A large (15kg) holiday decoration is suspended on either side of Hay Street in Perth by two cables stretching between buildings. The length of the first cable is
8m, the second is 6m and the buildings are 12m apart. The 8m rope hangs at an angle 26.4° below the horizontal while the 6m rope hangs at 36.3°.
a. Draw a vector diagram of this situation
b. Find the tension in each cable.
a. To draw a vector diagram of this situation, follow these steps:
1. Draw two horizontal lines to represent the buildings, with a distance of 12m between them.
2. On the left side of the diagram, draw a line extending downward at an angle of 26.4° below the horizontal, representing the 8m cable.
3. On the right side of the diagram, draw a line extending downward at an angle of 36.3° below the horizontal, representing the 6m cable.
4. Connect the bottom ends of the two lines with a horizontal line to represent the holiday decoration.
b. To find the tension in each cable, let's use basic trigonometry and force equilibrium.
For the 8m cable:
- Resolve the tension force into horizontal and vertical components.
- The vertical component is T₁ * sin(θ) (θ being the angle of 26.4°).
- The horizontal component is T₁ * cos(θ).
- The upward vertical component must balance the weight of the decoration, which is 15kg * g (where g is the acceleration due to gravity, approximately 9.8 m/s²).
- So, we have T₁ * sin(26.4°) = 15kg * g.
For the 6m cable:
- Resolve the tension force into horizontal and vertical components.
- The vertical component is T₂ * sin(θ) (θ being the angle of 36.3°).
- The horizontal component is T₂ * cos(θ).
- The upward vertical component must balance the weight of the decoration, which is 15kg * g.
- So, we have T₂ * sin(36.3°) = 15kg * g.
Since the two cables are connected at the same point, the horizontal components of the tension forces must cancel each other out. Thus, T₁ * cos(26.4°) = T₂ * cos(36.3°).
Now, you can solve these three equations simultaneously to find the values of T₁ and T₂.