Moment Force 02Db06

1. **State the problem:** We need to find the moment of the force $\mathbf{F} = -6\mathbf{i} + 4\mathbf{j} + 8\mathbf{k}$ kN about point $P$, expressed as a Cartesian vector. 2. **Identify position vectors:** Let $\mathbf{r}_A$ be the position vector of point $A$ where the force is applied, and $\mathbf{r}_P$ be the position vector of point $P$. From the given dimensions and coordinate system: - $O$ is the origin. - $P$ is located at $\mathbf{r}_P = 6\mathbf{i} + 3\mathbf{j} + 1\mathbf{k}$ m (using the given distances along $x$, $y$, and $z$ axes). - $A$ is located at $\mathbf{r}_A = 0\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$ m (assuming from the problem's description and dimensions). 3. **Calculate the position vector from $P$ to $A$:** $$\mathbf{r} = \mathbf{r}_A - \mathbf{r}_P = (0 - 6)\mathbf{i} + (4 - 3)\mathbf{j} + (2 - 1)\mathbf{k} = -6\mathbf{i} + 1\mathbf{j} + 1\mathbf{k}$$ 4. **Moment formula:** The moment $\mathbf{M}_P$ of force $\mathbf{F}$ about point $P$ is given by the cross product: $$\mathbf{M}_P = \mathbf{r} \times \mathbf{F}$$ 5. **Calculate the cross product:** $$\mathbf{M}_P = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -6 & 1 & 1 \\ -6 & 4 & 8 \end{vmatrix}$$ Calculate each component: - $M_{Px} = (1)(8) - (1)(4) = 8 - 4 = 4$ - $M_{Py} = -\left((-6)(8) - (1)(-6)\right) = -(-48 + 6) = -(-42) = 42$ - $M_{Pz} = (-6)(4) - (1)(-6) = -24 + 6 = -18$ 6. **Write the moment vector:** $$\mathbf{M}_P = 4\mathbf{i} + 42\mathbf{j} - 18\mathbf{k} \text{ kN}\cdot\text{m}$$ 7. **Interpretation:** The moment of the force about point $P$ is a vector with components $4$ in the $x$-direction, $42$ in the $y$-direction, and $-18$ in the $z$-direction, all in units of kN·m. **Final answer:** $$\boxed{\mathbf{M}_P = 4\mathbf{i} + 42\mathbf{j} - 18\mathbf{k} \text{ kN}\cdot\text{m}}$$