To solve this problem, we will use the principles of vector decomposition and equilibrium. Given:

• The force $F = 5000$ lb exerted by the cable on the pole,
• The angle $\theta = 15^\circ$ (assumed to be the angle between the cable and the horizontal).

Let’s break down the problem into parts:

Part (a) — The Horizontal and Vertical Components of the Cable Pull $F$:

The force $F$ has both horizontal and vertical components. These can be found using trigonometry:

• Horizontal component: $F_x = F \cos(\theta)$
• Vertical component: $F_y = F \sin(\theta)$

Given:

• $F = 5000 \, \text{lb}$
• $\theta = 15^\circ$

We can now calculate:

$F_x = 5000 \cos(15^\circ)$ $F_y = 5000 \sin(15^\circ)$

Using these values: $F_x = 5000 \times 0.9659 \approx 4829.5 \, \text{lb}$ $F_y = 5000 \times 0.2588 \approx 1294 \, \text{lb}$

So, the horizontal and vertical components of the cable pull are:

• $F_x \approx 4829.5 \, \text{lb}$
• $F_y \approx 1294 \, \text{lb}$

Part (b) — The Horizontal and Vertical Components of the Reaction in the Stay Wire:

Since the resultant force on the pole must be downward and collinear with the pole, the horizontal components of the forces must cancel out, and the vertical components must sum up to a force pointing downward.

Let’s assume the reaction force in the stay wire has components $R_x$ (horizontal) and $R_y$ (vertical). For equilibrium:

• The horizontal components must balance out: $F_x + R_x = 0$
• The vertical components must sum to a downward force: $F_y + R_y$

From the horizontal equilibrium condition: $R_x = -F_x = -4829.5 \, \text{lb}$

From the vertical equilibrium condition: $R_y + F_y = 0 \quad \Rightarrow \quad R_y = -F_y = -1294 \, \text{lb}$

Thus, the horizontal and vertical components of the reaction in the stay wire are:

• $R_x \approx -4829.5 \, \text{lb}$
• $R_y \approx -1294 \, \text{lb}$

Part (c) — The Net Force on the Pole:

Since the horizontal forces cancel each other out ($F_x + R_x = 0$), the only remaining force is the sum of the vertical components.

The net force on the pole is simply the downward force, which is:

$F_{\text{net}} = F_y + R_y = -1294 + (-1294) = -2588 \, \text{lb}$

Thus, the net force on the pole is a downward force of 2588 lb.

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Summary:

1. Horizontal and vertical components of the cable pull:

   • $F_x \approx 4829.5 \, \text{lb}$
   • $F_y \approx 1294 \, \text{lb}$

2. Horizontal and vertical components of the reaction in the stay wire:

   • $R_x \approx -4829.5 \, \text{lb}$
   • $R_y \approx -1294 \, \text{lb}$

3. Net force on the pole:

   • $F_{\text{net}} \approx 2588 \, \text{lb}$ downward

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Would you like further details or clarifications on any part of this solution?

Related Questions:

1. How would the forces change if the angle of the cable were increased to $30^\circ$?
2. What would happen if the force in the cable were increased to 6000 lb?
3. How does the equilibrium of forces work in three-dimensional systems?
4. How do you calculate the moment caused by a force acting at a distance from a point?
5. What is the difference between static and dynamic equilibrium in mechanical systems?

Tip:

When solving equilibrium problems, always start by ensuring that the sum of forces in each direction (horizontal and vertical) equals zero. This will help you set up correct equations.