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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 .

Ask by Potter Gray. in South Africa
Jan 28,2025

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The torque of the driver pulley is 242 N·m.

Solution

To calculate the **torque of the driver pulley** in the belt drive system, we'll follow these steps: ### **Given:** - **Tensions in the belt:** - \( T_1 = 2000 \, \text{N} \) (tight side) - \( T_2 = 900 \, \text{N} \) (slack side) - **Diameter of the driver pulley:** \( d = 0.44 \, \text{m} \) - **Radius of the driver pulley:** \( r = \frac{d}{2} = \frac{0.44}{2} = 0.22 \, \text{m} \) ### **Torque Calculation:** The torque (\( \tau \)) produced by the driver pulley due to the belt tension is given by the difference in tensions multiplied by the radius of the pulley: \[ \tau = (T_1 - T_2) \times r \] Plugging in the known values: \[ \tau = (2000 \, \text{N} - 900 \, \text{N}) \times 0.22 \, \text{m} = 1100 \, \text{N} \times 0.22 \, \text{m} = 242 \, \text{N} \cdot \text{m} \] ### **Answer:** Problem b Answer: The driver pulley torque is (2000 N – 900 N) × 0.22 m = 242 N·m.

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To solve this problem, we need to calculate several aspects of the belt drive system given the tensions, pulley diameter, and rotational frequency. ### a) Effective Tension (T_eff) The effective tension in the belt drive can be calculated using the formula: \[ T_{\text{eff}} = T_1 - T_2 \] where \(T_1\) is the tension on the tight side (2000 N) and \(T_2\) is the tension on the slack side (900 N). \[ T_{\text{eff}} = 2000 \, \text{N} - 900 \, \text{N} = 1100 \, \text{N} \] ### b) Torque of the Driver Pulley (τ) The torque exerted by the driver pulley can be calculated using the formula: \[ \tau = T_{\text{eff}} \times r \] First, we need the radius \(r\), which is half the diameter. Given the diameter is \(0.44 \, \text{m}\): \[ r = \frac{0.44 \, \text{m}}{2} = 0.22 \, \text{m} \] Now, substituting the values: \[ \tau = 1100 \, \text{N} \times 0.22 \, \text{m} = 242 \, \text{N m} \] ### c) Work Done During One Revolution (W) The work done can be calculated as: \[ W = T_{\text{eff}} \times \text{Distance moved by the belt in one revolution} \] The distance for one full revolution is the circumference of the pulley: \[ \text{Circumference} = \pi \times d = \pi \times 0.44 \, \text{m} \approx 1.385 \, \text{m} \] Thus, the work done during one revolution: \[ W = 1100 \, \text{N} \times 1.385 \, \text{m} \approx 1523.5 \, \text{J} \] ### d) Output Power of the Pulley after 45 seconds (P) To calculate power, we need to find the number of revolutions in 45 seconds. The frequency is given as \(800 \, \text{r/min}\), so we convert that to revolutions per second: \[ \text{Revolutions per second} = \frac{800 \, \text{r/min}}{60} \approx 13.33 \, \text{r/s} \] In 45 seconds, the number of revolutions is: \[ \text{Total revolutions} = 13.33 \, \text{r/s} \times 45 \, \text{s} \approx 600 \, \text{r} \] Now, the total work done would be: \[ \text{Total Work} = W \times \text{Total revolutions} = 1523.5 \, \text{J} \times 600 \approx 914100 \, \text{J} \] Finally, power output can be calculated by: \[ P = \frac{\text{Total Work}}{\text{Time}} = \frac{914100 \, \text{J}}{45 \, \text{s}} \approx 20220 \, \text{W} \approx 20.22 \, \text{kW} \] So, to sum up: - **Effective Tension:** \(1100 \, \text{N}\) - **Torque of the Driver Pulley:** \(242 \, \text{N m}\) - **Work Done During One Revolution:** \(1523.5 \, \text{J}\) - **Output Power of the Pulley after 45 Seconds:** \(20.22 \, \text{kW}\)

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