Answer :
To solve this problem, let's follow a step-by-step approach.
### Step 1: Determine the Forces
1. Total length of the rod [tex]\( L \)[/tex]:
[tex]\[ L = 2 \text{ meters} \][/tex]
2. Masses:
- Mass of the rod [tex]\( m_{\text{rod}} = 4 \text{ kg} \)[/tex]
- Mass of the load [tex]\( m_{\text{load}} = 6 \text{ kg} \)[/tex]
3. Weight Forces
- Weight of the rod [tex]\( W_{\text{rod}} \)[/tex]:
[tex]\[ W_{\text{rod}} = m_{\text{rod}} \times g = 4 \times 9.81 = 39.24 \text{ N} \][/tex]
- Weight of the load [tex]\( W_{\text{load}} \)[/tex]:
[tex]\[ W_{\text{load}} = m_{\text{load}} \times g = 6 \times 9.81 = 58.86 \text{ N} \][/tex]
### Step 2: Equilibrium of Forces
The system is in static equilibrium, so let's use the conditions for both vertical and rotational equilibrium.
#### Vertical Equilibrium
Sum of vertical forces must be zero. Let [tex]\( R_A \)[/tex] be the reaction force at [tex]\( A \)[/tex] and [tex]\( R_B \)[/tex] be the reaction force at [tex]\( B \)[/tex].
Given that [tex]\( R_B = 3 \times R_A \)[/tex], we have:
[tex]\[ R_A + R_B = W_{\text{rod}} + W_{\text{load}} \][/tex]
Substituting [tex]\( R_B = 3R_A \)[/tex]:
[tex]\[ R_A + 3R_A = W_{\text{rod}} + W_{\text{load}} \][/tex]
[tex]\[ 4R_A = W_{\text{rod}} + W_{\text{load}} \][/tex]
[tex]\[ 4R_A = 39.24 + 58.86 = 98.1 \text{ N} \][/tex]
[tex]\[ R_A = \frac{98.1}{4} = 24.525 \text{ N} \][/tex]
Now, calculating [tex]\( R_B \)[/tex]:
[tex]\[ R_B = 3 \times R_A = 3 \times 24.525 = 73.575 \text{ N} \][/tex]
### Step 3: Rotational Equilibrium
For rotational equilibrium around point [tex]\( A \)[/tex], the sum of moments about point [tex]\( A \)[/tex] should be zero.
Taking clockwise moments as positive:
[tex]\[ \sum M_A = 0 \][/tex]
[tex]\[ R_A \cdot 0 + R_B \cdot L - W_{\text{rod}} \cdot \left( \frac{L}{2} \right) - W_{\text{load}} \cdot x = 0 \][/tex]
Since [tex]\( R_A \cdot 0 = 0 \)[/tex], we simplify to:
[tex]\[ R_B \cdot L - W_{\text{rod}} \cdot \left( \frac{L}{2} \right) - W_{\text{load}} \cdot x = 0 \][/tex]
Substitute known values:
[tex]\[ 73.575 \cdot 2 - 39.24 \cdot 1 - 58.86 \cdot x = 0 \][/tex]
[tex]\[ 147.15 - 39.24 - 58.86x = 0 \][/tex]
[tex]\[ 107.91 = 58.86x \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{107.91}{58.86} \approx 1.833 \text{ meters} \][/tex]
### Final Results
- The distance [tex]\( x \)[/tex] at which the mass is placed from [tex]\( A \)[/tex]:
[tex]\[ x \approx 1.833 \text{ meters} \][/tex]
- The reactions at the supports are:
[tex]\[ R_A \approx 24.525 \text{ N} \][/tex]
[tex]\[ R_B \approx 73.575 \text{ N} \][/tex]
Therefore, the distance [tex]\( x \)[/tex] from point [tex]\( A \)[/tex] to the mass is approximately 1.833 meters, the reaction at support [tex]\( A \)[/tex] is approximately 24.525 N, and the reaction at support [tex]\( B \)[/tex] is approximately 73.575 N.
### Step 1: Determine the Forces
1. Total length of the rod [tex]\( L \)[/tex]:
[tex]\[ L = 2 \text{ meters} \][/tex]
2. Masses:
- Mass of the rod [tex]\( m_{\text{rod}} = 4 \text{ kg} \)[/tex]
- Mass of the load [tex]\( m_{\text{load}} = 6 \text{ kg} \)[/tex]
3. Weight Forces
- Weight of the rod [tex]\( W_{\text{rod}} \)[/tex]:
[tex]\[ W_{\text{rod}} = m_{\text{rod}} \times g = 4 \times 9.81 = 39.24 \text{ N} \][/tex]
- Weight of the load [tex]\( W_{\text{load}} \)[/tex]:
[tex]\[ W_{\text{load}} = m_{\text{load}} \times g = 6 \times 9.81 = 58.86 \text{ N} \][/tex]
### Step 2: Equilibrium of Forces
The system is in static equilibrium, so let's use the conditions for both vertical and rotational equilibrium.
#### Vertical Equilibrium
Sum of vertical forces must be zero. Let [tex]\( R_A \)[/tex] be the reaction force at [tex]\( A \)[/tex] and [tex]\( R_B \)[/tex] be the reaction force at [tex]\( B \)[/tex].
Given that [tex]\( R_B = 3 \times R_A \)[/tex], we have:
[tex]\[ R_A + R_B = W_{\text{rod}} + W_{\text{load}} \][/tex]
Substituting [tex]\( R_B = 3R_A \)[/tex]:
[tex]\[ R_A + 3R_A = W_{\text{rod}} + W_{\text{load}} \][/tex]
[tex]\[ 4R_A = W_{\text{rod}} + W_{\text{load}} \][/tex]
[tex]\[ 4R_A = 39.24 + 58.86 = 98.1 \text{ N} \][/tex]
[tex]\[ R_A = \frac{98.1}{4} = 24.525 \text{ N} \][/tex]
Now, calculating [tex]\( R_B \)[/tex]:
[tex]\[ R_B = 3 \times R_A = 3 \times 24.525 = 73.575 \text{ N} \][/tex]
### Step 3: Rotational Equilibrium
For rotational equilibrium around point [tex]\( A \)[/tex], the sum of moments about point [tex]\( A \)[/tex] should be zero.
Taking clockwise moments as positive:
[tex]\[ \sum M_A = 0 \][/tex]
[tex]\[ R_A \cdot 0 + R_B \cdot L - W_{\text{rod}} \cdot \left( \frac{L}{2} \right) - W_{\text{load}} \cdot x = 0 \][/tex]
Since [tex]\( R_A \cdot 0 = 0 \)[/tex], we simplify to:
[tex]\[ R_B \cdot L - W_{\text{rod}} \cdot \left( \frac{L}{2} \right) - W_{\text{load}} \cdot x = 0 \][/tex]
Substitute known values:
[tex]\[ 73.575 \cdot 2 - 39.24 \cdot 1 - 58.86 \cdot x = 0 \][/tex]
[tex]\[ 147.15 - 39.24 - 58.86x = 0 \][/tex]
[tex]\[ 107.91 = 58.86x \][/tex]
Solving for [tex]\( x \)[/tex]:
[tex]\[ x = \frac{107.91}{58.86} \approx 1.833 \text{ meters} \][/tex]
### Final Results
- The distance [tex]\( x \)[/tex] at which the mass is placed from [tex]\( A \)[/tex]:
[tex]\[ x \approx 1.833 \text{ meters} \][/tex]
- The reactions at the supports are:
[tex]\[ R_A \approx 24.525 \text{ N} \][/tex]
[tex]\[ R_B \approx 73.575 \text{ N} \][/tex]
Therefore, the distance [tex]\( x \)[/tex] from point [tex]\( A \)[/tex] to the mass is approximately 1.833 meters, the reaction at support [tex]\( A \)[/tex] is approximately 24.525 N, and the reaction at support [tex]\( B \)[/tex] is approximately 73.575 N.
