# Step-by-Step Solution: V-Belt Pulley Countershaft Problem Since the figure and specific belt tensions are not provided, I'll outline the **typical process** for solving a problem involving a countershaft with two V-belt pulleys, as commonly encountered in mechanical engineering studies. Please supply the figure or exact belt tensions for a tailored solution. --- ## Problem Overview - **Two V-belt pulleys (A and B) on a countershaft.** - **Pulley A**: Receives power from a motor (input side). - **Pulley B**: Delivers power via a belt to a load (output side). - **Given**: Belt tensions on pulley A (tight side \( T_1 \), slack side \( T_2 \)), pulley diameters, and rotational speed. ### Objective Find the **power transmitted** and **shaft torque**. --- ## Step 1: Understand Power Transmission by Belt Power transmitted by a belt: \[ P = (T_1 - T_2) \cdot v \] - \( T_1 \): Tension in the tight side (input). - \( T_2 \): Tension in the slack side (input). - \( v \): Linear speed of the belt over the pulley. --- ## Step 2: Find Belt Speed \[ v = \pi D N \] - \( D \): Pulley diameter (in meters). - \( N \): Rotational speed (in revolutions per second). If \( D \) is in mm and \( N \) in rpm: \[ v = \frac{\pi D N}{60} \] where \( D \) in meters, \( N \) in rpm, \( v \) in m/s. --- ## Step 3: Compute Power Transmitted Insert values into the power formula: \[ P = (T_1 - T_2) \cdot v \] - \( T_1 \) and \( T_2 \) in newtons (N) - \( v \) in meters per second (m/s) - \( P \) in watts (W) --- ## Step 4: Find Shaft Torque Torque transmitted by the pulley: \[ \tau = (T_1 - T_2) \cdot r \] - \( r = \frac{D}{2} \): Pulley radius (in meters) - \( \tau \): Torque in newton-meters (Nm) --- ## Step 5: Power at Shaft Alternatively, check via rotational power: \[ P = \tau \cdot \omega \] - \( \omega = 2\pi N \) (if \( N \) is in revolutions per second) --- ## Example Calculation Suppose: - Pulley A diameter \( D = .2 \) m - Rotational speed \( N = 900 \) rpm - \( T_1 = 120 \) N - \( T_2 = 400 \) N ### (a) Belt speed: \[ v = \frac{\pi D N}{60} = \frac{\pi \times .2 \times 900}{60} = \frac{.2\pi \times 900}{60} = \frac{180\pi}{60} = 3\pi \approx 9.42 \text{ m/s} \] ### (b) Power transmitted: \[ P = (T_1 - T_2) \times v = (120 - 400) \times 9.42 = 800 \times 9.42 = 7536 \text{ W} = 7.54 \text{ kW} \] ### (c) Shaft torque at pulley A: \[ \tau = (T_1 - T_2) \times \frac{D}{2} = 800 \times .1 = 80 \text{ Nm} \] --- ## Summary Table | Quantity | Formula | Value | |------------------|----------------------------------------|------------| | Belt Speed \( v \) | \( v = \frac{\pi D N}{60} \) | \( 9.42 \) m/s | | Power \( P \) | \( P = (T_1 - T_2) \cdot v \) | \( 7.54 \) kW | | Torque \( \tau \) | \( \tau = (T_1 - T_2) \cdot \frac{D}{2} \) | \( 80 \) Nm | --- ## Final Answers (for Example Values) - **Belt speed**: \( 9.42 \) m/s - **Power transmitted**: \( 7.54 \) kW - **Shaft torque**: \( 80 \) Nm --- > **Note:** Insert your actual belt tensions, pulley diameters, and speeds for your specific problem. --- **Image Explanation**: *Diagram of a countershaft with two pulleys, showing forces and directions of rotation. (Not included here as image not provided)* --- If you provide the actual figure or numbers, I can give a **precise solution** for your case.