Question

To hang a 6.20-kg pot of flowers, a gardener uses two wires—one attached horizontally
to a wall, the other sloping upward at an angle of and attached to the ceiling.
Find the tension in each wire.

Answer 1

To find the tension in the two wires supporting a 6.20-kg pot, we need to analyze the forces acting on the pot and apply Newton's second law of motion.

1. **Identify Forces**:
- The weight of the pot, \( W \), acts downward. It can be calculated using:
\[
W = mg
\]
where \( m = 6.20 \, \text{kg} \) and \( g \approx 9.81 \, \text{m/s}^2 \).

- The tensions in the two wires will be denoted as \( T_h \) (horizontal wire) and \( T_s \) (sloping wire).

2. **Draw the Free Body Diagram**:
- The vertical component of the tension in the sloping wire must balance the weight of the pot.
- The horizontal component of the tension in the sloping wire must balance the tension in the horizontal wire.

3. **Set Up the Equations**:
- For the vertical forces:
\[
T_s \sin(\theta) = W
\]
- For the horizontal forces:
\[
T_h = T_s \cos(\theta)
\]

4. **Calculate the Weight of the Pot**:
\[
W = 6.20 \, \text{kg} \times 9.81 \, \text{m/s}^2 \approx 60.82 \, \text{N}
\]

5. **Substitute the Weight into the Vertical Forces Equation**:
\[
T_s \sin(\theta) = 60.82 \, \text{N}
\]
Rearranging gives:
\[
T_s = \frac{60.82}{\sin(\theta)}
\]

6. **Substitute \( T_s \) into the Horizontal Forces Equation**:
\[
T_h = \left(\frac{60.82}{\sin(\theta)}\right) \cos(\theta) = 60.82 \frac{\cos(\theta)}{\sin(\theta)} = 60.82 \cot(\theta)
\]

At this point, we need the angle \( \theta \) to proceed further and calculate the specific values for \( T_h \) and \( T_s \).

### Example Calculation:

If we assume an angle of \( \theta = 30^\circ \):

1. Calculate \( T_s \):
- \( \sin(30^\circ) = 0.5 \)
\[
T_s = \frac{60.82}{0.5} = 121.64 \, \text{N}
\]

2. Calculate \( T_h \):
- \( \cos(30^\circ) = \sqrt{3}/2 \approx 0.866 \)
- The cotangent is:
- \( \cot(30^\circ) = \sqrt{3} \approx 1.732 \)
\[
T_h = 60.82 \cdot 1.732 \approx 105.39 \, \text{N}
\]

### Final Result (Example):
For \( \theta = 30^\circ \):
- \( T_s \approx 121.64 \, \text{N} \) (tension in the sloping wire)
- \( T_h \approx 105.39 \, \text{N} \) (tension in the horizontal wire)

Modify the angle \( \theta \) in the calculations to find a specific result as needed.