[tex] {⚠️ \: Don't \: Copy \: Like these < \ p >. < p > }[/tex]
______________________________________
### Step 1: Sum of Vertical Forces
[tex]\[
R_1 + R_2 + R_3 = 240 \text{ lb} + 960 \text{ lb}
\][/tex]
[tex]\[
R_1 + R_2 + R_3 = 1200 \text{ lb} \quad \text{(Equation 1)}
\][/tex]
### Step 2: Sum of Moments about \( R_2 \)
Taking moments about \( R_2 \):
[tex]\[
\sum M_{R_2} = 0 \implies R_1 \cdot 8 \text{ ft} - 240 \text{ lb} \cdot 9 \text{ ft} - 960 \text{ lb} \cdot (9 + 3) \text{ ft} + R_3 \cdot 12 \text{ ft} = 0
\]
[/tex]
Simplifying the distances for moments:
[tex]\[
R_1 \cdot 8 \text{ ft} - 240 \text{ lb} \cdot 9 \text{ ft} - 960 \text{ lb} \cdot 12 \text{ ft} + R_3 \cdot 12 \text{ ft} = 0
\][/tex]
[tex]\[
R_1 \cdot 8 - 240 \cdot 9 - 960 \cdot 12 + R_3 \cdot 12 = 0
\]
[/tex]
[tex]\[
8R_1 - 2160 - 11520 + 12R_3 = 0
\][/tex]
[tex]
\[
8R_1 + 12R_3 = 13680 \quad \text{(Equation 2)}
\][/tex]
### Step 3: Sum of Moments about \( R_3 \)
Taking moments about \( R_3 \):
[tex]\[
\sum M_{R_3} = 0 \implies R_1 \cdot 20 \text{ ft} - 240 \text{ lb} \cdot 21 \text{ ft} - 960 \text{ lb} \cdot 3 \text{ ft} + R_2 \cdot 12 \text{ ft} = 0
\][/tex]
Simplifying the distances for moments:
[tex]\[
R_1 \cdot 20 - 240 \cdot 21 - 960 \cdot 3 + R_2 \cdot 12 = 0
\][/tex]
[tex]\[
20R_1 - 5040 - 2880 + 12R_2 = 0
\]
[/tex]
[tex]\[
20R_1 + 12R_2 = 7920 \quad \text{(Equation 3)}
\][/tex]
### Step 4: Solving the Equations
We now have a system of three equations with three unknowns:
[tex]1. \( R_1 + R_2 + R_3 = 1200 \)
2. \( 8R_1 + 12R_3 = 13680 \)
3. \( 20R_1 + 12R_2 = 7920 \)[/tex]
Let's solve this system of equations.
[tex]From Equation 2:
\[
8R_1 + 12R_3 = 13680 \implies R_3 = \frac{13680 - 8R_1}{12}
\][/tex]
[tex]From Equation 3:
\[
20R_1 + 12R_2 = 7920 \implies R_2 = \frac{7920 - 20R_1}{12}
\][/tex]
Substitute \( R_3 \) and \( R_2 \) back into Equation 1:
[tex]\[
R_1 + \frac{7920 - 20R_1}{12} + \frac{13680 - 8R_1}{12} = 1200
\]
[/tex]
Simplifying the equation:
[tex]\[
R_1 + \frac{7920 - 20R_1 + 13680 - 8R_1}{12} = 1200
\][/tex]
[tex]\[
R_1 + \frac{21600 - 28R_1}{12} = 1200
\][/tex]
Multiply through by 12 to clear the denominator:
[tex]\[
12R_1 + 21600 - 28R_1 = 14400
\][/tex]
[tex]\[
-16R_1 = -7200
\][/tex]
[tex]\[
R_1 = 450 \text{ lb}
\][/tex]
Now, substitute \( R_1 = 450 \text{ lb} \) back into the equations for \( R_2 \) and \( R_3 \):
[tex]\[
R_2 = \frac{7920 - 20 \cdot 450}{12}
\][/tex]
[tex]\[
R_2 = \frac{7920 - 9000}{12} = \frac{-1080}{12} = -90 \text{ lb} \quad \text{(Negative value indicates a downward force)}
\][/tex]
[tex]\[
R_3 = \frac{13680 - 8 \cdot 450}{12}
\]
[/tex]
[tex]\[
R_3 = \frac{13680 - 3600}{12} = \frac{10080}{12} = 840 \text{ lb}
\]
[/tex]
### Summary of Reactions
[tex]
\[
R_1 = 450 \text{ lb}
\][/tex]
[tex]\[
R_2 = -90 \text{ lb} \quad \text{(Indicates downward force)}
\][/tex]
[tex]\[
R_3 = 840 \text{ lb}
\]
[/tex]
[tex]For \( F_{R_2} \):
\[
F_{R_2} = R_2 = -90 \text{ lb}
\][/tex]
The reactions are:
[tex]- \( R_1 = 450 \text{ lb} \)
- \( R_2 = -90 \text{ lb} \) (downward force)
- \( R_3 = 840 \text{ lb} \)
- \( F_{R_2} = -90 \text{ lb} \)
[/tex]
