Let's solve each part step by step: --- ### a) **Assuming no losses due to friction, what torque is being applied to the output shaft attached to Gear 4? What speed is it rotating?** #### **Step 1: Gear Ratios and Speed Calculation** - **Gear 1 (N1):** 20 teeth, input at 1100 rpm - **Gear 2 (N2):** 60 teeth - **Gear 3 (N3):** 20 teeth (mounted on same shaft as Gear 2, thus same speed as Gear 2) - **Gear 4 (N4):** 60 teeth **Gear train:** - Gear 1 drives Gear 2 - Gear 3 (on same shaft as 2) drives Gear 4 **First stage (1 drives 2):** \[ \text{Speed of Gear 2} = \text{Speed of Gear 1} \times \frac{N_1}{N_2} \] \[ = 1100 \times \frac{20}{60} = 1100 \times 0.333 = 366.7 \text{ rpm} \] **Second stage (3 drives 4):** \[ \text{Speed of Gear 4} = \text{Speed of Gear 3} \times \frac{N_3}{N_4} \] But Gear 3 and 2 are on same shaft, so speed of Gear 3 = speed of Gear 2 = 366.7 rpm \[ \text{Speed of Gear 4} = 366.7 \times \frac{20}{60} = 366.7 \times 0.333 = 122.2 \text{ rpm} \] --- #### **Step 2: Torque Calculation** **Input Power:** 1.5 HP = 1.5 × 746 = **1119 W** \[ \text{Power} = \text{Torque} \times \text{Angular Velocity} \] \[ T = \frac{P}{\omega} \] where \(\omega\) in rad/s: \(\omega = 2\pi \times n / 60\) **At Output Shaft (Gear 4):** \[ n_{4} = 122.2 \text{ rpm} \] \[ \omega_{4} = 2\pi \times 122.2 / 60 = 12.8 \text{ rad/s} \] \[ T_{4} = \frac{P}{\omega_{4}} = \frac{1119}{12.8} = 87.4 \text{ Nm} \] --- **Summary for (a):** - **Output Shaft Speed (Gear 4):** **122.2 rpm** - **Output Torque:** **87.4 Nm** --- ### b) **Assume 95% efficiency for each meshing gear pair. What is the output shaft speed and torque?** There are **2 gear meshes**, each at 95% efficiency. **Overall efficiency:** \[ \eta_{total} = 0.95 \times 0.95 = 0.9025 \] **Output Power:** \[ P_{out} = P_{in} \times \eta_{total} = 1119 \times 0.9025 = 1010 \text{ W} \] **Speed of output shaft remains the same:** **122.2 rpm** **Output Torque:** \[ T_{4, \,eff} = \frac{P_{out}}{\omega_{4}} = \frac{1010}{12.8} = 78.9 \text{ Nm} \] --- **Summary for (b):** - **Output Shaft Speed:** **122.2 rpm** - **Output Shaft Torque:** **78.9 Nm** --- ### c) **Redesign for Coaxial Shafts (Input and Output in Line)** To have input and output shafts **coaxial**, you can use an **idler gear** or a **planetary gear system**. Since you’re to use the same gears, you can use an **idler gear train** (see sketch below): **Sketch:** ``` Input Shaft (Gear 1) --- Gear 2 (on intermediate shaft) --- Idler Gear(s) --- Output Shaft (Gear 4) ``` - Place Gear 4 in line with Gear 1 (input and output on the same axis). - Use an idler gear between Gears 3 and 4 to transmit motion while aligning the shafts. **[You can draw a straight line with gears 1 and 4 on the ends, and 2, 3, and any idlers in the middle.]** --- ### d) **Calculating Stress in the Teeth of Gear 3** To calculate the **tooth stress** (bending stress) in Gear 3 (the pinion): 1. **Calculate the tangential force on Gear 3:** \[ F_t = \frac{2T}{d} \] where \(T\) is the torque on Gear 3 and \(d\) is the pitch diameter. 2. **Use the Lewis Bending Equation:** \[ \sigma = \frac{F_t}{b m Y} \] where: - \(F_t\): Tangential load - \(b\): Face width - \(m\): Module (pitch diameter/number of teeth) - \(Y\): Lewis form factor (depends on number of teeth and gear shape) 3. **Plug in values for Gear 3's torque, module, and geometry.** --- ## **Summary Table** | Part | Output Speed (rpm) | Output Torque (Nm) | |------|--------------------|--------------------| | (a) | 122.2 | 87.4 | | (b) | 122.2 | 78.9 | --- Let me know if you need a sketch or further explanation on any step!