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This problem involves equilibrium conditions for a rod and a disc in a setup with smooth walls and a horizontal plane, with forces acting on the system. Let's break it down.

### Setup Details:
1. **Rod \( AB \)**:
   - Weight: \( 2W \)
   - Length: \( 3a \)
   - It is hinged at point \( A \), resting horizontally between walls \( EF \) and \( CD \).
   
2. **Disc \( Q \)**:
   - Weight: \( W \)
   - It is in contact with the wall \( EF \) at point \( P \) and the rod \( AB \) at point \( Q \).

3. **Geometrical Condition**:
   - The angle \( \angle BAC \) is given as \( 60^\circ \).
   - \( AB \) and the disc are in equilibrium, implying that the sum of forces and moments acting on them are zero.

### Questions:
1. **(a) The Magnitude of the Reaction at \( B \)**:  
   The reaction force at \( B \) is given by \( \frac{7\sqrt{3}}{9} W \).

2. **(b) The Magnitude of the Reaction at \( A \)**:  
   We need to find the reaction at \( A \), correct to 3 significant figures.

---

### Solution Process:
We will use equilibrium conditions:
1. **Vertical Equilibrium**:  
   The sum of the vertical forces acting on the system (both rod and disc) must be zero.

2. **Horizontal Equilibrium**:  
   The sum of the horizontal forces acting on the system must be zero.

3. **Moment Equilibrium**:  
   The sum of the moments about a convenient point (often the hinge \( A \)) must also be zero.

#### Let's start with step-by-step analysis:

1. **Vertical Force Balance:**
   - Weight of rod = \( 2W \)
   - Weight of disc = \( W \)
   - Reactions at \( A \) and \( B \) provide vertical support.
   
   Sum of vertical forces = 0:
   \[
   R_A^V + R_B^V = 2W + W = 3W
   \]

2. **Horizontal Force Balance:**
   The smooth walls imply no friction, so the horizontal components of the reactions at \( A \), \( B \), and the contact with the wall \( P \) must balance.

3. **Moment Balance:**
   Taking moments about point \( A \) to eliminate \( R_A \), we can set up the equation for the moments caused by the weights of the rod and the disc, along with the reaction forces at \( B \).

We can proceed with calculations, or I can provide a more detailed walkthrough.

Would you like a full breakdown of the solution with calculations?

Two parallel vertical smooth walls EF and CD meet a horizontal plane at E and C, respectively. A uniform smooth rod AB, of weight 2W and length 3a, is freely hinged to the horizontal plane at the point A. Between EF and CD, the rod AB rests against CD. A uniform smooth disc of weight W is in contact with the wall EF at the point P and with the rod at the point Q. It is given that the angle BAC is 60° and that ABQ is a triangle. The rod and the disc are in equilibrium in the same vertical plane, which is perpendicular to both walls.

Solve a mechanics problem involving a rod and disc in equilibrium. Learn how to calculate the reaction forces at points A and B using statics principles, force balance, and moment equations. Suitable for engineering or physics students.

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