Math Problem Statement

A traffic light of weight of 139 N is hanged from a circular knot as shown with two cables as shown.

a. Write down two equations by balancing the forces in x and y directions.

b. Solve the simultaneous equations and determine T1 in N.

A traffic light of weight of 139 N is hanged from a circular knot as shown with two cables as shown. (a) Write down two equations by balancing the forces in x and y directions. (b) Solve the simultaneous equations and determine T1 in N.

Solution

To solve this problem, we’ll balance the forces in both the xx- and yy-directions and solve for the tension T1T_1.

Given Information

  • Weight of the traffic light, W=139NW = 139 \, \text{N}
  • Angle of T1T_1 with the horizontal: 4040^\circ
  • Angle of T2T_2 with the horizontal: 6060^\circ

Part (a): Writing Equilibrium Equations

Since the system is in equilibrium, the sum of the forces in both the xx- and yy-directions should be zero.

  1. In the xx-direction: T1cos(40)=T2cos(60)T_1 \cos(40^\circ) = T_2 \cos(60^\circ)

  2. In the yy-direction: T1sin(40)+T2sin(60)=WT_1 \sin(40^\circ) + T_2 \sin(60^\circ) = W Substituting W=139NW = 139 \, \text{N}, we get: T1sin(40)+T2sin(60)=139T_1 \sin(40^\circ) + T_2 \sin(60^\circ) = 139

Part (b): Solving the Simultaneous Equations

Now, let’s solve these equations to find T1T_1.

  1. From the xx-direction equation: T2=T1cos(40)cos(60)T_2 = \frac{T_1 \cos(40^\circ)}{\cos(60^\circ)}

  2. Substitute T2T_2 from this equation into the yy-direction equation: T1sin(40)+T1cos(40)sin(60)cos(60)=139T_1 \sin(40^\circ) + \frac{T_1 \cos(40^\circ) \sin(60^\circ)}{\cos(60^\circ)} = 139

  3. Simplify and solve for T1T_1.

Let's calculate the values using trigonometric approximations and substitutions for the angles.

Approximate Trigonometric Values:

  • sin(40)0.6428\sin(40^\circ) \approx 0.6428
  • cos(40)0.7660\cos(40^\circ) \approx 0.7660
  • sin(60)0.8660\sin(60^\circ) \approx 0.8660
  • cos(60)=0.5\cos(60^\circ) = 0.5

Substitute these into the equation:

T1×0.6428+T1×0.7660×0.86600.5=139T_1 \times 0.6428 + \frac{T_1 \times 0.7660 \times 0.8660}{0.5} = 139

Calculating T1T_1 from this equation will yield the tension in T1T_1.

Let me proceed with the computation.The tension T1T_1 in the cable is approximately 70.57N70.57 \, \text{N}.

Would you like more details on the steps or have any questions?

Here are some related questions that could further deepen your understanding:

  1. How would the tensions change if the weight of the traffic light increased?
  2. What effect would changing the angles of the cables have on the tension values?
  3. How can we derive the trigonometric relationships used for the angles here?
  4. What are some real-life scenarios where understanding force equilibrium is essential?
  5. How would the system behave if one of the cables was replaced with a stronger or weaker material?

Tip: When solving equilibrium problems, always start by resolving forces into their components and writing down the balance equations before solving.

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Math Problem Analysis

Mathematical Concepts

Statics
Trigonometry
Equilibrium of Forces

Formulas

Sum of forces in the x-direction: T1 * cos(40°) = T2 * cos(60°)
Sum of forces in the y-direction: T1 * sin(40°) + T2 * sin(60°) = W

Theorems

Equilibrium Theorem (sum of forces in each direction equals zero)

Suitable Grade Level

Grades 11-12