Math Problem Statement

DA 125 N sign is supported by two ropes. One rope pulls up © Macmillan Learn and to the right 01 = 35.5° above the horizontal with a tension T1, and the other rope pulls up and to the left 02 = 48.5° above the horizontal with a tension Tz, as shown in the figure. Find the tensions Ti and Tz.

Solution

To solve for the tensions T1T_1 and T2T_2 in the ropes supporting the sign, we need to apply the principles of equilibrium. Since the sign is in equilibrium, the sum of forces in both the horizontal and vertical directions must be zero.

Step-by-Step Solution:

  1. Breakdown of Forces: Each tension force can be split into horizontal and vertical components. Let’s denote the tension forces and their components as follows:

    • Tension T1T_1:

      • Horizontal component: T1cosθ1T_1 \cos \theta_1
      • Vertical component: T1sinθ1T_1 \sin \theta_1
      • Angle θ1=35.5\theta_1 = 35.5^\circ
    • Tension T2T_2:

      • Horizontal component: T2cosθ2T_2 \cos \theta_2
      • Vertical component: T2sinθ2T_2 \sin \theta_2
      • Angle θ2=48.5\theta_2 = 48.5^\circ
  2. Sum of Forces in the Horizontal Direction: The horizontal components must cancel each other out: T1cosθ1=T2cosθ2T_1 \cos \theta_1 = T_2 \cos \theta_2

  3. Sum of Forces in the Vertical Direction: The vertical components must add up to the weight of the sign, which is 125 N (assuming the sign is in equilibrium and stationary): T1sinθ1+T2sinθ2=125T_1 \sin \theta_1 + T_2 \sin \theta_2 = 125

  4. Solving the System of Equations: From the horizontal force balance equation: T1=T2cosθ2cosθ1T_1 = T_2 \frac{\cos \theta_2}{\cos \theta_1}

    Substitute this into the vertical force balance equation: (T2cosθ2cosθ1)sinθ1+T2sinθ2=125\left(T_2 \frac{\cos \theta_2}{\cos \theta_1}\right) \sin \theta_1 + T_2 \sin \theta_2 = 125

    Simplifying to solve for T2T_2: T2(cosθ2sinθ1cosθ1+sinθ2)=125T_2 \left( \frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1} + \sin \theta_2 \right) = 125

    T2=125cosθ2sinθ1cosθ1+sinθ2T_2 = \frac{125}{\frac{\cos \theta_2 \sin \theta_1}{\cos \theta_1} + \sin \theta_2}

  5. Calculating Tensions: Now we can plug in the angles and compute T2T_2. Once we have T2T_2, we can find T1T_1 using the horizontal force equation.

Let’s compute these values.The tensions in the two ropes are:

  • T1=83.28NT_1 = 83.28 \, \text{N} (up and to the right at 35.535.5^\circ)
  • T2=102.32NT_2 = 102.32 \, \text{N} (up and to the left at 48.548.5^\circ)

These values ensure that the sign remains in equilibrium.

Would you like more details or have any further questions? Here are some related questions you might find useful:

  1. How do you derive the formulas for horizontal and vertical force components?
  2. What are the conditions for equilibrium in a system?
  3. How would the tensions change if the angles were different?
  4. What if the weight of the sign changed to 150 N?
  5. How can you solve similar problems using a different approach, like vector diagrams?

Tip: Always ensure to break down forces into components when dealing with inclined forces and use trigonometric functions to analyze them accurately.

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Math Problem Analysis

Mathematical Concepts

Equilibrium
Trigonometry
Force Analysis

Formulas

Horizontal Force Balance: T1 * cos(θ1) = T2 * cos(θ2)
Vertical Force Balance: T1 * sin(θ1) + T2 * sin(θ2) = Weight
System of Equations

Theorems

Equilibrium Theorem (Sum of forces in a system is zero)

Suitable Grade Level

Grades 11-12