Short Answer
The scenario involves a vendor suspending an 8.0 kg sign with a cable supported by a pole angled at 60 degrees. By applying equilibrium equations for both vertical and horizontal forces, the tension in the cable is calculated to be approximately 156.96 Newtons.
Step 1: Understand the Setup
In this scenario, a vendor is suspending an 8.0 kg sign with a cable supported by a pole that pivots against a wall. The key to solving for the tension in the cable is recognizing the role of the angle. The pole makes an angle of 60 degrees with the cable and leans away from the wall, affecting the tension’s components.
Step 2: Establish Equations for Equilibrium
To find the tension in the cable, we must apply the conditions for equilibrium, identifying vertical and horizontal forces. The equations are established as:
- Vertical equilibrium: Tsin(60¬¨‚à û) – Fsin(150¬¨‚à û) – W = 0
- Horizontal equilibrium: Tcos(60¬¨‚à û) + Fcos(150¬¨‚à û) = 0
In these equations, T represents the tension, F is the force from the pole, and W is the weight of the sign (78.48 N).
Step 3: Solve for Tension
By substituting the horizontal equilibrium equation into the vertical one, we eliminate the force F to solve for T. Following through the calculations leads to:
- Substituting yields: 0.866T – (-0.5)(0.5T – 0.866F) – 78.48 = 0
- Simplifying results in T ‚Äöaa 156.96 N
This means the tension in the cable is approximately 156.96 Newtons, confirming the solution for the given setup.