Let's break down the problem step by step: --- ## **Step : Understand What's Given** - **Power to transmit** (\(P\)): 15 hp - **Shock factor** (\(K_s\)): 1.25 - **Minimum safety factor**: 1.1 - **Center distance between shafts**: 9.00 ft - **Driving pulley diameter**: 5.00 in - **Driven pulley diameter**: 18.00 in - **Driving pulley speed** (\(N_1\)): 185 rpm - **Belt type**: Polyamide A-3 flat belt - **Belt width to be found** (\(b\)), such that \( n_{fs} = n_d = 1.1 \) --- ## **Step 2: Convert Units Where Needed** - **Power**: \(1~\text{hp} = 33,000~\text{ft} \cdot \text{lb}/\text{min}\) - **Pulley diameters**: 5 in and 18 in (keep in inches for now). --- ## **Step 3: Find the Belt Velocity** Belt velocity, \(v\), (in ft/min): \[ v = \pi \times D_1 \times N_1 \] Where \(D_1\) is in **feet**: \[ D_1 = 5~\text{in} = \frac{5}{12}~\text{ft} \] \[ v = \pi \times \frac{5}{12} \times 185 \] \[ v = 3.1416 \times .4167 \times 185 \approx 2421.2~\text{ft/min} \] --- ## **Step 4: Calculate the Effective Tension Needed** Power transmitted (\(P\)): \[ P = (T_1 - T_2) \times v \] Where: - \(T_1\): Tight side tension (lbf) - \(T_2\): Slack side tension (lbf) - \(v\): Velocity in ft/min Convert power to ft-lb/min: \[ P = 15~\text{hp} \times 33,000 = 495,000~\text{ft-lb/min} \] \[ T_1 - T_2 = \frac{P}{v} = \frac{495,000}{2421.2} \approx 204.4~\text{lbf} \] --- ## **Step 5: Find the Tension Ratio** The ratio is: \[ \frac{T_1}{T_2} = e^{f \theta} \] Where: - \(f\): coefficient of friction (from data tables, for Polyamide A-3 flat belt, typically \(f = .3\)) - \(\theta\): arc of contact in radians (to be calculated) ### **Find Arc of Contact (\(\theta\))** \[ \theta = \pi - 2 \sin^{-1} \left( \frac{D_2 - D_1}{2C} \right) \] Where: - \(D_2 = 18~\text{in}\) - \(D_1 = 5~\text{in}\) - \(C = 9.00~\text{ft} \times 12 = 108~\text{in}\) \[ \sin^{-1} \left( \frac{18 - 5}{2 \times 108} \right) = \sin^{-1} \left( \frac{13}{216} \right) \approx \sin^{-1}(.0602) \approx .0602~\text{radians} \] \[ \theta = \pi - 2(.0602) = 3.1416 - .1204 = 3.0212~\text{radians} \] --- ## **Step 6: Find Maximum Allowable Tension per Inch Width** From standard tables (for polyamide A-3), the **allowable tension** (\(T_{allow}\)) is about **300 lbf/in width** under light shock and with safety factor applied. But since a safety factor is specified, we must divide by it: \[ T_{max} = \frac{T_{allow}}{n_{fs}} \] \[ T_{max} = \frac{300}{1.1} \approx 272.7~\text{lbf/in width} \] --- ## **Step 7: Calculate Belt Width (\(b\))** The total tension on the tight side is: \[ T_1 = \text{maximum tension per inch} \times b \] But also, from the tension ratio and effective tension: \[ T_1 - T_2 = 204.4~\text{lbf} \] \[ \frac{T_1}{T_2} = e^{f \theta} = e^{.3 \times 3.0212} = e^{.9064} \approx 2.475 \] \[ T_1 = 2.475 T_2 \] \[ T_1 - T_2 = 204.4 \] \[ 2.475 T_2 - T_2 = 204.4 \] \[ 1.475 T_2 = 204.4 \] \[ T_2 = \frac{204.4}{1.475} \approx 138.6~\text{lbf} \] \[ T_1 = 2.475 \times 138.6 \approx 343.1~\text{lbf} \] Now, solve for belt width \(b\): \[ T_1 = T_{max} \times b \] \[ 343.1 = 272.7 \times b \] \[ b = \frac{343.1}{272.7} \approx 1.258~\text{in} \] --- ## **Step 8: Report Final Answer** \[ \boxed{1.258~\text{in}} \] --- ### **Final Answer** The belt width \(b\) should be **1.258 in** (rounded to three decimal places). --- ## **Summary Table** | Quantity | Value | |---------------------------|-------------| | Power (\(P\)) | 15 hp | | Velocity (\(v\)) | 2421.2 ft/min| | Effective Tension | 204.4 lbf | | Tension Ratio | 2.475 | | Tight Side Tension (\(T_1\)) | 343.1 lbf | | Allowable Tension (per in)| 272.7 lbf | | **Belt Width (\(b\))** | **1.258 in** | --- If you need the derivation for any particular step, let me know!