Replace the loading by an equivalent resultant force and couple moment acting at point O. Assume F1 = {-420i + 210j + 280k} N. F2 = {-450k} N.
Understand the Problem
The question asks to replace the given loading system (forces F1 and F2) with an equivalent resultant force and couple moment acting at point O. This involves calculating the resultant force by summing the individual forces, and calculating the resultant moment about point O by considering the moments caused by each force.
Answer
$\mathbf{F_R} = \{-420\mathbf{i} + 210\mathbf{j} - 170\mathbf{k}\} \, \text{N}$ $\mathbf{M_R} = \{-550\mathbf{i} - 840\mathbf{j} + 1155\mathbf{k}\} \, \text{N} \cdot \text{m}$
Answer for screen readers
$\mathbf{F_R} = {-420\mathbf{i} + 210\mathbf{j} - 170\mathbf{k}} , \text{N}$
$\mathbf{M_R} = {-550\mathbf{i} - 840\mathbf{j} + 1155\mathbf{k}} , \text{N} \cdot \text{m}$
Steps to Solve
- Calculate the resultant force
To find the resultant force $\mathbf{F_R}$, we sum the individual forces $\mathbf{F_1}$ and $\mathbf{F_2}$.
$\mathbf{F_1} = {-420\mathbf{i} + 210\mathbf{j} + 280\mathbf{k}} , \text{N}$ $\mathbf{F_2} = {-450\mathbf{k}} , \text{N}$
$\mathbf{F_R} = \mathbf{F_1} + \mathbf{F_2} = (-420\mathbf{i} + 210\mathbf{j} + 280\mathbf{k}) + (-450\mathbf{k}) = {-420\mathbf{i} + 210\mathbf{j} - 170\mathbf{k}} , \text{N}$
- Determine the position vectors
We need the position vectors from point O to the points where the forces are applied (points A and B). From the diagram:
$\mathbf{r_A} = {2\mathbf{j}} , \text{m}$ $\mathbf{r_B} = {2\mathbf{j} + 1.5\mathbf{i} + 1\mathbf{k}} , \text{m} = {1.5\mathbf{i} + 2\mathbf{j} + 1\mathbf{k}} , \text{m}$
- Calculate the moment due to F1 about point O
The moment due to force $\mathbf{F_1}$ about point O is given by the cross product $\mathbf{M_1} = \mathbf{r_B} \times \mathbf{F_1}$.
$\mathbf{M_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1.5 & 2 & 1 \ -420 & 210 & 280 \end{vmatrix} = (2 \cdot 280 - 1 \cdot 210)\mathbf{i} - (1.5 \cdot 280 - 1 \cdot (-420))\mathbf{j} + (1.5 \cdot 210 - 2 \cdot (-420))\mathbf{k} = (560 - 210)\mathbf{i} - (420 + 420)\mathbf{j} + (315 + 840)\mathbf{k} = {350\mathbf{i} - 840\mathbf{j} + 1155\mathbf{k}} , \text{N} \cdot \text{m}$
- Calculate the moment due to F2 about point O
The moment due to force $\mathbf{F_2}$ about point O is given by the cross product $\mathbf{M_2} = \mathbf{r_A} \times \mathbf{F_2}$.
$\mathbf{M_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 2 & 0 \ 0 & 0 & -450 \end{vmatrix} = (2 \cdot (-450) - 0 \cdot 0)\mathbf{i} - (0 \cdot (-450) - 0 \cdot 0)\mathbf{j} + (0 \cdot 0 - 2 \cdot 0)\mathbf{k} = (-900)\mathbf{i} , \text{N} \cdot \text{m}$
- Calculate the resultant moment about point O
The resultant moment $\mathbf{M_R}$ is the sum of the individual moments $\mathbf{M_1}$ and $\mathbf{M_2}$.
$\mathbf{M_R} = \mathbf{M_1} + \mathbf{M_2} = (350\mathbf{i} - 840\mathbf{j} + 1155\mathbf{k}) + (-900\mathbf{i}) = {-550\mathbf{i} - 840\mathbf{j} + 1155\mathbf{k}} , \text{N} \cdot \text{m}$
$\mathbf{F_R} = {-420\mathbf{i} + 210\mathbf{j} - 170\mathbf{k}} , \text{N}$
$\mathbf{M_R} = {-550\mathbf{i} - 840\mathbf{j} + 1155\mathbf{k}} , \text{N} \cdot \text{m}$
More Information
The resultant force and moment represent the single force and moment that would produce the same effect on the body as the original loading system. This is a fundamental concept in statics, used to simplify complex force systems for analysis.
Tips
A common mistake is incorrectly calculating the cross product, especially the signs of the components. Double-checking the determinant calculation is crucial. Another mistake is using the wrong position vectors or mixing them up.
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