Let's break down the problem step by step: **Given Data:** - Major diameter (d): 3 in - Pitch (p): 0.2 in - Coefficient of friction for collar (f_c): 0.16 - Coefficient of friction for thread (f_t): 0.12 - Frictional radius of the collar (r_c): 2.5 in - Input power (P_in): 3.5 hp - Velocity of the nut (v): 2.5 in/s **Asked:** Calculate the overall **efficiency** of the power screw for raising the load. --- ### Step 1: Calculate Input Power (in consistent units) 1 hp = 746 W So, \( P_{in} = 3.5 \times 746 = 2611 \) W Convert to lb·in/s for easier matching with other units: \( 1 \text{ hp} = 550 \text{ ft·lbf/s} = 6600 \text{ in·lbf/s} \) So \( P_{in} = 3.5 \times 6600 = 23100 \text{ in·lbf/s} \) --- ### Step 2: Output Power Output power is \( \text{Load} \times \text{velocity of nut} \) Let load = W (to be found from input power and screw mechanics). --- ### Step 3: Power Screw Calculations #### Lead of the screw (l) Since it's a double square thread: Lead = 2 × pitch = 0.4 in #### Velocity relation: \( v = n \times \text{lead} \) \( n = \frac{v}{\text{lead}} = \frac{2.5}{0.4} = 6.25 \) rev/s --- ### Step 4: Efficiency Calculation Efficiency (\( \eta \)) is the ratio of output power to input power. \[ \eta = \frac{\text{Output Power}}{\text{Input Power}} \] #### Output Power: \[ \text{Output Power} = W \times v \] #### Input Power: The input power to raise the load = Total torque × angular velocity Total torque = Torque for thread + Torque for collar \[ T_{\text{total}} = T_{\text{thread}} + T_{\text{collar}} \] Where: - \( T_{\text{thread}} = W \frac{d_m}{2} \frac{\tan(\lambda) + f_t}{1 - f_t \tan(\lambda)} \) - \( T_{\text{collar}} = f_c \times W \times r_c \) Here, - \( d_m \) = mean diameter ≈ major diameter = 3 in (for square threads, minor difference) - \( \lambda \) = lead angle = \( \tan^{-1}\left(\frac{\text{lead}}{\pi d_m}\right) \) Calculate lead angle: \[ \lambda = \tan^{-1}\left(\frac{0.4}{\pi \times 3}\right) = \tan^{-1}\left(\frac{0.4}{9.4248}\right) = \tan^{-1}(0.04246) \approx 2.43^\circ \] \[ \tan(\lambda) \approx 0.0425 \] Now, \[ T_{\text{thread}} = W \frac{3}{2} \frac{0.0425 + 0.12}{1 - 0.12 \times 0.0425} = W \times 1.5 \frac{0.1625}{1 - 0.0051} = W \times 1.5 \frac{0.1625}{0.9949} = W \times 1.5 \times 0.1634 = W \times 0.2451 \] \[ T_{\text{collar}} = 0.16 \times W \times 2.5 = W \times 0.4 \] \[ T_{\text{total}} = W \times (0.2451 + 0.4) = W \times 0.6451 \] Angular velocity: \[ \omega = 2\pi n = 2\pi \times 6.25 = 39.27 \text{ rad/s} \] \[ \text{Input Power} = T_{\text{total}} \times \omega = W \times 0.6451 \times 39.27 = W \times 25.33 \text{ in·lbf/s} \] \[ \text{Output Power} = W \times 2.5 \text{ in·lbf/s} \] --- ### Step 5: Efficiency \[ \eta = \frac{W \times 2.5}{W \times 25.33} = \frac{2.5}{25.33} = 0.0986 \approx 0.10 \] --- ### **Final Answer** The overall efficiency of the power screw for raising the load is: \[ \boxed{0.10} \] So, the correct option is **0.10**.