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Statics — Section 4.6 In-Class Problem — Real Worl

Question

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Statics — Section 4.6 In-Class Problem — Real World Problem
The guardrail post shown is subjected to a code-prescribed load of 200 Ibs at the
4
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VIPSolutions · Accepted Answer

# Statics Problem Explanation

Let's break down the problem step by step and address each part:

---

## Problem Overview

A guardrail post is subjected to a 200 lb force at the top rail (3'6" above the base), applied at an angle θ from the horizontal. You need to:

1. Find the angle θ for **maximum moment** at the base (O).
2. Compute the **maximum moment**.
3. Determine the **force-couple components** at anchors A and B to create a couple that cancels the moment at O.
4. Find the **minimum distance between anchors** if their strength is limited to 600 lbs.

---

## Part (a) — **Angle θ for Maximum Moment**

The moment $ M_O $ at point O due to force F is:
$$
M_O = F \cdot d_\perp
$$
where $ d_\perp $ is the perpendicular distance from O to the line of action of F.

- The vertical height from O to where F is applied is 3'6" = 3.5 ft.
- The moment arm is maximized when the force is **perpendicular** to the moment arm.

For a force applied at angle θ from the horizontal:
$$
M_O = F \cdot h \cdot \sin(\theta)
$$
where $ h = 3.5 $ ft.

- **Maximum** value occurs when $ \sin(\theta) = 1 $ ($ \theta = 90^\circ $).

**Answer:**
- θ = 90°, i.e., the force is applied **horizontally**.

---

## Part (b) — **Maximum Moment Magnitude**

Substitute the maximum values:
$$
M_{O,\text{max}} = F \cdot h \cdot \sin(90^\circ) = 200 \text{ lb} \cdot 3.5 \text{ ft} \cdot 1 = 700 \text{ lb} \cdot \text{ft}
$$

---

## Part (c) — **Force-Couple at Anchors A and B**

We want to replace the moment $ M_O $ with a **couple** formed by forces at A and B.

- **Couple**: Two equal and opposite forces separated by a distance.

Let $ F_A $ and $ F_B $ be the forces at anchors A and B, separated by distance $ d $.

$$
M_O = F_{\text{couple}} \cdot d
$$

If the couple is formed by tension at A and compression at B (or vice versa), and both have the same magnitude $ F_c $:

$$
M_O = F_c \cdot d
$$

So, the **force at each anchor** ($ F_c $) for a given spacing $ d $ is:

$$
F_c = \frac{M_O}{d}
$$

---

## Part (d) — **Minimum Distance for Anchor Strength**

Given:
- Maximum anchor capacity $ F_{c, \text{max}} = 600 $ lbs

Set $ F_c \leq 600 $ lbs:

$$
F_c = \frac{700}{d} \leq 600
$$

Solve for $ d $:

$$
d \geq \frac{700}{600} = 1.17 \text{ ft}
$$

Convert to inches:

$$
1.17 \text{ ft} \times 12 = 14.0 \text{ in}
$$

---

## **Summary Table**

| Part | Equation / Reasoning | Answer |
|------|---------------------|--------|
| a | Max moment when force is perpendicular ($ \theta = 90^\circ $) | $ \theta = 90^\circ $ |
| b | $ M_O = 200 \times 3.5 $ | $ 700 $ lb·ft |
| c | $ F_c = \frac{M_O}{d} $ | $ F_c = \frac{700}{d} $ |
| d | $ F_c \leq 600 \implies d \geq 1.17 $ ft ($ 14 $ in) | $ d \geq 14 $ in |

---

### **Let me know if you want worked-out calculations for any step!**

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