Let’s solve the problem step by step: --- ## **Step 1: Understanding the Gear Train** This is a **reverted compound gear train**. Given: - Overall train value = \(\dfrac{\omega_{out}}{\omega_{in}} = \dfrac{1}{64}\) - \(N_2 = 36\) (teeth on gear 2) - Diametral pitch, \(P = 5\) teeth/in - Pressure angle, \(\phi = 20^\circ\) - Full depth teeth - No interference (minimum teeth requirement) - Power input = 30 hp - Input speed (gear 2) = 1500 rpm - Efficiency = 100% --- ## **Step 2: Gear Train Ratio** For a reverted compound gear train: - Gears 2 and 3 are on one shaft, gears 4 and 5 are on another. - 2 meshes with 3, 4 meshes with 5. - **Reverted:** Distance between centers of 2-3 = distance between centers of 4-5. Let: - \(N_2\), \(N_3\), \(N_4\), \(N_5\) = number of teeth on respective gears The total gear ratio: \[ \frac{\omega_{out}}{\omega_{in}} = \frac{N_2 \cdot N_4}{N_3 \cdot N_5} \] Given: \[ \frac{\omega_{out}}{\omega_{in}} = \frac{1}{64} \] So, \[ \frac{N_2 \cdot N_4}{N_3 \cdot N_5} = \frac{1}{64} \] \[ N_2 \cdot N_4 = \frac{N_3 \cdot N_5}{64} \] But more conveniently: \[ N_2 \cdot N_4 = N_3 \cdot N_5 / 64 \implies N_2 \cdot N_4 \cdot 64 = N_3 \cdot N_5 \] But actually, cross-multiplying: \[ \frac{N_2 \cdot N_4}{N_3 \cdot N_5} = \frac{1}{64} \implies N_2 \cdot N_4 = \frac{N_3 \cdot N_5}{64} \implies N_3 \cdot N_5 = 64 N_2 \cdot N_4 \] --- ## **Step 3: Reverted Condition** **Reverted:** Distance from input shaft to intermediate shaft = distance from intermediate shaft to output shaft. Let the pitch diameters be \(d_2, d_3, d_4, d_5\): So, \[ d_2 + d_3 = d_4 + d_5 \] But \(d = N / P\), so: \[ \frac{N_2}{P} + \frac{N_3}{P} = \frac{N_4}{P} + \frac{N_5}{P} \implies N_2 + N_3 = N_4 + N_5 \] --- ## **Step 4: List the Equations** From above, we have: 1. \(N_2 \cdot N_4 = \dfrac{N_3 \cdot N_5}{64}\) 2. \(N_2 + N_3 = N_4 + N_5\) 3. \(N_2 = 36\) --- ## **Step 5: Solve for Teeth Numbers** Let’s denote: - \(N_2 = 36\) (given) - \(N_3 = x\) - \(N_4 = y\) - \(N_5 = z\) From equation 1: \[ \frac{N_2 \cdot N_4}{N_3 \cdot N_5} = \frac{1}{64} \implies 64 N_2 N_4 = N_3 N_5 \implies 64 \times 36 \times y = x \cdot z \implies 2304y = xz \] From equation 2: \[ N_2 + N_3 = N_4 + N_5 \implies 36 + x = y + z \implies z = 36 + x - y \] Substitute \(z\) in the first equation: \[ xz = x(36 + x - y) = 2304y \] \[ x(36 + x - y) = 2304y \] \[ 36x + x^2 - xy = 2304y \] \[ 36x + x^2 = 2304y + xy \] \[ 36x + x^2 - xy = 2304y \] \[ x^2 + 36x - xy - 2304y = 0 \] \[ x^2 + 36x - y(x + 2304) = 0 \] \[ x^2 + 36x = y(x + 2304) \] \[ y = \frac{x^2 + 36x}{x + 2304} \] We need integer values for \(x\) and \(y\), and all gears must have enough teeth to avoid interference. For 20° full-depth, minimum teeth ≈ 18. Let’s try integer values for \(x\) (try \(x = 72\), \(x = 96\), etc.) Let’s try \(x = 96\): \[ y = \frac{96^2 + 36 \times 96}{96 + 2304} = \frac{9216 + 3456}{2400} = \frac{12672}{2400} = 5.28 \] Not integer. Try \(x = 144\): \[ y = \frac{144^2 + 36 \times 144}{144 + 2304} = \frac{20736 + 5184}{2448} = \frac{25920}{2448} = 10.59 \] No. Try \(x = 288\): \[ y = \frac{288^2 + 36 \times 288}{288 + 2304} = \frac{82944 + 10368}{2592} = \frac{93312}{2592} = 36 \] Bingo! So, \(x = 288\), \(y = 36\). Now, \(z = 36 + x - y = 36 + 288 - 36 = 288\). So, the numbers of teeth are: - \(N_2 = 36\) (given) - \(N_3 = 288\) - \(N_4 = 36\) - \(N_5 = 288\) --- ## **Step 6: Check for Interference** For 20° full-depth, minimum number is about 18. All gears have more than that—so **no interference**. --- ## **Step 7: Final Answers for Number of Teeth** \[ \boxed{ N_2 = 36, \quad N_3 = 288, \quad N_4 = 36, \quad N_5 = 288 } \] --- ## **Step 8: Free Body Diagram (F.B.D) of Gear 4 and Load Calculation** Given: - Power input = 30 hp - Input shaft speed (gear 2) = 1500 rpm - Efficiency = 100% ### **Torque on Gear 4** **Output speed**: \[ \omega_{out} = \omega_{in} \times \text{Train ratio} = 1500 \times \frac{1}{64} = 23.4375 \text{ rpm} \] **Output torque (since efficiency is 100%):** \[ P = 2\pi N T / 60 \] Convert 30 hp to Watts: \[ 1\,hp = 746\,W \implies 30\,hp = 22380\,W \] \[ P = 2\pi N_{out} T_{out} / 60 \implies T_{out} = \frac{P \cdot 60}{2\pi N_{out}} \] \[ T_{out} = \frac{22380 \times 60}{2\pi \times 23.4375} \] \[ T_{out} = \frac{1342800}{147.223} \] \[ T_{out} \approx 9123.7 \text{ Nm} \] This is the torque on gear 4. --- ### **Forces on Gear 4** Gear 4 meshes with gear 3 and gear 5. - **Tangential Force (\(F_t\))**: \[ F_t = \frac{2T}{d} \] - \(T\) = 9123.7 Nm = 9123700 Nmm - Pitch diameter \(d_4 = N_4 / P = 36 / 5 = 7.2\) in \(= 182.88\) mm \[ F_t = \frac{2 \times 9123700}{182.88} = \frac{18247400}{182.88} \approx 99784\, \text{N} \] - **Radial Force (\(F_r\))**: \[ F_r = F_t \tan \phi \] \(\phi = 20^\circ\) \[ F_r = 99784 \times \tan(20^\circ) \approx 99784 \times 0.3640 \approx 36313\, \text{N} \] --- ### **Free Body Diagram (F.B.D) for Gear 4** Draw gear 4 with: - Tangential force \(F_t\) at the pitch point (due to mesh with gear 3 and 5, acting in opposite directions) - Radial force \(F_r\) perpendicular to \(F_t\) - Shaft reaction (at center) - Applied torque (\(T_{out}\)) --- ## **Summary Table** | Gear | Number of Teeth | |------|----------------| | 2 | 36 | | 3 | 288 | | 4 | 36 | | 5 | 288 | ### **Loads Acting on Gear 4** - Tangential force: \(\approx 99784\, \text{N}\) - Radial force: \(\approx 36313\, \text{N}\) - Torque: \(9123.7\, \text{Nm}\) --- ### **Final Answers** ### **(a) Number of teeth:** \[ \boxed{N_2 = 36,\ N_3 = 288,\ N_4 = 36,\ N_5 = 288} \] ### **(b) F.B.D. of gear 4 and loads:** - Torque: \(9123.7\) Nm - Tangential force: \(99784\) N - Radial force: \(36313\) N --- Let me know if you need the actual drawing for the FBD or further clarifications!