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 = -20\\) n
\\(a_y = -60\\) n
\\(a_z = 65\\) n

\\(m_{ax} =\\) number \\(n \cdot m\\)
\\(m_{ay} =\\) number \\(n \cdot m\\)
\\(m_{az} =\\) number \\(n \cdot m\\)

Explanation:

🆕 New Concept Discovered: 3D Rigid Body Equilibrium
*Balancing forces and moments in three dimensions*

Step 1: Identify Coordinates of Key Points

Let the origin of the coordinate system be at the fixed support \( A \). Based on the geometry shown in the diagram:
- Point \( A \) is at the origin:
\[ A = (0, 0, 0) \]
- Segment \( AB \) of length \( L_Y = 5\text{ m} \) lies along the positive \( Y \)-axis:
\[ B = (0, 5, 0) \]
- Segment \( BC \) of length \( L_X = 6\text{ m} \) is parallel to the positive \( X \)-axis:
\[ C = (6, 5, 0) \]
- Segment \( CD \) of length \( L_Z = 5\text{ m} \) goes vertically downwards parallel to the negative \( Z \)-axis:
\[ D = (6, 5, -5) \]

Step 2: Define Position Vectors

We need the position vectors from the reference point \( A \) to the points where the forces are applied (\( C \) and \( D \)):
- Position vector to point \( C \) (where \( \vec{F}_2 \) acts):
\[ \vec{r}_C = \vec{r}_{C/A} = 6\hat{i} + 5\hat{j} + 0\hat{k}\text{ m} \]
- Position vector to point \( D \) (where \( \vec{F}_1 \) acts):
\[ \vec{r}_D = \vec{r}_{D/A} = 6\hat{i} + 5\hat{j} - 5\hat{k}\text{ m} \]

Step 3: Calculate the Reaction Forces at Support A

For static equilibrium, the sum of all forces must equal zero:
\[ \sum \vec{F} = \vec{A} + \vec{F}_1 + \vec{F}_2 = \vec{0} \]

Given:
\[ \vec{F}_1 = 75\hat{i} + 10\hat{j} - 85\hat{k}\text{ N} \]
\[ \vec{F}_2 = -55\hat{i} + 50\hat{j} + 20\hat{k}\text{ N} \]

Sum of applied forces:
\[ \vec{F}_1 + \vec{F}_2 = (75 - 55)\hat{i} + (10 + 50)\hat{j} + (-85 + 20)\hat{k} = 20\hat{i} + 60\hat{j} - 65\hat{k}\text{ N} \]

Therefore, the reaction force vector \( \vec{A} \) is:
\[ \vec{A} = -(\vec{F}_1 + \vec{F}_2) = -20\hat{i} - 60\hat{j} + 65\hat{k}\text{ N} \]

The components are:
- \( A_X = -20\text{ N} \)
- \( A_Y = -60\text{ N} \)
- \( A_Z = 65\text{ N} \)

Step 4: Calculate the Reaction Moments at Support A

For static equilibrium, the sum of all moments about point \( A \) must equal zero:
\[ \sum \vec{M}_A = \vec{M}_A + \vec{r}_C \times \vec{F}_2 + \vec{r}_D \times \vec{F}_1 = \vec{0} \]

First, calculate the moment due to \( \vec{F}_2 \) about \( A \):
\[ \vec{M}_{C} = \vec{r}_C \times \vec{F}_2 = \det \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 5 & 0 \\ -55 & 50 & 20 \end{pmatrix} \]
\[ \vec{M}_{C} = \hat{i}(5 \cdot 20 - 0 \cdot 50) - \hat{j}(6 \cdot 20 - 0 \cdot (-55)) + \hat{k}(6 \cdot 50 - 5 \cdot (-55)) \]
\[ \vec{M}_{C} = 100\hat{i} - 120\hat{j} + (300 + 275)\hat{k} = 100\hat{i} - 120\hat{j} + 575\hat{k}\text{ N}\cdot\text{m} \]

Next, calculate the moment due to \( \vec{F}_1 \) about \( A \):
\[ \vec{M}_{D} = \vec{r}_D \times \vec{F}_1 = \det \begin{pmatrix} \hat{i} & \hat{j} & \hat{k} \\ 6 & 5 & -5 \\ 75 & 10 & -85 \end{pmatrix} \]
\[ \vec{M}_{D} = \hat{i}(5 \cdot (-85) - (-5) \cdot 10) - \hat{j}(6 \cdot (-85) - (-5) \cdot 75) + \hat{k}(6 \cdot 10 - 5 \cdot 75) \]
\[ \vec{M}_{D} = \hat{i}(-425 + 50) - \hat{j}(-510 + 375) + \hat{k}(60 - 375) \]
\[ \vec{M}_{D} = -375\hat{i} - (-135)\hat{j} - 315\hat{k} = -375\hat{i} + 135\hat{j} - 315\hat{k}\text{ N}\cdot\text{m} \]

Sum the moments from the applied forces:
\[ \vec{M}_{\text{applied}} = \vec{M}_{C} + \vec{M}_{D} = (100 - 375)\hat{i} + (-120 + 135)\hat{j} + (575 - 315)\hat{k} \]
\[ \vec{M}_{\text{applied}} = -275\hat{i} + 15\hat{j} + 260\hat{k}\text{ N}\cdot\text{m} \]

The reaction moment \( \vec{M}_A \) at the support must balance this sum:
\[ \vec{M}_A = -\vec{M}_{\text{applied}} = 275\hat{i} - 15\hat{j} - 260\hat{k}\text{ N}\cdot\text{m} \]

The components are:
- \( M_{AX} = 275\text{ N}\cdot\text{m} \)
- \( M_{AY} = -15\text{ N}\cdot\text{m} \)
- \( M_{AZ} = -260\text{ N}\cdot\text{m} \)

Answer:

• \( A_X = -20\text{ N} \)
• \( A_Y = -60\text{ N} \)
• \( A_Z = 65\text{ N} \)
• \( M_{AX} = 275\text{ N}\cdot\text{m} \)
• \( M_{AY} = -15\text{ N}\cdot\text{m} \)
• \( M_{AZ} = -260\text{ N}\cdot\text{m} \)