First, calculate the power available at the workpiece, $P_{workpiece}$.

$$P_{workpiece} = P_{in} \eta_1 \eta_2$$

Next, determine the rotational speed of the workpiece, $n_2$. Since the belt speed is constant, the speeds of the pulley rims are equal: $\pi d_1 n_1 = \pi d_2 n_2$.

$$n_2 = n_1 \frac{d_1}{d_2}$$

Then, calculate the cutting speed $v$, which is the tangential speed at the surface of the workpiece.

$$v = \pi d n_2(\text{in rps}) = \pi d \frac{n_2(\text{in rpm})}{60} = \frac{\pi d n_1}{60} \frac{d_1}{d_2}$$

Finally, the cutting force $F_{cut}$ is related to power and speed by $P_{workpiece} = F_{cut} v$.

$$F_{cut} = \frac{P_{workpiece}}{v} = \frac{P_{in} \eta_1 \eta_2}{\frac{\pi d n_1 d_1}{60 d_2}} = \frac{60 P_{in} \eta_1 \eta_2 d_2}{\pi n_1 d_1 d}$$

Substitute the given values into the derived formula (using SI units: $P_{in} = 7500$ W, $d_1 = 0.060$ m, $d_2 = 0.240$ m, $d = 0.100$ m, $n_1 = 1200$ r/min):

$$F_{cut} = \frac{60 \cdot 7500 \text{ W} \cdot 0.90 \cdot 0.95 \cdot 0.240 \text{ m}}{\pi \cdot 1200 \text{ min}^{-1} \cdot 0.060 \text{ m} \cdot 0.100 \text{ m}} \approx 4.1 \times 10^3 \text{ N}$$