Question

Determine the moment about point A of each of the three forces acting on the beam.

Answer 1

Moment due to F1: M1 = -3000 lb⋅ft (or 3000 lb⋅ft clockwise)

Moment due to F2: M2 = -5600 lb⋅ft (or 5600 lb⋅ft clockwise)

Moment due to F3: M3 = (40√3 - 1520) lb⋅ft ≈ -1450.72 lb⋅ft (or 1450.72 lb⋅ft clockwise)

Answer 2

To determine the moment about point A for each force, we use the formula $M = F \times d$, where $F$ is the force and $d$ is the perpendicular distance from point A to the line of action of the force. We will use the convention that counter-clockwise moments are positive and clockwise moments are negative.

1. Moment due to Force $F_1 = 375 \text{ lb}$

• Force: $F_1 = 375 \text{ lb}$ (acting downwards).
• Perpendicular distance from A: $d_1 = 8 \text{ ft}$.
• Direction of rotation about A: Clockwise.
• Moment: $M_1 = -F_1 \times d_1 = -375 \text{ lb} \times 8 \text{ ft} = -3000 \text{ lb} \cdot \text{ft}$

2. Moment due to Force $F_2 = 500 \text{ lb}$

First, resolve $F_2$ into its horizontal ($F_{2x}$) and vertical ($F_{2y}$) components. The force is defined by a 3-4-5 triangle, where the vertical component corresponds to 4 and the horizontal to 3.

• Vertical component: $F_{2y} = F_2 \times \frac{4}{5} = 500 \text{ lb} \times \frac{4}{5} = 400 \text{ lb}$ (acting downwards).
• Horizontal component: $F_{2x} = F_2 \times \frac{3}{5} = 500 \text{ lb} \times \frac{3}{5} = 300 \text{ lb}$ (acting to the left).

Now, calculate the moment for each component about A:

• Moment due to $F_{2y}$:
  • Perpendicular distance from A: $d_{2y} = 8 \text{ ft} + 6 \text{ ft} = 14 \text{ ft}$.
  • Direction of rotation about A: Clockwise.
  • Moment: $M_{2y} = -F_{2y} \times d_{2y} = -400 \text{ lb} \times 14 \text{ ft} = -5600 \text{ lb} \cdot \text{ft}$
• Moment due to $F_{2x}$:
  • The horizontal component $F_{2x}$ acts along the beam's axis (assuming A is on the axis). Therefore, its line of action passes through the same vertical level as point A.
  • Perpendicular distance from A: $d_{2x} = 0 \text{ ft}$.
  • Moment: $M_{2x} = F_{2x} \times d_{2x} = 300 \text{ lb} \times 0 \text{ ft} = 0 \text{ lb} \cdot \text{ft}$
• Total moment due to $F_2$: $M_2 = M_{2x} + M_{2y} = 0 - 5600 \text{ lb} \cdot \text{ft} = -5600 \text{ lb} \cdot \text{ft}$

3. Moment due to Force $F_3 = 160 \text{ lb}$

First, resolve $F_3$ into its horizontal ($F_{3x}$) and vertical ($F_{3y}$) components. The force acts at $30^\circ$ below the horizontal.

• Horizontal component: $F_{3x} = F_3 \cos(30^\circ) = 160 \text{ lb} \times \frac{\sqrt{3}}{2} = 80\sqrt{3} \text{ lb}$ (acting to the right).
• Vertical component: $F_{3y} = F_3 \sin(30^\circ) = 160 \text{ lb} \times \frac{1}{2} = 80 \text{ lb}$ (acting downwards).

The force $F_3$ is applied at point B, which is $0.5 \text{ ft}$ below the beam's axis and at a total horizontal distance from A: $d_h = 8 \text{ ft} + 6 \text{ ft} + 5 \text{ ft} = 19 \text{ ft}$.

Now, calculate the moment for each component about A:

• Moment due to $F_{3x}$:
  • The horizontal component $F_{3x}$ acts to the right at a vertical distance of $0.5 \text{ ft}$ below A.
  • Perpendicular distance from A: $d_{3x} = 0.5 \text{ ft}$.
  • Direction of rotation about A: Counter-clockwise.
  • Moment: $M_{3x} = F_{3x} \times d_{3x} = 80\sqrt{3} \text{ lb} \times 0.5 \text{ ft} = 40\sqrt{3} \text{ lb} \cdot \text{ft} \approx 69.28 \text{ lb} \cdot \text{ft}$
• Moment due to $F_{3y}$:
  • The vertical component $F_{3y}$ acts downwards at a horizontal distance of $19 \text{ ft}$ from A.
  • Perpendicular distance from A: $d_{3y} = 19 \text{ ft}$.
  • Direction of rotation about A: Clockwise.
  • Moment: $M_{3y} = -F_{3y} \times d_{3y} = -80 \text{ lb} \times 19 \text{ ft} = -1520 \text{ lb} \cdot \text{ft}$
• Total moment due to $F_3$: $M_3 = M_{3x} + M_{3y} = 40\sqrt{3} \text{ lb} \cdot \text{ft} - 1520 \text{ lb} \cdot \text{ft} \approx 69.28 - 1520 = -1450.72 \text{ lb} \cdot \text{ft}$

Summary of Results:

• Moment due to $F_1$: $M_1 = -3000 \text{ lb} \cdot \text{ft}$ (or $3000 \text{ lb} \cdot \text{ft}$ clockwise)
• Moment due to $F_2$: $M_2 = -5600 \text{ lb} \cdot \text{ft}$ (or $5600 \text{ lb} \cdot \text{ft}$ clockwise)
• Moment due to $F_3$: $M_3 = (40\sqrt{3} - 1520) \text{ lb} \cdot \text{ft} \approx -1450.72 \text{ lb} \cdot \text{ft}$ (or $1450.72 \text{ lb} \cdot \text{ft}$ clockwise)