To solve this problem, we need to consider the forces and torques acting on the system. Here's how we can approach it:
1. **Total Weight:** The total weight of the board and the block acts downward, which is the combined weight of both. This is given by \( W = (m_{\text{block}} + m_{\text{board}}) \times g \), where \( m_{\text{block}} \) and \( m_{\text{board}} \) are the masses of the block and board, respectively, and \( g \) is the acceleration due to gravity.
2. **Torque Balance:** Since the board is in equilibrium, the torques about any point must sum to zero. We can choose the point under the block (5.64 m from the left) as the pivot point to simplify calculations, as this will eliminate the torque due to the block's weight.
3. **Calculating Torques:** We need to calculate the torque due to the weight of the board and the forces exerted by the scales. The torque due to the board's weight can be considered to act at its center of mass, which is at the midpoint of its length.
4. **Forces by Scales:** Let \( F_{\text{left}} \) and \( F_{\text{right}} \) be the forces exerted by the left and right scales, respectively. These forces also create torques about the pivot point.
5. **Equations of Equilibrium:** We set up the equations for the sum of forces and the sum of torques to be zero.
Given:
- \( m_{\text{block}} = 2.2 \, \text{kg} \)
- \( m_{\text{board}} = 33 \, \text{kg} \)
- Length of the board, \( L = 44.4 \, \text{m} \)
- Distance of block from left, \( d = 5.64 \, \text{m} \)
- \( g \approx 9.81 \, \text{m/s}^2 \)
Let's calculate the forces and torques to find the reading of the scale on the right.
The scale on the right reads approximately 138.31 kg. This value represents the force exerted by the right scale to balance the torques and keep the system in equilibrium.