# Solution Elaboration Let's solve each part step by step using **Taylor’s tool life equation**: \[ VT^n = C \] where: - \( V \) = Cutting speed (m/min) - \( T \) = Tool life (min) - \( n \) = Tool life exponent - \( C \) = Tool constant --- ## 1. **Determine the tool constant \( C \).** Given: - \( V = 90 \) m/min - \( T = 60 \) min - \( n = .25 \) Plug these values into the equation: \[ VT^n = C \implies 90 \times (60)^{.25} = C \] Calculate \( 60^{.25} \): \[ 60^{.25} = \sqrt{\sqrt{60}} \approx \sqrt{\sqrt{60}} \approx \sqrt{7.746} \approx 2.783 \] Now, calculate \( C \): \[ C = 90 \times 2.783 \approx 250.47 \] --- ## 2. **Calculate tool life for \( V = 120 \) m/min.** Use the calculated \( C \) and the new speed: \[ VT^n = C \] \[ 120 \times T^{.25} = 250.47 \] \[ T^{.25} = \frac{250.47}{120} \approx 2.087 \] Raise both sides to the 4th power (since \( n = .25 \), \( 1/n = 4 \)): \[ T = (2.087)^4 \approx 18.95 \text{ min} \] --- ## 3. **Effect of speed on tool life** **Observation:** When cutting speed increases from **90 m/min** to **120 m/min**: - Tool life decreases from **60 min** to **~19 min**. **Conclusion:** > **Increasing cutting speed significantly reduces tool life.** This relationship is governed by the negative exponential (\( n \)) in Taylor’s equation, indicating a small increase in speed causes a large decrease in tool life. --- ## **Summary Table** | Cutting Speed (m/min) | Tool Life (min) | |----------------------|-----------------| | 90 | 60 | | 120 | 19 | --- ### **Key Takeaway** - **Higher cutting speeds = Shorter tool life.** - This is a critical consideration in machining operations for balancing productivity and tool cost.