Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 9 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension (in N). (Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.)


To solve this problem, we need to use the principles of static equilibrium. The decoration is in equilibrium, meaning the net forces in both the horizontal and vertical directions are zero. We will break down the tension in each wire into horizontal and vertical components.

### Step 1: Label the tensions

- Let \( T_1 \) be the tension in the 3 m rope, which makes an angle of 52° with the horizontal.
- Let \( T_2 \) be the tension in the 5 m rope, which makes an angle of 40° with the horizontal.

The mass of the decoration is 9 kg, so the gravitational force acting on it is:
\[
F_g = mg = 9 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 88.2 \, \text{N}
\]

### Step 2: Set up equilibrium equations

Since the decoration is in equilibrium, the sum of the forces in the horizontal direction must be zero, and the sum of the forces in the vertical direction must also be zero.

1. **Horizontal direction:**
   \[
   T_1 \cos(52^\circ) = T_2 \cos(40^\circ)
   \]
   This equation expresses the fact that the horizontal components of the two tensions must cancel each other out.

2. **Vertical direction:**
   \[
   T_1 \sin(52^\circ) + T_2 \sin(40^\circ) = 88.2 \, \text{N}
   \]
   This equation ensures that the vertical components of the two tensions must balance the weight of the decoration.

### Step 3: Solve the system of equations

From the first equation, solve for \( T_2 \) in terms of \( T_1 \):
\[
T_2 = T_1 \frac{\cos(52^\circ)}{\cos(40^\circ)}
\]

Now substitute this expression for \( T_2 \) into the second equation:
\[
T_1 \sin(52^\circ) + \left( T_1 \frac{\cos(52^\circ)}{\cos(40^\circ)} \right) \sin(40^\circ) = 88.2
\]

Now let's solve for \( T_1 \) and \( T_2 \).The tensions in the ropes are as follows:

- Tension in the 3 m rope (\( T_1 \)) is approximately \( 67.61 \, \text{N} \).
- Tension in the 5 m rope (\( T_2 \)) is approximately \( 54.33 \, \text{N} \).

These values ensure that the forces in both the horizontal and vertical directions are balanced, keeping the decoration in equilibrium.

Do you want more details or have any further questions?

Here are 5 related questions to deepen your understanding:
1. How would the tensions change if the mass of the decoration were increased?
2. What happens to the tensions if the angles of the ropes change?
3. Can you explain how the components of forces work in two-dimensional equilibrium problems?
4. How does the angle of the rope affect the horizontal and vertical components of the tension?
5. If the gravitational acceleration was different, how would that impact the tensions?

**Tip:** Always break forces into components when solving equilibrium problems to simplify the calculations.

Ropes 3 m and 5 m in length are fastened to a holiday decoration that is suspended over a town square. The decoration has a mass of 9 kg. The ropes, fastened at different heights, make angles of 52° and 40° with the horizontal. Find the tension in each wire and the magnitude of each tension (in N). (Use g = 9.8 m/s2 for the acceleration due to gravity. Round your answers to two decimal places.)

Solve the physics problem involving a 9 kg holiday decoration suspended by two ropes of different lengths (3 m and 5 m) at angles of 52° and 40° with the horizontal. This problem explores force equilibrium and uses trigonometry to find the tensions in each rope. The detailed step-by-step solution is ideal for college-level students in physics or engineering.

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