Let's break down the problem step by step: ### **Given:** - Weight of cylinder \( B \): \( 16.5 \) lb - Weight of wedge \( A \): \( 7.5 \) lb - Incline angle (\( \theta \)): \( 20^\circ \) - Cylinder radius (\( r \)): \( 3 \) in \( = 0.25 \) ft - The cylinder rolls without slipping on the wedge. - Neglect friction between wedge and ground. ### **Step 1: Free Body Diagrams** #### **Cylinder B:** For the cylinder: - Weight acts vertically downward (\( W_B = 16.5 \) lb \( = 16.5/32.2 = 0.512 \) slugs). - Normal force \( N \) from the wedge, perpendicular to the wedge. - Friction force \( f \), parallel to the wedge (required for rolling). #### **Wedge A:** For the wedge: - Weight acts vertically downward (\( W_A = 7.5 \) lb \( = 7.5/32.2 = 0.233 \) slugs). - Normal force from the ground (vertical). - Normal and friction forces from the cylinder (reaction). ### **Step 2: Equations of Motion** #### **Cylinder:** Let \( a_B \) be the acceleration of the cylinder's center **relative to the ground**. Let \( a_A \) be the acceleration of the wedge **relative to the ground**. Let’s take **rightward as positive**. Forces on cylinder \( B \) along the incline: \[ m_B a_{B/A} = m_B g \sin\theta - f \] But \( a_{B/A} \) is the acceleration of the cylinder **relative to the wedge**. But, the **absolute acceleration** of the cylinder (relative to ground) is: \[ a_B = a_A + a_{B/A} \] #### **Wedge:** Horizontal forces on wedge: \[ m_A a_A = f \] #### **Rolling Condition:** Since the cylinder rolls **without slipping**: \[ a_{B/A} = r \alpha \] where \( \alpha \) is the angular acceleration. Sum of moments about center of cylinder: \[ I \alpha = f r \] For a solid cylinder, \( I = \frac{1}{2} m_B r^2 \): \[ \frac{1}{2} m_B r^2 \alpha = f r \implies \frac{1}{2} m_B r \alpha = f \] But \( a_{B/A} = r \alpha \implies \alpha = \frac{a_{B/A}}{r} \): \[ f = \frac{1}{2} m_B a_{B/A} \] ### **Step 3: Express All in Terms of \( a_A \) and \( a_{B/A} \)** From wedge equation: \[ m_A a_A = f \] From above, \( f = \frac{1}{2} m_B a_{B/A} \): \[ m_A a_A = \frac{1}{2} m_B a_{B/A} \implies a_A = \frac{1}{2} \frac{m_B}{m_A} a_{B/A} \implies a_{B/A} = \frac{2 m_A}{m_B} a_A \] From cylinder equation along incline: \[ m_B (a_B - a_A) = m_B g \sin\theta - f \] Recall, \( a_B = a_A + a_{B/A} \): \[ m_B (a_A + a_{B/A} - a_A) = m_B g \sin\theta - f \] \[ m_B a_{B/A} = m_B g \sin\theta - f \] But \( f = \frac{1}{2} m_B a_{B/A} \): \[ m_B a_{B/A} = m_B g \sin\theta - \frac{1}{2} m_B a_{B/A} \] \[ m_B a_{B/A} + \frac{1}{2} m_B a_{B/A} = m_B g \sin\theta \] \[ \frac{3}{2} m_B a_{B/A} = m_B g \sin\theta \] \[ a_{B/A} = \frac{2}{3} g \sin\theta \] Now substitute into earlier wedge equation: \[ a_A = \frac{1}{2} \frac{m_B}{m_A} a_{B/A} \] Plugging in \( a_{B/A} \): \[ a_A = \frac{1}{2} \frac{m_B}{m_A} \left(\frac{2}{3} g \sin\theta\right) \] \[ a_A = \frac{m_B}{m_A} \frac{g \sin\theta}{3} \] ### **Step 4: Plug in Values** - \( m_B = \frac{16.5}{32.2} = 0.512 \) slugs - \( m_A = \frac{7.5}{32.2} = 0.233 \) slugs - \( g = 32.2 \) ft/s² - \( \sin(20^\circ) = 0.342 \) \[ a_A = \frac{0.512}{0.233} \cdot \frac{32.2 \times 0.342}{3} \] \[ a_A = 2.197 \cdot \frac{11.0124}{3} \] \[ a_A = 2.197 \cdot 3.6708 \] \[ a_A = 8.06 \text{ ft/s}^2 \] --- ### **Final Answer:** \[ \boxed{8.06} \text{ ft/s}^2 \] **The acceleration of the wedge immediately after the cord is cut is \( 8.06 \) ft/s².**

Immediately after the cord is cut, the wedge accelerates horizontally with an acceleration determined by the ratio of the masses, gravity, and the incline angle, resulting in approximately 8.06 ft/s², due to the combined effects of the rolling cylinder and the wedge's reaction forces.