Math Problem Statement

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.)

Solution

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 T1T_1 be the tension in the 3 m rope, which makes an angle of 52° with the horizontal.
  • Let T2T_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: Fg=mg=9kg×9.8m/s2=88.2NF_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: T1cos(52)=T2cos(40)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: T1sin(52)+T2sin(40)=88.2NT_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 T2T_2 in terms of T1T_1: T2=T1cos(52)cos(40)T_2 = T_1 \frac{\cos(52^\circ)}{\cos(40^\circ)}

Now substitute this expression for T2T_2 into the second equation: T1sin(52)+(T1cos(52)cos(40))sin(40)=88.2T_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 T1T_1 and T2T_2.The tensions in the ropes are as follows:

  • Tension in the 3 m rope (T1T_1) is approximately 67.61N67.61 \, \text{N}.
  • Tension in the 5 m rope (T2T_2) is approximately 54.33N54.33 \, \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.

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Math Problem Analysis

Mathematical Concepts

Statics
Force Equilibrium
Trigonometry

Formulas

Horizontal force equilibrium: T1 * cos(52°) = T2 * cos(40°)
Vertical force equilibrium: T1 * sin(52°) + T2 * sin(40°) = mg
Gravitational force: Fg = mg

Theorems

Equilibrium of Forces
Trigonometric Decomposition of Forces

Suitable Grade Level

College Level (Physics/Engineering)