We have three forces acting at point O: T1, T2, and T3. The tension T1 is given as 5000 lbs. The angles of the cables are given by their slopes:
- T1: slope 4/3
- T2: slope 5/12
- T3: slope 4/3
Resolve each force into its horizontal (x) and vertical (y) components:
For T1 (5000 lbs):
- \( T1_x = T1 \cdot \frac{3}{5} = 5000 \cdot \frac{3}{5} = 3000 \) lbs
- \( T1_y = T1 \cdot \frac{4}{5} = 5000 \cdot \frac{4}{5} = 4000 \) lbs
For T2:
- \( T2_x = T2 \cdot \frac{12}{13} \)
- \( T2_y = T2 \cdot \frac{5}{13} \)
For T3:
- \( T3_x = T3 \cdot \frac{3}{5} \)
- \( T3_y = T3 \cdot \frac{4}{5} \)
Since the system is in equilibrium, the sum of the forces in both the x and y directions must be zero.
\[ T1_x + T2_x - T3_x = 0 \]
\[ 3000 + T2 \cdot \frac{12}{13} - T3 \cdot \frac{3}{5} = 0 \]
\[ T1_y + T2_y + T3_y = 10,000 \text{ lbs (downward force)} \]
\[ 4000 + T2 \cdot \frac{5}{13} + T3 \cdot \frac{4}{5} = 10,000 \]
We have two equations with two unknowns (T2 and T3):
- \( 3000 + T2 \cdot \frac{12}{13} - T3 \cdot \frac{3}{5} = 0 \)
- \( 4000 + T2 \cdot \frac{5}{13} + T3 \cdot \frac{4}{5} = 10,000 \)
From the first equation:
\[ T2 \cdot \frac{12}{13} = T3 \cdot \frac{3}{5} - 3000 \]
\[ T2 = \left( T3 \cdot \frac{3}{5} - 3000 \right) \cdot \frac{13}{12} \]
Substitute \( T2 \) into the second equation:
\[ 4000 + \left( \left( T3 \cdot \frac{3}{5} - 3000 \right) \cdot \frac{13}{12} \right) \cdot \frac{5}{13} + T3 \cdot \frac{4}{5} = 10,000 \]
\[ 4000 + \left( T3 \cdot \frac{3}{5} - 3000 \right) \cdot \frac{5}{12} + T3 \cdot \frac{4}{5} = 10,000 \]
\[ 4000 + T3 \cdot \frac{15}{60} - 3000 \cdot \frac{5}{12} + T3 \cdot \frac{48}{60} = 10,000 \]
\[ 4000 + T3 \cdot \frac{63}{60} - 1250 + T3 \cdot \frac{48}{60} = 10,000 \]
\[ 2750 + T3 \cdot \frac{111}{60} = 10,000 \]
\[ T3 \cdot \frac{111}{60} = 7250 \]
\[ T3 = 7250 \cdot \frac{60}{111} \]
\[ T3 \approx 3914.41 \text{ lbs} \]
Now, substitute \( T3 \) back into the equation for \( T2 \):
\[ T2 = \left( 3914.41 \cdot \frac{3
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