Question: Replace the loading by an equivalent resultant force and couple moment acting at point O (Figure 1). Assume $F_1 = 200$ N, $F_2 = 400$ N, and $M_c = 95$ N\cdot m.
Part A
Determine the resultant force.
Express the $x$, $y$ and $z$ components of the resultant force in newtons to three significant figures separated by commas.
Part B
Determine the couple moment acting at point O.
Express the $x$, $y$ and $z$ components of the couple moment in newton-meters to three significant figures separated by commas.
1. **Problem Statement:**
We need to find the equivalent resultant force and couple moment at point $O$ for the given forces $F_1 = 200$ N, $F_2 = 400$ N, and couple moment $M_c = 95$ N\cdot m.
2. **Given Data and Coordinates:**
- Force $F_1 = 200$ N acts at a point located $0.5$ m along the $x$-axis and $0.3$ m along the $z$-axis.
- Force $F_2 = 400$ N acts at a point located $0.4$ m along the $y$-axis.
- Couple moment $M_c = 95$ N\cdot m acts at the far right end.
3. **Step 1: Determine the Resultant Force $\mathbf{F}_R$**
The resultant force is the vector sum of all forces:
$$\mathbf{F}_R = \mathbf{F}_1 + \mathbf{F}_2$$
Assuming directions:
- $F_1$ acts vertically upward along $z$-axis: $\mathbf{F}_1 = (0, 0, 200)$ N
- $F_2$ acts horizontally along $x$-axis: $\mathbf{F}_2 = (400, 0, 0)$ N
Therefore,
$$\mathbf{F}_R = (0 + 400, 0 + 0, 200 + 0) = (400, 0, 200) \text{ N}$$
4. **Step 2: Determine the Couple Moment $\mathbf{M}_O$ at Point $O$**
The couple moment at $O$ is the sum of the original couple moment $M_c$ plus the moments generated by the forces about $O$:
$$\mathbf{M}_O = \mathbf{M}_c + \mathbf{r}_1 \times \mathbf{F}_1 + \mathbf{r}_2 \times \mathbf{F}_2$$
Where:
- $\mathbf{M}_c = (0, 0, 95)$ N\cdot m (assuming along $z$-axis)
- Position vectors:
- $\mathbf{r}_1 = (0.5, 0, 0.3)$ m
- $\mathbf{r}_2 = (0, 0.4, 0)$ m
Calculate cross products:
$$\mathbf{r}_1 \times \mathbf{F}_1 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0.5 & 0 & 0.3 \\ 0 & 0 & 200 \end{vmatrix} = (0 \cdot 200 - 0.3 \cdot 0)\mathbf{i} - (0.5 \cdot 200 - 0.3 \cdot 0)\mathbf{j} + (0.5 \cdot 0 - 0 \cdot 0)\mathbf{k} = (0, -100, 0)$$
$$\mathbf{r}_2 \times \mathbf{F}_2 = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 0.4 & 0 \\ 400 & 0 & 0 \end{vmatrix} = (0.4 \cdot 0 - 0 \cdot 0)\mathbf{i} - (0 \cdot 0 - 0 \cdot 400)\mathbf{j} + (0 \cdot 0 - 0.4 \cdot 400)\mathbf{k} = (0, 0, -160)$$
5. **Sum the moments:**
$$\mathbf{M}_O = (0, 0, 95) + (0, -100, 0) + (0, 0, -160) = (0, -100, -65) \text{ N\cdot m}$$
6. **Final Answers:**
- Resultant force components:
$$(F_R)_x, (F_R)_y, (F_R)_z = 400, 0, 200 \text{ N}$$
- Couple moment components at $O$:
$$(M_O)_x, (M_O)_y, (M_O)_z = 0, -100, -65 \text{ N\cdot m}$$