Let's solve the problem step by step and include all necessary diagrams and explanations. --- ## **Problem Restatement** - There are two pulleys connected by a belt. - The speed of the belt is **12 ft/s**. - The acceleration of point **A** is **600 ft/s²**. - The radii of the pulleys are: - For **A**: 9 in (radius) - For **B**: 15 in (radius) - We are to find the **acceleration of point B**. --- ## **Step 1: Convert Units (if necessary)** The radii are given in **inches**, but speeds and accelerations are in **feet**. Let's convert all radii to **feet**: - 9 in = 9/12 = **0.75 ft** (radius of A) - 15 in = 15/12 = **1.25 ft** (radius of B) --- ## **Step 2: Relationship Between Angular Velocity and Belt Speed** The **tangential velocity** at the rim of each pulley is equal to the belt speed: \[ v = r \cdot \omega \] Where: - \( v \) = belt speed - \( r \) = radius - \( \omega \) = angular velocity So, - For A: \( v = r_A \omega_A \) - For B: \( v = r_B \omega_B \) Thus, \[ \omega_A = \frac{v}{r_A} \] \[ \omega_B = \frac{v}{r_B} \] --- ## **Step 3: Relationship Between Angular Acceleration and Linear Acceleration** The **tangential acceleration** at the rim is: \[ a_t = r \alpha \] Where: - \( a_t \) = tangential acceleration - \( \alpha \) = angular acceleration For point A: \[ a_A = r_A \alpha_A \] Given \( a_A = 600 \) ft/s², \( r_A = 0.75 \) ft, \[ \alpha_A = \frac{a_A}{r_A} = \frac{600}{0.75} = 800 \text{ rad/s}^2 \] --- ## **Step 4: Find the Angular Acceleration of Pulley B** When the pulleys are connected by a belt, the **tangential acceleration** at the rims must be equal in magnitude but opposite in direction (since the belt doesn't slip): \[ a_A = a_B \] But, since the radii are different: \[ r_A \alpha_A = r_B \alpha_B \] So, \[ \alpha_B = \frac{r_A \alpha_A}{r_B} \] Plug in the values: \[ \alpha_B = \frac{0.75 \times 800}{1.25} = \frac{600}{1.25} = 480 \text{ rad/s}^2 \] --- ## **Step 5: Find the Linear Acceleration at Point B** The **linear (tangential) acceleration** of point B is: \[ a_B = r_B \alpha_B \] Plug in the values: \[ a_B = 1.25 \times 480 = 600 \text{ ft/s}^2 \] --- ## **Final Answer** \[ \boxed{600\ \text{ft/s}^2} \] The acceleration of point B is **600 ft/s²**. --- ### **Diagram** Here is a labeled diagram for clarity: ``` _________ / \ | 24 in | | | (| |) | 9 in |---- Belt ----| 15 in | (| |) | | | | | | \_________/ \_________/ Pulley A Pulley B (r_A) (r_B) Given: - Belt speed: 12 ft/s - a_A = 600 ft/s² - r_A = 0.75 ft, r_B = 1.25 ft Find: - a_B = ? ``` --- ## **Summary Table** | Pulley | Radius (ft) | Angular Accel (rad/s²) | Linear Accel (ft/s²) | |--------|-------------|------------------------|----------------------| | A | 0.75 | 800 | 600 | | B | 1.25 | 480 | 600 | --- If you have any questions about any step, let me know!