Let's solve the problem step by step: --- ## **Step 1: Understand What’s Given** - Pulley A receives power from a motor through a belt. - Power is transmitted through the shaft and delivered to the belt on pulley B. - Tension on the loose side of B's belt is 8% of the tension on the tight side. - Given: - **B (diameter of pulley B):** 7.00 in - **T(A1) (tight side tension on A):** 560. lb - **T(A2) (loose side tension on A):** 93.3 lb --- ## **Step 2: Find Torque Transmitted by the Shaft** The **net belt tension** on pulley A = difference between tight and loose side tensions: \[ T_{A1} - T_{A2} = 560.\ \text{lb} - 93.3\ \text{lb} = 466.7\ \text{lb} \] The **torque** delivered by pulley A (also transmitted to pulley B, since the shaft is rigid and speed is constant): \[ \text{Torque} = (\text{Net belt tension}) \times (\text{radius of pulley A}) \] But we aren’t given the diameter of pulley A, so let's use the fact that the same torque is applied to pulley B. --- ## **Step 3: Belt Tensions on Pulley B** Let: - \( T_{B1} \) = tension on the **tight** side of pulley B - \( T_{B2} \) = tension on the **loose** side of pulley B We're told: \[ T_{B2} = .08 \times T_{B1} \] --- ## **Step 4: Torque Delivered to Pulley B** \[ \text{Torque at B} = (T_{B1} - T_{B2}) \times r_B \] Where \( r_B = \frac{B}{2} = \frac{7.00}{2} = 3.5\ \text{in} \) The torque must be the **same** as that delivered by pulley A: \[ \text{Torque} = (T_{A1} - T_{A2}) \times r_A \] But since we don’t have \( r_A \), but we do have \( (T_{A1} - T_{A2}) \), and since torque is transmitted through the shaft, we can **equate the torques**: \[ (T_{A1} - T_{A2}) \times r_A = (T_{B1} - T_{B2}) \times r_B \] But since the shaft speed is constant and there are no losses, **the net belt force difference is the same:** \[ T_{A1} - T_{A2} = T_{B1} - T_{B2} \] \[ 466.7 = T_{B1} - T_{B2} \] --- ## **Step 5: Solve for the Belt Tensions on Pulley B** We already have: \[ T_{B2} = .08 T_{B1} \] \[ T_{B1} - T_{B2} = 466.7 \] Substitute: \[ T_{B1} - .08 T_{B1} = 466.7 \] \[ .92 T_{B1} = 466.7 \] \[ T_{B1} = \frac{466.7}{.92} = 507.28~\text{lb} \] Now, find \( T_{B2} \): \[ T_{B2} = .08 \times 507.28 = 40.58~\text{lb} \] --- ## **Step 6: Final Answers** ### **Tension on the loose side of the pulley B:** \[ \boxed{40.6~\text{lb}} \] ### **Tension on the tight side of the pulley B:** \[ \boxed{507.3~\text{lb}} \] --- ## **Summary Table** | Quantity | Value | |------------------|-----------| | Loose side \(T_{B2}\) | 40.6 lb | | Tight side \(T_{B1}\) | 507.3 lb | --- **That's the step-by-step solution!**