Let's break down the problem given in the image: **Question:** A double square threaded power screw has the following specifications: - Major diameter: 3 in - Pitch: 0.2 in - Coefficient of friction for collar: 0.16 - Coefficient of friction for thread: 0.12 - Frictional radius of the collar: 2.5 in - Input power: 3.5 hp - Velocity of the nut: 2.5 in/s **Asked:** Calculate the overall efficiency of the power screw for raising the load. --- ## Step-by-Step Solution ### 1. **Calculate the Lead of the Screw** Double-threaded means the lead \( l \) = 2 × pitch = 2 × 0.2 in = **0.4 in** --- ### 2. **Calculate the Axial Velocity** The nut moves at 2.5 in/s. --- ### 3. **Calculate the Axial Force (\( F \))** Power (\( P \)) = Force × Velocity But first, convert power to consistent units: - 1 hp = 550 ft-lb/s = 550 × 12 in-lb/s = 6600 in-lb/s - Input power = 3.5 × 6600 = **23100 in-lb/s** --- ### 4. **Calculate Output Power (\( P_{out} \))** \( P_{out} = F \times v \) --- ### 5. **Calculate the Torque Required to Raise Load (\( T \))** The total torque required to overcome both thread and collar friction: \[ T = T_{thread} + T_{collar} \] Where: - For thread: \[ T_{thread} = F \times \left[ \frac{d_m}{2} \left( \frac{l + \pi \mu_t d_m}{\pi d_m - \mu_t l} \right) \right] \] - For collar: \[ T_{collar} = F \times \mu_c \times r_c \] Where: - \( d_m \) = mean diameter = major diameter - (0.5 × pitch) = 3 - (0.5 × 0.2) = **2.9 in** - \( \mu_t \) = thread friction = 0.12 - \( \mu_c \) = collar friction = 0.16 - \( r_c \) = collar radius = 2.5 in - \( l \) = lead = 0.4 in Let's compute: #### a) Compute \( T_{thread} \): \[ T_{thread} = F \times \frac{d_m}{2} \left( \frac{l + \pi \mu_t d_m}{\pi d_m - \mu_t l} \right) \] Numerator: \( l + \pi \mu_t d_m = 0.4 + \pi \times 0.12 \times 2.9 \approx 0.4 + 1.094 = 1.494 \) Denominator: \( \pi d_m - \mu_t l = \pi \times 2.9 - 0.12 \times 0.4 \approx 9.11 - 0.048 = 9.062 \) So, \[ T_{thread} = F \times \frac{2.9}{2} \times \frac{1.494}{9.062} = F \times 1.45 \times 0.165 = F \times 0.239 \] #### b) Compute \( T_{collar} \): \[ T_{collar} = F \times 0.16 \times 2.5 = F \times 0.4 \] #### c) Total Torque: \[ T = F \times (0.239 + 0.4) = F \times 0.639 \] --- ### 6. **Input Power (\( P_{in} \))** The torque multiplied by the angular velocity (\( \omega \)) gives the input power: \[ P_{in} = T \times \omega \] The angular velocity (\( \omega \)): \[ \omega = \frac{v}{lead} \times 2\pi \] \[ \omega = \frac{2.5}{0.4} \times 2\pi = 6.25 \times 2\pi = 39.27 \text{ rad/s} \] But in power screws, usually use rev/s: \[ \text{rev/s} = \frac{2.5}{0.4} = 6.25 \text{ rev/s} \] \[ \omega = 6.25 \times 2\pi = 39.27 \text{ rad/s} \] So, \[ P_{in} = T \times \omega = (F \times 0.639) \times 39.27 \] --- ### 7. **Set \( P_{in} \) Equal to Given Input Power** \[ P_{in} = 23100 \text{ in-lb/s} \] \[ 23100 = F \times 0.639 \times 39.27 \] \[ F = \frac{23100}{0.639 \times 39.27} = \frac{23100}{25.09} \approx 920 \text{ lb} \] --- ### 8. **Calculate Output Power (\( P_{out} \))** \[ P_{out} = F \times v = 920 \times 2.5 = 2300 \text{ in-lb/s} \] --- ### 9. **Calculate Efficiency (\( \eta \))** \[ \eta = \frac{P_{out}}{P_{in}} = \frac{2300}{23100} \approx 0.0995 \approx 0.10 \] --- ## **Final Answer:** **The overall efficiency of the power screw for raising the load is approximately 0.10 (10%).** So, the closest option in the list is **0.10**. --- ### **Summary Table** | Step | Value/Formula | Result | |----------------------|--------------------------|------------------| | Lead | 2 × 0.2 in | 0.4 in | | Mean diameter (\(d_m\)) | 3 - 0.1 | 2.9 in | | \(T_{thread}\) | \(F \times 0.239\) | | | \(T_{collar}\) | \(F \times 0.4\) | | | Total Torque | \(F \times 0.639\) | | | Angular velocity | 39.27 rad/s | | | Input Power | 23100 in-lb/s | | | Force (\(F\)) | \( \approx 920 \) lb | | | Output Power | 2300 in-lb/s | | | Efficiency | \( \approx 0.10 \) | | If you need further clarification or a simplified version, let me know!