In a grinding operation of a metal, specific energy consumption is 15 J/mm³ . If a grinding wheel with a diameter of 200 mm is rotating at 3000 rpm to obtain a material removal rate of 6000 mm³ /min, then the³ tangential force on the wheel is ______________ N (round off to two decimal places).

The correct answer is 47.74.

Step-by-Step Explanation:

To find the tangential force on the grinding wheel, we can relate the energy consumed in the grinding process to the mechanical work done by the wheel.

1. Identify the given parameters and convert them to standard units:
- Specific energy consumption (^e): 15 J/mm³
- Material removal rate (MRR): 6000 mm³/min
- Grinding wheel diameter (D): 200 mm
- Rotational speed (N): 3000 rpm

2. Convert the Material Removal Rate (MRR) to mm³/s:
$$\text{MRR}=\frac{6000}{60}{\text{mm}}^3\text{/s}=100{\text{mm}}^3\text{/s}$$

3. Calculate the total power consumption (P) of the grinding process:
Power is the product of specific energy consumption and the material removal rate:
$$P=e\times \text{MRR}$$
Substituting the values:
$$P=15{\text{J/mm}}^3\times 100{\text{mm}}^3\text{/s}=1500\text{W}$$

4. Calculate the tangential velocity (V) of the grinding wheel:
The linear speed at the periphery of the wheel is given by:
$$V=\frac{\pi \times D\times N}{60\times 1000}$$
Note: The factor of 1000 in the denominator converts the diameter from millimeters to meters so that velocity is obtained in m/s.
$$V=\frac{\pi \times 200\times 3000}{60000}=10\pi \text{m/s}\approx 31.4159\text{m/s}$$

5. Calculate the tangential force (F_c):
Power is also equal to the product of tangential force and tangential speed:
$$P=F_c\times V$$
Rearranging the formula to solve for the tangential force:
$$F_c=\frac{P}{V}=\frac{1500}{10\pi }=\frac{150}{\pi }\approx 47.746\text{N}$$
Rounding off to two decimal places yields approximately 47.74 N.