This physics problem tests your understanding of fundamental physical laws and their applications. The step-by-step solution below breaks down the problem using relevant equations and physical reasoning.
ScanSolve AI Answer
Here are the calculations for Question 3: Given: Pulley diameter D = 500 mm = 0.5 m Rotational speed N = 450 r/min Tight side tension T_1 = 3 × slack side tension T_2 Power transmitted P = 18 kW = 18000 W Maximum allowable tension in the belt = 14 N/mm width Step 1: Calculate the angular velocity () of the pulley. = (2 N)/(60) = 2 × 450 r/min60 s/min = 15 rad/s Step 2: Calculate the linear belt speed (v). The radius of the pulley is r = (D)/(2) = 0.5 m2 = 0.25 m. v = × r = 15 rad/s × 0.25 m = 3.75 m/s Step 3: Calculate the tight side tension (T_1). The power transmitted by a belt drive is given by P = (T_1 - T_2) × v. We are given T_1 = 3 T_2, so T_1 - T_2 = 3 T_2 - T_2 = 2 T_2. Substituting this into the power equation: P = 2 T_2 × v 18000 W = 2 T_2 × 3.75 m/s 18000 = 7.5 T_2 T_2 = (18000)/(7.5 ) N Now, calculate T_1: T_1 = 3 T_2 = 3 × (18000)/(7.5 ) = (54000)/(7.5 ) = (7200)/() N T_1 ≈ (7200)/(3.14159) ≈ 2291.83 N Step 4: Calculate the width of the belt. The maximum allowable tension in the belt is 14 N/mm width. This means the total tight side tension T_1 must be equal to or less than the allowable tension per unit width multiplied by the belt width. Width = (T_1)/(Allowable tension per mm width) Width = 2291.83 N14 N/mm ≈ 163.70 mm The width of the belt required is 163.70 mm. That's 2 down. 3 left today — send the next one.