Question

find the components of the reaction at the fixed wall support, a, given:
\\(\vec{f}_1 = \langle 75, 10, -85 \
angle\\) n, \\(\vec{f}_2 = \langle -55, 50, 20 \
angle\\) n, \\(l_x = 6\\) m, \\(l_y = 5\\) m, \\(l_z = 5\\) m

\\(a_x =\\) number n
\\(a_y =\\) number n
\\(a_z =\\) number n

Explanation:

🆕 New Concept Discovered: 3D Static Equilibrium of Rigid Bodies
Finding support reactions by balancing forces and moments in three dimensions.

Step 1: Identify the Equilibrium Conditions

For a rigid body in static equilibrium, the sum of all external forces acting on the system must equal zero:

$$ \sum \vec{F} = \vec{0} $$

This vector equation can be broken down into three scalar equations for the coordinate axes:

$$ \sum F_x = 0 $$

$$ \sum F_y = 0 $$

$$ \sum F_z = 0 $$

Step 2: Set up the Force Balance Equations

The external forces acting on the pipe assembly are the reaction forces at the fixed support \( A \) (represented by the vector \(\vec{A} = \langle A_x, A_y, A_z
angle\)) and the applied forces \(\vec{F}_1\) and \(\vec{F}_2\).

Given:

$$ \vec{F}_1 = \langle 75, 10, -85 angle \text{ N} $$

$$ \vec{F}_2 = \langle -55, 50, 20 angle \text{ N} $$

The equilibrium equation for the forces is:

$$ \vec{A} + \vec{F}_1 + \vec{F}_2 = \vec{0} $$

This means:

$$ A_x + F_{1x} + F_{2x} = 0 $$

$$ A_y + F_{1y} + F_{2y} = 0 $$

$$ A_z + F_{1z} + F_{2z} = 0 $$

Step 3: Calculate the Reaction Components

Substitute the given force components into the equilibrium equations:

For the x-component:

$$ A_x + 75 + (-55) = 0 $$

$$ A_x + 20 = 0 $$

$$ A_x = -20 \text{ N} $$

For the y-component:

$$ A_y + 10 + 50 = 0 $$

$$ A_y + 60 = 0 $$

$$ A_y = -60 \text{ N} $$

For the z-component:

$$ A_z + (-85) + 20 = 0 $$

$$ A_z - 65 = 0 $$

$$ A_z = 65 \text{ N} $$

Answer:

\( A_x = -20 \) N
\( A_y = -60 \) N
\( A_z = 65 \) N