Let's solve the problem step by step as requested: --- ### **Step 1: Find the Length of the Belt** Given: - Diameter of smaller pulley, \( d_1 = 400 \) mm \( = 0.4 \) m - Diameter of larger pulley, \( d_2 = 900 \) mm \( = 0.9 \) m - Center distance, \( C = 4.5 \) m For an open belt drive, the length of the belt is given by: \[ L = \pi \left( \frac{d_1 + d_2}{2} \right) + 2C + \frac{(d_2 - d_1)^2}{4C} \] Plug the values: \[ L = \pi \left( \frac{0.4 + 0.9}{2} \right) + 2 \times 4.5 + \frac{(0.9 - 0.4)^2}{4 \times 4.5} \] \[ = \pi \times 0.65 + 9 + \frac{0.25}{18} \] \[ = 2.042 \text{ m} + 9 \text{ m} + 0.0139 \text{ m} \] \[ = 11.056 \text{ m} \] **Explanation:** We have used the standard formula for the length of an open belt drive, substituting all given values. --- ### **Step 2: Find the Maximum Force Before the Belt Breaks** Given: - Permissible initial load per unit width = 17.5 kN/m - Width \( b = 200 \) mm = 0.2 m Maximum force before the belt breaks: \[ F_{\text{max}} = \text{Permissible load per unit width} \times \text{Width} \] \[ = 17.5 \text{ kN/m} \times 0.2 \text{ m} = 3.5 \text{ kN} = 3500 \text{ N} \] **Explanation:** The maximum force is simply the permissible load per meter width multiplied by the width of the belt. --- ### **Step 3: Find the Maximum Transmitted Power** Given: - Maximum tension \( T_1 = 3500 \) N (from above) - Belt thickness \( t = 4 \) mm = 0.004 m - Belt width \( b = 0.2 \) m - Weight per unit volume = 11.4 kN/m³ = 11400 N/m³ - Coefficient of friction \( \mu = 0.8 \) - Velocity \( v = 10 \) m/s #### **First, Find Centrifugal Tension (\( T_c \)):** \[ \text{Mass per unit length} = \text{Volume per unit length} \times \text{Density} \] But we have weight per unit volume, so: \[ \text{Weight per m} = \text{Volume per m} \times \text{Weight per unit volume} \] \[ = b \times t \times 1 \times 11400 = 0.2 \times 0.004 \times 11400 = 9.12 \text{ N/m} \] \[ \text{Mass per m} = \frac{9.12}{9.81} = 0.929 \text{ kg/m} \] \[ T_c = mass/m \times v^2 = 0.929 \times (10)^2 = 92.9 \text{ N} \] #### **Find Effective Tension for Power Transmission** \[ T_1 = 3500 \text{ N} \] \[ \text{Effective } T_1 = T_1 - T_c = 3500 - 92.9 = 3407.1 \text{ N} \] #### **Find Tension Ratio (\( \frac{T_1}{T_2} \)):** The angle of lap on the smaller pulley for open belt: \[ \theta = \pi - 2 \sin^{-1} \left( \frac{d_2 - d_1}{2C} \right) \] \[ = \pi - 2 \sin^{-1} \left( \frac{0.9 - 0.4}{2 \times 4.5} \right) = \pi - 2 \sin^{-1} \left( \frac{0.5}{9} \right) \] \[ = \pi - 2 \sin^{-1} (0.0556) \] \[ \sin^{-1}(0.0556) \approx 0.0556 \text{ radians} \] \[ \theta = \pi - 2 \times 0.0556 = 3.1416 - 0.1112 = 3.0304 \text{ radians} \] Now, \[ \frac{T_1}{T_2} = e^{\mu \theta} = e^{0.8 \times 3.0304} = e^{2.424} \approx 11.29 \] \[ T_2 = \frac{T_1}{11.29} = \frac{3407.1}{11.29} = 301.9 \text{ N} \] #### **Power Transmitted** \[ P = (T_1 - T_2) \times v \] \[ = (3407.1 - 301.9) \times 10 = 3105.2 \times 10 = 31052 \text{ W} = 31.05 \text{ kW} \] **Explanation:** We calculated the centrifugal tension, effective tight and slack side tensions, and used the velocity and tension difference to find the maximum transmitted power. --- ## **Summary and Final Answers** 1. **Length of the belt:** \( \boxed{11.06 \ \text{m}} \) 2. **Maximum force before the belt breaks:** \( \boxed{3500 \ \text{N}} \) 3. **Maximum transmitted power:** \( \boxed{31.1 \ \text{kW}} \) (rounded to 3 significant figures) --- ### **Explanations Recap:** - **Step 1:** Used the formula for open belt length. - **Step 2:** Used width and permissible load per width for max force. - **Step 3:** Accounted for centrifugal tension and friction, calculated tension ratio, and then power. If you need clarification on any step or formula, feel free to ask!