric values:
- $\cos ({40}^{\circ })\approx 0.766044$
- $\cos ({65}^{\circ })\approx 0.422618$
Now plug these values in:
$$F_c=1419.72\times \frac{0.766044}{0.422618}$$
$$F_c\approx 1419.72\times 1.81261\approx 2573.40\text{N}$$

Thus, the cutting force is 2573.40 N.