Let's solve for the **moment about point A** due to the 150-N force. --- # **Step-by-Step Solution** ## **Step 1: Find Location of Force Application (Point B)** Given: - \( AB = 12 \) units (horizontal segment) - \( BC = 5 \) units (vertical segment) - The triangle at \( C \) is a right triangle with sides 3, 4, and 5. Coordinates (relative to A): - \( A = (0, 0) \) - \( B = (12, 0) \) - To find \( C \): From \( B \), move 3 units right and 4 units up (based on the 3-4-5 triangle at C). Therefore: - \( C = (12 + 3, 0 + 4) = (15, 4) \) ## **Step 2: Find Perpendicular Distance from A to Line of Action of Force** The force is vertical (downward) at C. Moment arm = perpendicular distance from A to the vertical line through C. - \( x_A = 0 \) - \( x_C = 15 \) **So, perpendicular distance = \( 15 \) units.** But the figure says the total horizontal distance from A to C is **260 mm**. So, **1 unit = 260/15 = 17.33 \text{ mm/unit}**. Thus, the perpendicular distance in mm is: \[ d = 260 \text{ mm} \] ## **Step 3: Calculate the Moment** The moment about A due to a force F at a perpendicular distance d is: \[ M = F \cdot d \] Where: - \( F = 150 \) N (downward) - \( d = 0.260 \) m (convert 260 mm to meters) \[ M_A = 150 \text{ N} \times 0.260 \text{ m} = 39.0 \text{ N}\cdot\text{m} \] **Direction:** - The force tends to rotate the bar **clockwise** about A (right-hand rule). - **Moment is negative by convention (clockwise):** \[ \boxed{M_A = -39.0\ \text{N}\cdot\text{m}} \] --- ## **Final Answer** \[ \boxed{ M_A = -39.0~\text{N}\cdot\text{m} \quad \text{(clockwise about point A)} } \]

# **1. Introduction** The concept of moments in physics, particularly in statics, is crucial for understanding how forces interact with structures or objects. A moment (or torque) about a point is defined as the product of the force applied and the perpendicular distance from the point of rotation to the line of action of the force. This concept is foundational when analyzing mechanical systems, as it helps predict rotational effects caused by forces. In this problem, we are tasked with calculating the moment about point A due to a downward force of 150 N applied at point C. Understanding the geometry of the system and the relationship between forces and distances is essential for accurately calculating the moment. **Explanation:** Grasping the foundational principles of moments provides insight into how forces can produce rotational effects. By applying these concepts, the solution to the problem becomes clearer, highlighting the importance of geometry and force application in static equilibrium. --- # **2. Presentation of Relevant Formulas and Representation of Given Data** ## **Relevant Formulas** 1. **Moment About a Point:** \[ M = F \cdot d \] - Where \( M \) is the moment, \( F \) is the force, and \( d \) is the perpendicular distance from the point to the line of action of the force. ## **Given Data** - Force, \( F = 150 \) N (downward) - Distances: - Horizontal distance from A to B, \( AB = 12 \) units - Vertical distance from B to C, \( BC = 5 \) units - The right triangle at C has sides 3, 4, and 5. **Explanation:** The moment formula is essential to derive the rotational effect of the force about point A. The given data includes both the force magnitude and the necessary distances to facilitate a correct calculation of the moment. --- # **3. A Detailed Step-by-Step Solution** ## **Step 1: Determine Coordinates of Points** 1. **Coordinates of Point A:** \[ A = (0, 0) \] 2. **Coordinates of Point B:** \[ B = (12, 0) \] 3. **Coordinates of Point C:** - From point B, moving right 3 units and up 4 units gives: \[ C = (12 + 3, 0 + 4) = (15, 4) \] ## **Step 2: Calculate Perpendicular Distance from A to the Force Line** - The force is applied downward at point C, which is at \( (15, 4) \). - The perpendicular distance \( d \) from point A to the line of action of the force at C is simply the x-coordinate of C: \[ d = 15 \text{ units} \] ## **Step 3: Convert Distance to Meters** - Given that 1 unit corresponds to 260 mm, convert to meters: \[ d = 15 \text{ units} \times \frac{260 \text{ mm}}{1 \text{ unit}} = 3900 \text{ mm} = 3.9 \text{ m} \] ## **Step 4: Calculate the Moment About Point A** Using the moment formula: \[ M_A = F \cdot d \] Substituting the values: \[ M_A = 150 \text{ N} \times 3.9 \text{ m} = 585.0 \text{ N}\cdot\text{m} \] **Direction:** The force causes a clockwise rotation about point A, which by convention is considered a negative moment: \[ M_A = -585.0 \text{ N}\cdot\text{m} \] **Explanation:** Each step contributes to calculating the moment by first determining the force's application point and its distance from the reference point. The conversion from units ensures compatibility in calculations, leading to an accurate moment evaluation. --- # **Conclusion** The moment about point A due to the 150-N force applied at point C is calculated as follows: \[ \boxed{M_A = -585.0~\text{N}\cdot\text{m} \quad \text{(clockwise about point A)}} \] This solution demonstrates the process of determining moments by using geometrical relationships and mechanics principles to arrive at the final outcome effectively.