Let's analyze the problem step by step: ### Given Data: - Major diameter of the screw, \( D = 3 \) inches - Pitch, \( p = .2 \) inches - Coefficient of friction for collar, \( \mu_c = .15 \) - Coefficient of friction for thread, \( \mu_t = .12 \) - Frictional radius of the collar, \( r_c = 2.5 \) inches - Input power, \( P = 3.5 \) hp - Voltage of the nut, \( v = 2.5 \) inches (though this seems to be the velocity of the nut, not voltage) ### Step 1: Convert power to consistent units: - 1 hp = 746 W - \( P = 3.5 \times 746 = 2611 \) W ### Step 2: Calculate the efficiency: The efficiency (\( \eta \)) of a power screw can be approximated using the formula: \[ \eta = \frac{\text{Work output per unit time}}{\text{Work input per unit time}} \] But in practical problems involving friction in screw mechanisms, efficiency is derived from the following relation: \[ \eta = \frac{\text{Mechanical advantage}}{\text{Effort advantage}} \] or \[ \eta = \frac{\text{Useful work done}}{\text{Input power}} \] Given the complexity, an approximate formula for efficiency considering friction is: \[ \eta = \frac{\tan \lambda}{\tan (\lambda + \phi)} \] where: - \( \lambda = \arctan \left(\frac{p}{\pi D}\right) \) (lead angle) - \( \phi = \arctan(\mu) \) (angle of friction) ### Step 3: Calculating the lead angle, \( \lambda \): \[ \lambda = \arctan \left(\frac{p}{\pi D}\right) = \arctan \left(\frac{.2}{\pi \times 3}\right) \] \[ \lambda = \arctan \left(\frac{.2}{9.4248}\right) \approx \arctan(.0212) \approx 1.21^\circ \] ### Step 4: Calculating the angle of friction, \( \phi \): \[ \phi = \arctan(\mu) = \arctan(.15) \approx 8.53^\circ \] ### Step 5: Calculate the efficiency: \[ \eta \approx \frac{\tan \lambda}{\tan (\lambda + \phi)} = \frac{\tan 1.21^\circ}{\tan (1.21^\circ + 8.53^\circ)} = \frac{.0212}{\tan 9.74^\circ} \] \[ \text{Since} \quad \tan 9.74^\circ \approx .171 \] \[ \eta \approx \frac{.0212}{.171} \approx .124 \] ### **Final answer:** \[ \boxed{\eta \approx .125 \text{ or } 12.5\%} \] ### **Closest option:** - .15 - .10 - .20 - .05 The closest option to .125 is **.15**. --- ## **Correct Option:** ### **.15**