4. Tensions are 2000 N and 900 N in a belt drive. The diameter of the driver pulley is \( 0,44 \mathrm{~m} \) 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 .

To solve the problem, let’s break it down into each part step by step! ### a) Effective Tension The effective tension \( T_{e} \) in the belt can be calculated using the formula: \[ T_{e} = T_1 - T_2 \] Where \( T_1 \) is the tension on the tight side (2000 N) and \( T_2 \) is the tension on the slack side (900 N). So, \[ T_{e} = 2000 \, \text{N} - 900 \, \text{N} = 1100 \, \text{N} \] ### b) Torque of the Driver Pulley The torque \( \tau \) can be calculated using the formula: \[ \tau = T_{e} \times r \] Where \( r \) is the radius of the pulley. Given the diameter of the pulley is 0.44 m, the radius \( r = \frac{0.44}{2} = 0.22 \, \text{m} \). Thus, \[ \tau = 1100 \, \text{N} \times 0.22 \, \text{m} = 242 \, \text{N m} \] ### c) Work Done During One Revolution The work done \( W \) in one revolution can be calculated using the formula: \[ W = T_{e} \cdot d \] Where \( d \) is the circumference of the pulley: \[ d = 2\pi r = 2\pi \times 0.22 \approx 1.38 \, \text{m} \] So, \[ W = 1100 \, \text{N} \times 1.38 \, \text{m} \approx 1518 \, \text{J} \] ### d) Output Power After 45 Seconds First, calculate the number of revolutions in 45 seconds. The rotational frequency is given as \( 800 \, \text{r/min} \): \[ \text{Revolutions in 45 seconds} = \frac{800}{60} \times 45 = 600 \, \text{revolutions} \] The total work done in 600 revolutions is: \[ \text{Total Work} = 600 \times 1518 \, \text{J} = 910800 \, \text{J} \] The power \( P \) is then calculated by dividing the work done by time: \[ P = \frac{\text{Total Work}}{\text{Time}} = \frac{910800 \, \text{J}}{45 \, \text{s}} \approx 20240 \, \text{W} \] Thus, the output power is approximately \( 20.24 \, \text{kW} \).