4. Tensions are 2000 N and 900 N in a belt drive. The diameter of the driver pull \( 0,44 \mathrm{~m} \) and its rotational frequency is \( 800 \mathrm{r} / \mathrm{min} \). Calculate: a) The effective tension. b) The torque of the driver pulley. c) The work done during one revolution. d) The output power of the pulley after 45 s . 5. The wheel of a belt drive rotates at \( 1200 \mathrm{r} / \mathrm{min} \). The belt is subjected to an effec force of 600 N . If the power transmitted by the belt is \( 12,4 \mathrm{~kW} \), calculate the

**Problem 4:** - **Given:** - Tight side tension (\( T_1 \)) = 2000 N - Slack side tension (\( T_2 \)) = 900 N - Diameter of driver pulley (\( D \)) = 0.44 m - Rotational frequency (\( n \)) = 800 rpm - **a) Effective Tension (\( T \))**: \[ T = T_1 - T_2 = 2000\, \text{N} - 900\, \text{N} = 1100\, \text{N} \] **Answer:** 1100 N - **b) Torque of the Driver Pulley (\( \tau \))**: \[ \tau = T \times \frac{D}{2} = 1100\, \text{N} \times \frac{0.44\, \text{m}}{2} = 1100 \times 0.22 = 242\, \text{N·m} \] **Answer:** 242 N·m - **c) Work Done During One Revolution (\( W \))**: \[ W = \tau \times 2\pi r = 242\, \text{N·m} \times 2\pi \times \frac{0.44}{2} = 242 \times 2\pi \times 0.22 \approx 242 \times 1.382 = 334.6\, \text{J} \] **Answer:** Approximately 334.6 J - **d) Output Power of the Pulley After 45 Seconds (\( P \))**: \[ P = \frac{W}{t} = \frac{334.6\, \text{J}}{45\, \text{s}} \approx 7.43\, \text{W} \] **Answer:** Approximately 7.43 W --- **Problem 5:** - **Given:** - Rotational speed of the wheel (\( n \)) = 1200 rpm - Effective force (\( F \)) = 600 N - Power transmitted by the belt (\( P \)) = 12.4 kW - **To Find:** - (Assuming the question is to find the diameter of the pulley or another related parameter, but it's not specified. For the sake of this example, let's assume it's to find the diameter of the pulley.) - **Assuming the question is to find the diameter of the pulley:** First, convert rotational speed from rpm to radians per second (\( \omega \)): \[ \omega = \frac{2\pi n}{60} = \frac{2\pi \times 1200}{60} = 40\pi\, \text{rad/s} \] Power (\( P \)) is related to torque (\( \tau \)) and angular velocity (\( \omega \)) by: \[ P = \tau \times \omega \] Solving for torque: \[ \tau = \frac{P}{\omega} = \frac{12400\, \text{W}}{40\pi\, \text{rad/s}} \approx \frac{12400}{125.66} \approx 98.7\, \text{N·m} \] Torque is also related to the force (\( F \)) and the radius (\( r \)) of the pulley: \[ \tau = F \times r \] Solving for radius: \[ r = \frac{\tau}{F} = \frac{98.7\, \text{N·m}}{600\, \text{N}} \approx 0.1645\, \text{m} = 16.45\, \text{cm} \] Therefore, the diameter (\( D \)) of the pulley is: \[ D = 2r = 2 \times 0.1645\, \text{m} = 0.329\, \text{m} = 32.9\, \text{cm} \] **Answer:** The diameter of the pulley is approximately 32.9 cm. --- **Note:** If the question in Problem 5 is different, please provide the specific details to offer a more accurate solution.