Let's analyze and solve Question 3 step-by-step. --- ## Question 3: **Given Data:** - Tool: Ge-head lathe - Material: AISI 1055, BH 250 - Cutting speed: 400 ft/min - Metal removal rate: 2.4 cubic inches per minute - Character of the tool: -7-7-7-15-15-1/32 --- ## Step 1: Understand what is asked We need to estimate: a) Energy consumed per unit time b) Power required at the motor c) Tangential component of the cutting force --- ## Step 2: Convert units where necessary ### Convert cutting speed from ft/min to inches/min: \[ \text{Cutting speed} = 400 \text{ ft/min} \times 12 \text{ in/ft} = 480 \text{ in/min} \] --- ## Step 3: Find the **cutting length per minute** The **material removal rate** is 2.4³/min. The **chip cross-sectional area** (\(A\)) is the product of chip thickness and chip. But since the problem states the **character** of the tool with the code "-7-7-7-15-15-1/32", this indicates the **chip thickness**. The **average chip thickness** is given as .007 inches (from the '7' in the code), and the **width of cut** is typically taken as the same as the tool's width, which is not explicitly given. However, for the estimation, assume the **chip width** to be the same as the tool's width, which is not specified; hence, we'll use the **material removal rate** to find the **cutting length**. ### Step 4: Calculate **cutting length per minute** \[ \text{Volume} = \text{Area} \times \text{Length} \] \[ \Rightarrow \text{Length} = \frac{\text{Volume}}{\text{Area}} \] The **area of the chip**: \[ A = \text{chip thickness} \times \text{width} \] Assuming the **width of cut** is 1 inch (common for such calculations): \[ A = .007 \text{ in} \times 1 \text{ in} = .007 \text{ in}^2 \] Now, the **total length** of cut per minute: \[ L = \frac{2.4 \text{ in}^3/\text{min}}{.007 \text{ in}^2} \approx 342.86 \text{ in/min} \] --- ## Step 5: Calculate **power required** Power is related to the **cutting force** and **cutting speed**: \[ \text{Power} = \text{Cutting force} \times \text{Cutting speed} \] But first, we need to find the **cutting force**. ### Step 6: Find **cutting force** The **cutting force** (\(F_c\)) can be estimated from the **specific cutting energy** (\(U\)): \[ U = \frac{\text{Work done}}{\text{Volume removed}} \] The **specific cutting energy** for steel can be approximated around **10 to 20 ft-lbf/in³**. Let's take **U = 15 ft-lbf/in³** as an average. \[ \text{Work per minute} = U \times \text{Volume per minute} = 15 \text{ ft-lbf/in}^3 \times 2.4 \text{ in}^3 = 36 \text{ ft-lbf} \] Since power is work per unit time: \[ \text{Power} = \text{Work per minute} \div 60 \text{ sec} = \frac{36 \text{ ft-lbf}}{60} = .6 \text{ ft-lbf/sec} \] Convert ft-lbf/sec to watts: \[ 1 \text{ ft-lbf/sec} \approx 1.356 \text{ W} \] \[ \Rightarrow \text{Power} \approx .6 \times 1.356 \approx .814 \text{ W} \] **Alternatively**, since the earlier calculations are approximate, the power required is roughly **about .8 W**. --- ## **Final Answers:** ### a) Energy consumed per unit time: Approximately **.8 watts**. ### b) Power required at the motor: Approximately **.8 watts**. ### c) Tangential component of the cutting force: The **tangential force** can be computed from: \[ F_t = \frac{\text{Power}}{\text{Cutting speed}} \] Expressing power in consistent units: \[ \text{Power} = .8 \text{ W} \approx .8 \text{ J/sec} \] \[ \text{Cutting speed} = 480 \text{ in/min} = 80 \text{ in/sec} \] Convert inches to meters: \[ 80 \text{ in/sec} \times .0254 \text{ m/in} \approx 2.04 \text{ m/sec} \] Now: \[ F_t = \frac{.8 \text{ J/sec}}{2.04 \text{ m/sec}} \approx .392 \text{ N} \] --- ## **Summary:** - **Energy per unit time:** ~.8 W - **Motor power:** ~.8 W - **Tangential force:** ~.392 N --- Let me know if you'd like a more detailed explanation on any step!