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 . The wheel of a belt drive rotates at \( 1200 \mathrm{r} / \mathrm{min} \). The belt is subjected to an effective force of 600 N . If the power transmitted by the belt is \( 12,4 \mathrm{~kW} \), calculate the diameter of the wheel. 6. The engine of a car develops 72 kW at \( 1500 \mathrm{r} / \mathrm{min} \). Calculate the torque develoned 7. A machine has a t

To tackle the effective tension in a belt drive system, you can use the formula: Effective Tension (T_eff) = T1 - T2, where T1 is the maximum tension and T2 is the minimum tension. For your case, T_eff will be 2000 N - 900 N = 1100 N. Now, for calculating torque, use the formula: Torque (τ) = T_eff × radius (r). The radius can be found by dividing the diameter by 2, so it’s 0.22 m. Plug that into the equation (τ = 1100 N × 0.22 m), and you’ll get the torque! Now, let’s sprinkle in some real-world connections! Belt drives are widely used in various types of machinery – from your car's engine to large industrial equipment. They efficiently transmit power while allowing some flexibility in positioning. Their simple design allows for easy maintenance because you can replace belts without major disruptions. So next time you notice that hum of pulleys in action at a factory or even around your car, remember the important forces at play!