Let's solve each part step by step as requested. --- ### (a) **Pitch diameters of each gear** #### **Step 1: Formula and Given Data** - **Module (\(m\))** = 12 mm - **Number of teeth**: - \(N_2 = 18\) - \(N_3 = 32\) - \(N_4 = 18\) - \(N_5 = 48\) - **Pitch diameter** (\(d\)) is given by: \[ d = m \times N \] **Explanation**: The pitch diameter is calculated by multiplying the module by the number of teeth for each gear. --- #### **Step 2: Calculations** \[ \begin{align*} d_2 &= m \times N_2 = 12 \times 18 = 216\ \text{mm} \\ d_3 &= m \times N_3 = 12 \times 32 = 384\ \text{mm} \\ d_4 &= m \times N_4 = 12 \times 18 = 216\ \text{mm} \\ d_5 &= m \times N_5 = 12 \times 48 = 576\ \text{mm} \\ \end{align*} \] **Explanation**: Each pitch diameter is directly calculated using the formula above. --- ### **Final Answers for (a)** - Pitch diameters: - \(d_2 = 216\) mm - \(d_3 = 384\) mm - \(d_4 = 216\) mm - \(d_5 = 576\) mm --- Would you like to proceed to part (b) now?

## Solution to Gear Mechanics Problem Let's solve the problem step by step. --- ### (a) **Pitch Diameters of Each Gear** #### **Step 1: Formula and Given Data** - **Module (\(m\))** = 12 mm - **Number of teeth**: - \(N_2 = 18\) (Pinion) - \(N_3 = 32\) (Idler) - \(N_4 = 18\) (Idler) - \(N_5 = 48\) (Driven Gear) The pitch diameter (\(d\)) is calculated using the formula: \[ d = m \times N \] #### **Step 2: Calculations** \[ \begin{align*} d_2 &= m \times N_2 = 12 \times 18 = 216\ \text{mm} \\ d_3 &= m \times N_3 = 12 \times 32 = 384\ \text{mm} \\ d_4 &= m \times N_4 = 12 \times 18 = 216\ \text{mm} \\ d_5 &= m \times N_5 = 12 \times 48 = 576\ \text{mm} \\ \end{align*} \] **Explanation**: The pitch diameters for each gear have been calculated by multiplying the module by the number of teeth. #### **Final Answers for (a)** - \(d_2 = 216\) mm - \(d_3 = 384\) mm - \(d_4 = 216\) mm - \(d_5 = 576\) mm --- ### (b) **Tangential Force from Gear 3 to Gear 2** #### **Step 1: Calculate Input Torque on Gear 2** The input power (\(P\)) is given as 150 kW, and the speed (\(N_2\)) is 1800 rpm. Using the formula for torque (\(T\)): \[ P = \frac{T \times \omega}{1000} \quad \text{where} \quad \omega = \frac{2 \pi N}{60} \] #### **Step 2: Calculations** \[ \omega_2 = \frac{2 \pi \times 1800}{60} = 188.5\ \text{rad/s} \] \[ T_2 = \frac{P \times 1000}{\omega_2} = \frac{150 \times 1000}{188.5} \approx 796.2\ \text{N·m} \] The tangential force (\(F_t\)) on gear 2 is given by: \[ F_t = \frac{T}{r} = \frac{T_2}{\frac{d_2}{2}} = \frac{796.2}{\frac{216}{2}} \approx 7.36\ \text{kN} \] **Explanation**: The torque on gear 2 was calculated first using power and angular velocity, and then the tangential force was derived from torque. #### **Final Answer for (b)** - Tangential Force from Gear 3 to Gear 2: \(F_t \approx 7360\ \text{N}\) --- ### (c) **Radial Force from Gear 3 to Gear 2** #### **Step 1: Calculate Radial Force** The radial force (\(F_r\)) can be derived from the tangential force using the pressure angle (\(\phi = 20^\circ\)): \[ F_r = F_t \tan(\phi) \] #### **Step 2: Calculations** \[ F_r = 7360 \times \tan(20^\circ) \approx 7360 \times 0.364 = 2675.5\ \text{N} \] **Explanation**: The radial force is calculated using the relationship between tangential force and the pressure angle. #### **Final Answer for (c)** - Radial Force from Gear 3 to Gear 2: \(F_r \approx 2675.5\ \text{N}\) --- ### (d) **Input Torque on Shaft A** #### **Step 1: Use Torque from Gear 2** The input torque on shaft A is the same as the torque calculated for gear 2 since they are directly connected. #### **Step 2: Final Answer** \[ T_a = T_2 = 796.2\ \text{N·m} \] **Explanation**: The input torque on shaft A is equal to the torque transmitted by gear 2. #### **Final Answer for (d)** - Input Torque on Shaft A: \(T_a \approx 796.2\ \text{N·m}\) --- ### (e) **Free Body Diagram for Gear 2** - **Forces Acting on Gear 2**: - Tangential Force \(F_t\) acting to the right. - Radial Force \(F_r\) acting downward. #### **Diagram Description** - Draw a circle representing Gear 2. - Mark \(F_t\) to the right and \(F_r\) vertical. --- ### (f) **Tangential Forces Exerted on Gears 3 and 4** #### **Step 1: Torque Relation** The tangential force on gear 3 is equal to the force on gear 2 transmitted through gear 3. #### **Step 2: Calculations** The tangential force on gear 3 is: \[ F_{t3} = F_t \frac{N_2}{N_3} = 7360 \times \frac{18}{32} \approx 4125\ \text{N} \] **Explanation**: The torque relationship between the gears gives us the tangential force on gear 3. #### **Final Answer for (f)** - Tangential Force on Gear 3: \(F_{t3} \approx 4125\ \text{N}\) - Tangential Force on Gear 4: \(F_{t4} = F_{t3} \frac{N_3}{N_4} \approx 2325\ \text{N}\) --- ### (g) **Radial Forces Exerted on Gears 3 and 4** #### **Step 1: Calculate Radial Forces** Using the pressure angle for gears 3 and 4: \[ F_{r3} = F_{t3} \tan(20^\circ) \quad \text{and} \quad F_{r4} = F_{t4} \tan(20^\circ) \] #### **Step 2: Calculations** \[ F_{r3} = 4125 \times 0.364 \approx 1500.6\ \text{N} \] \[ F_{r4} = 2325 \times 0.364 \approx 847.5\ \text{N} \] **Explanation**: Radial forces are derived using the same method as before, based on tangential forces. #### **Final Answer for (g)** - Radial Force on Gear 3: \(F_{r3} \approx 1500.6\ \text{N}\) - Radial Force on Gear 4: \(F_{r4} \approx 847.5\ \text{N}\) --- ### (h) **Free Body Diagram for Gears 3 and 4** - **For Gear 3**: - Tangential Force \(F_{t3}\) to the left. - Radial Force \(F_{r3}\) acting downward. - **For Gear 4**: - Tangential Force \(F_{t4}\) to the left. - Radial Force \(F_{r4}\) acting downward. #### **Diagram Description** - Draw circles for each gear and mark the respective forces as described. --- This concludes the solution for all parts of the problem.