Question

find the reactions at pin a and the magnitude of the compressive force in member bc, given:
m = 15 kg, l_1 = 2.4 m, l_2 = 2 m, l_3 = 3.2 m

bc = number n (enter a positive number)
a_x = number n
a_y = number n

Explanation:

🆕 New Concept Discovered: Static Equilibrium of Rigid Bodies
Using conditions of equilibrium to find unknown forces in pinned structures.

Step 1: Identify forces and geometry

The mass \( M = 15\text{ kg} \) exerts a downward gravitational force at point \( D \):

$$ W = M \cdot g = 15\text{ kg} \times 9.81\text{ m/s}^2 = 147.15\text{ N} $$

Member \( BC \) is a two-force member pinned at both ends, meaning it only carries an axial force along the line \( BC \).
The horizontal distance of \( BC \) is \( L_1 = 2.4\text{ m} \), and the vertical distance is \( L_3 = 3.2\text{ m} \).

The length of member \( BC \) is:

$$ L_{BC} = \sqrt{L_1^2 + L_3^2} = \sqrt{2.4^2 + 3.2^2} = \sqrt{5.76 + 10.24} = \sqrt{16} = 4\text{ m} $$

The force in member \( BC \), denoted as \( F_{BC} \), acts along the line from \( B \) to \( C \). Since it is in compression, it pushes outward on the pins, meaning at joint \( C \), the force \( F_{BC} \) pushes upward and to the right:

$$ F_{BC, x} = F_{BC} \cdot \frac{L_1}{L_{BC}} = F_{BC} \cdot \frac{2.4}{4} = 0.6 F_{BC} $$

$$ F_{BC, y} = F_{BC} \cdot \frac{L_3}{L_{BC}} = F_{BC} \cdot \frac{3.2}{4} = 0.8 F_{BC} $$

Step 2: Apply moment equilibrium about pin A

We isolate the horizontal beam \( ACD \). The forces acting on this beam are:

• Reaction forces at pin \( A \): \( A_x \) and \( A_y \)
• Force from member \( BC \) acting at point \( C \): \( F_{BC, x} \) (pointing right) and \( F_{BC, y} \) (pointing up)
• Downward load \( W \) acting at point \( D \)

Taking the sum of moments about pin \( A \) to eliminate the reaction forces \( A_x \) and \( A_y \):

$$ \sum M_A = 0 $$

$$ (F_{BC, y} \cdot L_1) - (W \cdot (L_1 + L_2)) = 0 $$

$$ (0.8 F_{BC} \cdot 2.4) - (147.15 \cdot (2.4 + 2)) = 0 $$

$$ 1.92 F_{BC} = 147.15 \cdot 4.4 $$

$$ 1.92 F_{BC} = 647.46 $$

$$ F_{BC} = \frac{647.46}{1.92} \approx 337.21875\text{ N} $$

Step 3: Solve for reaction forces at pin A

Using the equations of static equilibrium for the horizontal beam \( ACD \):

For the horizontal forces:

$$ \sum F_x = 0 $$

$$ A_x + F_{BC, x} = 0 $$

$$ A_x + 0.6 F_{BC} = 0 $$

$$ A_x = -0.6 \cdot 337.21875 = -202.33125\text{ N} $$

For the vertical forces:

$$ \sum F_y = 0 $$

$$ A_y + F_{BC, y} - W = 0 $$

$$ A_y + 0.8 F_{BC} - 147.15 = 0 $$

$$ A_y + (0.8 \cdot 337.21875) - 147.15 = 0 $$

$$ A_y + 269.775 - 147.15 = 0 $$

$$ A_y = -122.625\text{ N} $$

Answer:

• BC = 337.22
• \(A_x\) = -202.33
• \(A_y\) = -122.63