Let's break down the information given and solve for the overall efficiency of the power screw. ## **Given Data** - Major diameter, \( d = 3 \) in - Pitch, \( p = .2 \) in - Coefficient of friction for collar, \( f_c = .16 \) - Coefficient of friction for thread, \( f_t = .12 \) - Frictional radius of collar, \( r_c = 2.5 \) in - Input power, \( P_{in} = 3.5 \) hp - Velocity of nut, \( v = 2.5 \) in/s Let's convert input power to watts: \[ 1 \text{ hp} = 746 \text{ W} \] \[ P_{in} = 3.5 \times 746 = 2611 \text{ W} \] ## **Step 1: Output Power** Output power is the work done in raising the load: \[ P_{out} = \text{Load} \times \text{velocity} \] But the load is not given directly. Let's express efficiency in terms of available data. ## **Step 2: Efficiency Formula** Efficiency (\(\eta\)) is: \[ \eta = \frac{\text{Output Power}}{\text{Input Power}} \] ### **Output Power** \[ P_{out} = F \times v \] Where \( F \) is the axial load. ### **Input Power** Input power is calculated using torque and angular velocity: \[ P_{in} = T \cdot \omega \] Where \( T \) is the total torque required to raise the load and \( \omega \) is the angular speed. ### **Torque Required for Power Screw (with Collar Friction):** For a double square threaded screw: \[ T = F \left[ \frac{d_m}{2} \cdot \frac{\tan \lambda + f_t}{1 - f_t \tan \lambda} + f_c r_c \right] \] Where: - \( d_m \) = mean diameter = major diameter - (pitch/2) (since it's square threaded) - \( \tan \lambda = \frac{\text{lead}}{\pi d_m} \) - For double thread, **lead = 2 × pitch** #### Calculate terms: - Pitch = .2 in, so lead = .4 in - \( d_m = 3 - .2/2 = 2.9 \) in \[ \tan \lambda = \frac{.4}{\pi \times 2.9} = \frac{.4}{9.11} \approx .044 \] Now, plug values into torque equation: \[ T = F \left[ \frac{2.9}{2} \cdot \frac{.044 + .12}{1 - .12 \times .044} + .16 \times 2.5 \right] \] Calculate denominator: \[ 1 - .12 \times .044 = 1 - .00528 = .99472 \] Numerator: \[ .044 + .12 = .164 \] So, \[ \frac{.164}{.99472} = .16498 \] \[ \frac{2.9}{2} = 1.45 \] \[ 1.45 \times .16498 = .239 \] \[ .16 \times 2.5 = .4 \] \[ T = F (.239 + .4) = F \times .639 \] ### **Angular Speed (\(\omega\)):** \[ \text{Lead} = .4 \text{ in/rev} \] \[ \text{Velocity} = 2.5 \text{ in/s} \] So, \[ \text{Revolutions per second} = \frac{2.5}{.4} = 6.25 \text{ rev/s} \] \[ \omega = 2\pi \times 6.25 = 39.27 \text{ rad/s} \] ### **Input Power:** \[ P_{in} = T \cdot \omega = .639F \times 39.27 = 25.09F \] ### **Output Power:** \[ P_{out} = F \times 2.5 \] ### **Efficiency:** \[ \eta = \frac{P_{out}}{P_{in}} = \frac{F \times 2.5}{25.09F} = \frac{2.5}{25.09} = .0996 \approx .10 \] --- ## **Final Answer** ### **The correct option is:** \[ \boxed{.10} \]