1. **Problem Statement:** We have a force $F$ of magnitude 300 N acting on a plane at point $B$ directed diagonally downwards towards point $A$. Points are located as follows: $O$ at origin, $C$ at (1 m, 0), $B$ at (0, 1 m), $A$ at (1 m, 1 m), and $D$ at (3 m, 1 m). We need to find: - a. Moment about point $O$ ($M_O$) - b. Shortest distance $d$ of $F$ from $O$ - c. Moment about point $C$ ($M_C$) - d. Moment about point $D$ ($M_D$) 2. **Formulas and Rules:** - Moment $M$ about a point is given by $M = r \times F$, where $r$ is the position vector from the point to the line of action of the force. - The magnitude of the moment is $M = F \times d$, where $d$ is the perpendicular (shortest) distance from the point to the line of action of the force. - The direction of the moment is determined by the right-hand rule. 3. **Step 1: Define vectors and force direction** - Position vectors: - $\vec{O} = (0,0)$ - $\vec{B} = (0,1)$ - $\vec{A} = (1,1)$ - $\vec{C} = (1,0)$ - $\vec{D} = (3,1)$ - Force vector $\vec{F}$ acts from $B$ to $A$: $$\vec{F} = 300 \times \frac{\vec{A} - \vec{B}}{|\vec{A} - \vec{B}|} = 300 \times \frac{(1,1) - (0,1)}{\sqrt{(1-0)^2 + (1-1)^2}} = 300 \times (1,0) = (300,0)$$ 4. **Step 2: Moment about $O$ ($M_O$)** - Position vector from $O$ to $B$: $$\vec{r}_O = (0,1)$$ - Moment vector: $$\vec{M}_O = \vec{r}_O \times \vec{F} = (0,1) \times (300,0) = 0 \times 0 - 1 \times 300 = -300$$ - The moment magnitude is 300 Nm, direction is into the plane (negative z-direction). 5. **Step 3: Shortest distance $d$ from $O$ to line of action of $F$** - The line of action passes through $B(0,1)$ with direction vector $\vec{u} = (1,0)$. - Distance from $O(0,0)$ to line: $$d = \frac{|(\vec{O} - \vec{B}) \times \vec{u}|}{|\vec{u}|} = \frac{|(0,0)-(0,1) \times (1,0)|}{1} = \frac{|(0,-1) \times (1,0)|}{1}$$ - Cross product magnitude in 2D: $$|(x_1,y_1) \times (x_2,y_2)| = |x_1 y_2 - y_1 x_2|$$ - Calculate: $$|0 \times 0 - (-1) \times 1| = |0 + 1| = 1$$ - So, $d = 1$ m. 6. **Step 4: Moment about $C$ ($M_C$)** - Position vector from $C$ to $B$: $$\vec{r}_C = \vec{B} - \vec{C} = (0,1) - (1,0) = (-1,1)$$ - Moment: $$\vec{M}_C = \vec{r}_C \times \vec{F} = (-1,1) \times (300,0) = (-1) \times 0 - 1 \times 300 = -300$$ - Moment magnitude is 300 Nm, direction into the plane. 7. **Step 5: Moment about $D$ ($M_D$)** - Position vector from $D$ to $B$: $$\vec{r}_D = \vec{B} - \vec{D} = (0,1) - (3,1) = (-3,0)$$ - Moment: $$\vec{M}_D = \vec{r}_D \times \vec{F} = (-3,0) \times (300,0) = (-3) \times 0 - 0 \times 300 = 0$$ - Moment about $D$ is zero because $\vec{r}_D$ and $\vec{F}$ are parallel. **Final answers:** - a. $M_O = -300$ Nm (clockwise) - b. $d = 1$ m - c. $M_C = -300$ Nm (clockwise) - d. $M_D = 0$ Nm